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APPEARED IN BULLETIN OF THE

arXiv:math/9207210v1 [math.NT] 1 Jul 1992APPEARED IN BULLETIN OF THEAMERICAN MATHEMATICAL SOCIETYVolume 27, Number 1, July 1992, Pages 68-133GALOIS THEORY ON THE LINE IN NONZEROCHARACTERISTICShreeram S. AbhyankarDedicated to Walter Feit, J-P. Serre, and e-mail1. What is Galois theory?Originally, the equation Y 2 + 1 = 0 had no solution.

Then the two solutions iand −i were created. But there is absolutely no way to tell who is i and who is−i.1 That is Galois Theory.Thus, Galois Theory tells you how far we cannot distinguish between the rootsof an equation.

This is codified in the Galois Group.2. Galois groupsMore precisely, consider an equationY n + a1Y n−1 + · · · + an = 0and let α1, .

. .

, αn be its roots, which are assumed to be distinct. By definition,the Galois Group G of this equation consists of those permutations of the rootswhich preserve all relations between them.

Equivalently, G is the set of all thosepermutations σ of the symbols {1, 2, . .

., n} such that φ(ασ(1), . .

. , ασ(n)) = 0 forevery n-variable polynomial φ for which φ(α1, .

. .

, αn) = 0. The coefficients of φare supposed to be in a field K which contains the coefficients a1, .

. .

, an of thegiven polynomialf = f(Y ) = Y n + a1Y n−1 + · · · + an.We call G the Galois Group of f over K and denote it by GalY (f, K) or Gal(f, K).This is Galois’ original concrete definition.According to the modern abstract definition, the Galois Group of a normal ex-tension L of a field K is defined to be the group of all K-automorphisms of L andis denoted by Gal(L, K). Note that a normal extension L of a field K is a fieldobtained by adjoining to K all the roots of a bunch of univariate polynomials with1991 Mathematics Subject Classification.

Primary 12F10, 14H30, 20D06, 20E22.This work was partly supported by NSF grant DMS 88-16286Received by the editors December 19, 1990, and, in revised form, on July 31, 19911Only the physicist can tell the difference by declaring that i is up and −i is down. But then,what is up and what is down?

What I am saying is that intrinsically there is no way to distinguish,in an abstractly given copy of the complex numbers, between the two square-roots of minus one.However, practically everyone has their own favorite concrete model of C, perhaps R × R, perhapsR[Y ]/(Y 2 + 1), and then there is no problem in pointing to say (0, 1) in the first case or [Y ] inthe second and calling that i.c⃝1992 American Mathematical Society0273-0979/92 $1.00 + $.25 per page1

2S. S. ABHYANKARcoefficients in K. To relate the two definitions, let L = K(α1, .

. .

, αn) and note thatwe get an isomorphism of Gal(L, K) onto Gal(f, K) by sending any τ ∈Gal(L, K)to that σ ∈Gal(f, K) for which τ(αi) = ασ(i) for 1 ≤i ≤n.3. Permutation groupsThe above concrete definition brings out the close connection between grouptheory and theory of equations.To wit, the Galois Group Gal(f, K) is now asubgroup of Sn where, as usual, Sn denotes the symmetric group of degree n, i.e.,the group of all permutations of n symbols; note that the order of Sn is n!.

Quitegenerally, a subgroup of Sn is called a (permutation) group of degree n.Herethe use of the word degree is meant to remind us that potentially it comes froman equation of degree n.To convert this potentiality into actuality, in varioussituations, constitutes Inverse Galois Theory. To further bring out the parallelismbetween group theory and the theory of equations, we note that.

(1) f is irreducible iffGal(f, K) is transitive.Here, a permutation group G “acting”2 on the set Ω= {1, 2, . .

., n} is transitive iffor all i, j in Ω, there exists σ ∈G such that σ(i) = j. Likewise, G is 2-transitive(or doubly transitive) if for all i ̸= i′ and j ̸= j′ in Ω, there exists σ ∈G suchthat σ(i) = j and σ(i′) = j′.

Quite generally, G is l-transitive for a positive integerl ≤n, if for all pairwise distinct elements i1, i2, . .

. , il in Ωand pairwise distinctelements j1, j2, .

. .

, jl in Ω, there exists σ ∈G such that σ(ie) = je for 1 ≤e ≤l.This brings us to MTR, i.e., the method of4. Throwing away rootsAssuming f to be irreducible in K[Y ], let us “throw away” a root of f, say α1,and getf1 = f1(Y ) =f(Y )(Y −α1) = Y n−1 + b1Y n−2 + · · · + bn−1 ∈K(α1)[Y ].In continuation of (1), we see that.

(2) f and f1 are irreducible in K[Y ] and K(α1)[Y ] respectively iffGal(f, K) is2-transitive.It may be noted that, assuming f to be irreducible, it does not matter whichroot of f we throw away; for instance, the irreducibility of f1 in K(α1)[Y ] and, upto isomorphism, the Galois group Gal(f1, K(α1)) are independent of which root wecall α1.Likewise, by throwing away s roots of f0 = f we getfs = fs(Y ) =f(Y )(Y −α1) · · · (Y −αs)= Y n−s + d1Y n−s−1 + · · · + dn−s ∈K(α1, . .

. , αs)[Y ]and then:(3) fs is irreducible in K(α1, .

. .

, αs)[Y ] for 0 ≤s < l iffGal(f, K) is l-transitive.2Self-advice: Don’t be so scared of the term “acting.” It is simply the modern substitute for“permuting.”

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC35. Classification theoremsNow you would have thought that you could (easily or possibly) construct apolynomial f = f0, say of degree 20, such that f0, f1, .

. .

, f9 are irreducible whereasf10 is reducible. But No!

And that is the surprise of the century. You cannot!!

Sosays the CT, i.e., the recently established Classification Theorem of Finite SimpleGroups, which was a magnificent piece of “team work”. According to the staggeringstatistics as reported by Coach Gorenstein [G3], the CT took 30 years (1950–1980),100 authors, 500 papers, and 15000 pages!

Yet several hundred more pages arerequired to prove the implicationsCT ⇒CDT ⇒CTT ⇒CQT ⇒CFT ⇒CSTwhere CDT (resp: CTT, CQT, CFT, and CST) stands for the Classification The-orem of Doubly (resp: Triply, Quadruply, Fivefold, and Sixfold) transitive permu-tation groups.3 Promising to come back to CDT to CFT in a moment, let us stateCST. It simply says that the symmetric group Sn for n ≥6 and the alternatinggroup An for n ≥8 are the only sixfold transitive groups!!

Vis-a-vis equations,what we are saying is that if f is an irreducible polynomial of degree n > 7 suchthat f1, f2, f3, f4 and f5 are irreducible then so are f6, f7, . .

. , fn−3.4To take a first shot at CT, in addition to permutation groups, we should alsoconsider groups of matrices over the (Galois) Field GF(q) of q elements where q isa power of a prime.

So letGL(m, q) = the general linear group of degree m over GF(q)=the group of all nonsingular m by m matriceswith entries in GF(q).Here the multiplicative group GF(q)∗comes in two ways.Firstly, thinking ofscalar matrices, GF(q)∗becomes a normal subgroup (and, in fact, the center)of GL(m, q).Secondly, taking determinants we get a surjective homomorphismGL(m, q) →GF(q)∗. This motivates the definitionsPGL(m, q) =the projective general linear group of degree mover GF(q)= GL(m, q)/ GF(q)∗andSL(m, q) = the special linear group of degree m over GF(q)= ker GL(m, q) →GF(q)∗.3This is an expanded version of a lecture given at Walter Feit’s 60th birthday conference inOxford, England.In addition to Walter Feit, the audience included the group theorists PeterCameron, Michael Collins, Sandy Green, Graham Higman, Peter Neumann, Ron Solomon, andJohn Thompson; my talking group theory in this meet of topnotch group theorists was like carryingcoal to Newcastle, or bringing holy water to the Ganges!

!4Self-Challenge = challenge to the extollers of high school algebra: prove that by high schoolalgebra if you can! Of course we can simply decree CT be high school algebra!

!

4S. S. ABHYANKARCombining these two roles of GF(q)∗we getPSL(m, q) = the projective special linear group of degree mover GF(q)= SL(m, q)/(SL(m, q) ∩GF(q)∗).In group theory parlance5Lm(q) = the linear group of degree m over GF(q)= PSL(m, q).In yet another notationAn(q) = Ln+1(q) = PSL(n + 1, q).With these preliminaries,CT essentially says that An and An(q) are the only (finite) simple groups.Here we have to exclude “small” cases; namely, from the alternating group Anexclude n ≤4, and from An(q) exclude n = 1 and q ≤3 (we define An(q) onlyfor n ≥1).

Moreover, “essentially” means that with An(q) we have to includeits relatives and incarnations, to be discussed later. Finally, in addition to theseinfinite families, there are 26 “sporadics”, again to be discussed later.66.

Brief thirty year historyFirst, in the fifties, there was the fundamental work of Brauer [B] and Chevalley[Ch]. In 1962 this was followed by the path-breaking odd order paper of Feit andThompson [FT].

Then came the large team coached by Gorenstein [G4]. At anyrate, we are meeting here to felicitate our friend Walter Feit on the occasion of hisforthcoming sixtieth birthday.77.

Primitive groupsAs another example of the parallelism between group theory and theory of equa-tions, let us note that a permutation group G is said to be primitive if it is transitive,and the one-point stabilizer G1 of G is a maximal subgroup of G. Here we are as-suming G to be a subgroup of the symmetric group Sn acting on {1, 2, . .

., n}, andthen by definition, G1 = G ∩Sn−1 with Sn−1 = {σ ∈Sn : σ(1) = 1}. Just as it didnot matter which root of the irreducible equation we threw away, so in the presentsituation, if G is transitive then we may replace G1 by any Gi = {σ ∈G : σ(i) = i},which is called the stabilizer of i in G.

Clearly. (4) If f is irreducible and G = Gal(f, K) then G1 = Gal(f1, K(α1)).5In learning group theory, I am following the traditional Indian method: memorize things byheart and the meaning will eventually be revealed to you.

Moreover, every subject has its lingo.Thus GL, SL and PSL are the Tom, Dick, and Harry of group theory.6Does the number 26 vindicate the spread of the English language which has exactly thatmany letters?7My fondest memory of Walter is that in 1957, when we were both at Cornell, we decided togo on a diet together and the one who lost more weight was to get a quarter. At the end of onemonth, Walter gained one pound and I gained two.

Who lost and who won?

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC5Moreover;(5) Gal(f, K) is primitive ifff is irreducible and there is no field between K andK(α1).Not to get completely lost in group theory, let us revert to algebraic geometryby talking about8. Fundamental groupsIn my 1957 paper on “Coverings of Algebraic Curves” in the American Journal[A3], I considered the algebraic fundamental group πA(C) of a nonsingular curve C.Here C is allowed to be “open”, i.e., it may consist of a projective algebraic curveminus a finite number of points.

We assume that C is irreducible and defined over analgebraically closed ground field k. By πA(C) we mean the family of Galois groupsGal(L/k(C)) as L varies over all finite normal extensions of the function field k(C)of C such that no point of C (equivalently, no valuation of k(C)/k having centeron C) is ramified in L.In case C is the (affine) line Lk over k, or more generally if C = Lk,r = theline Lk minus r points λ1, . .

. , λr (with λi ̸= λj in k for 1 ≤i < j ≤r) then thisamounts to considering Gal(F, k(X)) whereF = F(X, Y ) = Y n + φ1(X)Y n−1 + · · · + φn(X)is a bivariate polynomial with coefficients φ1(X), .

. .

, φn(X) in k[X] such that F isunramified at all (finite) values of X other than λ1, . .

. , λr, i.e., such that for everyλ in k different from λ1, .

. .

, λr we haveF(X + λ, Y ) =nYi=1(Y −η(i)λ (X))with η(i)λ (X) in the (formal) power series ring k[[X]]. We may call F an unramifiedcovering of Lk,r.For any group G, let Gh denote the family of finite homomorphic images of G.Let Fr be the free group on r generators, and let Jr be the family of all finitegroups generated by r generators, and note that then Jr = Frh.

By the RiemannExistence Theorem etc., we see that if k = the field of complex numbers C, thenfor any (irreducible) nonsingular curve C over k we have πA(C) = π1(C)h where,as usual, π1(C) denotes the (topological) fundamental group of C; see Serre [S1].Hence in particular πA(LC,r) = π1(LC,r)h, and clearly π1(LC,r) = Fr. ThereforeπA(LC,r) = Jr.By taking r = 0 or 1 in the last equation, we get πA(LC) = J0 = {1} andπA(LC,1) = J1 = the family of all finite cyclic groups.

These two facts can alsobe proved purely algebraically, by the genus formula due to Hurwitz–Riemann–Zeuthen; see Serre [S2]. The said genus formula actually shows thatπ∗A(Lk) = J0 = {1}andπ∗A(Lk,1) =J1[char k]∗= the family of all finite cyclic groups with ordernondivisible by char k,

6S. S. ABHYANKARwhere char k is the characteristic of k and where for any (irreducible) nonsingularcurve C over any algebraically closed field k we defineπ∗A(C) = the globally tame fundamental group of C= the family of all the members of πA(C) whose order isnondivisible by char kand for any nonnegative integer m and any family of finite groups J we putJ[m]∗= the family of all the members of J whose order is nondivisible by m.9.

Quasi p-groupsGiven any prime number p we putQ(p) = the family of all quasi p-groups,where by a quasi p-group we mean a finite group which is generated by all of itsp-Sylow subgroups, and for any finite group G we putp(G) = the (normal) subgroup of G generated by all of its p-Sylow subgroupsand for any family of finite groups J we putJ(p) = the family of all finite groups G such that G/p(G) ∈Jand we note that then J0(p) = Q(p) and for every nonnegative integer r we haveJr(p) =Jr[p]∗(p)= the family of all finite groups G such that G/p(G) is generatedby r generatorsand{p(G) : G ∈Jr(p)} = Q(p).10. ConjecturesFor any algebraically closed field k of characteristic p ̸= 0, in the above cited1957 paper, I conjectured that πA(Lk,r) = πA(LC,r)(p), i.e., equivalently,General Conjecture.

For every nonnegative integer r we have πA(Lk,r) = Jr(p).Hence in particularQuasigroup Conjecture. πA(Lk) = Q(p).Now a (finite) simple group whose order is divisible by p is obviously a quasip-group, and therefore QC (= the Quasigroup Conjecture) subsumes theSimple Group Conjecture.

πA(Lk) contains every simple group whose order isdivisible by p.In particular

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC7Alternating Group Conjecture. For every integer n ≥p we have An ∈πA(Lk)except when p = 2 < n < 5.Here, as usual, An denotes the alternating group of degree n, i.e., the group ofall even permutations on n symbols, and we note that the order of An is n!/2 or1 according as n > 1 or n = 1.

As proved by Galois, An is a simple group forevery n > 4, and hence AGC (= the Alternating Group Conjecture) is a specialcase of SGC (= the Simple Group Conjecture) except when (p, n) = (3,4); in thisexceptional case, An is obviously a quasi p-group and so we can directly fall backupon QC. The symmetric group Sn of degree n is obviously a quasi 2-group forevery n, and hence QC subsumes theEven Prime Symmetric Group Conjecture.

If p = 2 then for every integern ≥2 we have Sn ∈πA(Lk).To match up with EPSGC (= the Even Prime Symmetric Group Conjecture),let us divide AGC into EPAGC and OPAGC, i.e., into the following two conjecturesrespectively.Even Prime Alternating Group Conjecture. If p = 2 then for every integern ≥2 except when 2 < n < 5 we have An ∈πA(Lk).Odd Prime Alternating Group Conjecture.

If p > 2 then for every integern ≥p we have An ∈πA(Lk).11. Again some historyLet 0 ̸= a ∈k = an algebraically closed field of characteristic p ̸= 0, and let n, s, tbe positive integers such that t ̸≡0(p).

Now in support of the above conjectures,in the above cited 1957 paper, I had written down the following two examples ofunramified coverings of Lk:bFn = Y n −aXsY t + 1with n = p + tandeFn = Y n −aY t + Xs with t < n ≡0(p) and GCD(n, t) = 1 and s ≡0(t)and had suggested that their Galois groups bGn = Gal( bFn, k(X)) and eGn = Gal( eFn, k(X))should be calculated.Now, after a gap of thirty years, with Serre’s encouragement and with the helpof CT, I can calculate these Galois groups, and the answers are as follows. (I)(I.1) If t = 1, then bGn = PSL(2, p) = PSL(2, n −1).

(I.2) If t = 2 and p = 7, then bGn = PSL(2, 8) = PSL(2, n −1). (I.3) If t = 2 and p ̸= 7, then bGn = An.

(I.4) If t > 2 and p ̸= 2, then bGn = An. (I.5) If p = 2, then bGn = Sn.

8S. S. ABHYANKAR(II)(II.1) If 1 < t < 4 and p ̸= 2, then eGn = An.

(II.2) If 1 < t < n −3 and p ̸= 2, then eGn = An. (II.3) If 1 < t = n −3 and p ̸= 2 and 11 ̸= p ̸= 23, then eGn = An.

(II.4) If 1 < t < 4 < n and p = 2, then eGn = An or Sn. (II.5) If 1 < t < n −3 and p = 2, then eGn = An or Sn.Actually, bFn is a slight generalization of the original equationF n = Y n −XY t + 1with n = p + twritten down in the 1957 paper.

This equation F n was discovered by taking asection of a surface extracted from my 1955 paper “Ramification of Algebraic Func-tions” in the American Journal [A1], which was the second part of my Ph.D. Thesiswritten under the able guidance of Oscar Zariski. In the 1956 paper in the Annalsof Mathematics [A2], which was the first part of my Ph.D Thesis, I proved reso-lution of singularities of algebraic surfaces in nonzero characteristic.

In the 1955American Journal paper, I was showing why Jung’s method of surface resolutionin the complex case does not generalize to nonzero characteristic, because the localfundamental group, above a normal crossing of the branch locus, in the former caseis abelian whereas in the latter case it can even be unsolvable. It was a surfaceconstructed for this purpose whose section I took in the 1957 paper.Although the second equation eFn is also a slight generalization of an equationoccuring in the 1957 paper but, amusingly, it got rediscovered in 1989 as a variationof the first equation bFn.Now (I.1) was originally proved by Serre and when he told me about it in Sep-tember 1988, that is what started offmy calculations after a thirty year freeze!

Asa slight generalization of (I.1), in the case of t = 1, I can also calculate the Galoisgroup bGn,q = Gal( bFn,q, k(X)) of the unramified covering of Lk given bybFn,q = Y n −aX−sY t + 1with n = q + t,where q is any positive power of p, and it turns out that(III)(III.1) If t = 1, then bGn,q = PSL(2, q) = PSL(2, n −1).Again, bFn,q is a slight generalization of the equationF n,q = Y n −XY t + 1with n = q + t,which also occurs in the 1957 paper. Note that thenF n = F n,p.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC912. Using MRTApplying MRT (= method of removing tame ramification through cyclic com-positums = so called Abhyankar’s Lemma) to the one-point stabilizer of bFn+1 weget the monic polynomial of degree n in Y with coefficients in k(X) given bybF ′n = h(Y )(Y + b)p −aX−sY twith 0 ̸= b ∈k,where h(Y ) is the monic polynomial of degree n−p in Y with coefficients in k givenbyh(Y ) = (Y + n + 1)n+1−p −Y n+1−p(n + 1)2and we let bG′n = Gal( bF ′n, k(X)).

As an immediate consequence of (I) we now get(IV)Assuming that n + 1 ̸≡0(p), in the following cases bF ′n gives an unramifiedcovering of Lk with the indicated Galois group. (IV.1) If n + 1 −p = t > 2 ̸= p and b = t and s ≡0(p −1) and s ≡0(t), thenbG′n = An.

(IV.2) If n + 1 −p = t = 2 and p ̸= 7 and b = t and s ≡0(p −1), then bG′n = An. (IV.3) If n = p + 1 and p > 5, then t can be chosen so that 1 < t < p+12andGCD(p+1, t) = 1, and for any such t, upon assuming b =tt−1 and s ≡0(t(p+1−t)),we have bG′n = An.

(IV.4) If n + 1 −p = t and p = 2 and b = t and s ≡0(t), then bG′n = Sn.13. Unramified coveringsWe have the following four corollaries of calculations (I) through (IV).In Calculations (I) through (IV), we use a lot of Ramification Theory, or, equiv-alently, CS (= Cycle Structure).

In addition to CS + MTR + MRT, in the proofsof Calculations (I) to (IV) we also use CT. In our original version of these proofs,the use of CT was heavy.

Gradually the use of CT decreased, but could not beremoved completely. However, by traversing a delicate path through calculations(I) through (IV), we have arranged a proof of the First and the Second Corollariesindependent of CT.First Corollary.

OPAGC is true. Equivalently, for any n ≥p > 2, there existsan unramified covering of the affine line in characteristic p whose Galois group isthe alternating group An of degree n.Second Corollary.

EPSGC is true. Equivalently, for any n ≥p = 2, there existsan unramified covering of the affine line in characteristic p whose Galois group isthe symmetric group Sn of degree n.Third Corollary.

Unramified coverings of the affine line in characteristic p witha few more Galois groups have been constructed.

10S. S. ABHYANKARDefinition-Remark.

By the minimal index of a finite group G we mean thesmallest number d such that G has a subgroup of index d which does not containany nonidentity normal subgroup of G.8 Now QC is obviously equivalent to sayingthat for every integer d ≥p the following is true.QC(d). Every quasi p-group of minimal index d belongs to πA(Lk), i.e., occurs asthe Galois group of an unramified covering of the affine line in characteristic p.Therefore it is interesting to point out that as a corollary of the above resultswe have the following:Fourth Corollary.

QC(p+1) is true for every p which is not a Mersenne prime9and which is different from 11 and 23. More generally, if G is a quasi p-groupcontaining a subgroup H of index p+1 such that H does not contain any nonidentitynormal subgroup of G, and if p is not a Mesenne prime and p is different from 11and 23, then there exists an unramified covering of the affine line in characteristicp having G as the Galois group.The proofs of the above four Corollaries and the four claims (I) to (IV) will becompletely given in this paper with the exception that the proof of claim (I.2) willbe completed in my forthcoming paper [A7].

In connection with (II.4), it may benoted that the case of p = 11 or 23 is still open.1014. History of a pilgrimageWhen I said that “Now...I can calculate these Galois groups”, what I reallymeant was that, from September 1988 to August 1989, I undertook a pilgrimage(physical as well as mental)11 to seek the help of lots of mathematicians, and then Isimply collated the help so obtained.

In chronological order, these mathematicianswere. Serre (oh yes, very much Serre), Kantor, Feit, Cameron, Sathaye, Eakin,Stennerson, Gorenstein, O’Nan, Mulay, and Neumann.The pilgrimage started when in September and October of 1988, Serre sent meone after another four long letters briefly saying that“In your 1957 paper you suggested that the Galois group of F n should be calcu-lated.

I can now prove that for t = 1 it is PSL(2, p). Can you calculate it for othervalues of t?

Also, the conjectures in your paper include AGC. Can you now proveAGC?”Fortunately, in his last letter, Serre added a sentence saying that “my e-mailis... .”8In other words, we are minimizing the index over subgroups of G which do not contain anyminimal normal subgroups of G, where we recall that a minimal normal subgroup of a group Gis a nonidentity normal subgroup N of G such that N does not contain any nonidentity normalsubgroup of G other than N itself.9A Mersenne prime is a prime p of the form p = 2µ −1 for some positive integer µ.

Note thatthen µ is necessarily prime, because otherwise by factoring µ = µ′µ′′ with µ′ > 1 and µ′′ > 1,we would get a factorization 2µ −1 = l′l′′ with l′ = 2µ′ −1 > 1 and l′′ = 1 + 2µ′ + 22µ′ + · · · +2µ′(µ′′−1) > 1.10As will become apparent later, the reason for this, as well as for the exclusion of these valuesof p from the Fourth Corollary, is the existence of the “Mathieu Groups”.11mental = e-mail + s-mail. s-mail = snail mail = usual mail.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC1115. Multiply transitive groupsHaving already commented on the significance of transitivity for Galois theory,before proceeding further with calculations of Galois groups, let us give a briefreview of multiply transitive groups.So let G be a permutation group, say of degree n, i.e., let G act on Ω={1, 2, .

. ., n}.

In analogy with the concept of transitivity introduced in §3, we saythat G is λ-antitransitive (or λ-fold antitransitive) for a positive integer λ ≤n, if forall pairwise distinct elements i1, i2, . .

. , iλ in Ωwe have that the identity is the onlymember of G which keeps them fixed.12 Moreover, for positive integers l ≤λ ≤n,we say that G is (l, λ)-transitive if G is l-transitive and λ-antitransitive; we mayexpress this by simply saying that G is (l, λ).13 Finally, G is sharply l-transitivemeans G is (l, l).

Now if G is l-transitive, with l > 1, then the one-point stabilizerof G is obviously (l−1)-transitive as a permutation group of degree n−1, acting onthe “remaining” n −1 elements; conversely, if G is transitive and its one-point sta-bilizer is (l −1)-transitive then G is l-transitive. Similarly, if G is λ-antitransitive,with λ > 1, then the one-point stabilizer of G is (λ −1)-antitransitive as a permu-tation group of degree n −1, acting on the “remaining” n −1 elements; conversely,if G is transitive and its one-point stabilizer is (l −1)-antitransitive then G is l-antitransitive.

Thus, to classify all (l, λ) groups, we can make induction and eachtime increase l, λ, and n by one. So we start with (1, 1).By definition, G is regular means G is (1, 1).

Now the classification of (1, 1)groups is either obvious or impossible. Obvious because it is so easy to define whatis (1, 1).

Indeed, the usual proof of the usual theorem which says that every finitegroup is a permutation group, amounts to representing the given group as a regularpermutation group.14 Hence impossible because it would amount to classifying allfinite groups.G is Frobenius means G is (1, 2) but not (1, 1). A prototype of a Frobenius groupis the group of all affine linear transformations ax + b with a, b in a finite field, ormore generally in a finite near-field.

Zassenhaus [Z2], in his 1936 Thesis writtenunder Artin, proved that the converse is true for (2, 2) groups.Zassenhaus’ Theorem. G is (2, 2) ⇔G = AGLNF(1, Ψ) for some finite near-field Ψ.Here, by AGLNF(1, Ψ) we are denoting the group of all affine linear transfor-mations of degree 1 over the near-field Ψ, where “near-field” is a generalizationof “field” obtained by weakening the distributive law.

Namely, a near-field is anadditive abelian group Ψ in which the nonzero elements form a multiplicative groupsuch that for all a, b, c in Ψ we have a(b + c) = ab + ac. Thus we are not assumingthe other distributive law (b + c)a = ba + bc which is not a consequence of the firstdistributive law because the multiplication is not required to be commutative.

Itcan easily be seen that, as in the case of a field, the number of elements in anyfinite near-field is a power of a prime number. By an affine linear transformation12That is, if σ ∈G is such that σ(ie) = ie for 1 ≤e ≤λ then we must have σ = 1.13Note that if G is l-transitive and m-antitransitive for positive integers l ≤n and m ≤n,then automatically l ≤m.

Also note that if G is l-transitive for some positive integer l ≤n thenG is l′-transitive for every positive integer l′ ≤l. Likewise, if G is λ-antitransitive for a positiveinteger λ then G is λ′-antitransitive for every positive integer λ′ ≥λ with λ′ ≤n.14By making G act on itself by right or left (but not both) multiplication.

12S. S. ABHYANKARof degree 1 over the near-field Ψ we mean a map Ψ →Ψ given by x 7→ax + b witha, b in Ψ and a ̸= 0.

It can easily be seen that distinct (a, b) give distinct mapsΨ →Ψ. A proof of Zassenhaus’ Theorem is given in 9.10 on page 424 of volume IIIof Huppert-Blackburn [HB].15 In the proof of this Theorem, as well as in the proofsof various other theorems on multitransitive groups, an important role is played bythe following Theorem of Frobenius (1901) [Fr] for a proof of which we refer to 8.2on page 496 of volume I of Huppert–Blackburn [HB].Frobenius’ Theorem.

A Frobenius group G always has a (1, 1) normal subgroup.More precisely, the subset of G consisting of the identity together with those elementsσ which fix no letter (i.e., σ(i) ̸= i for i = 1, 2, . .

., n) forms a regular normalsubgroup of G.In the notation AGLNF(1, Ψ), the letters NF are meant to remind us of a near-field. In case Ψ = a field Φ, we may write AGL(1, Φ) instead of AGLNF(1, Ψ).To put the notation AGL(1, Φ) in proper perspective, first we remark that fora field Φ and a positive integer m the groups GL(m, Φ), PGL(m, Φ), SL(m, Φ), andPSL(m, Φ) are defined by replacing GF(q) by Φ in §5.16 Note that Z(GL(m, Φ)) =the set of all scalar matrices, and Z(SL(m, Φ)) = SL(m, Φ) ∩Z(GL(m, Φ)), wherethe center of any group Γ is denoted by Z(Γ), i.e., Z(Γ) is the normal subgroupof Γ given by putting Z(Γ) = {a ∈Γ : ab = ba for all b ∈Γ}.

Now a nonsingularm by m matrix α ∈GL(m, Φ) corresponds to the bijection Φm →Φm which sendsany 1 by m matrix ξ ∈Φm to the matrix product ξα ∈Φm.In this manner,the group GL(m, Φ), and hence also the subgroup SL(m, Φ), may be regarded as apermutation group on Φm. Let P(Φm) be the (m −1)-dimensional projective spaceover Φ, where we think of P(Φm) as the set of all one-dimensional subspaces of Φm.Now the bijection Φm →Φm corresponding to any α ∈GL(m, Φ) clearly induces abijection P(Φm) →P(Φm); moreover, if α∗∈GL(m, Φ) differs from α by a scalarmatrix, then α and α∗induce the same bijection P(Φm) →P(Φm).

Thus the groupPGL(m, Φ), and hence also the subgroup PSL(m, Φ), becomes a permutation groupon P(Φm). Members of PGL(m, Φ) are called projective transformations of P(Φm).Note that for any γ ∈PGL(m, Φ) and ζ ∈P(Φm) we have γ(ζ) ∈P(Φm).Now the affine general linear group AGL(m, Φ) of degree m over Φ may beintroduced as the semidirect product Φm ⋊GL(m, Φ) of Φm by GL(m, Φ) withthe obvious action of GL(m, Φ) on Φm, where we recall that a group Γ is saidto be the (internal) semidirect product of a normal subgroup Θ by a subgroup ∆provided Γ = Θ∆and Θ ∩∆= 1, and we note that in this case ∆acts on Θ byconjugation.

Concretely, AGL(m, Φ) may be regarded as the set of all m by m + 1matrices whose entries are in the Φ and whose m by m piece is nonsingular; in otherwords, we think of AGL(m, Φ) as the set of all pairs (α, β) with α ∈GL(m, Φ) andβ ∈Φm, where multiplication is defined by (α, β)(α′, β′) = (αα′, βα′ + β′).17 To(α, β) ∈AGL(m, Φ) there corresponds the bijection Φm →Φm which sends every15Although Dickson found all finite near-fields, it was left to Zassenhaus to prove that therewere no more. E. H. Moore [Mo] of the newly opened University of Chicago classified finite fields,around 1895, and then, around 1905, his students Wedderburn [W] and Dickson [D2] studied skewfields and near-fields respectively.16Writing q for GF(q) is justified because for any prime power q, up to isomorphism, there isexactly one field with q elements.17This is an example of an “external” semidirect product.

For further elucidation see Suzuki[Su2], Huppert-Blackburn [HB], and Wielandt [Wi]. A lot of the group theory background, re-quired in this paper, I learned in the last two years from these nice books.

I highly recommend

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC13ξ ∈Φm to ξα + β ∈Φm.18 Thus AGL(m, Φ) also becomes a permutation group onΦm. Members of AGL(m, Φ) are called affine transformations of Φm.19To make another example of a semidirect product, let Aut Φ be the group of allautomorphisms of the field Φ.

For every matrix γ with entries in Φ and for every g ∈Aut Φ, let γg be the matrix obtained by applying g to each entry of γ. This gives anaction of Aut Φ on GL(m, Φ).

The semidirect product GL(m, Φ)⋊Aut Φ is denotedby Γ L(m, Φ) and members of Γ L(m, Φ) are called semilinear transformations ofΦm. A member of Γ L(m, Φ) may be thought of as a pair (g, α) with g ∈Aut Φ andα ∈GL(m, Φ), and the corresponding bijection Φm →Φm sends every ξ ∈Φm toξgα ∈Φm.

The multiplication in Γ L(m, Φ) is given by (g, α)(g′, α′) = (gg′, αg′α′).Thus Γ L(m, Φ), which may be called the semilinear group of degree m over Φ,acts on Φm. With this action we can form the semidirect product A Γ L(m, Φ) =Φm ⋊Γ L(m, Φ) and call it the affine semilinear group of degree m over Φ.Amember of A Γ L(m, Φ) is called an affine semilinear transformation of Φm and itmay be represented as a triple (g, α, β) with g ∈Aut Φ, α ∈GL(m, Φ), β ∈Φm; thecorresponding bijection Φm →Φm sends every ξ ∈Φm to ξgα + β ∈Φm, and themultiplication in A Γ L(m, Φ) is given by (g, α, β)(g′, α′, β′) = (gg′, αg′α′, βg′α′+β′).Thus A Γ L(m, Φ) is a permutation group on Φm, and in a natural manner we haveSL(m, Φ) ⊳GL(m, Φ) ⊳Γ L(m, Φ) < A Γ L(m, Φ)andGL(m, Φ) < AGL(m, Φ) < A Γ L(m, Φ),where < and ⊳denote subgroup and normal subgroup respectively.To construct one more example of a semidirect product, we note that the ac-tion of Aut Φ on GL(m, Φ) obviously induces an action of Aut Φ on the factorgroup PGL(m, Φ).With this induced action we form the semidirect productP Γ L(m, Φ) = PGL(m, Φ) ⋊Aut Φ and call it the projective semilinear group ofdegree m over Φ.

A member of P Γ L(m, Φ) is called a projective semilinear transfor-mation of P(Φm) and it may be represented by a pair (γ, g) with γ ∈PGL(m, Φ) andg ∈Aut Φ; the corresponding bijection P(Φm) →P(Φm) sends every ζ ∈P(Φm) toγ(ζg) ∈P(Φm) where ζg ∈P(Φm) is given by ζg = {ξg : ξ ∈ζ}. Thus P Γ L(m, Φ)becomes a permutation group on P(Φm), and in a natural manner we havePSL(m, Φ) ⊳PGL(m, Φ) ⊳P Γ L(m, Φ).Although we have spoken of SL(m, Φ), GL(m, Φ), Γ L(m, Φ), AGL(m, Φ), A Γ L(m, Φ)as permutation groups on Φm, and PSL(m, Φ), PGL(m, Φ), P Γ L(m, Φ) as permu-tation groups on P(Φm), this is relevant mainly when Φ = GF(q) for some primethem.

It may be noted that a Frobenius group G is the semidirect product of the Frobenius kernelof G by a Frobenius complement of G, where by the Frobenius kernel of G we mean the (1, 1)normal subgroup of G, and by a Frobenius complement of G we mean a 1-point stabilizer of G.18In the above case of m = 1, this reduces to x 7→ax + b by taking x = ξ ∈Φ, 0 ̸= a = α ∈Φ,and b = β ∈Φ.19Transformations, or substitutions, of the type x′ = ax + by + c and y′ = a′x + b′y + c′ arefamiliar to us from high school. By adding “points at infinity” to the ordinary plane we get theprojective plane.

To distinguish between the ordinary plane and the projective plane, the ordinaryplane is called the affine plane and the above transformations are called affine transformations ofthe plane Φ2. To know that they can be redefined in terms of semidirect products should help tomake this notion friendly.

14S. S. ABHYANKARpower q and in that case we may write SL(m, q),GL(m, q), Γ L(m, q), AGL(m, q),Γ L(m, q), PSL(m, q), PGL(m, q), and P Γ L(m, q) for SL(m, Φ), GL(m, Φ), Γ L(m, Φ),AGL(m, Φ), A Γ L(m, Φ), PSL(m, Φ), PGL(m, Φ), and P Γ L(m, Φ) respectively.Henceforth by a permutation group we shall again mean a permutation group ona finite set.16.

Zassenhaus groupsHaving talked about (1, 1), (1, 2), and (2, 2) groups, let us now discuss (2, 3)groups which are not (2, 2). Basically they fall into the following three classes.

(i) A Feit group is defined to be a (2, 3) group which is not (2, 2) but has a (1, 1)normal subgroup. As a prototype we have the group A Γ L(1, 2p) where p is a primenumber.

This consists of all transformations x 7→axg + b with 0 ̸= a ∈GF(2p) andb ∈GF(2p) and g ∈Aut GF(2p). Now | GF(2p)| = 2p and |Aut GF(2p)| = p where| | denotes cardinality.

Thus A Γ L(1, 2p) is a permutation group of degree 2p andorder 2p(2p −1)p.(ii) A sharp Zassenhaus group is defined to be a (3, 3) group; such a group isclearly a (2, 3) group; moreover, it is a (2, 2) group only when its degree is 3 and inthat case it is simply S3. The Fundamental Theorem of Projective Geometry saysthat, on the projective line over a field Φ, any three points can be sent to any otherthree points by one and only one projective transformation.

In case, Φ = GF(q),where q is a prime power,20 this amounts to saying that the group PGL(2, q) is a(3, 3) group, where we regard PGL(2, q) as a permutation group of degree q + 1;21it is easily seen that PGL(2, q) has no (1, 1) normal subgroup; clearly the orderof PGL(2, q) is (q + 1)q(q −1).22 It can be shown that, if q is an even power ofan odd prime, P Γ L(2, q) has exactly one 3-transitive subgroup, which we denoteby PML(2, q), such that PML(2, q) ̸= PGL(2, q) and PSL(2, q) is a subgroup ofPML(2, q) of index 2. Now PML(2, q) is also a (3, 3) group, where we again regardit as a permutation group of degree q+1; it is easily seen that PML(2, q) has no (1, 1)normal subgroup provided q ≥4; clearly the order of PML(2, q) is (q+1)q(q−1); wecall PML(2, q) the projective mock linear group of degree 2 over GF(q).

For furtherdiscussion about PML(2, q) see page 163 of volume III of Huppert-Blackburn [HB]where it is denoted by M(q). (iii) A strict Zassenhaus group is defined to be a (2, 3) group which is neither(2, 2) nor (3, 3) and does not have any (1, 1) normal subgroup.23 It can easily beseen that, for any odd prime power q, the group PSL(2, q) is a strict Zassenhausgroup of degree q + 1.

To find the order of this group, we might as well start bycalculating the order of GL(m, q) for any positive integer m and any prime powerq which need not be odd. Now the number of ways of choosing the first column of20That is q is a positive integral power of a prime number.

Clearly then q is an odd primepower or an even prime power according as the corresponding prime number is even or odd. Whenq is an odd prime power, it can be an even power of an odd prime or an odd power of an oddprime.21PGL(2, q) acts on the projective line over GF(q) which has q + 1 points on it, out of whichq are at “finite distance” and one is the point at “infinity”.22Obviously the order of any (l, l) group of degree n is n(n −1) · · · (n −l + 1).23In Gorenstein’s book [G1] every (2, 3) group which is not (2, 2) is called a Zassenhaus group.In Huppert and Blackburn’s book [HB] every (2, 3) group which is not (2, 2) and does not haveany (1, 1) normal subgroup is called a Zassenhaus group.

We are calling the groups mentioned in(i) Feit groups because they were completely characterized by Feit [F] in 1960.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC15an element of GL(m, q) equals | GF(q)m| −1 = qm −1. The number of multiplesof the first column is q and hence, having chosen the first column, the number ofways of choosing the second column equals | GF(q)m| −| GF(q)| = qm −q.

Moregenerally, the first i columns generate an i-dimensional vector space over GF(q)and hence, having chosen the first i columns, the number of ways of choosing the(i + 1)th columns equals | GF(q)m| −| GF(q)i| = qm −qi. Therefore| GL(m, q)| = (qm −1)(qm −q) · · · (qm −qm−1).Consequently| AGL(m, q)| = | GF(q)m|| GL(m, q)| = qm(qm −1)(qm −q) · · · (qm −qm−1).Now | GF(q)∗| = q−1, where GF(q)∗is the multiplicative group of nonzero elementsof GF(q), and we have obvious exact sequences of finite groups 1 →SL(m, q) →GL(m, q) →GF(q)∗→1 and 1 →GF(q)∗→GL(m, q) →PGL(m, q) →1, andhence| PGL(m, q)| = | SL(m, q)| = | GL(m, q)|/(q −1)= (qm −1)(qm −q) · · · (qm −qm−2)qm−1.LetZν =the cyclic group of order ν where ν is either a positiveinteger or ∞.Now clearlyAut GF(q) = Zµ where q = pµ with p = char kand hence|Γ L(m, q)|/| GL(m, q)| = | A Γ L(m, q)|/| AGL(m, q)|= | P Γ L(m, q)|/ PGL(m, q)| = µ.Let SL(m, q)∗be the group of all m by m scalar matrices whose entries are in GF(q)and whose determinant is 1.

Then SL(m, q)∗is isomorphic to the group of all mthroots of 1 in GF(q) and hence | SL(m, q)∗| = GCD(m, q −1). Also we have anobvious exact sequence of finite groups 1 →SL(m, q)∗→SL(m, q) →PSL(m, q) →1 and hence| PSL(m, q)| = | SL(m, q)|/ GCD(m, q −1)= | PGL(m, q)|/ GCD(m, q −1).So in particular| PSL(2, q)| = (q + 1)q(q −1)/2or(q + 1)q(q −1)according as q is odd or even i.e., according as q −1 is or is not divisible by 2.

Ifq is even then we cannot divide q −1 by 2, and so we do the best we can; namely,

16S. S. ABHYANKARassuming q to be a square of a proper odd power24 of 2, in the expression (q+1)q(q−1)/2 we replace (q −1)/2 by (q1/2 −1) to get the expression (q + 1)q(q1/2 −1),and now to get rid of the fractional power we write q in place of q1/2.

For every qwhich is a proper odd power of 2, the resulting expression (q2 + 1)q2(q −1) is theorder of a certain strict Zassenhaus group Sz(q) of degree q2 + 1; this group Sz(q)is isomorphic25 to a certain subgroup of GL(4, q) and, since it was discovered bySuzuki [Su1] in 1962, it is called the Suzuki group over GF(q). We may think ofSz(q) as ersatz PSL(2, q2); we have just given a heuristic reason for its existenceand a mnemonic device for remembering its order; to recapitulate| Sz(q)| = (q2 + 1)q2(q −1)if q is any odd power of 2.As hinted in the above order formula, the definition of the Suzuki group Sz(q) canbe extended so as to include the case of q = 2; in this case we still get a 2-transitivepermutation group of degree q2 + 1 and of the above order, which is however nota strict Zassenhaus group; indeed, as a permutation group, Sz(2) is isomorphic tothe (2, 2) group AGL(1, 5).The following theorem of Zassenhaus [Z1], Feit [F], and Suzuki [Su1] says thatthe above examples of (2, 3) groups which are not (2, 2) are exhaustive; for a proofsee 1.1 and 11.16 on pages 161 and 286 of volume III of Huppert-Blackburn [HB].Zassenhaus-Feit-Suzuki Theorem.

For a permutation group G we have thefollowing. (1) G is Feit ⇔G = A Γ L(1, 2p) for some prime p.(2) G is sharp Zassenhaus ⇔G = PGL(2, q) for some prime power q, or G =PML(2, q) for some even power q of an odd prime.

(3) G is strict Zassenhaus ⇔G = PSL(2, q) for some odd prime power q, orG = Sz(q) for some proper odd power q of 2.Moreover, this gives an exhaustive and mutually exclusive listing of (2, 3) groupswhich are not (2, 2), with the proviso that the group PSL(2, 2) = S3 is included initem (2) even though it is a (2, 2) group in addition to being a (3, 3), and hence a(2, 3), group.17. More about classification theoremsBy analyzing the Suzuki group Sz(q), a certain analogous 2-transitive permuta-tion group R1(q) of degree q3 + 1, for every odd power q of 3, was discovered byRee [R] in 1964; although R1(q) is 2-transitive, it is not 3-antitransitive; the groupR1(q) is defined in terms of some 7 by 7 matrices over GF(q) and is called the Reegroup over GF(q); the order of R1(q) is given by|R1(q)| = (q3 + 1)q3(q −1)if q is any odd power of 3.For every odd power q of 2, in terms of certain matrices over GF(q), Ree defined agroup R2(q) which is also called the Ree group over GF(q); the order is now givenby|R2(q)| = q12(q6 + 1)(q4 −1)(q3 + 1)(q −1)24That is with an odd exponent > 1.25Not as a permutation group.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC17if q is any odd power of 2. All the Suzuki groups and all the Ree groups turn outto be a simple groups except: Sz(2) is solvable; R1(3) is isomorphic to P Γ L(2, 8)and hence the simple group PSL(2, 8) may be regarded as a normal subgroup ofR1(3) of index 3 and then PSL(2, 8) is the unique minimal normal subgroup ofR1(3); the commutator subgroup R2(2)′ of R2(2) is a (normal) subgroup of R2(2)of index 2, and the said commutator subgroup R2(2)′ is actually a simple groupcalled the Tits group.

Thus we have the following three infinite families of finitesimple groups deduced from matrices over GF(q).Suzuki and Ree groups. Sz(q) for every proper odd power q of 2.

R1(q) forevery proper odd power q of 3. R2(q) for every proper odd power q of 2, togetherwith the commutator subgroup R2(2)′.Just before Suzuki and Ree found these groups, Steinberg [St], in 1959, discoveredthat some known groups together with some further ones suggested by them couldbe organized into four other infinite families of finite simple groups which are definedin terms of matrices over GF(q2) and which are now labeled as follows.Steinberg groups.2An(q), 2Dn(q), 3D4(q), and 2E6(q).

Here n is any positiveinteger and q is any prime power except:in case of2An(q) exclude (n, q) =(1, 2), (1, 3), (2, 2), and in case of 2Dn(q) exclude n = 1.These 3+4 = 7 families are the “twisted incarnations” of the original nine infinitefamilies of “relatives” of An(q). These nine infinite families of finite simple groupsdefined in terms of matrices over GF(q) were systematized by Chevalley [Ch] in1955, and they are labeled as follows.26Chevalley groups.

An(q), Bn(q), Cn(q), Dn(q), E6(q), E7(q), E8(q), F4(q), and G2(q).Again here n is any positive integer and q is any prime power except: in case ofAn(q) exclude (n, q) = (1, 2), (1, 3); in case of Bn(q) and Cn(q) exclude (n, q) =(1, 2), (1, 3), (2, 2); in case of Dn(q) exclude n < 4; and in case of G2(q) excludeq = 2.Thus we have 7 + 9 = 16 infinite families of finite simple groups defined in termsof matrices over finite fields. It may be noted that, just as the group An(q) =PSL(n + 1, q) is obtained by projectivizing the special linear group SL(n + 1, q),the group Cn(q) = PSp(2n, q) is obtained by projectivizing the “symplectic group”Sp(2n, q), and the groups Bn(q) = P Ω(2n + 1, q) and Dn(q) = P Ω+(2n, q) areobtained by projectivizing the “commutator groups” Ω(2n + 1, q) and Ω+(2n, q)of the “orthogonal groups” O(2n + 1, q) and O+(2n, q) respectively.

The groupsAn(q), Bn(q), Cn(q),Dn(q) are collectively called “classical groups”, and the re-maining Chevalley groups E6(q), E7(q), E8(q), F4(q), G2(q) are collectively called“exceptional groups.”Likewise, amongst the Steinberg groups, the group 2An(q) = PSU(n + 1, q)is obtained by projectivizing the “special unitary group” SU(n + 1, q), and thegroup 2Dn(q) = P Ω−(2n, q) is obtained by projectivizing the “commutator group”Ω−(2n, q) of the “orthogonal group” O−(2n, q). The Suzuki and Ree groups havethe alternative labels Sz(q) = 2B2(q), R1(q) = 2G2(q), and R2(q) = 2F4(q).

Thegroups 2An(q), 2B2(q), 2Dn(q), 3D4(q) may collectively be called the “twisted26When q = a prime p, many of these were already studied by Jordan [J1] in the last century.For general q, some of them were discussed by Dickson [D1] at the turn of the century. Likewisethe first two families of Steinberg groups were already known to Jordan and Dickson, while thelast two were independently found by Tits.

18S. S. ABHYANKARclassical groups”, and the groups 2E6(q), 2F4((q), 2G2(q) may collectively be calledthe “twisted exceptional groups.”Now the projective mock linear group PML(2, q) may also be called the pre-mathieu linear group of degree 2 over GF(q).

The reason for this nomenclature isthat, around 1865, Mathieu [Mat] found a transitive extension of PML(2, 9) whichis denoted by M11. By a transitive extension of a permutation group of degree nwe mean a transitive permutation group of degree n + 1 having the given group asa one-point stabilizer.

Note that M11 is a (4, 4) group of degree 11 and hence itsorder is 11 · 10 · 9 · 8. Mathieu also found a transitive extension of M11 which wedenote by M12.

Clearly M12 is a (5, 5) group of degree 12 and hence its order is12·11·10·9·8. The permutation groups M11 and M12 are called Mathieu groups ofdegree 11 and 12 respectively.

It can be shown that M12 has a noninner automor-phism α of order 2 such that α(M11) is 3-transitive of degree 12. The permutationgroup α(M11) may be regarded as an incarnation of the Mathieu group M11 andwe may denote it by cM11.

To recapitulate|M12| = 12 · 11 · 10 · 9 · 8and|cM11| = |M11| = 11 · 10 · 9 · 8.To introduce the remaining three groups discovered by Mathieu, let us first notethat for any positive integer m and any prime power q we have | GF(q)m| = qm andhence for the corresponding (m −1)-dimensional projective space we have|P(GF(q)m)| = (qm −1)/(q −1) = qm−1 + qm−2 + · · · + q + 1.So in particular |P(GF(4)3)| = 21 and hence PSL(3, 4), which acts onP(GF(4)3),is a permutation group of degree 21; it can easily be seen that it is 2-transitive butnot 3-transitive; by the above order formula we also get| PSL(3, 4)| = (43 −1)(43 −4)42/ GCD(3, 3) = 21 · 20 · 48.Mathieu obtained a transitive extension M22 of PSL(3, 4), a transitive extensionM23 of M22, and a transitive extension M24 of M23. Clearly M22 is a 3-transitivebut not 4-transitive permutation group of degree 22, M23 is a 4-transitive but not5-transitive permutation of degree 23, and M24 is a 5-transitive but not 6-transitivepermutation group of degree 24.

These groups are called Mathieu groups of degree22, 23, 24 respectively, and obviously their orders are27 |M22| = 22 · 21 · 20 · 48;|M23| = 23 · 22 · 21 · 20 · 48, and |M24| = 24 · 23 · 22 · 21 · 20 · 48. Note the strikingsimilarity between the numbers (2, 9) occuring in the “parent group” PML(2, 9) ofM11 and M12, and the numbers (3, 4) occuring in the “parent group” PSL(3, 4) ofM22, M23, and M24.

All of the five Mathieu groups M11, M12, M22, M23, and M24turn out to be simple groups. One hundred years after their discovery, during 1965to 1975, twenty-one other finite simple sporadic groups, i.e., those which do notnaturally fit in any infinite family, were discovered by various people; the largest ofthe 21 + 5 = 26 sporadic groups is called the monster and its order is246 · 320 · 59 · 76 · 112 · 133 · 17 · 19 · 23 · 29 · 31 · 41 · 47 · 59 · 71;for details see Gorenstein [G1].

We are now ready to state, of course without proof,the Classification Theorem of Finite Simple Groups.2827For a transitive permutation group G of degree n we clearly have |G| = n|G1| where G1 isa one-point stabilizer of G.28See Gorenstein [G2] or Aschbacher [As].At least one part of this extremely long proof,namely Mason’s paper on quasi-thin groups [Mas], is still to see the light of day!

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC19CT. The following is a complete list of finite simple groups.

(1) The cyclic group Zp for every prime p.(2) The alternating group An for every integer n ≥5. (3) The sixteen infinite families of “matrix” groups mentioned above.29(4) The twenty-six sporadics including the five Mathieus.Just as PSL(m, q) is the typical example of a finite simple group, soPGL(2, q)is the quintessential example of a 3-transitive permutation group.

As obvious vari-ations of this, additional 3-transitive permutation groups are obtained by takinginto account all the groups between PGL(2, q) and P Γ L(2, q) for every prime powerq, and all the groups between PML(2, q) and P Γ L(2, q) for every even power q ofan odd prime; both these types have degree q + 1. All these arise out of the factthat, by a projective transformation, any 3 points of a projective line can be sentto any other 3.

Going to higher dimensions, by a projective transformation, any4 points of a projective plane can be sent to any other 4, and any 5 points of aprojective 3-space can be sent to any other 5, and so on. This would give rise to4-transitives, 5-transitives, and so on.

But there is a flaw. 3 collinear points in aprojective plane, or in a projective 3-space, cannot be sent to 3 noncollinear points.4 coplanar points in a projective 3-space cannot be sent to 4 noncoplanar points.And so on.

Thus, because of questions of linear independence, for every integerm > 1 and every prime power q, the group PGL(m, q), instead of being (m + 1)-transitive, is only 2-transitive, unless every line contains only 2 distinct points, inwhich case it would be 3-transitive. Well, for q = 2, a line should contain only 2points.

But even that is so only in the affine case because then we don’t have thepoint at infinity. Thus, it is not PGL(m, 2) which is 3-transitive, but AGL(m, 2).We can see that AGL(m, 2) is, however, not 4-transitive unless m = 2 in which casewe actually have AGL(2, 2) = S4.NowAGL(m, 2) = GF(2)m ⋊GL(m, 2) = 2m · Lm(2),where a dot stands for the semidirect product symbol ⋊, and 2m stands for (Z2)m,i.e., for the direct product Z2 × Z2 × · · · × Z2 of m copies of Z2.

By the orderformula we have|L4(2)| = (24 −1)(24 −2)(24 −4)(24 −8) = 8!/2 = |A8|.Hence, by the philosophical principle that two finite simple groups of equal order areusually isomorphic, we expect that L4(2) ≈A8 where ≈stands for isomorphism,and this can, in fact, be easily proved. Note that L4(2) = GL(4, 2) is a 1-pointstabilizer of AGL(4, 2) and hence in this incarnation A8 is only 2-transitive;30 letus denote this incarnation by bA8.31In a natural manner, A7 may be regardedas a subgroup of A8 and then it turns out that the image of A7 under the said29I am using the more friendly term “matrix groups” instead of the awe inspiring “Lie typegroups.”30By taking the stabilizer at the origin, GL(m, q) becomes the 1-point stabilizer of AGL(m, q)for every integer m > 1 and every prime power q.

Likewise, by taking the stabilizer at the pointat infinity, AGL(1, q) may be regarded as the 1-point stabilizer of PGL(2, q) for every prime powerq.31That is, as a permutation group, bA8 = GL(4, 2) = PGL(4, 2) ⊂S15.

20S. S. ABHYANKARisomorphism is also only 2-transitive; let us denote the said image by bA7.32 Thecorresponding subgroup of 24 · bA8 = AGL(4, 2) may be denoted by 24 · bA7; this is a3-transitive but not 4-transitive permutation group of degree 24 and order 23 · 7!.33As a consequence of CT, it can be shown that there are no more 3-transitivepermutation groups other than those we have already listed.

In other words wehave the following detailed version of CTT, i.e., the Classification Theorem of TriplyTransitive Permutation Groups; this theorem was compiled from conversations withCameron, Neumann, and O’Nan.CTT or Refined Fundamental Theorem of Projective Geometry. Thefollowing is a complete list of 3-transitive permutation groups.

(1) For every prime power q, each group between PGL(2, q) andP Γ L(2, q) is a3-transitive permutation group of degree q + 1, and we have | PGL(2, q)| =| P Γ L(2, q)|/µ = (q +1)q(q −1) where q = pµ with p = char GF(q). Amongthese, PGL(2, 3) and P Γ L(2, 4) are the only groups which are 4-transitive,and for them we actually have PGL(2, 3) = S4 and P Γ L(2, 4) = S5.

(2) For every even power q of an odd prime, each group between PML(2, q) andP Γ L(2, q) is a 3-transitive but not 4-transitive permutation group of degreeq + 1, and we have | PML(2, q)| = | P Γ L(2, q)|/µ = (q + 1)q(q −1) whereq = pµ with p = char GF(q). (3) For every integer m > 1, the group AGL(m, 2) is a 3-transitive permutationgroup of degree 2m and order 2m(2m−1)(2m−2) · · · (2m−2m−1).

This is 4-transitive only for m = 2, and in that case we actually have AGL(2, 2) = S4. (4) The group 24 · bA7 is a 3-transitive but not 4-transitive permutation group ofdegree 24 and order 24 · 7!, and as a permutation group it is a subgroup of24 · bA8 = AGL(4, 2).

(5) The reincarnated Mathieu group cM11 is a 3-transitive but not 4-transitivepermutation group of degree 12 and order 11 · 10 · 9 · 8 = 12 · 11 · 10 · 6. (6) The Mathieu group M22 and its automorphism group Aut M22 are 3-transitivebut not 4-transitive permutation groups of degree 22 with |M22| = |Aut M22|/2 =22 · 21 · 20 · 48.

(7) The Mathieu groups M11 and M23 are 4-transitive but not 5-transitive per-mutation groups of degree 11 and 23 and order 11·10·9·8 and 23·22·21·20·48respectively. (8) The Mathieu groups M12 and M24 are 5-transitive but not 6-transitive per-mutation groups of degree 12 and 24 and order 12 · 11 · 10 · 9 · 8 and24 · 23 · 22 · 21 · 20 · 48 respectively.

(9) For every integer n ≥5, the alternating group An is an (n −2)-transitivebut not (n −1)-transitive permutation group of degree n and order n!/2. (10) For every integer n ≥3, the symmetric group Sn is an n-transitive but not(n + 1)-transitive permutation group of degree n and order n!.The above formulation of CTT obviously subsumes CQT, CFT, and CST.

Inturn the CTT is of course subsumed under the CDT which is given by Cameron [C]32In other words, bA7 is the image of A7 under some injective group homomorphism A7 →bA8 = GL(4, 2) ⊂S15, and bA7 is a 2-transitive but not 3-transitive permutation group of degree15.33As permutation groups, bA7 and 24· bA7 are independent of the injective group homomorphismA7 →bA8 we choose for defining bA7.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC21and Kantor [K2] and which, in addition to heavily using the CT, is based on theprevious work of Curtis-Kantor-Seitz [CKS], O’Nan [O], and others. The followingweaker version of CDT, dealing mainly with the degrees of 2-transitive permutationgroups which are not 3-transitive, was communicated to me by Cameron.Weak CDT.

Concerning the degrees of 2-transitive permutation groups we havethe following. (1) For every integer m > 1 and every prime power q, each group betweenPSL(m, q) and P Γ L(m, q) is a 2-transitive permutation group of degree|P(GF(q)m)| = (qm −1)/(q −1).

Out of these, only the groups listed initems (1) and (2) of CTT are 3-transitive. In case of m > 2, for each groupbetween PSL(m, q) and P Γ L(m, q), by considering the action on “hyper-planes” in P(GF(q)m), we get a second representation as a 2-transitive butnot 3-transitive permutation group of degree (qm −1)/(q −1).

(2) For every integer m > 2, the group Sp(2m, 2) has 2-transitive but not 3-transitive permutation representations of degrees 22m−1+2m−1 and 22m−1−2m−1. (3) For every prime power q = pµ > 2 with prime p, each group betweenPSU(3, q) and its automorphism group Aut PSU(3, q) has a 2-transitive butnot 3-transitive permutation representation of degree q3 + 1, and moreoverPSU(3, q) is a normal subgroup of index [GCD(3, q+1)]2µ in Aut PSU(3, q).34(4) For every proper odd power q = 2µ of 2, each group between the Suzukigroup Sz(q) and its automorphism group Aut Sz(q) has a 2-transitive butnot 3-transitive permutation group of degree q2 + 1, and moreover Sz(q) isa normal subgroup of index µ in Aut Sz(q).

(5) For every odd power q = 3µ of 3, each group between the Ree group R1(q)and its automorphism group Aut R1(q) has a representation as a 2-transitivebut not 3-transitive permutation group of degree q3 +1, and moreover R1(q)is a normal subgroup of index µ in Aut R1(q). (6) The group PSL(2, 11) has two distinct 2-transitive but not 3-transitive per-mutation representations of degree 11.35(7) The alternating group A7 has two distinct 2-transitive but not 3-transitivepermutation representations of degree 15; both are equivalent to isomor-phisms A7 →bA7 ⊂S15.

(8) The “Higman-Sims” sporadic group HS has two different 2-transitive butnot 3-transitive permutation representations of degree 176. (9) The “third Conway” sporadic group Co3 has a 2-transitive but not 3-transitivepermutation representation of degree 276.The above items (1) to (9) contain a complete list of 2-transitive but not 3-transitivepermutation groups having a nonabelian minimal normal subgroup.

The degree ofa 2-transitive but not 3-transitive permutation group G having an abelian minimalnormal subgroup is necessarily some power pm of some prime p; the said minimal34A centerless group G, i.e., a group G whose center is the identity, may be identified with itsinner automorphism group and hence may be regarded as a normal subgroup of its automorphismgroup Aut G.35Two permutation representations G →Sn and G →Sn′ of a (finite) group G are equivalentif they differ by an isomorphism Sn →Sn′ induced by a bijection between the underlying sets ofSn and Sn′; note that then automatically n = n′. Two representations are distinct if they are notequivalent.

22S. S. ABHYANKARnormal subgroup is isomorphic to (Zp)m and a 1-point stabilizer of the group Gitself is isomorphic to a subgroup of GL(m, p); moreover: in the case m = 1 wemust have G = AGL(1, p), whereas, in the case m = a prime number and p = 2we must have G = AGL(1, 2m) or A Γ L(1, 2m), and finally, in the case m = 2 andp = 3 we must have G = AGL(1, 9) or A Γ L(1, 9) or AML(1, 9) or AGL(2, 3) orASL(2, 3), where we have put AML(1, 9) = the 1-point stabilizer of PML(2, 9),36and ASL(2, 3) = (GF(3))2 ⋊SL(2, 3).In connection with the above statement, we note the following Theorem of Burn-side which is really the starting point of the classification of 2-transitive permutationgroups.

Although most modern proofs of this make use of Frobe- nius’ Theorem(1901) [Fr],37 it is clear that Burnside’s original proof did not, since it is alreadygiven as Theorem IX on page 192 of the first edition of [Bu] published in 1897. Inthe second edition of [Bu] published in 1911, it occurs as Theorem XIII on page202, and there Burnside gives two proofs of it, one using Frobenius and the otherwithout.38 Thus Burnside gives an “elementary” proof of the following Theoremwithout using “character theory.”39Burnside’s Theorem.

A 2-transitive permutation group has a unique minimalnormal subgroup. The said subgroup is either an elementary abelian group40 or anonabelian simple group.As a consequence of CTT and Weak CDT, and in view of a simple numericallemma, we have the following.Special CDT.

Given any prime p and any positive integer µ, concerning 2-transitive but not 3-transitive permutation groups of degree q + 1, where q = pµ, wehave the following. If µ = 1 and p is a Mersenne prime, then PSL(2, p), AGL(1, p+1) and A Γ L(1, p + 1) are the only such groups.

If µ = 1 but p is not a Mersenneprime, then PSL(2, p) is the only such group, except that for p = 2 this group is“accidentally” 3-transitive because it coincides with S3. If µ > 1 then, in addi-tion to the relevant groups listed in items (1), (3), (4), (5) of Weak CDT, the onlyother such groups are the groups AGL(1, 9), A Γ L(1, 9), AML(1, 9), AGL(2, 3),and ASL(2, 3), which occur when (µ, p) = (3, 2), and the group AGL(1, q +1) whichoccurs when q + 1 is a Fermat prime.41Here is the said36In other words, AML(1, 9) is the 2-point stabilizer of the Mathieu group M11.37For Frobenius’ Theorem there is no “character free” proof.

As examples of modern proofsof Burnside’s Theorem which seem to use Frobenius’ Theorem, see 12.4 on page 32 of Wielandt[Wi] and 7.12 on page 233 of volume III of Huppert-Blackburn [HB].38In Burnside’s classical style of writing, Theorem x means Theorem x together with thediscussion around it. In other words, although everything is proved, only some of the conclusionsare called theorems.

This “classical” style is quite different from the so called “Landau Style” ofSatz-Beweis-Bemerkung. In the classical style, you first discuss things and then suddenly say thatyou have proved such and such; in other words, the proof precedes the statement of a theorem.39And certainly without using CT!40A group is elementary abelian if it is isomorphic to (Zp)m for some positive integer m andsome prime p.41A Fermat prime is a prime of the form 2µ + 1 for some positive integer µ.

It follows thenthat µ must be a power of 2, because otherwise µ = µ′µ′′ where µ′ is even and µ′′ > 1 is odd andthis would give the nontrivial factorization 2µ + 1 = (2µ′ + 1)(2µ′(µ′′−1) −2µ′(µ′′−2) + · · · + 1).

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC23Simple Numerical Lemma. Given any primes p and π and any positive integersµ and u, such that pµ + 1 = πu, we have the following.

If p > 2, then µ = 1 andπ = 2 and u = a prime number, and so p must be a Mersenne prime. If p = 2 andu > 1, then µ = 3 and π = 3 and u = 2.

If p = 2 and u = 1, then µ = a power of2, and so π must be a Fermat prime.To see this, first suppose that p > 2; now p is odd and hence pµ + 1 is even andhence π = 2; since p is odd, we have p ≡1(4) or p ≡3(4) and hence p2 ≡1(4), andtherefore if µ is even then 2u = pµ + 1 ≡2(4) and this would imply u = 1 whichwould be absurd; on the other hand, if µ is odd then 2u = pµ + 1 = (p + 1)(pµ−1 −pµ−2 + · · · + 1) where the second parenthesis contains an odd number of odd termsand hence its value is odd, but since that value divides 2u, it must be 1, and thisgives pµ+1 = p+1 which implies µ = 1. Next suppose that p = 2 and u > 1; now πmust be odd, and if u is also odd then 2µ = πu −1 = (π −1)(πu−1 + πu−2 + · · ·+ 1)where the second parenthesis consists of u positive odd terms and hence its valueis an odd number ≥u and, since it divides 2µ, it must be 1 which is absurd; on theother hand, if u is even then 2µ = πu −1 = (πu/2 −1)(πu/2 + 1) and hence πu/2 −1and πu/2 + 1 are both powers of 2 whose difference is 2 and therefore they must beequal to 2 and 4, and this gives µ = 3 and π = 3 and u = 2.

Finally, if p = 2 andu = 1, then by the last footnote we see that µ must be a power of 2, and π mustbe a Fermat prime.Here is another consequence of CDT.Uniqueness Theorem for Transitive Extensions. Any two transitive exten-sions of a transitive permutation group are isomorphic as permutation groups, withonly one exception.42 The exception is that PSL(2, 7) and A Γ L(1, 8) have a com-mon 1-point stabilizer; note that both these are 2-transitive but not 3-transitivepermutation groups of degree 8.As an immediate corollary of the above theorem we have the following.Uniqueness Theorem for Transitive Extensions of 2-Transitive Groups.

Anytwo transitive extensions of a 2-transitive permutation group are isomorphic as per-mutation groups. In particular, the Mathieu groups M11, M12, M22, M23, and M24are the unique transitive extensions of PML(2, 9), M11, PSL(3, 4), M22, and M23 re-spectively.To end this review of group theory, we note that by the rank of a transitivepermutation group is meant the number of orbits of its 1-point stabilizer; the lengthsof these orbits, excluding the obvious one point orbit, are called subdegrees of thegroup; so the number of subdegrees is one less than the rank, and the sum ofthe subdegrees is one less than the degree.

Thus a 2-transitive group is simply atransitive group of rank 2. Now CT has also been used by Kantor-Liebler [KL],Liebeck [L], and others, to give CR3 = classification of transitive groups of rank 3,which although much longer than CDT, should be quite useful for Galois theory.Here is an amusing sample from CR3 which does not use CT and which can befound in Kantor [K1].42If we don’t assume the given group to be transitive, then there are numerous exceptions.For example every finite group, in its standard representation as a regular permutation group,is a transitive extension of the identity group.

Since for increasing l, there are fewer and fewerl-transitive permutation groups, it follows that “most” transitive permutation groups have notransitive extensions.

24S. S. ABHYANKARSample from CR3.

For any integer n > 1 and any prime power q, the groupsPSp(2n, q) and O(2n + 1, q) are the only transitive permutation groups of rank 3whose subdegrees are q(q2n−2−1)/(q−1) and q2n−1. A rank 3 transitive permutationgroup G with subdegrees q(q + 1)2 and q4 for a prime power q > 1, is a subgroup ofAut PSL(4, q); moreover, if q > 2 then G contains PSL(4, q).18.

A type of derivativeTo continue with the calculation of Galois groups, let me explain how to throwaway a root α = α1 of a polynomialf = f(Y ) = Y n + a1Y n−1 + · · · + an =nYi=1(Y −αi)by using a type of derivative. Now the coefficients a1, .

. .

, an belong to a field K,and we want to find the polynomialf1 = f1(Y ) =f(Y )(Y −α) = Y n−1 + b1Y n−2 + · · · + bn−1 ∈K(α)[Y ].To this end, first recall the three basic transformations of equations described inany old book. For instance, we may quote the following three relevant articles (=sections) from Burnside-Panton’s 1904 book on the theory of equations [BP].43Art 31.To multiply the roots by a given quantity.

For any u ̸= 0, thepolynomial g = g(Y ) whose roots are u times the roots of f is given byg(Y ) = unfYu= Y n + c1Y n−1 + · · · + cn =nYi=1(Y −uαi)with ci = uiai.Art 32. To reciprocate the roots.

In case αi ̸= 0 for 1 ≤i ≤n, i.e., in casean ̸= 0, the polynomial g = g(Y ) whose roots are the reciprocals of the roots of fis given byg(Y ) = Y nanf 1Y= 1an(1 + a1Y + · · · + anY n) =nYi=1Y − 1α i.Art 33. To decrease the roots by a given quantity.

For any u, the polynomialg = g(Y ) whose roots are −u plus the roots of f is given byg(Y ) = f(Y + u) = Y n + c1Y n−1 + · · · + cn =nYi=1(Y −(αi −u))43For the last forty years I had happily assumed that this Burnside of the theory of equations[BP] was the same as the Burnside of the theory of groups of finite order [Bu]. To my dismay, atthe Oxford Conference in April 1990, Peter Neumann told me that, although both were namedWilliam and both obtained a D.Sc.

from Dublin around 1890, the equations Burnside was WilliamSnow whereas the group theory Burnside was simply William. Strangely, I first learnt group theoryfrom William Snow’s book on the theory of equations.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC25with c1 = a1 + nu, . .

. , cn = f(u).Now in the first and the third cases provided u ∈K, and in the second casewithout any proviso, we have g(Y ) ∈K[Y ] and, assuming the roots α1, .

. .

, αn tobe pairwise distinct, we have Gal(g, K) = Gal(f, K) as permutation groups, and sofor Galois theory purposes we may conveniently modify f by one or more of thesethree transformations.For example, sometimes it may be easier to compute the polynomial g1(Y ) ob-tained by decreasing the roots of f1 by α. In view of what we have just said, we getGal(g1, K(α)) = Gal(f1, K(α)) and hence, assuming f to be irreducible in K[Y ],for the one-point stabilizer G1 of G = Gal(f, K) we get G1 = Gal(g1, K(α)).Clearly g1 can also be obtained by first decreasing the roots of f by α to get thepolynomial g = g(Y ) = f(Y + α), and then throwing away the root Y = 0 of g;this gives g1(Y ) = g(Y )/Y , and now remembering that f(α) = 0 we getg1(Y ) = f(Y + α) −f(α)Y.According to the calculus definition, by taking the “limit” of the RHS as Y tendsto 0, we get f ′(α).

This motivates the following definition according to which g1turns out to be the twisted Y -derivative of f at α.Definition. For any polynomial θ = θ(Y ) in an indeterminate Y with coefficientsin a field L and for any element β in L, we call (θ(Y + β) −θ(β))/Y the twistedY -derivative of θ at β.For a moment let us denote the twisted Y -derivative of θ at β by θ′.

Then clearlyθ′ = θ′(Y ) is a polynomial in Y with coefficients in L, and if θ ∈L then θ′ = 0,whereas: if θ /∈L then θ′ ̸= 0 and the Y -degree of θ′ is 1 less than the Y -degree ofθ, and the two polynomials θ and θ′ have the same leading coefficient, and hencein particular, if θ is monic then so in θ′.Next we note that this is L-linear because for any δ = δ(Y ) ∈L[Y ] and λ, µ ∈Lwe have(λδ + µθ)′ = λδ(Y + β) + µθ(Y + β) −λδ(β) −µθ(β)Y= λδ′ + µθ′.However, the usual product rule is to be replaced by a twisted product rule becauseby the standard trick of adding and subtracting the same quantity we get(δθ)′ = δ(Y + β)θ(Y + β) −δ(Y + β)θ(β)Y+ δ(Y + β)θ(β) −δ(β)θ(β)Y= δ∗θ′ + δ′θ♯where δ∗is the Y -translation of δ by β and θ♯is the evaluation of θ at β, i.e.,δ∗= δ∗(Y ) = δ(Y + β) and θ♯= θ(β).Finally, for any positive integer m we have the power rule(Y m)′ = Y m−1 + mβY m−2 + · · · +miβiY m−i−1 + · · · + mβm−1

26S. S. ABHYANKARand, in case char L ̸= 0, for any power q of char L we have the prime power rule(Y q)′ = Y q−1 and combining this with the product rule, we get the power productrule[Y qθ(Y )]′ = (Y + β)qθ′(Y ) + Y q−1θ(β).The reason for explicitly mentioning Y in all this is that there may be other indeter-minates present; for instance, if ψ = ψ(X, Y ) is a polynomial in indeterminates Xand Y , and β is an element in a field which contains X as well as all the coefficientsof ψ, then the twisted Y -derivative of ψ at β is given by (ψ(X, Y ) −ψ(X, β))/Y .Reverting to the original situation by taking β = α and L = K(α) in the aboveset-up, we conclude with the following.Summary about the twisted derivative.

If f = f(Y ) is a nonconstant monicirreducible polynomial in an indeterminate Y with coefficients in a field K suchthat f has no multiple root in any overfield of K, and if α is a root of f in someoverfield of K, then by letting f ′ = f ′(Y ) to be the twisted Y -derivative of f atα we have that the Galois group Gal(f ′, K(α)) is the one-point stabilizer of theGalois group Gal(f, K).Now without assuming f to be irreducible and without any precondition aboutmultiple roots, suppose the degree of f is n > 2 and suppose for every root α off(Y ) in a splitting field of K we have that the twisted Y -derivative of f(Y ) at αis irreducible in K(α)[Y ], then f(Y ) must be devoid of multiple roots; namely, iff(Y ) = Qni=1(Y −αi) and α1 = α2 and f ′(Y ) is the twisted Y -derivative of f(Y )at α = α1, then f ′(Y ) is reducible in K(α)[Y ] because its degree is n −1 > 1 andit has (Y −α2) as a factor in K(α)[Y ]. Thus we have the following.Twisted Derivative Criterion.

If f(Y ) is a nonconstant monic polynomial ofdegree > 2 in an indeterminate Y with coefficients in a field K such that for everyroot α of f(Y ) in a splitting field of K we have that the twisted Y -derivative of f(Y )at α is irreducible in K(α)[Y ], then f(Y ) has no multiple roots in any overfield ofK.19. Cycle lemmaAs another tool for calculating Galois groups, let us make note of a “cyclelemma”.Let K be a field and consider a monic polynomialf = f(Y ) = Y n + a1Y n−1 + · · · + an =nYi=1(Y −αi)of degree n in an indeterminate Y with coefficients a1, .

. .

, an in K having pairwisedistinct roots α1, . .

. , αn in some overfield of K. Now by conveniently enlarging thesaid overfield and moving it by a K-isomorphism, it can be construed to containany preassigned overfield K∗of K, and this gives us the following obvious but basicprinciple of computational Galois theory.Basic Extension Principle.

For any given overfield K∗of K, the Galois groupGal(f, K∗), as a permutation group of degree n, acting on the roots {α1, . .

. , αn},may be regarded as a subgroup of the Galois group Gal(f, K).

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC27Given any overfield K∗of K and any factorizationf(Y ) =mYj=1ej(Y )where ej(Y ) = Y nj + aj1Y nj−1 + · · · + ajnjwith aj1, . .

. , ajnj in K∗we can relabel the roots α1, .

. .

, αn as α11, . .

. , α1,n1, .

. .

,αm1, . .

. , αmnm so thatej = ej(Y ) =njYi=1(Y −αji)for 1 ≤j ≤mand we can identify the direct product Sn1 × · · ·× Snm, where Snj is the symmetricgroup acting on αj1, .

. .

, αjnj, with a subgroup of the symmetric group Sn actingon α1, . .

. , αn.

As a second obvious but basic principle we then have the following.Basic Projection Principle. For the Galois group Gal(f, K∗) ⊂Sn we haveGal(f, K∗) ⊂Sn1 × · · · × Snm, and for 1 ≤j ≤m, the Galois group Gal(f, K∗)maps onto the Galois group Gal(ej, K∗) ⊂Snj under the natural projection Sn1 ×· · · × Snm →Snj.Recall that a ν-cycle is a permutation σ, say in Sn, such that for some ν dis-tinct elements αi1, .

. ., αiν in {α1, .

. ., αn} we have σ(αi1) = αi2, .

. ., σ(αiν−1)=αiν, σ(αiν) = αi1 and σ(αj) = αj for all j /∈{i1, .

. .

, iν}. Now if e1(Y ) is irreduciblein K∗[Y ] and if either n1 is prime or Gal(e1, K∗) is cyclic, then clearly Gal(e1, K∗)contains an n1-cycle τ1, and by the Projection Principle τ1 is the projection of someτ ∈Gal(f, K∗), and if also | Gal(ej, K∗)| and n1 are coprime for 2 ≤j ≤m thenupon letting µ to be the product of | Gal(e2, K∗)|, .

. .

, | Gal(em, K∗)| we see thatτ µ ∈Gal(f, K∗) is an n1-cycle. Therefore in view of the Extension Principle weget the following.Cycle Prelemma.

If | Gal(ej, K∗)| and n1 are coprime for 2 ≤j ≤m, ande1(Y ) is irreducible in K∗[Y ], and either n1 is prime or Gal(e1, K∗) is cyclic, thenGal(f, K) contains an n1-cycle.To convert the Cycle Prelemma into the Cycle Lemma, let v be a (real discrete)valuation44 of K, i.e., v is a map of K onto the set of all integers together withthe symbol ∞such that for all a, b in K we have v(a) = ∞⇔a = 0, and v(ab) =v(a) + v(b), and v(a + b) ≥min(v(a), v(b)). Recall that {a ∈K : v(a) ≥0} iscalled the valuation ring of v, and this ring modulo the unique maximal ideal{a ∈K : v(a) > 0} in it is called the residue field of v. Also recall that v is said tobe trivial on a subfield k of K, or v is said to be a valuation of K/k, if v(a) = 0 forall 0 ̸= a ∈k.

Let bK be a finite algebraic field extension of K and let ˆv1, . .

. , ˆvh bethe extensions of v to bK, i.e., ˆv1, .

. .

, ˆvh are those valuations of bK whose valuationrings intersected with K give the valuation ring of v; we may also say that v splitsin bK into ˆv1, . .

. , ˆvh.

By ¯r(ˆvj : v) we denote the reduced ramification exponent45 ofˆvj over v, i.e., ¯r(ˆvj : v) is the unique positive integer such that for all a ∈K wehave ˆvj(a) = ¯r(ˆvj : v)v(a). By ¯d(ˆvj : v) we denote the residue degree of ˆvj over v,44In this paper, by a valuation we shall mean a real discrete valuation.45Also called the reduced ramification index.

28S. S.

ABHYANKARi.e., ¯d(ˆvj : v) is the field degree of the residue field of ˆvj over the residue field ofv. Note that if either bK/K is separable, or v is trivial over a subfield k of K suchthat K/k is finitely generated of transcendence degree 1, then(†)hXj=1¯r(ˆvj : v) ¯d(ˆvj : v) = [ bK : K].Also note that ˆvj is unramified over v, or over K, means that ¯r(ˆvj : v) = 1 andthe residue field of ˆvj is separable over the residue field of v; ˆvj is ramified overv, or over K, means that ˆvj is not unramified over v; v is unramified in bK meansthat ˆvj is unramified over v for 1 ≤j ≤h; and finally, v is ramified in bK meansthat v is not unramified in bK.

Now it is well known that if f(Y ) is irreduciblein K[Y ], bK = K(α1), K∗= the completion of K with respect to v, and ej(Y ) isirreducible in K∗[Y ] for 1 ≤j ≤m, then h = m and, after a suitable relabelling,¯r(ˆvj : v) ¯d(ˆvj : v) = nj for 1 ≤j ≤h; for instance see §2 of [A2].Moreover,by Newton’s Theorem, if the residue field of v is an algebraically closed field ofthe same characteristic as K, and if nj ̸≡0(char K) for some j, then for that jthe Galois group Gal(ej, K∗) is cyclic; for a proof of Newton’s Theorem based onShreedharacharya’s method of completing the square, see my new book on algebraicgeometry for scientists and engineers [A6]. Therefore by the Cycle Prelemma weget theCycle Lemma.

If f(Y ) is irreducible in K[Y ] and there exists a valuation v of Ksuch that the residue field of v is an algebraically closed field of the same character-istic as K and such that for the extensions ˆv1, . .

. , ˆvh of v to a root field46 of f(Y )over K we have that ¯r(ˆvj : v) and ¯r(ˆv1 : v) are coprime and ¯r(ˆvj : v) ̸≡0(char K)for 1 < j ≤h, and either ¯r(ˆv1 : v) is prime or ¯r(ˆv1 : v) ̸≡0(char K), then theGalois group Gal(f, K) contains an ¯r(ˆv1 : v)-cycle.The Basic Extension Principle can be refined thus.Refined Extension Principle.

Given any field extensions K ⊂K′ ⊂K∗, by theBasic Extension Principle we may regard Gal(f, K∗) < Gal(f, K′) < Gal(f, K)

. , αn) and L∗= K∗(α1, .

. .

, αn). Now L is a (finite)Galois extension of K, and given any σ ∈Gal(f, K), we view σ as a permutation of{1, 2, .

. ., n} such that for some (actually unique) τ ∈Gal(L, K) we have τ(αi) =ασ(i) for 1 ≤i ≤n.

Likewise, L∗is a (finite) Galois extension of K∗, and givenany σ∗∈Gal(f, K∗), we view σ∗as a permutation of {1, 2, . .

., n} such that forsome (actually unique) τ ∗∈Gal(L∗, K∗) we have τ ∗(αi) = ασ∗(i) for 1 ≤i ≤n.Obviously, σ = σ∗⇔τ = τ ∗|L where τ∗|L denotes the restriction of τ ∗to L. Thus46A root field of f(Y ) over K is a field obtained by adjoining a root of f(Y ) to K, for instancethe field K(α1).47That is, Gal(f, K)/ Gal(f, K∗) ≈Gal(K∗, K)/N for some normal subgroup N of Gal(K∗, K).Note that for any finite normal extension K∗of a field K, without assuming K∗to be separableover K, the Galois group Gal(K∗, K) is defined to be the group of all K-automorphisms of K∗.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC29we get the commutative diagramGal(f, K∗)−−−−→Gal(f, K)yyGal(L∗, K∗)δ−−−−→Gal(L, K)where the left arrow is the isomorphism σ∗7→τ ∗, the right arrow is the isomorphismσ 7→τ, the top arrow is the inclusion Gal(f, K∗) ⊂Gal(f, K), and the bottom arrowδ is the injection τ ∗7→τ ∗|L. Therefore our assertion is equivalent to saying thatim δ is a normal subgroup of Gal(L, K) and Gal(L, K)/im δ ≈Gal(K∗, K)/N forsome normal subgroup N of Gal(K∗, K).

To prove this new version of the assertion,let K0 = L ∩K∗, let K∗0 be the maximal separable algebraic field extension of K0in K∗, let L∗0 = K∗0(α1, . .

. , αn), and let us depict all this in the following Hassediagram.L∗•L∗0 •L••K∗•K∗0K0••KReferring to the lower quadrilateral in the above diagram, L∗0/K0 is a finite Galoisextension, the field L∗0 is a compositum of the fields L and K∗0 with L∩K∗0 = K0, andthe four sides of the said quadrilateral represent the finite Galois extensions L/K0,K∗0/K0, L∗0/L, and L∗0/K∗0; hence by the Fundamental Theorem of Galois Theory,the group Gal(L∗0, K0) is the internal direct product of the two normal subgroups48Gal(L∗0, L) and Gal(L∗0, K∗0), and λ∗7→λ∗|L gives a surjection Gal(L∗0, K0) →Gal(L, K0) whose kernel is Gal(L∗0, L) and whose restriction to Gal(L∗0, K∗0) is anisomorphismGal(L∗0, K∗0)θ−→Gal(L, K0).

By applying the Fundamental Theoremof Galois Theory to the left triangle in the above diagram, i.e., by noting thatL/K, L/K0, and K0/K are Galois extensions, we see that Gal(L, K0) is a normalsubgroup of Gal(L, K) and Gal(L, K)/ Gal(L, K0) ≈Gal(K0, K).Upon lettingGal(L, K0)θ∗−→Gal(L, K) be the natural inclusion Gal(L, K0) ⊂Gal(L, K), we seethat the composition Gal(L∗0, K∗0)θ−→Gal(L, K0)θ∗−→Gal(L, K) coincides withthe injection Gal(L∗0, K∗0)δ0−→Gal(L, K) given by λ∗7→λ∗|L, and henceim δ0 = Gal(L, K0) ⊳Gal(L, K)andGal(L, K)/im δ0 ≈Gal(K0, K).48That is, the intersection of the two normal subgroups is the identity and they generate thewhole group.

30S. S. ABHYANKARBy applying the Fundamental Theorem of Galois Theory to the right triangle inthe above diagram, i.e., by noting that K∗0/K, K∗0/K0, and K0/K are finite Galoisextensions, we see thatGal(K∗0, K0) ⊳Gal(K∗0, K)andGal(K∗0, K)/ Gal(K∗0, K0) ≈Gal(K0, K).Therefore, upon letting N = Gal(K∗0, K0), we conclude thatim δ0 ⊳Gal(L, K)andGal(L, K)/im δ0 ≈Gal(K∗0, K)/Nfor someN ⊳Gal(K∗0, K).Finally, referring to the modified diagram obtained by deleting the three lines ema-nating from K0 in the above diagram, L/K and K∗0/K are finite Galois extensions;K∗/K is a finite normal extension; K∗/K∗0 is pure inseparable; L∗0 is a compositumof L and K∗0; and L∗is a compositum of L∗0 and K∗.

Consequently λ 7→λ|K∗0gives an isomorphism Gal(K∗, K) →Gal(K∗0, K), whereas λ′ 7→λ′|L∗0 gives anisomorphismGal(L∗, K∗)δ′−→Gal(L∗0, K∗0)such that δ0(im δ′) = im δ; therefore by the next to last display we conclude thatim δ ⊳Gal(L, K)andGal(L, K)/im δ ≈Gal(K∗, K)/Nfor some N ⊳Gal(K∗, K).For applying to specific situations, here are someCorollaries of the Refined Extension Principle. Given any finite algebraicfield extension K′ of K, by the Basic Extension Principle we may regard Gal(f, K′)

(1.1) There exists N ⊳Gal(K∗, K) and M ⊳Gal(f, K) with M < Gal(f, K′) suchthat Gal(f, K)/M ≈Gal(K∗, K)/N. (1.2) If Gal(K∗, K) is solvable, then there exists M ⊳Gal(f, K) with M

(1.3) If Gal(K∗, K) is solvable, and Gal(f, K) = Sn, and 3 ̸= n ̸= 4, thenGal(f, K′) = Sn or An. (1.4) If Gal(K∗, K) is cyclic and Gal(f, K) = Sn, then Gal(f, K′) = Sn or An.

(1.5) If Gal(K∗, K) is cyclic of odd order and Gal(f, K) = Sn, thenGal(f, K′) =Sn. (1.6) If there is no nonidentity group which is a homomorphic image ofGal(f, K)as well as Gal(K∗, K), then Gal(f, K′) = Gal(f, K).

(1.7) If Gal(f, K) is a simple group which is not a homomorphic image ofGal(K∗, K), then Gal(f, K′) = Gal(f, K). (1.8) If Gal(K∗, K) is solvable and Gal(f, K) is nonabelian simple, then Gal(f, K′) =Gal(f, K).

(1.9) If Gal(K∗, K) is solvable, and Gal(f, K) = An, and 3 ̸= n ̸= 4, thenGal(f, K′) = An. (1.10) If Gal(K∗, K) is cyclic, and Gal(f, K) = An, and n = 4, then Gal(f, K′) =An or (Z2)2.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC31(1.11) If Gal(K∗, K) is cyclic, and Gal(f, K) = An, and n = 3, then Gal(f, K′) =An or Z1. (1.12) If Gal(K∗, K) is cyclic of order nondivisible by 3 and Gal(f, K) = An,then Gal(f, K′) = An.

(1.13) If Gal(K∗, K) is solvable and Gal(f, K) = PSL(2, q) for a prime powerq > 3, then Gal(f, K′) = PSL(2, q). (1.14) If Gal(K∗, K) is cyclic of order nondivisible by char k and Gal(f, K) =PSL(2, q) with 1 < q = n −1 = a power of char k, then Gal(f, K′) = PSL(2, q).

(1.1) follows by taking M = Gal(f, K∗) in the Refined Extension Principle. Theimplication (1.1) ⇒(1.2) follows from the fact that a homomorphic image of afinite solvable group is solvable.

In view of (1.1), the implication (1.2) ⇒(1.3)follows from the facts that if n ≥5 then Sn is nonsolvable, and An and Sn arethe only nonidentity normal subgroups of Sn, and there are no other subgroups ofSn between An and Sn, whereas if n ≤2 then An and Sn are the only subgroupsof Sn. In view of (1.1), the implication (1.3) ⇒(1.4) follows from the fact thatS3 and S4 are noncyclic; A3 is the only nonidentity normal subgroup of S3, forthe factor group we have S3/A3 = Z2, and the only nonidentity normal subgroupof S4 other than A4 is the Klein group (Z2)2 consisting of the four permutations(1), (12)(34), (13)(24), (14)(23), and the factor group of S4 by the Klein group isisomorphic to S3.

In view of (1.1), the implication (1.4) ⇒(1.5) follows from thefact that Sn/An = Z2 or Z1. The implications (1.1) ⇒(1.6) ⇒(1.7) are obvious.The implication (1.7) ⇒(1.8) follows from the fact that a homomorphic image ofa finite solvable group is solvable.

In view of (1.1), the implication (1.8) ⇒(1.9)follows from the facts that if n ≥5 then An is nonabelian simple, whereas if n ≤2then An = Z1. The implication (1.1) ⇒(1.10) follows from the facts that A4 is notcyclic, and the Klein group (Z2)2 is the only nonidentity normal subgroup of A4, andthe factor group by the Klein group is Z3.

The implication (1.1) ⇒(1.11) followsfrom the fact that A3 = Z3. In view of (1.1), (1.10), and (1.11), the implication(1.9) ⇒(1.12) follows from the fact that A4/(Z2)2 = Z3 = A3.

The implication(1.8) ⇒(1.13) follows from the fact that PSL(2, q) is nonabelian simple for everyprime power q > 3. Finally, in view of (1.1), and what we have said about S3 andA4, the implication (1.13) ⇒(1.14) follows from the facts that PSL(2, 2) = S3 andPSL(2, 3) = A4.As a consequence of the Refined Extension Principle we have theSubstitutional Principle.

Assume that K = the field k(X) of rational functionsin an indeterminate X with coefficients in a field k, i.e., ai = ai(X) ∈k(X) for1 ≤i ≤n. Given any χ(X) ∈k(X) \ k, letfχ = fχ(Y ) = Y n + a1(χ(X))Y n−1 + · · · + an(χ(X)) ∈k(X)[Y ] = K[Y ]and let K′ = k(V ) where V is an indeterminate.Then the k-homomorphismX 7→χ(V ) gives an embedding K = k(X) ⊂k(V ) = K′, and by sending Y toY it gives an embedding K[Y ] = k(X)[Y ] ⊂k(V )[Y ] = K′[Y ], and this sends f tofχ with X changed to V .

Therefore, fχ has no multiple roots in any field exten-sion of k(X) and, upon letting K∗= a least normal extension of k(X) containingk(V ), by the Basic Extension Principle we may regard Gal(f, K∗) < Gal(f, k(V )) =Gal(fχ, k(X)) < Gal(f, k(X)) < Sn, and now by the Refined Extension Principle

32S. S. ABHYANKARGal(f, K∗) ⊳Gal(f, k(X)) and Gal(f, k(X))/ Gal(f, K∗) ≈Gal(K∗, k(X))/N forsome N ⊳Gal(K∗, k(X)).49Here are the correspondingCorollaries of the Substitutional Principle.

Letting the situation be as in theSubstitutional Principle, and remembering that K∗= a finite normal extension ofk(X) and Gal(fχ, k(X)) < Gal(f, k(X)) < Sn, we have the following. (2.1) There exists N⊳Gal(K∗, k(X)) and M⊳Gal(f, k(X)) with M < Gal(fχ, k(X))such that Gal(f, k(X))/M ≈Gal(K∗, k(X))/N.

(2.2) If Gal(K∗, k(X)) is solvable, then there exists M ⊳Gal(f, k(X)) with M

(2.4) If Gal(K∗, k(X)) is cyclic and Gal(f, k(X)) = Sn, then Gal(fχ, k(X)) = Snor An. (2.5) If Gal(K∗, k(X)) is cyclic of odd order and Gal(f, k(X)) = Sn, thenGal(fχ, k(X)) = Sn.

(2.6) If there is no nonidentity group which is a homomorphic image ofGal(f, k(X))as well as Gal(K∗, k(X)), then Gal(fχ, k(X)) = Gal(f, k(X)). (2.7) If Gal(f, k(X)) is a simple group which is not a homomorphic image ofGal(K∗, k(X)), then Gal(fχ, k(X)) = Gal(f, k(X)).

(2.8) If Gal(K∗, k(X)) is solvable and Gal(f, k(X)) is nonabelian simple, thenGal(fχ, k(X)) = Gal(f, k(X)). (2.9) If Gal(K∗, k(X)) is solvable, and Gal(f, k(X)) = An, and 3 ̸= n ̸= 4, thenGal(fχ, k(X)) = An.

(2.10) If Gal(K∗, k(X)) is cyclic, and Gal(f, k(X)) = An, and n = 4, thenGal(fχ, k(X)) = An or (Z2)2. (2.11) If Gal(K∗, k(X)) is cyclic, and Gal(f, k(X)) = An, and n = 3, thenGal(fχ, k(X)) = An or Z1.

(2.12) If Gal(K∗, k(X)) is cyclic of order nondivisible by 3 and Gal(f, k(X)) =An, then Gal(fχ, k(X)) = An. (2.13) If Gal(K∗, k(X)) is solvable and Gal(f, k(X)) = PSL(2, q) for a primepower q > 3, then Gal(fχ, k(X)) = PSL(2, q).

(2.14) If Gal(K∗, k(X)) is cyclic of order nondivisible by char k andGal(f, k(X)) =PSL(2, q) with 1 < q = n −1 = a power of char k, thenGal(fχ, k(X)) = PSL(2, q).The proof of (2.1) to (2.14) follows from the above proof of (1.1) to (1.14) bychanging “Refined Extension Principle” to “Substitutional Principle” and by chang-ing (1.i) to (2.i) for 1 ≤i ≤14.Stated in a form more suitable for applying to the specific equations describedearlier, here are some furtherCorollaries of the Substitutional Principle. Letting the situation be as in theSubstitutional Principle, and remembering that Gal(fχ, k(X)) < Gal(f, k(X))

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC33(3.1) There exists M⊳Gal(f, k(X)) with M < Gal(fχ, k(X)) such that Gal(f, k(X))/Mis a cyclic group whose order is nondivisible by char k but divides r.(3.2) If Gal(f, k(X)) = Sn, then Gal(fχ, k(X)) = Sn or An. (3.3) If Gal(f, k(X)) = Sn and char k = 2, then Gal(fχ, k(X)) = Sn.

(3.4) If Gal(f, k(X)) is nonabelian simple, then Gal(fχ, k(X)) =Gal(f, k(X)). (3.5) If Gal(f, k(X)) = An and 3 ̸= n ̸= 4, then Gal(fχ, k(X)) = An.

(3.6) If Gal(f, k(X)) = An and n = 4, then Gal(fχ, k(X)) = An or (Z2)2. (3.7) If Gal(f, k(X)) = An and n = 3, then Gal(fχ, k(X)) = An or Z1.

(3.8) If Gal(f, k(X)) = An and char k = 3, then Gal(fχ, k(X)) = An. (3.9) If Gal(f, k(X)) = PSL(2, q) for a prime power q > 3, thenGal(fχ, k(X)) =PSL(2, q).

(3.10) If Gal(f, k(X)) = PSL(2, q) with 1 < q = n −1 = a power of char k, thenGal(fχ, k(X)) = PSL(2, q).Namely, Gal(K∗, K), i.e., the Galois group of the splitting field of Y |r|− Xcr/|r|over k(X) is a cyclic group whose order is nondivisible by char k but divides r, andhence (3.1), (3.2), (3.3) follow from (2.1), (2.4), (2.5) respectively, and (3.i) followsfrom (2.i + 4) for 4 ≤i ≤10.Here are still some moreCorollaries of the Substitutional Principle. Letting the situation be as in theSubstitutional Principle, and remembering that Gal(fχ, k(X)) < Gal(f, k(X))

(4.1) If for each i with 1 ≤i ≤d we have that there is no nonidentity groupwhich is a homomorphic image of Gal(f, k(X)) as well as Gal(K∗i , k(X)), thenGal(fχ, k(X)) = Gal(f, k(X)). (4.2) If Gal(f, k(X)) is a simple group which is not a homomorphic image ofGal(K∗i , k(X)) for any i with 1 ≤i ≤d, then Gal(fχ, k(X)) = Gal(f, k(X)).

(4.3) If Gal(K∗i , k(X)) is solvable for each i with 1 ≤i ≤d, andGal(f, k(X)) isnonabelian simple, then Gal(fχ, k(X)) = Gal(f, k(X)). (4.4) If Gal(K∗i , k(X)) is solvable for each i with 1 ≤i ≤d, andGal(f, k(X)) =An, and 3 ̸= n ̸= 4, then Gal(fχ, k(X)) = An.Namely, (4.1), (4.2), (4.3), and (4.4) follow by repeatedly applying (2.6), (2.7),(2.8), and (2.9) respectively.Although we shall not use it in this paper, here is a third basic principle ofcomputational Galois theory; for a proof see §61 of volume I of van der Waerden’sbook [V], and for some applications see the 1958 follow-up [A4] of my 1957 paper.

34S. S. ABHYANKARBasic Homomorphism Principle.

If f(Y ) ∈Rv[Y ] where Rv is the valuationring of a valuation v of K and if ¯f(Y ) has no multiple roots in any overfield ofK, where K is the residue field of v and ¯f(Y ) ∈K(Y ) is obtained by applying thecanonical epimorphism Rv →K to the coefficients of f(Y ), then, as a permutationgroup, the Galois group Gal( ¯f, K) may be regarded as a subgroup of the Galoisgroup Gal(f, K).To give some concrete examples of valuations, assume that K = k(x) where x isa transcendental over a field k. Then for every nonconstant irreducible φ(x) ∈k[x]we get a valuation vφ of k(x)/k by takingvφπ(x)φ(x)lǫ(x)= lfor any integer l and any nonzero π(x) and ǫ(x) in k[x] which are nondivisible byφ(x); we may call this the φ(x) = 0 valuation of k(x)/k; if φ(x) = x −c with c ∈kthen we may also call it the x = c valuation of k(x)/k; note that for any othernonconstant irreducible φ∗(x) ∈k[x] we have vφ = vφ∗⇔φ and φ∗are constantmultiples of each other. In addition to these valuations, there is exactly one morevaluation v∞of k(x)/k given by takingv∞π(x)ǫ(x)= deg ǫ(x) −deg π(x)for all nonzero π(x) and ǫ(x) in k[x]; we may call this the x = ∞valuation ofk(x)/k.Let λ > µ ≥0 be integers, and consider the polynomial ξ(Z)Y λ + η(Z)Y µ inindeterminates Y and Z where ξ(Z) and η(Z) are nonzero coprime polynomials inZ with coefficients in k. Let y be an element in an overfield of K = k(x) such thatξ(x)yλ + η(x)yµ = 0.

Now, upon letting ˆv1, . .

. , ˆvh be the extensions of v = vφ tobK = K(y), for 1 ≤j ≤h we have¯r(ˆvj : v)v(ξ(x)η(x)) =¯r(ˆvj : v)v−η(x)ξ(x) = |ˆvj(yλ−µ)| = (λ −µ)|ˆvj(y)|and hence¯r(ˆvj : v) ≡0λ −µGCD(λ −µ, v(ξ(x)η(x)))and therefore by (†) we get(††)[ bK : K] ≡0λ −µGCD(λ −µ, v(ξ(x)η(x))).From this we deduce the followingFirst Irreducibility Lemma.

Let λ > µ ≥0 be integers, and consider the poly-nomial ξ(Z)Y λ + η(Z)Y µ in indeterminates Y and Z where ξ(Z) and η(Z) arenonzero coprime polynomials in Z with coefficients in a field k. Let y be an ele-ment in an overfield of k(Z) such that ξ(Z)yλ + η(Z)yµ = 0. Assume that thereexists a finite number of nonconstant irreducible polynomials φ1(Z), .

. .

, φm(Z) in

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC35Z with coefficients in k such that, upon letting νi be the largest integer for whichφi(Z)νi divides ξ(Z)η(Z) in k[Z], we have GCD(λ −µ, ν1, . .

. , νm) = 1.Then[k(y, Z) : k(Z)] = λ −µ, and the polynomial ξ(Z)Y λ + η(Z)Y µ is irreducible ink(Y )[Z].50Namely, by (††) we see that[k(y, Z) : k(Z)] ≡0λ −µGCD(λ −µ, νi)for 1 ≤i ≤mand clearlyλ −µGCD(λ −µ, ν1, .

. .

, νm) = LCMλ −µGCD(λ −µ, ν1), . .

. ,λ −µGCD(λ −µ, νm)and hence[k(y, Z) : k(Z)] ≡0λ −µGCD(λ −µ, ν1, .

. .

, νm).Since GCD(λ −µ, ν1, . .

. , νm) = 1, we get[k(y, Z) : k(Z)] ≡0(λ −µ)and since yλ−µ = −η(Z)/ξ(Z), we conclude that[k(y, Z) : k(Z)] = λ −µand the polynomial Y λ−µ + η(Z)/ξ(Z) is irreducible in k(Z)[Y ].Therefore bythe Gauss Lemma,51 the polynomial ξ(Z)Y λ−µ +η(Z) is irreducible in k[Y, Z], andhence again by the Gauss Lemma, the polynomial ξ(Z)Y λ−µ+η(Z) is irreducible ink(Y )[Z], and therefore the polynomial ξ(Z)Y λ + η(Z)Y µ is irreducible in k(Y )[Z].Let us now convert the above lemma into the followingSecond Irreducibility Lemma.

Let λ > µ ≥0, and let ξλ(Y, Z) and ηµ(Y, Z)be nonzero homogeneous polynomials of respective degrees λ and µ in (Y, Z) withcoefficients in a field k. Assume that the polynomials ξλ(Y, Z) and ηµ(Y, Z) areregular in Z,52 and the polynomials ξλ(1, Z) and ηµ(1, Z) have no nonconstantcommon factor in k[Z]. Also assume that there exists a finite number of nonconstantirreducible polynomials φ1(Z), .

. .

, φm(Z) in Z with coefficients in k such that, uponletting νi to be the largest integer for which φi(Z)νi divides ξλ(1, Z)ηµ(1, Z) in k[Z],we have GCD(λ −µ, ν1, . .

. , νm) = 1.

Then the polynomial ξλ(Y, Z) + ηµ(Y, Z) isirreducible in k(Y )[Z].To prove this, note that Z 7→Y Z gives a k(Y )-automorphism of k(Y )[Z] whichsends the polynomial ξλ(Y, Z)+ηµ(Y, Z) to the polynomial ξλ(1, Z)Y λ+ηµ(1, Z)Y µ.50That is, ξ(Z)Y λ + η(Z)Y µ is either a nonzero element of k(y) or a nonconstant irreduciblepolynomial in Z with coefficients in k(y).51The Gauss Lemma says that a nonzero polynomial in indeterminates Y and Z with coef-ficients in a field k is irreducible in k[Y, Z] if and only if it is irreducible in k(Z)[Y ] and, as apolynomial in Y , its coefficients have no nonconstant common factor in k[Z].52That is, their degrees in Z coincide with their degrees in (Y, Z).

36S. S. ABHYANKARBy the First Irreducibility Lemma, the second polynomial is irreducible in k(Y )[Z]and hence so is the first.As an illustration, let n > t > 1 be integers such that GCD(n, t) = 1.

Now ifu is any element in an algebraic closure ¯k of k such that un = 1 = ut, then, sinceGCD(n, t) = 1, we must have u = 1; therefore 1 is the only common root of Zn −1and Zt −1 in ¯k. Again since GCD(n, t) = 1, either n or t is nondivisible by thecharacteristic of k, and hence either Zn −1 or Zt −1 is devoid of multiple rootsin ¯k.

Therefore the polynomials (Zn −1)/(Z −1) and (Zt −1)/(Z −1) have nocommon root in ¯k and at least one of them has no multiple root in ¯k, and hencethey have no nonconstant common factor in k[Z] and at least one of them has nononconstant multiple irreducible factor in k[Z]; since n > t > 1, we conclude thattheir product has at least one nonconstant nonmultiple irreducible factor in k[Z].53By applying the k-automorphism Z 7→Z + 1 of k[Z] and by multiplying the secondpolynomial by −a, where a is any nonzero element of k, we see that the polynomials(Z + 1)n −1)/Z and −a[(Z + 1)t −1]/Z have no nonconstant common factor ink[Z] and their product has at least one nonconstant nonmultiple irreducible factorin k[Z]. Consequently by taking (Z + Y )n −Y n/Z and −a[(Z + Y )t −Y t]/Z forξλ(Y, Z) and ηµ(Y, Z), and by letting φ1(Z), .

. .

, φm(Z) to be the distinct monicnonconstant irreducible factors of ξλ(1,Z)ηµ(1, Z) in k[Z], and by letting νi to bethe largest integer for which φi(Z)νi divides ξλ(1, Z)ηµ(1, Z) in k[Z], we see thatξλ(Y, Z) and ηµ(Y, Z) in k[Z] are nonzero homogeneous polynomials of respectivedegrees λ = n −1 and µ = t −1 in (Y, Z) with coefficients in k such that thepolynomials ξλ(1, Z) and ηµ(1, Z) have no nonconstant common factor in k[Z], andfor some i we have νi = 1, and hence trivially we have GCD(λ −µ, ν1, . .

. , νm) = 1.Therefore by the Second Irreducibility Lemma we get the followingThird Irreducibility Lemma.

Let n > t > 1 be integers such that GCD(n, t) =1. Let Y and Z be indeterminates over a field k, and let 0 ̸= a ∈k.

Then thepolynomial ((Z + Y )n −Y n/Z −a[(Z + Y )t −Y t]/Z is irreducible in k(Y )[Z].20. The tilde polynomial and borrowing cyclesLet us now apply the twisted derivative method to the polynomialeFn,s = eFn,s(X, Y ) = Y n −aY t + Xswith 1 ≤t < n and GCD(n, t) = 1, where s is a positive integer, a is a nonzeroelement in a field k of characteristic p,54 and X and Y are indeterminates over k.We want to calculate the Galois group eGn,s = Gal( eFn,s, k(X)).55Let ¯k be an algebraic closure of k, and let(i)d = n −tewhere e = 1if p = 0,max pµ with n −t ≡0(pµ)if p ̸= 0.53That is, there exists a nonconstant irreducible member of k[Z] which divides the said productbut whose square does not divide the said product.54Here p may or may not be zero.

Later on we shall specialize to the case of p ̸= 0; with thefurther assumption that n ≡0(p) and s ≡0(t), the polynomial eFn,s reduces to the polynomialeFn considered in §11.55In a moment we shall show that eFn,s has no multiple roots in any overfield of k(X), andhence the Galois group Gal( eFn,s, k(X)) is defined.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC37Now(ii)Y n −aY t = Y tdYi=1(Y −βi)ewith pairwise distinct nonzero elements β1, . .

. , βd in ¯k and GCD(e, t) = 1, andhence by the First Irreducibility Lemma we see that eFn,s is irreducible in ¯k(X)[Y ],and therefore eFn,s is irreducible in k(X)[Y ].56Let us put x = X and let y be a root of eFn,s in an overfield of k(x).

Then(iii)xs = ayt −ynand hence y is transcendental over k.Let E(y, Z) be the polynomial in an in-determinate Z such that E(y, Y ) is the twisted Y -derivative of eFn,s(x, Y ) at y.ThenE(y, Z) = (Z + y)n −ynZ−a[(Z + y)t −yt]Zand hence by the Third Irreducibility Lemma we see that E(y, Z) is irreducible ink(y)[Z].For a moment suppose that s = 1 < t; then by (iii) we have x ∈k(y), andhence E(y, Z) is irreducible in k(x, y)[Z], and therefore by the Twisted DerivativeCriterion we see that eFn,1(X, Y ) has no multiple roots in any overfield of k(X),and so Gal( eFn,1, k(X)) is defined; since E(y, Z) is irreducible in k(x, y)[Z], it alsofollows that Gal( eFn,1, k(X)) is 2-transitive. Thus, if s = 1 < t then eFn,1 is devoidof multiple roots and eGn,1 is 2-transitive.Reverting to general s but still assuming t > 1, since eFn,s(X, Y ) = eFn,1(Xs, Y )and eFn,1(X, Y ) has no multiple roots in any overfield of k(X), it follows thateFn,s(X, Y ) has no multiple roots in any overfield of k(X).

Thus, if t > 1 theneFn,s is devoid of multiple roots and hence eGn,s is defined.To give a direct proof of eFn,s being devoid of multiple roots, let us calculate itsY -discriminant.So first recall that the Y -discriminant of a monic polynomial f = f(Y ) of degreen > 0 in Y with coefficients in a field K is denoted by DiscY (f) and is defined byputtingDiscY (f) = ResY (f, fY ) = the Y -resultant of f and fY ,where fY is the (ordinary) Y -derivative of f. Upon letting m be the degree of fY ,we note that ResY (f, fY ) is the determinant of a certain n + m by n + m matrix;also note that m = n −1 ⇔n ̸≡0(char K); finally note that if fY = 0, i.e., ifchar K = p ̸= 0 and f ∈K[Y p], then we take DiscY (f) = 0. For calculationalpurposes, upon letting f = Qi=1(Y −αi) we observe thatDiscY (f) =nYi=1fY (αi)56For s = 1 this also follows by noting that eFn,1 is monic of degree 1 in X.

38S. S. ABHYANKARand upon assuming fY ̸= 0 and upon letting fY = ǫ Qmi=1(Y −ǫi) we observe thatDiscY (f) = (−1)nmǫnmYi=1f(ǫi).Finally, upon lettingDisc∗Y (f) = the modified Y -discriminant of F =Yi

If char K ̸= 2 and f is devoid of multiple roots, then:Gal(f, K) ⊂An ⇔Disc∗Y (f) is a square in K.Now for the (ordinary) Y -derivative of eFn,s we have( eFn,s)Y = nY n−1 −taY t−1= −taY t−1 ̸= 0if n ≡0(p),nY t−1 Qn−ti=1 (Y −bωi) ̸= 0if n ̸≡0(p),with b, ω1, . .

. , ωn−t in ¯k such that nbn−t = ta and Qn−ti=1 (Y −ωi) = Y n−t −1 andhenceDiscY ( eFn,s) =(−1)n(t−1)(−ta)n(Xs)t−1if n ≡0(p),(−1)n(n−1)nn(Xs)t−1 Qn−ti=1 [Xs −n−1(n −t)abtωti]if n ̸≡0(p),and therefore, because GCD(n −t, t) = 1, we get(iv)DiscY ( eFn,s) = (−1)nttnanXs(t−1)if n ≡0(p),Xs(t−1)[nnXs(n−t) −(n −t)n−tttan]if n ̸≡0(p),and, observing that if n ≡0(p) then in k we have nn = 0 and −(n −t)n−ttt =(−1)n−t+1tn = (−1)nttn,58 we conclude that in both the cases we haveDiscY ( eFn,s) = nnXs(n−1) −(n −t)n−tttanXs(t−1).Thus always DiscY ( eFn,s) ̸= 0 and hence eFn,s is devoid of multiple roots.If s = 1 and n −t ̸≡0(p) then by (i), (ii), (iii) we see that the valuationx = 0 of ¯k(x)/¯k splits in ¯k(x, y) = ¯k(y) into the n −t + 1 = d + 1 valuationsy = 0, y = β1, .

. .

, y = βd with reduced ramification exponents t, 1, . .

. , 1, and henceif also t ̸≡0(p) then by the Cycle Lemma we can find a t-cycle in Gal( eFn,1, ¯k(X)),and therefore, since by the Basic Extension Principle we have Gal( eFn,1, ¯k(X))

Thus, if n −t ̸≡0(p) andt ̸≡0(p) then eGn,1 contains a t-cycle.Let us note the following Corollary of an 1871 Theorem of Jordan [J2]; a proofcan also be found in 13.3 on page 34 of Wielandt [Wi], and in 4.4 on page 171 ofVolume I of Huppert [HB].57On pages 82–87 of volume I of van der Waerden’s book [V], it seems to be wrongly assertedthat Disc∗Y (f) = DiscY (f). Some authors call Disc∗Y (f) the discriminant of f.58Note that if n and t are both odd then n −t + 1 and nt are both odd, whereas if one outof n and t is odd and the other even then n −t + 1 and nt are both even, and finally, sinceGCD(n, t) = 1, the remaining possibility of n and t both being even cannot occur.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC39Jordan’s Corollary. An and Sn are the only primitive permutation groups ofdegree n containing a 3-cycle.Let us also note the following two 1892 Theorems of Marggraff[Mar]; these aregiven as Theorems 13.5 and 13.8 on pages 35 and 38 of Wielandt [Wi] respectively.59Marggraff’s First Theorem.

An and Sn are the only primitive permutationgroups of degree n having a (n −ν)-point stabilizer, with 1 < ν < n/2, which actstransitively on the remaining ν symbols.Marggraff’s Second Theorem. A primitive permutation group of degree n con-taining a ν-cycle, with 1 < ν < n, is (n −ν + 1)-transitive.For a moment suppose that n −t ̸≡0(p) and 1 < t ̸≡0(p).60 Now, in view ofwhat we have proved above, eGn,1 is a 2-transitive permutation group containing at-cycle and hence if t = 2 then obviously61 eGn,1 = Sn; if t = 3 then by Jordan’sCorollary62 we have eGn,1 = An or Sn; if t < n2 then by Marggraff’s First Theoremwe have eGn,1 = An or Sn; if t < n −4 then by Marggraff’s Second Theorem andCTT we have eGn,1 = An or Sn; if t = n −4 then automatically 12 ̸= n ̸= 24 andhence by Marggraff’s Second Theorem and CTT we have eGn,1 = An or Sn; and ift = n−3 then again automatically 12 ̸= n ̸= 24 and hence if also 11 ̸= n ̸= 23, thenby Marggraff’s Second Theorem and CTT we have eGn,1 = An or Sn.

In all thesecases, if t is even then we must have eGn,1 = Sn because An cannot have a cycle ofeven length, and if t is odd and p ̸= 2 then, in view of the Discriminant Criterion,by (iv) we can unambiguously decide between An and Sn; in particular, if k isalgebraically closed and n ≡0(p) and p ̸= 2 and t is odd then we get An becausein that case Disc∗Y ( eFn,1) is a square in k(x). Again, in all these cases, assumingk to be algebraically closed, in view of Corollaries (3.2) to (3.8) of the Substitu-tional Principle, we see that (1) t is even ⇒eGn,1 = Sn ⇒eGn,s = An or Sn; (2)t is odd and n = 4 ⇒eGn,1 = An or Sn with n = 4 ⇒eGn,s = (Z2)2 or An or Sn;(3) t is odd and n > 4 and p = 2 ⇒eGn,1 = An or Sn with n > 4 ⇒eGn,s =An or Sn; (4) t is odd and n = 4 and p ̸= 2 and Disc∗Y ( eFn,1) is a square in k(x) ⇒eGn,1 = An with n = 4 ⇒eGn,s = (Z2)2 or An; (5) t is odd and n > 4 and p ̸=2 and Disc∗Y ( eFn,1) is a square in k(x) ⇒eGn,1 = An with n > 4 ⇒eGn,s = An; andfinally (6) t is odd and p ̸= 2 and Disc∗Y ( eFn,1) is not a square in k(x) ⇒eGn,1 =Sn ⇒eGn,s = An or Sn.Now for a moment suppose that n ≡0(p).

Then t ̸≡0(p) and hence by (iv) wesee that x = 0 and x = ∞are the only valuations of k(x)/k which are possiblyramified in k(x, y). Moreover, if s ≡0(t) then, in view of (i), (ii), (iii), either bydirect reasoning we see that the valuation x = 0 of k(x)/k is unramified in k(x, y),or alternatively, first, upon letting x∗= xs we see that the valuation x∗= 0 ofk(x∗)/k splits in k(x∗, y) = k(y) into several valuations out of which y = 0 is59Apparently, Marggraff’s Second Theorem can also be found in [J2]; see Neumann [N].60Note that if n ≡0(p) then automatically n −t ̸≡0(p) and t ̸≡0(p) because by assumptionGCD(n, t) = 1.61A 2-cycle is simply a transposition, and a 2-transitive permutation group containing a trans-position must contain all transpositions and hence must be the symmetric group.62A 2-transitive permutation group is automatically primitive.

40S. S. ABHYANKARthe only valuation which is possibly ramified over k(x∗) and for it the reducedramification index is t and the residue degree is 1, and now upon lettingx′ = xif p = 0,xs′ where s′ = max pλ with s ≡0(pλ)if p ̸= 0,we see that x′s/s′ = x∗and s/s′ is a positive integer which is divisible by t butnot divisible by p and hence by MRT63 we see that the valuation x′ = 0 of k(x′) isunramified in k(x′, y), and finally from this we deduce that the valuation x = 0 ofk(x)/k is unramified in k(x, y).64Thus, if n ≡0(p) and s ≡0(t), then x = ∞is the only valuation of k(x)/k whichis possibly ramified in k(x, y), and hence65 x = ∞is the only valuation of k(x)/kwhich is possibly ramified in a least Galois extension of k(x) containing k(x, y).It only remains to note that, by Result 4 on page 841 of the 1957 paper [A3], asa consequence of the genus formula, every member of the algebraic fundamentalgroup of the affine line over an algebraically closed field k of characteristic p ̸= 0is a quasi p-group; therefore in our case, if eGn,s = An or Sn and p ̸= 2 then wemust have eGn,s = An, because clearly Sn is not a quasi p-group for any n ≥p > 2;likewise, if eGn,s = (Z2)2 or An or Sn with n = 4 and p = 2, then we must haveeGn,s = (Z2)2 or Sn because A4 is not a quasi 2-group.Let us put all this together in the followingSummary about the tilde polynomial.

Let k be an algebraically closed fieldk of characteristic p ̸= 0, and consider the polynomial eFn = Y n −aY t + Xs inindeterminates X and Y over k, where a is a nonzero element of k, and n, s, t arepositive integers with GCD(n, t) = 1 and t < n ≡0(p) and s ≡0(t). Then eFngives an unramified covering of the affine line over k, and for the Galois groupeGn = Gal( eFn, k(X)) we have the following.

(II.1) If 1 < t < 4 and p ̸= 2, then eGn = An. (II.2) If 1 < t < n −3 and p ̸= 2, then eGn = An.

(II.3) If 1 < t = n −3 and p ̸= 2 and 11 ̸= p ̸= 23, then eGn = An. (II.4) If 1 < t < 4 < n and p = 2, then eGn = An or Sn.

(II.5) If 1 < t < n −3 and p = 2, then eGn = An or Sn.Here CT was not used in the proofs of (II.1) and (II.4). Moreover,66 the followingspecial cases of (II.2) and (II.5) were proved without using CT.(II.2*) If 1 < t < n/2 and p ̸= 2, then eGn = An.

(II.5*) If 1 < t < n/2 and p = 2, then eGn = An or Sn.Note. In the next section we shall consider the unramified covering of the affine linegiven by the polynomial F n = Y n −XY t + 1 with n = p + t and t ̸≡0(p) which63MRT=Abhyankar’s Lemma = pages 181–186 of [A5].64This last deduction follows from the easy to prove fact which says that if a valuation v of afield K is unramified in a finite separable algebraic field extension L of K then the unique extensionof v to a finite purely inseparable field extension K′ of K is unramified in the compositum of Land K′.65Say by Proposition 1 of [A1].66In view of the above discussion, without using CT we see that if 1 < t < 4 = n and p = 2,then eGn = (Z2)2 or Sn, and using CT we see that if 1 < t = n −3 and eGn ̸= An, then eithern = p = 11 and eGn = M11, or n = p = 23 and eGn = M23.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC41was introduced in §11; there the calculation of the Galois group of F n will be basedon the fact that it contains a p-cycle because the valuation X = ∞splits into thevaluations Y = 0 and Y = ∞with reduced ramification exponents t and p respec-tively.67 In the present section we considered the unramified covering of the affineline given by the polynomial eFn = Y n −aY t + Xs with n ≡0(p) and GCD(n, t) =1 and s ≡0(t) which was also introduced in §11; contrary to the Galois group ofthe polynomial F n, it is not easy to find any cycle in the Galois group of the poly-nomial eFn because now the valuation X = ∞has the valuation Y = ∞as the onlyextension and for it the reduced ramification exponent is n which is a multiple ofp; so what we did was to “borrow” a t-cycle by going down to the subfield k(Xs) ofk(X), i.e., by embedding the Galois group over k(X) as a subgroup of the Galoisgroup over k(Xs) where the latter does contain a t-cycle. This method of borrow-ing cycles can already be found in Hilbert’s 1892 paper [H] where he “borrows” atransposition to embed An in Sn, thereby constructing An coverings of the rationalnumber field Q.21.

The bar polynomialLet us now turn to the polynomialF n,q = Y n −XY t + 1with n = q + t and positive integer t ̸≡0(p) mentioned in §11. This is a polynomialin indeterminates X and Y with coefficients in a field k of characteristic p ̸= 0, andq is a positive power of p.The (ordinary) Y -derivative of F n,q is given by(F n,q)Y = n(Y n−1 −XY t−1)and so we getF n,q = Yn (F n,q)Y + 1and hence for the Y -discriminant of F n,q we haveDiscY (F n,q) = nn = a nonzero element of k.Therefore this gives an unramified covering of the affine line over k. At any rate,F n,q, as a polynomial in Y , is devoid of multiple roots, and so we can talk aboutits Galois group over k(X).We are interested in calculating this Galois groupGn,q = Gal(F n,q, k(X)).Now F n,q, as a polynomial in X, is linear and in it the coefficient of X has nocommon factor with the terms devoid of X.

Therefore F n,q is irreducible in k[Y ][X]and hence in k(X)[Y ]. Consequently, Gn,q, as a permutation group of degree n, istransitive.Regarding F n,q as a polynomial in Y and reciprocating its roots we get thepolynomialΘ = Θ(Y ) = Y q+t −xY q + 1,67Actually, by reciprocating the roots of F n we shall get the polynomial Y n −XY p + 1 forwhich the valuation X = ∞splits into the valuations Y = 0 and Y = ∞with reduced ramificationexponents p and t respectively.

42S. S. ABHYANKARwhere we have put x = X.

Let y be a root of this polynomial in some overfield ofk(x). Then solving Θ(y) = 0 for x we get the equation x = yt + y−q.

The q = pcase of this equation was really the starting point of this paper and it was originallyobtained by taking h = c1 = 1 in Proposition 1 of the 1957 paper [A3]. The case ofgeneral q can also be obtained by taking h = q/p and cq/p = 1 and ci = 0 for 1 ≤i < q/p in that Proposition.

By the said Proposition, or directly by looking atthe above equation, we see that (1) the simple transcendental extension k(y) ofk is a separable algebraic field extension of the simple transcendental extensionk(x) of k with field degree [k(y) : k(x)] = q + t; (2) v∞: x = ∞is the onlyvaluation of k(x)/k which is ramified in k(y); (3) v∞splits in k(y) into the valuationsw0 : y = 0 and w∞: y = ∞; and (4) for the residue degrees and reduced ramificationexponents we have ¯d(w0 : v∞) = 1 and ¯r(w0 : v∞) = q, and ¯d(w∞: v∞) = 1 and¯r(w∞: v∞) = t. Now Gn,q = Gal(F n,q, k(X)) = Gal(Θ, k(x)), and hence in view of(3) and (4), by the Cycle Lemma we see that if k is algebraically closed and q = pthen Gn,p contains a p-cycle; by the Basic Extension Principle, as a permutationgroup, Gal(F n,q, ¯k(X)) is a subgroup of Gal(F n,q, k(X)) where ¯k is an algebraicclosure of k; therefore, without assuming k to be algebraically closed, we have thatif q = p then Gn,p contains a p-cycle.Let prime denote the twisted Y -derivative at y. ThenΘ′(Y ) = [Y q(Y t −x)]′ + (1)′(by linearity)= [Y q(Y t −x)]′(because constant′ = 0)= (Y + y)q(Y t −x)′ + Y q−1(yt −x)(by power product rule)= (Y + y)q(Y t)′ + Y q−1(yt −x)(because constant′ = 0)= (Y + y)q(Y t)′ −y−qY q−1(because x = yt + y−q)and hence by the definition of (Y t)′ we getΘ′(Y ) = [(Y + y)q](Y + y)t −ytY−y−qY q−1and thereforeΘ′(0) = tyq+t−1 ̸= 0.Let ∆(Y ) be the polynomial obtained by reciprocating the roots of Θ′(Y ). Then∆(Y ) is a monic polynomial of degree q + t −1 in Y with coefficients in k(y) andwe have∆(Y ) = Y q+t−1tyq+t−1Θ′ 1Y=Y qyq 1Y + yq Y tyt h 1Y + yt −ytitYy 1Y− Y q+t−1tyq+t−1 y−qY q−1=[(Y + y−1)q]y[(Y + y−1)t −Y t]t−Y tty2q+t−1 .Let Λ(Z) be the polynomial obtained by multiplying the roots of ∆(Z) by ty.Then Λ(Z) is a monic polynomial of degree q + t −1 in an indeterminate Z with

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC43coefficients in k(y) and we haveΛ(Z) =(ty)q+t−1∆ Zty=(ty)q Zty + 1yq (ty)t−1yZty + 1yt−Ztytt−(ty)q+t−1 Ztytty2q+t−1=[(Z + t)q](Z + t)t −Ztt2−tq−2Ztyq+tand henceΛ(Z) = (Z + t)qγ(Z) −tq−2y−q−tZt,where γ(Z) is the monic polynomial of degree t −1 in Z with coefficients in theprime field kp ⊂k given byγ(Z) = t−2[(Z + t)t −Zt] =t−1Xi=0γiZt−1−i,where γi = ti−1ti+1∈kp with γ0 = 1. Now γ(0) = tt−2 and γ(−t) = (−1)t+1tt−2and hence γ(0) ̸= 0 ̸= γ(−t).Let z be a root of Λ(Z) in some overfield of k(y).

Then solving Λ(z) = 0 for ywe get(∗)yq+t =tq−2zt(z + t)qγ(z)and so in particular we see that z is transcendental over k. Consider the valuationκ : z = 0 of k(z)/k and let ζ be an extension of it to k(y, z). Since t ̸= 0 ̸= γ(0),by the above equation we see that(q + t)ζ(y) = ζ(yq+t) = ζtq−2zt(z + t)qγ(z)= ¯r(ζ : κ)κtq−2zt(z + t)qγ(z)= ¯r(ζ : κ)t.Since q + t and t are coprime, we must have ¯r(ζ : κ) ≥q + t. Therefore [k(y, z) :k(z)] ≥q+t and hence by the above equation for yq+t we get [k(y, z) : k(z)] = q+t.Since [k(y, z) : k(z)] = q + t, by the above equation for yq+t we see that thepolynomial (z + t)qγ(z)Y q+t −tq−2zt is irreducible in k(z)[Y ].

Since the nonzeropolynomials (Z+t)qγ(Z) and tq−2Zt in Z with coefficients in k have no nonconstantcommon factor in k[Z], we conclude that the polynomial(Z + t)qγ(Z)Y q+t −tq−2Ztis irreducible in k[Y, Z]. Consequently the polynomial (Z + t)qγ(Z)yq+t −tq−2Ztis irreducible in k(y)[Z], and hence the polynomial Λ(Z) is irreducible in k(y)[Z].68Therefore the polynomials Λ(Y ) and ∆(Y ) are irreducible in k(y)[Y ], and hence68This also follows from the First Irreducibility Lemma.In fact, the above argument is aspecial case of the argument used in the proof of the said lemma.

44S. S. ABHYANKARGal(Λ, k(y)) = Gal(∆, k(y)) = a transitive permutation group of degree n−1.

Sincethis group is the one-point stabilizer of Gn,q = Gal(F n,q, k(x)), we conclude that:Gn,q is 2-transitive.As we have noted above, if q = p then Gn,p contains a p-cycle, and hence by a1873 Theorem of Jordan we see that if also t > 2, then Gn,p = An or Sn. In Result4 on page 841 of the 1957 paper [A3] we have noted that, as a consequence of thegenus formula, every member of the algebraic fundamental group of the affine lineover an algebraically closed ground field of characteristic p ̸= 0 is a quasi p-group.Also obviously, for n > 1 and p ̸= 2, the symmetric group Sn is not a quasi p-group.Therefore if k is algebraically closed and q = p ̸= 2 and n = p + t with t ̸≡0(p) andt > 2, then Gn,p = the alternating group An.Here is the 1873 Theorem of Jordan [J3] we spoke of; proofs can also be found in13.9 on page 39 of Wielandt [Wi] and in 3.7 on page 331 of volume III of Huppert-Blackburn [HB].Jordan’s Theorem.

If a primitive permutation group G of degree n = p + tcontains a p-cycle, where p is prime and t > 2, then G = An or Sn.69In the case p = 2, a p-cycle is simply a transposition, and obviously Sn is theonly 2-transitive permutation group of degree n which contains a transposition.70Therefore if q = p = 2 and n = p + t with t ̸≡0(p) and t > 0, then Gn,p = thesymmetric group Sn.To throw away a second root of F n,q, or more precisely to throw away a root ofΛ(Z), this time around let prime denote the twisted Z-derivative at z. ThenΛ′(Z) = [(Z + t)qγ(Z)]′ −tq−2y−q−t[Zt]′(by linearity)= (Z + z + t)qγ′(Z) + Zq−1γ(z) −tq−2y−q−t[Zt]′(by power product rule)= (Z + z + t)qγ′(Z) + Zq−1γ(z) −γ(z)(z + t)qz−t[Zt]′(by (*))= (Z + z + t)q¯γ(Z) + γ(z)Zq−1 −γ(z)(z + t)qz−t¯ρt−1(Z),where ¯ρj(Z) is the monic polynomial of degree j ≥0 in Z with coefficients in theprime field kp ⊂k given by¯ρj(Z) = (Z + z)j+1 −zj+1Zand¯γ(Z) =t−2Xj=0γt−2−j ¯ρj(Z).Let Φ(W) be the monic polynomial of degree q + t −2 in an indeterminate W69A 2-transitive permutation group is automatically primitive.70A 2-transitive permutation group which contains a transposition, must automatically containall transpositions, and hence must be the symmetric group.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC45with coefficients in k(z) obtained by multiplying the roots of Λ′(W) by z. ThenΦ(W) =zq+t−2Λ′Wz=zqWz + z + tq zt−2¯γWz+ γ(z)zt−1"zq−1Wzq−1#−γ(z)(z + t)qzq−t−1zt−1¯ρt−1Wzand henceΦ(W) = (W + z2 + tz)qˆγ(W) + γ(z)zt−1W q−1 −γ(z)(z + t)qzq−t−1ˆρt−1(W)where ˆρj(W) is the monic polynomial of degree j ≥0 in W with coefficients in k[z]given byˆρj(W) = (W + z2)j+1 −z2j+2Wandˆγ(W) =t−2Xj=0zt−2−jγt−2−j ˆρj(W).Note that if t = 1 then ˆγ(W) = 0,71 whereas if t > 1 then ˆγ(W) is a monicpolynomial of degree t −2 in W with coefficients in k[z], and hence, in particular,if t = 2 then ˆγ(W) = 1.For a moment suppose that t = 1; then γ(Z) is a monic polynomial of degreet −1 = 0 in Z and hence γ(Z) = 1 and therefore γ(z) = 1; also as noted aboveˆγ(W) = 0; finally ˆρt−1(W) is a monic polynomial of degree t −1 = 0 in W andhence ˆρt−1(W) = 1. Thus,(1*)t = 1 ⇒(Λ(Z) = (Z + 1)q −y−q−1ZandΦ(W) = W q−1 −(z + 1)qzq−2.Next, for a moment suppose that t = 2; thenγ(Z) = 2−2[(Z + 2)2 −Z2] = Z + 1and hence γ(z) = z +1; also as noted above ˆγ(W) = 1 and clearly (W +z2 +tz)q =W q + (z + 2)qzq; finally ˆρt−1(W) is the monic polynomial of degree 1 in W givenby ˆρt−1(W) = W + 2z2 and henceΦ(0) =(z + 2)qzq −(z + 1)(z + 2)qzq−3(2z2)=(z + 2)qzq−1(z −2z −2) = −(z + 2)q+1zq−1.Thus,(2*)t = 2 ⇒Λ(Z) = (Z + 2)q(Z + 1) −2q−2y−q−2Z2andΦ(W) = W q + (z + 1)zW q−1 −(z + 1)(z + 2)qzq−3W−(z + 2)q+1zq−1.71Because in that case the summation in the above expression for ˆγ(W ) is empty.

46S. S. ABHYANKARIn the case of general t, by looking at the equation (∗), we get the following“ramification diagram” for the field extensions k(x) ⊂k(y) ⊂k(y, z), where thesquare bracketed numbers are the reduced ramification exponents.→ζi : ψi(z) = 0[1] for 1 ≤i ≤t∗−1→w∞: y = ∞[t]—→ζt∗: ψt∗(z) = 0[q]v∞: x = ∞—→ζ0 : z = 0[t]→w0 : y = 0[q] —–→ζ∞,j : z = ∞and ǫj(yq+tt′ zq−1t′ ) = 0[ q−1t′ ]for 1 ≤j ≤t′′.To explain the top half of the extreme right hand side in the above diagram, firstnote that, since t ̸≡0(p), the Z-discriminant of (Z + t−1)t −t−t ∈k[Z] is a nonzeroconstant; now 0 is a root of this monic polynomial and by reciprocating the rootsof ((Z + t−1)t −t−t/Z we get γ(Z); therefore γ(Z) is free from multiple factors;since γ(−t) ̸= 0, we conclude that γ(Z) = ψ1(Z)ψ2(Z) · · · ψt∗−1(Z), where, uponletting ψt∗(Z) = Z + t, we have that ψ1(Z), ψ2(Z), .

. .

, ψt∗(Z) are pairwise distinctnonconstant monic irreducible polynomials in Z with coefficients in k; let κi be thevaluation of k(z)/k given by ψi(z); since GCD(q + t, q) = 1, by (∗) we see that κihas a unique extension ζi to k(y, z), and ζ1, ζ2, . .

. , ζt∗are exactly all the extensionsof w∞to k(y, z), and we have ¯r(ζi : κi) = q + t for 1 ≤i ≤t∗, ¯r(ζi : w∞) = 1 for1 ≤i ≤t∗−1, and ¯r(ζt∗: w∞) = q.Turning to the bottom half, since GCD(q + t, t) = 1, by (∗) we see that thevaluation κ0 : z = 0 of k(z)/k has a unique extension ζ0 to k(y, z); since t ̸≡0(p),upon letting t′ = GCD(q + t, q −1) we have Zt′ −tq−2 = ǫ1(Z)ǫ2(Z) .

. .

ǫt′′(Z),where ǫ1(Z), ǫ2(Z), . .

. , ǫt′′(Z) are pairwise distinct nonconstant monic irreduciblepolynomials in Z with coefficients in k; now by (∗) we see that the valuationκ∞: z = ∞of k(z)/k has t′′ extensions ζ∞,1, ζ∞,2, .

. .

, ζ∞,t′′ to k(y, z) whichare characterized by saying that ζ∞,j(ǫj(y(q+t)/t′z(q−1)/t′)) > 0 for 1 ≤j ≤t′′, andmoreover ζ0, ζ∞,1, ζ∞,2, . .

. , ζ∞,t′′ are exactly all the extensions of w0 to k(y, z), andwe have¯r(ζ0 : κ0) = q + t, ¯r(ζ0 : w0) = t, ¯r(ζ∞,j : κ∞) = (q + t)/t′ for 1 ≤j ≤t′′,and ¯r(ζ∞,j : w0) = (q −1)/t′ for 1 ≤j ≤t′′.Finally note that, since v∞is the only valuation of k(x) which is ramified ink(y), no valuation of k(y), other than w0 and w∞, can be ramified in k(y, z), andno valuation of k(y, z), other than ζ1, ζ2, .

. .

, ζt∗, ζ0, ζ∞,1, ζ∞,2, . .

. , ζ∞,t′′, can beramified in the splitting field of Φ(W) over k(y, z).Now, as noted above, if t = 1 then γ(Z) = 1, whereas if t = 2 then γ(Z) = Z +1,and hence, in connection with the top half of the above diagram, we have that(1′)t = 1 ⇒yq+1 =z(z + 1)q , and t∗= 1 and ψ1(z) = z + 1whereas(2′) t = 2 ⇒yq+2 =2q−2z2(z + 2)q(z + 1), t∗= 2, ψ1(z) = z + 1, and ψ2(z) = z + 2.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC47Concerning the bottom half of the above diagram we note thatt = 2 ⇒LCM(¯r(ζ0 : w0), ¯r(ζ∞,1 : w0), . .

. , ¯r(ζ∞,t′′ : w0))¯r(ζ0 : w0)=( q−16if q ≡1(3),q−12if q ̸≡1(3).

(2′′)Recall that a permutation group is said to be semiregular if its stabilizer, at anypoint in the permuted set, is the identity; in our terminology this is equivalent to1-antitransitive. By analogy, let us say that a univariate monic polynomial, withcoefficients in some field, is semiregular over that field if by adjoining any one of itsroots we get the splitting field.

Note that then, assuming the roots to be distinct,the polynomial is semiregular if and only if its Galois group is semiregular.For a moment suppose that t = 1 and k contains all the (q −1)th roots of 1.Then by (1∗) we see that Φ(W) is semiregular over k(y, z) and hence its Galoisgroup is 1-antitransitive; since this group is the 2-point stabilizer of the 2-transitivepermutation group Gn,q, we conclude that Gn,q is a (2,3) group. Now if p = 2then ζ1((z + 1)qzq−2) = q(q + 1) and GCD(q −1, q(q + 1)) = 1, and therefore by(1∗) we see that Φ(W) is irreducible in k(y, z)[W] and its Galois group is a cyclicgroup of order q −1, and hence in particular the said Galois group is a (1,1) group;consequently, Gn,q is a (3,3) group.

On the other hand, if p ̸= 2 then by (1′) wehave (z + 1)qzq−2 = zq−1y−(q+1) and hence by (1∗) we getΦ(W) = (W (q−1)/2 −z(q−1)/2y−(q+1)/2)(W (q−1)/2 + z(q−1)/2y−(q+1)/2)and alsoζ1(z(q−1)/2y−(q+1)/2) = 12ζ1((z + 1)qzq−2) = q(q + 1)2andGCDq −12, q(q + 1)2= 1,and hence the above two factors of Φ(W) are irreducible in k(y, z)[W] and theGalois group of Φ(W) over k(y, z) is a cyclic group of order q−12 . Thus, if t = 1and k contains all the (q −1)th roots of 1, then Gn,q is a (2,3) group, and moreoverif also p = 2 then Gn,q is a (3,3) group and its order is (q + 1)q(q −1), whereas ifp ̸= 2 then Gn,q is not a (3,3) group and its order is (q+1)q(q−1)2.Therefore by an obvious corollary of the Zassenhaus-Feit-Suzuki Theorem we seethat if t = 1 and k contains all the (q −1)th roots of 1, then Gn,q = PSL(2, q) withthe possible exception that for q = 7 we may have Gn,q = A Γ L(1, 8).The said corollary may be formulated in the following manner.

For deducingthis corollary from the Zassenhaus-Feit-Suzuki Theorem, the only thing we need tocheck is that if the (degree, order) pair (2l, 2l(2l −1)l) of a Feit Group A Γ L(1, 2l),where l is a prime, equals (q + 1, (q+1)q(q−1)2) then l = (2l −2)/2 = 2l−1 −1 and bydirect calculation we see that this is not possible for l = 2, but is possible for l = 3,and is never possible for l ≥4 because then 2l−1 −1 = 1 + 2 + 22 + · · · + 2l−2 ≥1 + 2(l −2) = l + (l −3) > l.

48S. S. ABHYANKARCorollary of the Zassenhaus-Feit-Suzuki-Theorem.

If p = 2, i.e., if q is apositive power of 2, then PSL(2, q) is the only (3, 3) group of degree q + 1. If p ̸= 2,i.e., if q is a positive power of an odd prime p, then PSL(2, q) is the only (2, 3) groupof degree q + 1 and order (q+1)q(q−1)2with the exception that, in the case q = 7, thegroup A Γ L(1, 8) also satisfies this description.Now, on the one hand, by Result 4 on page 841 of [A3] we know that, in casek is algebraically closed, Gn,q is a quasi p-group, and, on the other hand, for thegroup A Γ L(1, 8) = GF(8) ⋊Γ L(1, 8) we have Γ L(1, 8) = GL(1, 8) ⋊Aut (GF(8))and hence Aut (GF(8)) = Z3 is a homomorphic image of A Γ L(1, 8) and thereforeA Γ L(1, 8) is not a quasi 7-group.72 Consequently, in case k is algebraically closed,the said exception cannot occur and hence if k is algebraically closed and t = 1,then Gn,q = PSL(2, q) = PSL(2, n −1).Before proceeding further, let us record the followingFourth Irreducibility Lemma.

Remember that p is a prime number, and t isa positive integer.73 For 0 ≤i ≤t, let fi(Y ) and gi(Y ) be monic polynomials ofdegree p + t −i in Y with coefficients in a field Ki, such that gi(Y ) is obtainedfrom fi(Y ) by making one or more of the three operations of multiplying the rootsby a nonzero quantity in Ki, reciprocating the roots,74 and decreasing the roots bya quantity in Ki. Assume that, for 1 ≤i ≤t, the polynomial fi(Y ) is obtained bythrowing away a root αi of gi−1(Y ), and for the field Ki we have Ki = Ki−1(αi).Also assume that f0(Y ) has no multiple roots.75 Finally assume that there existvaluations ˆu(i) and u(i) of Ki for 0 ≤i ≤t such that ˆu(0) = u(0); for 1 ≤i ≤t, thevaluation ˆu(i) is an extension of ˆu(i−1) to Ki; for 1 ≤i ≤t, the valuation u(i) is anextension of u(i−1) to Ki; for every i with 1 ≤i ≤t we have ¯r(ˆu(i) : ˆu(i−1)) ̸≡0(p);and for some j with 1 ≤j ≤t we have ¯r(u(j) : u(j−1)) ≡0(p).Let u′ be an extension of u(t) to a splitting field K′ of gt(Y ) over Kt, and letr∗= Qti=1 ¯r(u(i) : u(i−1)) and r′ = ¯r(u′ : u(t)).

Let u(t)1= u(t), u(t)2 , . .

. , u(t)δbe allthe extensions of u(t−1) to Kt, and letr′′ = LCM(¯r(u(t)1: u(t−1)), .

. .

, r(u(t)δ: u(t−1)))¯r(u(t) : u(t−1)).Finally let ¯u(0), . .

. , ¯u(t), ¯u′ be any valuations of K0, .

. .

, Kt, K′ respectively, with¯u(0) = u(0), such that ¯u(i) is an extension of ¯u(i−1) for 1 ≤i ≤t, and ¯u′ is anextension of ¯u(t) for 1 ≤i ≤t, and let ¯r∗= Qti=1 ¯r(¯u(i) : ¯u(i−1)) and ¯r′ = ¯r(¯u′ :¯u(t)).Then the polynomial gt(Y ) is irreducible in Kt[Y ], and we have | Gal(gt, Kt)| ≡0(r′) and r′ ≡0(r′′) and ¯r∗¯r′ = r∗r′. Moreover, if gt−1(Y ) is irreducible in Kt−1[Y ],and the residue field of u(0) is an algebraically closed field of the same characteristicas K0, and ¯r(uj : u(t−1)) ̸≡0(char K0) for 1 ≤j ≤δ, then r′ = r′′.To see this, take an extension ˆu′ of ˆu(t) to K′.

By assumption¯r(u(j) : u(j−1)) ≡0(p)for some j with 1 ≤j ≤t72A quasi p-group can be characterized as a finite group having no homomorphic image whoseorder is prime to p and greater than 1.73In this lemma, t is allowed to be divisible by p.74In case of reciprocating the roots, we of course assume that zero is not a root.75In any field extension of K0.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC49and clearly ¯r(u′ : u(0)) = r∗r′, and hence we must have ¯r(u′ : u(0)) ≡0(p). Nowboth the valuations ˆu′ and u′ are extensions of ˆu(0) = u(0) to K′ which is a Galoisextension of K0,76 and hence ¯r(ˆu′ : ˆu(0)) = ¯r(u′ : u(0)), and therefore ¯r(ˆu′ : ˆu(0)) ≡0(p).

By assumption ¯r(ˆu(i) : ˆu(i−1)) ̸≡0(p) for 1 ≤i ≤t and clearly ¯r(ˆu′ : ˆu(0)) =¯r(ˆu(1) : ˆu(0)) . .

. ¯r(ˆu(t) : ˆu(t−1))¯r(ˆu′ : ˆu(t)), and hence we must have ¯r(ˆu′ : ˆu(t)) ≡0(p).

By assumption K′ is a splitting field of gt(Y ) over Kt and hence | Gal(gt, Kt)|must be divisible by ¯r(ˆu′ : ˆu(t)), and therefore | Gal(gt, Kt)| ≡0(p); since gt(Y ) is amonic polynomial of degree p in Y with coefficients in Kt, we conclude that gt(Y )is irreducible in Kt[Y ]. Since K′ is a splitting field of gt(Y ) over Kt, we also get| Gal(gt, Kt)| ≡0(r′).

Now K′ is a Galois extension of Kt−1, and hence r′ ≡0(r′′).Since K′ is a Galois extension of K0, and ¯u′ and u′ are extensions of ¯u(0) = u(0) toK′, we also get ¯r∗¯r′ = ¯r(¯u′ : ¯u(0)) = ¯r(u′ : u(0)) = r∗r′. If gt−1(Y ) is irreducible inKt−1[Y ] then K′ is a least Galois extension of Kt−1 containing Kt, and hence thelast assertion follows from Proposition 7 on page 845 of [A3].Note that this lemma gives an alternative proof of the irreducibility of Λ(Y ) incase of q = p and t = 1.

In fact we have the followingCorollary of the Fourth Irreducibility Lemma. If k is algebraically closedand F(Y ) is a monic irreducible polynomial of degree p+1 in Y with coefficients ink(X) such that no valuation of k(X)/k, other than the valuation given by X = ∞,is ramified in the root field of F(Y ) over k(X), then the Galois group Gal(F, k(X))is 2-transitive.To deduce this from the lemma, it suffices to note that, in view of Proposition 6on page 835 of [A3], in the root field of F(Y ) over k(X), the valuation X = ∞mustsplit into two valuations with reduced ramification exponents p and 1 respectively.77More interestingly, in view of the ramification diagram and implication (2′′),by the above lemma we see that if q = p and n = p + t with t = 2 ̸≡0(p),then Φ(W) is irreducible in k(y, z)[W] and so Gn,p is 3-transitive, and the orderof its 3-point stabilizer Gal(Φ, k(y, z)) is divisible by τ, where τ = p−16or τ = p−12according as p ≡1(3) or p ̸≡1(3), and hence the order of Gn,p is divisible by (p +2)(p + 1)pτ, and moreover the reduced ramification exponent of any extension of ζ0(resp.

ζ1, ζ2, ζ∞,1, . .

. , ζ∞,t′′) to a splitting field of Φ(W) over k(y, z) is divisible by τ(resp.

pτ, τ, 1, . .

. , 1), and, in case k is algebraically closed, the reduced ramificationexponent of any extension of ζ0 (resp.

ζ1, ζ2, ζ∞,1, . .

. , ζ∞,t′′) to a splitting field ofΦ(W) over k(y, z) equals τ (resp.

pτ, τ, 1, . .

. , 1).78Therefore by the following Corollary of CTT we see that if q = p and n = p + twith t = 2 ̸≡0(p), and if Gn,p is neither equal to An nor equal to Sn,79 then eitherp = 7 and Gn,p = PSL(2, 8), or p = 7 and Gn,p = P Γ L(2, 8), or p = 31 andGn,p = P Γ L(2, 32).Corollary of CTT.

Let G be a 3-transitive permutation group of degree n = p+2with an odd prime p, such that G is neither equal to An nor equal to Sn. Then76Because it is a splitting field of f0(Y ) over K0.77Alternatively, in view of Proposition 6 on page 835 of [A3], this Corollary can be deducedfrom the fact that if G is a transitive permutation group of degree p + 1 of order divisible by pthen G must contain a p-cycle and hence it must be 2-transitive.78In the present case, by (2′) we have t∗= 2.79That is if, as a permutation group, it is not isomorphic to either of these.

50S. S. ABHYANKARp = 2µ −1 for some prime µ,80 and either G = PSL(2, 2µ) or G = P Γ L(2, 2µ).Moreover, if the order of G is divisible by (p + 2)(p + 1)pτ, where τ = p−16or p−12according as p ≡1(3) or p ̸≡1(3), then either p = 7 and G = PSL(2, 8), or p = 7and G = P Γ L(2, 8), or p = 31 and G = P Γ L(2, 32).To deduce the above Corollary from CTT, by inspection we see that no grouplisted in items (3) to (8) of CTT has degree of the form p + 2 with an odd primep.

Moreover the degree of every group listed in items (1) and (2) of CTT is of theform πµ + 1, with prime π and positive integer µ, and if we have πµ + 1 = p + 2then, in case of odd π, we would get the contradiction p = πµ −1 ≡0(2). Thereforeby CTT we conclude that p = 2µ −1 for some prime µ, and G is a group betweenPSL(2, 2µ) and P Γ L(2, 2µ).

Since µ is prime and PSL(2, 2µ) is a normal subgroupof P Γ L(2, 2µ) of index µ, there are no groups strictly between PSL(2, 2µ) andP Γ L(2, 2µ). Therefore G = PSL(2, 2µ) or G = P Γ L(2, 2µ).

Now if the order ofG is divisible by (p + 2)(p + 1)pτ, then (p + 2)(p + 1)pµ = | P Γ L(2, 2µ)| ≥|G| ≥(p + 2)(p + 1)pτ and hence µ ≥τ ≥p−16= (2µ −2)/6 and therefore 2µ−1 −1 ≤3µ;on the other hand, if µ > 5 then 2µ−1 −1 = (1 + 2 + 22) + (23 + 24 + · · · + 2µ−2) ≥7 + 8(µ −4) = 3µ + 5(µ −5) > 3µ. Therefore if the order of G is divisible by(p + 2)(p + 1)pτ, then we must have µ = 2 or 3 or 5, i.e., p = 3 or 7 or 31; now thecase p = 3 is ruled out because then PSL(2, 2µ) = An and P Γ L(2, 2µ) = Sn withµ = 2 and n = 5; also, in case of p = 31, we cannot have G = PSL(2, 32) because| PSL(2, 32)| = 33 · 32 · 31 < 33 · 32 · 31 · 5 = (p + 2)(p + 1)pτ.Once again by Result 4 on page 841 of [A3] we know that, in case k is algebraicallyclosed, the group Gn,q is a quasi p-group, and obviously, in the case p ̸= 2, the groupSn is not a quasi p-group because it has Z2 as a homomorphic image, and likewise,in the case of a Mersenne prime p = 2µ −1, with a prime µ, the group P Γ L(2, 2µ)is not a quasi p-group because it has Zµ as a homomorphic image.

Therefore bythe above italicized conclusion we see that if k is algebraically closed and q = p andn = p + t with t = 2 ̸≡0(p), then in the case p ̸= 7 we have Gn,p = An, whereas inthe case p = 7 we have Gn,p = An or PSL(2, n −1).Now let ω be an element in an overfield of k(y, z) such that(∗∗)Φ(ω) = 0,and let Ψ(U) be the monic polynomial of degree q −1 in an indeterminate U withcoefficients in k(y, z, ω) obtained by throwing away the root ω of Φ(U). Actually,Φ(U) ∈k(z)[U] and hence Ψ(U) ∈k(z, ω)[U].To compute Ψ(U), let prime stand for the twisted U-derivative at ω; then wehave Ψ(U) = Φ′(U), and by the prime power rule we have (U q)′ = U q−1, and by80Recall that a prime p is called a Mersenne prime if it is of the form 2µ −1 for some positiveinteger µ, and then automatically µ is itself a prime number.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC51the power rule we have (U 0)′ = 0 and (U)′ = 1, and by direct calculation we have(U q−1)′ =(U + ω)q−1 −ωq−1U=ωq−1 h−1 +1 + Uωq 1 + Uω−1iU=ωq−1 h−1 +1 + Uqωq 1 −Uω + U2ω2 −U3ω3 + · · ·iU=ωq−1 h−1 +1 −Uω + U2ω2 −· · · + Uq−1ωq−1iU=U q−2 −ωU q−3 + w2U q−4 −· · · + ωq−3U −wq−2,and therefore, in view of (2∗), by linearity we see that(2∗∗)t = 2 ⇒(Ψ(U) = U q−1 + (z + 1)z[U q−2 −ωU q−3 + ω2U q−4 −· · · + ωq−3U]−(z + 1)z[ωq−2 + (z + 2)qzq−4].For a moment suppose that k is algebraically closed and q = p = 7 and t = 2.Since A9 is 4-transitive and PSL(2, 8) is sharply 3-transitive, by the above italicizedassertion it follows that Gn,p ̸= An ⇔Gn,p = PSL(2, n −1) ⇔Ψ(U) is reduciblein k(y, z, ω)[U] ⇔Ψ(U) completely factors (into linear factors) in k(y, z, ω)[U]. Toconvert the question of reducibility of Ψ(U) over the field k(y, z, ω) to its reducibilityover the simpler field k(z, ω), first note that k(z, ω) ⊂k(y, z, ω) and hence Ψ(U)is reducible in k(z, ω)[U] ⇒Ψ(U) is reducible in k(y, z, ω)[U]; next, by (2′) we seethat k(y, z, ω) is a cyclic extension of k(z, ω) of degree 1 or 3 or 9, and hence Ψ(U)completely factors in k(y, z, ω)[U] ⇒the degrees of the irreducible factors of Ψ(U) ink(z, ω)[U] are (3,3) or (3,1,1,1) or (1,1,1,1,1,1).

Therefore if k is algebraically closedand q = p = 7 and n = p + t with t = 2 then Gn,p ̸= An ⇔Gn,p = PSL(2, n −1) ⇔Ψ(U) is reducible in k(y, z, ω)[U] ⇔Ψ(U) completely factors (into linear factors)in k(y, z, ω)[U] ⇔Ψ(U) is reducible in k(z, ω)[U] ⇔the degrees of the irreduciblefactors of Ψ(U) in k(z, ω)[U] are (3, 3) or (3, 1, 1, 1) or (1, 1, 1, 1, 1, 1).To further simplify the question of reducibility of Ψ(U), we proceed to eliminatethe last two possibilities of (3, 1, 1, 1) and (1, 1, 1, 1, 1, 1) for the degrees of theirreducible factors of Ψ(U) in k(z, ω). Afterwards, we shall also indicate how thequestion of reducibility of Ψ(U) in k(z, ω)[U] can be converted to its reducibility inkp(z, ω)[U].Recall that γ(Z) is a monic polynomial of degree t −1 in Z with coefficients inthe prime field kp ⊂k.

Lety∗= (z + t)qγ(z)ztandΛ∗(Z) = (Z + t)qγ(Z) −y∗Zt ∈kp(y∗)[Z].Now the element y∗is transcendental over k, the polynomial Λ∗(Z) is irreduciblein k(y∗)[Z], the element z is a root of Λ∗(Z), the polynomial Λ′(Z) ∈kp(z)[Z] isobtained by throwing away the root z of Λ∗(Z), the polynomial Φ(W) ∈kp(z)[W]

52S. S. ABHYANKARis obtained by multiplying the roots of Λ′(W) by z, the element ω is a root ofΦ(W), the polynomial Ψ(U) ∈kp(z, ω)[U] is obtained by throwing away the root ωof Φ(U), and finally if q = p and t = 2 then the polynomial Φ(W) is irreducible ink(z)[W] and hence Gal(Λ∗, k(y∗)) and Gal(Λ∗, kp(y∗)) are 2-transitive permutationgroups of degree p + 1.For t = 2 we have γ(Z) = Z + 1 and so(2∗∗∗)t = 2 ⇒y∗= (z + 2)q(z + 1)z2and Λ∗(Z) = (Z + 2)q(Z + 1) −y∗Z2.If q = p and t = 2 then either directly by the above expression for y∗, orindirectly by the formulae (∗) and (2′) and the ramification diagram of the fieldextension k(y) ⊂k(y, z), we see that w∗0 : y∗= 0 and w∗∞: y∗= ∞are the onlyvaluations of k(y∗)/k which are ramified in k(z), the valuation w∗0 splits into thevaluations κ1 : z + 1 = 0 and κ2 : z + 2 = 0 of k(z)/k with reduced ramificationexponents ¯r(κ1 : w∗0) = 1 and ¯r(κ2 : w∗0) = p, and the valuation w∗∞splits intothe valuations κ0 : z = 0 and κ∞: z = ∞of k(z)/k with reduced ramificationexponents ¯r(κ0 : w∗∞) = 2 and ¯r(κ∞: w∗∞) = p −1.

Thus, in case of q = p andt = 2 we have the special ramification diagram for the field extension k(y∗) ⊂k(z)where the square bracketed numbers are the reduced ramification exponents:→κ1 : z + 1 = 0[1]w∗0 : y∗= 0 —–→κ2 : z + 2 = 0[p]→κ0 : z = 0[2]w∗∞: y∗= ∞—→κ∞: z = ∞[p −1].For a moment suppose that q = p and t = 2. Since the degree of Φ(W) is p, bythe above diagram we see that κ1 has a unique extension λ1 to k(z, ω), for λ1 wehave ¯r(λ1 : κ1) = p, and the reduced ramification exponent of any extension of λ1to a splitting field of Ψ(U) over k(z, ω) equals the reduced ramification exponentof any extension of κ2 to the said splitting field.

By direct calculation with theexpression of Φ(W) given in (2∗), it can be shown that κ2 has 3 extensions tok(z, ω) with reduced ramification exponents (1, p−12 , p−12 ). Therefore, say by theFourth Irreducibility Lemma, the reduced ramification exponent of any extensionof λ1 to the said splitting field must be divisible by p−12 .

Consequently the degreeof some irreducible factor of Ψ(U) in k(z, ω)[U] must be ≥p−12 . The said directcalculation will not be given here,81 and hence, for deducing the last consequence,at least in the case when k is algebraically closed, let us indirectly argue in thefollowing manner.So for a moment suppose that k is algebraically closed and q = p and t = 2.Now in view of MRT,82 by implication (2′) and by the paragraphs “Finally note81But it will be given in my forthcoming paper [A7].82MRT = Abhyankar’s Lemma = Pages 181–186 of [A5].

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC53that...” and “More interestingly...”, we see that κ1, κ2, κ0 are the only valuationsof k(z)/k which are ramified in a splitting field of Φ(W) over k(z), and the reducedramification exponent of any extension of κ1 (resp. κ2, κ0) to the said splitting fieldis p(p−1)2(resp.

p−12 , p−12 ). Since the degree of Φ(W) is p, it follows that κ1 hasa unique extension λ1 to k(z, ω), for λ1 we have ¯r(λ1 : κ1) = p, and the reducedramification index of any extension of λ1 to a splitting field of Ψ(U) over k(z, ω) isp−12 .

Therefore the degree of some irreducible factor of Ψ(U) in k(z, ω)[U] must be≥p−12 .Finally for a moment suppose that k is algebraically closed and q = p = 7and t = 2. Then by the above paragraph and the paragraph following (2∗∗) wesee that Ψ(U) is reducible in k(z, ω)[U] ⇔the degrees of the irreducible fac-tors of Ψ(U) in k(z, ω)[U] are (3,3) or (3,1,1,1).Thus, Ψ(U) is reducible ink(z, ω)[U] ⇔Gal(Λ∗, k(y∗)) is 2-transitive but neither 3-transitive nor sharply 2-transitive.

Therefore, since AGL(1, 8) is sharply 2-transitive, by the Special CDTwe conclude that Ψ(U) is reducible in k(z, ω)[U] ⇔Gal(Λ∗, k(y∗)) = PSL(2, 7) orA Γ L(1, 8); by the Zassenhaus-Feit-Suzuki Theorem, both of these are (2,3) groups,and hence Ψ(U) is reducible in k(z, ω)[U] ⇔the degrees of the irreducible factorsof Ψ(U) in k(z, ω)[U] are (3,3) and they have a common splitting field over k(z, ω)with Galois group Z3. It follows that Ψ(U) is reducible in kp(z, ω)[U] ⇔the de-grees of the irreducible factors of Ψ(U) in kp(z, ω)[U] are (3, 3) ⇔Gal(Λ∗, kp(y∗))is 2-transitive but neither 3-transitive nor sharply 2-transitive ⇔Gal(Λ∗, kp(y∗))is not 3-transitive; therefore by CTT, Special CDT, and the Zassenhaus-Feit-Suzuki Theorem, we see that Ψ(U) is reducible in kp(z, ω)[U] ⇔Gal(Λ∗, kp(y∗)) =PSL(2, 7) or A Γ L(1, 8) ⇔Gal(Λ∗, kp(y∗)) ̸∈{S8, A8,AGL(3, 2), PGL(2, 7)}.Let us close this long section with aSummary about the bar polynomial.

Let k be an algebraically closed fieldof characteristic p ̸= 0. Let q be a positive power of p and let n = q + t wheret is a positive integer with t ̸≡0(p).

Let F n,q = Y n −XY t + 1 ∈k[X, Y ] andlet Gn,q = Gal(F n,q, k(X)).Without using CT we have shown that if t = 1then Gn,q = PSL(2, q) = PSL(2, n −1), whereas if q = p and t > 2 ̸= p thenGn,q = An, and finally if q = p and t > 2 = p then Gn,q = Sn.83 Using CT we haveshown that if q = p ̸= 7 and t = 2 then Gn,q = An. Using CT, and referring to(2∗), (2′), (∗∗), (2∗∗), (2∗∗∗) for notation and remembering that z is transcendentalover k, we have also shown that if q = p = 7 and t = 2 then Gn,q ̸= An ⇔Gn,q =PSL(2, n−1) ⇔Ψ(U) is reducible in k(y, z, ω)[U] ⇔Ψ(U) completely factors (intolinear factors) in k(y, z, ω)[U] ⇔Ψ(U) is reducible in k(z, ω)[U] ⇔the degrees of theirreducible factors of Ψ(U) in k(z, ω)[U] are (3, 3) and they have a common splittingfield over k(z, ω) with Galois group Z3 ⇔Gal(Λ∗, k(y∗)) = A Γ L(1, 8) ⇔Ψ(U)is reducible in kp(z, ω)[U] ⇔the degrees of the irreducible factors of Ψ(U) inkp(z, ω)[U] are (3, 3) ⇔Gal(Λ∗, kp(y∗)) = A Γ L(1, 8).Remark.

To establish the above long chain of equivalences, we still need to provethat(′)G9,7 = PSL(2, 8) ⇒Gal(Λ∗, k(y∗)) ̸= PSL(2, 7)83Since PSL(2, 2) = S3, it follows that if q = p = 2 then Gn,q = Sn.

54S. S. ABHYANKARandGal(Λ∗, k(y∗))(′′)= A Γ L(1, 8) ⇒Gal(Λ∗, kp(y∗)) ̸∈{S8, A8, AGL(3, 2),PGL(2, 7), PSL(2, 7)}.Now Gal(Λ, k(y)) and AGL(1, 8) are the 1-point stabilizers of G9,7 and PSL(2, 8)respectively, and hence to prove (′) it suffices to show that(′′′)Gal(Λ, k(y)) = AGL(1, 8) ⇒Gal(Λ∗, k(y∗)) ̸= PSL(2, 7)Let f be a nonconstant univariate monic polynomial with coefficients in a field Khaving no multiple roots in any field extension of K, and let K∗be a (finite) Galoisextension of K. By (2′) and (2∗∗∗) we see that, in the case q = p = 7 and t = 2, thefield k(y) is a Galois extension of the field k(y∗) and hence, by taking f = Λ = Λ∗and (K, K∗) = (k(y∗), k(y)), implication (′′′) follows from the implication(♯)Gal(f, K∗) = AGL(1, 8) ⇒Gal(f, K) ̸= PSL(2, 7).Also clearly for some finite algebraic field extension k∗of kp we have that Gal(Λ∗, k∗(y∗)) =Gal(Λ∗, k(y∗)) and k∗(y∗) is a Galois extension of kp(y∗) and hence, by takingf = Λ∗and (K, K∗) = (kp(y∗), k∗(y∗)), implication (′′) follows from the implica-tionGal(f, K∗) = A Γ L(1, 8)⇒Gal(f, K) ̸∈{S8, A8, AGL(3, 2), PGL(2, 7), PSL(2, 7)}.

(♯♯)Now AGL(1, 8) is a nonidentity solvable group and PSL(2, 7) is a nonabelian sim-ple group, and hence AGL(1, 8) is not isomorphic to a normal subgroup of PSL(2, 7),and therefore (♯) follows from the Refined Extension Principle. Likewise, A Γ L(1, 8)is a nonidentity solvable group but all the nonidentity normal subgroups of {S8, A8,AGL(3, 2), PGL(2, 7), PSL(2, 7)} are nonsolvable because they respectively containthe nonabelian simple group {A8, A8, PSL(3, 2), PSL(2, 7), PSL(2, 7)} as a normalsubgroup, and hence A Γ L(1, 8) cannot be isomorphic to a normal subgroup of anyone of the groups {S8, A8, AGL(3, 2), PGL(2, 7), PSL(2, 7)}, and therefore (♯♯) alsofollows from the Refined Extension Principle.Note.

Assuming q = p = 7 and t = 2, in my forthcoming paper [A7], first byresolving the singularities of the curve Φ(ω) = 0 in the (z, ω)-plane I calculate itsgenus in terms of infinitely near singularities, and then, seeing that the genus is 2and hence the curve is hyperelliptic, by means of adjoints I express it as a doublecovering of the line, and finally, using the resulting square-root parametrization ofthe curve I show that the polynomial Ψ(U) factors in k(z, ω)[U] into two factors ofdegree 3, and so I conclude that G9,7 = PSL(2, 8). Initially I used the square-rootparametrization to take the “norm” of Ψ(U) which is a monic polynomial Ψ♯(T, U)of degree 12 in U with coefficients which are polynomials in T over the prime fieldk7.

I used REDUCE and MACSYMA to calculate Ψ♯(T, U); the largest coefficientdegree in T turned out to be 216, and it took 10 pages to print out the exactexpressions of the coefficients. From what I have said above, it follows that Ψ(U)

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC55is reducible in k(z, ω)[U] ⇔Ψ♯(T, U) is reducible in k7[T, U] ⇔Ψ♯(T, U) factorsin k7[T, U] into two polynomials of degree 3 in U. Although the last two tasksare finitistic in nature, the computer algebra packages REDUCE and MACSYMArefused to factor bivariate polynomials over a finite field!

So I turned back to handcalculation in a hyperelliptic function field, using MACSYMA only for checkingordinary polynomial operations ! !22.

The roof polynomial and decreasing inductionLet X and Y be indeterminates over an algebraically closed field k of charac-teristic p ̸= 0, let 0 ̸= a ∈k, let s and t be positive integers with t ̸≡0(p), andconsider the polynomialbFn = Y n −aXsY t + 1with n = p + tmentioned in §11. This corresponds to the q = p case of the more general polynomialbFn,q = Y n −aXsY t + 1with n = q + t and q = a positive power of p which was also mentioned in §11.

Thepolynomials bFn and bFn,q in turn may be obtained by substituting aXs for X in thepolynomialsF n,q = Y n −XY t + 1with n = q + t and q = a positive power of p andF n = Y n −XY t + 1with n = p + t.As noted in the beginning of §21, the Y -discriminant of the polynomials F n and F n,qis the nonzero element nn of k, and hence the Y -discriminant of the polynomialsbFn and bFn,q is also the nonzero element nn of k, and therefore each one of the fourpolynomials F n, F n,q, bFn, bFn,q gives an unramified covering of the affine line overk.84In the beginning of §21, we noted that the polynomials F n and F n,q are irre-ducible in k(X)[Y ], and considering the Galois groups Gn = Gal(F n, k(X)) andGn,q = Gal(F n,q, k(X)), as summarized in the Summary and the Note at the endof that section, in the rest of that section we proved the following.85(I.1*) If t = 1, then Gn = PSL(2, p) = PSL(2, n −1). (I.2*) If t = 2 and p = 7, then Gn = PSL(2, 8) = PSL(2, n −1).

(I.3*) If t = 2 and p ̸= 7, then Gn = An. (I.4*) If t > 2 and p ̸= 2, then Gn = An.

(I.5*) If p = 2, then Gn = Sn. (III.1*) If t = 1, then Gn,q = PSL(2, q) = PSL(2, n −1).84That is, say in view of Proposition 1 of [A1], X = ∞is the only valuation of k(X)/k whichis possibly ramified in the splitting field of F n (resp.

F n,q, bFn, bFn,q) over K(X).85As said earlier, my work on this paper started when, in September 1988, Serre told me thathe could prove (I.1*). As he now tells me, his proof, which also applies to (III.1*), uses “descendingGalois theory” which is different from my method which he calls the “ascending” method.

WithSerre’s kind permission, his letter to me, dated 15 November 1990, describing his “descending”proof, is appended herewith.

56S. S. ABHYANKARAs noted in the said Summary and Note, on the one hand, CT was not used inthe proofs of (I.1*), (I.4*), (I.5*), and (III.1*), and on the other hand, the proof(I.2*) was complete only modulo the reducibility of the polynomial Ψ(U) whichwill be established in [A7].

Actually, CT was used only in the sense that, assumingt = 2, we first showed Gn to be a 3-transitive permutation group of degree n andfrom this by CTT we deduced that Gn = An or Sn. Now the 3-transitivity tellsus that |Gn| ≥n(n −1)(n −2) and hence if p = 3 then obviously Gn = An orSn.

Likewise, for p = 5 we need not invoke CT because again in that case, say inview of the following elementary theorem which occurs as item IV on page 148 ofCarmichael’s book [Ca],86 the 3-transitivity directly tells us that Gn = An or Sn.Carmichael’s Theorem. An and Sn are the only l-transitive groups of degree nfor which l > [ n3 + 1] where [ n3 + 1] is the greatest integer not exceeding n3 + 1.By Corollaries (3.5) to (3.10) of the Substitutional Principle, the Galois groupsbGn,q = Gal( bFn,q, k(X)) and bGn = Gal( bFn, k(X)) have the same description asthe above description of the Galois groups Gn = Gal(F n, k(X)) and Gn,q =Gal(F n,q, k(X)).

Let us summarize this in the followingSummary about the roof polynomial. Let k be an algebraically closed fieldof characteristic p ̸= 0, let 0 ̸= a ∈k, and let s and t be positive integers witht ̸≡0(p).

Then the polynomials bFn = Y n −aXsY t + 1 with n = p + t and bFn,q =Y n −aXsY t + 1 with n = q + t and q = a positive power of p, give unramifiedcoverings of the affine line over k, and for their Galois groups bGn = Gal( bFn, k(X))and bGn,q = Gal( bFn,q, k(X)) we have the following. (I.1) If t = 1, then bGn = PSL(2, p) = PSL(2, n −1).

(I.2) If t = 2 and p = 7, then bGn = PSL(2, 8) = PSL(2, n −1). (I.3) If t = 2 and p ̸= 7, then bGn = An.

(I.4) If t > 2 and p ̸= 2, then bGn = An. (I.5) If p = 2, then bGn = Sn.

(III.1) If t = 1, then bGn,q = PSL(2, q) = PSL(2, n −1).Here, CT is not used in the proofs of (I.1), (I.4), (I.5) and (III.1); likewise, itis not used in the proof of (I.3) for p < 7. Moreover, referring to the polynomialΨ(U) obtained by taking q = p = 7 in item (2**) of the §21, the proof of (I.2) iscomplete only modulo the reducibility of Ψ(U) in k(z, ω) to be established in [A7].Now what does the “decreasing induction” in the title of this section refer to?Roughly speaking, it says that if we can find an unramified covering of the affineline with Galois group An then we can find one with Galois group An−1.

If thiswere so without any qualification, then an An covering for large n would yield anAn covering for all smaller n. But there is a qualification! More precisely, we havethe followingMethod of decreasing induction.

Assuming that we are working over an al-gebraically closed field k of characteristic p ̸= 0, let there be given an irreducible86The statement of Charmichael’s Theorem given on page 154 of Volume I of Huppert [HB]says that An and Sn are the only l-transitive groups of degree n for which l > n3 . This seemsincorrect beacause for (l, n) = (4, 11) we have 4 > 113 but the Mathieu group M11 is a 4-transitivegroup of degree 11, and for (l, n) = (5, 12) we have 5 >123but the Mathieu group M12 is a5-transitive group of degree 12.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC57n-fold unramified covering of the affine line by the affine line,87 such that the pointat infinity splits exactly into two points, say the origin and the point at infinity,with ramification exponents p and t = n −p ̸≡0(p), and let G be the Galois groupof the given covering.88 Now if G = An with n > 5 then we can find an unramifiedcovering of the affine line with Galois group An−1. More generally, without anycondition on n or G, we can find an unramified covering of the affine line withGalois group G∗⊳G1 = the 1-point stabilizer of G such that G1/G∗is cyclic oforder nondivisible by p.To prove this informally, say the original line is the X-axis, and the coveringline is the Y -axis, and let C be the least Galois covering of the X-axis “containing”(or, “dominating”) the Y -axis.

Then the Galois group of C over the Y -axis is theone-point stabilizer G1 of G. Moreover, the origin of the Y -axis is tamely ramifiedin C, say with reduced ramification exponent t∗,89 and the point at infinity of theY -axis is the only other point which is possibly ramified in C. Now consideringthe Y ∗-axis with a∗Y ∗r = Y with 0 ̸= a∗∈k and r ≡0(t∗), and passing to the“compositum” C∗of C and the Y ∗-axis, by MRT90 we see that C∗is an unramifiedcovering of the affine line = the Y ∗-axis, and upon letting G∗to be the Galois groupof C∗over the Y ∗-axis, by Corollary (3.1) of the Substitutional Principle we seethat G∗⊳G1 = the 1-point stabilizer of G, and G1/G∗is cyclic of order nondivisibleby p.To apply this method to the polynomial F n, from the initial material91 of §21we recall that92 by putting X = x and reciprocating the roots of F n we get thepolynomialΘ(Y ) = Y p+t + xY p + 1and by letting y be a root of Θ(Y ) in an overfield of k(x) we have k(x, y) = k(y),and by letting Θ′(Y ) be the twisted Y -derivative of Θ(Y ) at y, and letting ∆(Y )be the polynomial obtained by reciprocating the roots of Θ′(Y ), and letting Λ(Z)be the polynomial obtained by multiplying the roots of ∆(Z) by ty, we getΛ(Z) = γ(Z)(Z + t)p −tp−2y−p−tZt,whereγ(Z) = t−2[(Z + t)t −Zt].Now Λ(Z) is irreducible in k(y)[Z] and has no multiple roots in any overfield ofk(y), and upon taking 1-point stabilizers, by (I.3*), (I.4*), and (I.5*) we get thefollowing where CT is used only in the p > 5 case of (I.3**). (I.3**) If t = 2 and p ̸= 7, then Gal(Λ(Z), k(y)) = An−1.

(I.4**) If t > 2 and p ̸= 2, then Gal(Λ(Z), k(y)) = An−1. (I.5**) If p = 2, then Gal(Λ(Z), k(y)) = Sn−1.87That is, a covering of the projective line by the projective line, which is unramified over theaffine line.88That is, G is the Galois group of the associated least Galois covering.89By the argument on pages 843–845 of [A3] we see that t∗divides LCM(t, (p −1)!) and hencethe origin of the Y -axis is tamely ramified in C.90MRT = pages 181–186 of [A5].91Up to formula (∗).92Until further notice q = p and n = p + t with positive integer t ̸≡0(p).

58S. S. ABHYANKARRemembering that s is any positive integer and a is any nonzero element of k,we letΛs,a(Z) = γ(Z)(Z + t)p −aY −sZtand we note that if s ≡0(p+t) then Λs,a(Z) can be obtained from Λ(Z) by replacingy by a′Y s/(p+t), where a′ ∈k is such that tp−2a′−p−t = a.

In view of the LCMTheorem,93 by the first ramification diagram94 in §21, we see that LCM(p −1, t) isdivisible by the reduced ramification exponent of every extension of the valuationy = 0 of k(y)/k to a splitting field of Λ(Z) over k(y), and no valuation of k(y)/k,other than the valuations y = 0 and y = ∞, is ramified in the said splitting field.Therefore, if s ≡0((p + t) LCM(p −1, t)) then by MRT we see that no valuation ofk(Y )/k, other than the valuation Y = ∞, is ramified in a splitting field of Λs,a(Z)over k(Y ), and obviously Λs,a(Z) has no multiple roots in any overfield of k(Y ), andin view of Corollaries (3.3), (3.5), and (3.8) of the Substitution Principle, by (I.3**),(I.4**), and (I.5**) we see that (3′) if t = 2 and p ̸= 7, then Gal(Λs,a(Z), k(Y )) =An−1, whereas (4′) if t > 2 and p ̸= 2, then Gal(Λs,a(Z), k(Y )) = An−1, and finally(5′) if p = 2, then Gal(Λs,a(Z), k(Y )) = Sn−1. Thus we have the following whereCT is used only in the p > 5 case of (I.3′).

(I.0′) If s ≡0((p + t) LCM(p −1, t)) then no valuation of k(Y )/k, other than thevaluation Y = ∞, is ramified in a splitting field of Λs,a(Z) over k(Y ), and Λs,a(Z)has no multiple roots in any overfield of k(Y ). (I.3′) If t = 2 and p ̸= 7 and s ≡0((p+t) LCM(p−1, t)), thenGal(Λs,a(Z), k(Y )) =An−1.

(I.4′) If t > 2 and p ̸= 2 and s ≡0((p+t) LCM(p−1, t)), thenGal(Λs,a(Z), k(Y )) =An−1. (I.5′) If p = 2 and s ≡0((p + t) LCM(p −1, t)), then Gal(Λs,a(Z), k(Y )) = Sn−1.Actually, the above four assertions remain valid if we replace the assumption thats ≡0((p + t) LCM(p −1, t)) by the weaker assumption that s ≡0(LCM(p −1, t)).To see this, first note that Λ(Z) can be obtained from Λ1,1(Z) by replacing Y byt2−pyp+t; now since no valuation of k(y)/k, other than the valuations y = 0 andy = ∞, is ramified in a splitting field of Λ(Z) over k(y), it follows that no valuationof k(Y )/k, other than the valuations Y = 0 and Y = ∞, is ramified in a splittingfield of Λ1,1(Z) over k(Y ); moreover since Λ(Z) is irreducible in k(y)[Z] and hasno multiple roots in any overfield of k(y), it follows that Λ1,1(Z) is irreduciblein k(Y )[Z] and has no multiple roots in any overfield of k(Y ); finally, in view ofCorollary (3.1) the Substitutional Principle, by (I.3**) to (I.5**) we see that (3***)if t = 2 and p ̸= 7, then Gal(Λ1,1(Z), k(Y )) = An−1 or Sn−1, and (4***) if t > 2and p ̸= 2, then again Gal(Λ1,1(Z), k(Y )) = An−1 or Sn−1, and (5***) if p = 2,then Gal(Λ1,1(Z), k(Y )) = Sn−1.

Again, Λs,a(Z) can be obtained from Λ1,1(Z)by substituting aY s for Y and hence, in view of Corollaries (3.2) to (3.8) of theSubstitutional Principle, we get the following where CT is used only in the p > 5case of (I.3′′). (I.0′′) No valuation of k(Y )/k, other than the valuations Y = 0 and Y = ∞,is ramified in a splitting field of Λs,a(Z) over k(Y ), and Λs,a(Z) is irreducible95 ink(Y )[Z] and has no multiple roots in any overfield of k(Y ).93That is, Proposition 7 on page 845 of [A3].94Together with its explanation in the three paragraphs following it.95The irreducibility of Λs,a follows from the First Irreducibility Lemma.Alternatively, itfollows from the fact that a polynomial is irreducible if and only if its Galois group is transitive.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC59(I.3′′) If t = 2 and p ̸= 7, then Gal(Λs,a(Z), k(Y )) = An−1 or Sn−1. (I.4′′) If t > 2 and p ̸= 2, then Gal(Λs,a(Z), k(Y )) = An−1 or Sn−1.

(I.5′′) If p = 2, then Gal(Λs,a(Z), k(Y )) = Sn−1.Now γ(0) ̸= 0 ̸= t and, upon letting Z′ to be a root of Λ1,1(Z) in an overfield ofk(Y ), we haveY =Z′tγ(Z′)(Z′ + t)pand hence the valuation Y = 0 of k(Y )/k splits in k(Y, Z′) = k(Z′) into thevaluations Z′ = 0 and Z′ = ∞with reduced ramification exponents t ̸≡0(p)and p −1 ̸≡0(p) respectively, and therefore by the LCM Theorem we see thatLCM(p −1, t) is divisible by the reduced ramification exponent of every extensionof the valuation Y = 0 of k(Y )/k to a splitting field of Λ(Z) over k(Y ), and henceby MRT we see that if s ≡0(LCM(p −1, t)) then no valuation of k(Y )/k, otherthan the valuation Y = ∞, is ramified in a splitting field of Λs,a(Z) over k(Y ), andtherefore by Result 4 on page 841 of [A3] we know that Gal(Λs,a(Z), k(Y )) is aquasi p-group. Thus, in view of (I.0′′), (I.3′′), (I.4′′), and (I.5′′), we conclude withthe following where CT is used only in the p > 5 case of (IV.2′).

(IV.0′) If s ≡0(LCM(p −1, t)), then no valuation of k(Y )/k, other than thevaluation Y = ∞, is ramified in a splitting field of Λs,a(Z) over k(Y ), and Λs,a(Z)is irreducible in k(Y )[Z] and has no multiple roots in any overfield of k(Y ). (IV.1′) If t > 2 and p ̸= 2 and s ≡0(LCM(p −1, t)), thenGal(Λs,a(Z), k(Y )) =An−1.

(IV.2′) If t = 2 and p ̸= 7 and s ≡0(LCM(p −1, t)), thenGal(Λs,a(Z), k(Y )) =An−1. (IV.4′) If p = 2 and s ≡0(LCM(p −1, t)), then Gal(Λs,a(Z), k(Y )) = Sn−1.As we have said, CT is used only when t = 2; now in the case t = 2 we haveγ(Z) = Z + 1 and hence Λs,a(Z) = (Z + 1)(Z + 2)p −aY −sZt; therefore by (I.0′′)we see that if t = 2 then Λs,a(Z) = (Z + 1)(Z + 2)p −aY −sZt is unramified outsideY = 0 and Y = ∞.

For getting hold of a variation of (I.3′′′) without CT, let usreprove the said unramifiedness in a more general context. So letE(Z) = (Z + 1)(Z + b)p −Y Zτwith 1 < τ ≤p and 0 ̸= b ∈k.

Now for the (ordinary) Z-derivative we haveEZ(Z) = (Z + b)p −τY Zτ−1and henceE(Z) = (Z + 1)EZ(Z) + E∗(Z),whereE∗(Z) = (τ −1)Y Zτ−1Z +ττ −1

60S. S. ABHYANKARand therefore for the Z-discriminant we haveDiscZ(E(Z)) = ResZ(E(Z), EZ(Z))= ResZ(E∗(Z), EZ(Z))= (τ −1)pY pEZ(0)τ−1EZ −ττ −1= (τ −1)pY pbp(τ−1)EZ −ττ −1= (τ −1)pY pbp(τ−1)(−τY ) −ττ −1τ−1if b =ττ −1= (−τ)τ(τ −1)p−τ+1bp(τ−1)Y p+1if b =ττ −1and hence if b =ττ−1 then no valuation of k(Y )/k, other than the valuations Y = 0and Y = ∞, is ramified in a splitting field of E(Z) over k(Y ), and E(Z) has norepeated roots in any overfield of k(Y ).

Therefore, upon remembering that a is anynonzero element of k and s and t are any positive integers with t ̸≡0(p), and uponlettingEs,a(Z) = (Z + 1)(Z + b)p −aY −sZtwe get the following.96(IV.0**) If 1 < t < p and b =tt−1 then no valuation of k(Y )/k, other than thevaluations Y = 0 and Y = ∞, is ramified in a splitting field of Es,a(Z) over k(Y ),and Es,a(Z) is irreducible97 in k(Y )[Z] and has no multiple roots in any overfieldof k(Y ).Before proceeding further, let us make note of the followingAlternate Corollary of the Fourth Irreducibility Lemma. If F(Z) is amonic irreducible polynomial of degree p + 1 in Z with coefficients in k(Y ) suchthat some valuation of k(Y )/Y , say the valuation Y = ∞, splits in a root field ofF(Z) over k(Y ) into two valuations with reduced ramification exponents p and 1,then the Galois group Gal(F(Z), k(Y )) is 2-transitive.For a moment suppose that 1 < t < p and b =tt−1.

Now upon letting Z∗to bea root of E1,1(Z) in some overfield of k(Y ), we haveY =Z∗t(Z∗+ 1)(Z∗+ b)pand hence the valuation Y = ∞of k(Y )/k splits in k(Y, Z∗) = k(Z∗) into twovaluations with reduced ramification exponents p and 1 and therefore, by the aboveAlternate Corollary, Gal(E1,1(Z), k(Y )) is 2-transitive. By the above equation forY we also see that the valuation Y = 0 of k(Y )/k splits in k(Z∗) into the valuationsZ∗= 0 and Z∗= ∞with reduced ramification exponents t ̸≡0(p) and p + 1 −t ̸≡0(p) respectively.

Therefore on the one hand, by the LCM Theorem and MRT wesee that if s ≡0(t(p+ 1 −t)) then no valuation of k(Y )/k, other than the valuation96For a while there will be no reference to n.97The irreducibility of Es,a,b follows from the First Irreducibility Lemma.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC61Y = ∞, is ramified in a splitting field of Es,a(Z) over k(Y ), and hence by Result4 on page 841 of [A3] we see that Gal(Es,a(Z), k(Y )) is a quasi p-group, and so ifwe already knew that Gal(Es,a(Z), k(Y )) = Ap+1 or Sp+1 then we would concludethat Gal(Es,a(Z), k(Y )) = Ap+1. On the other hand, by the Cycle Lemma we seethat if GCD(p + 1 −t, t) = 1 then Gal(E1,1(Z), k(Y )) contains a t-cycle, and henceby Marggraff’s First Theorem we see that if also t < p+12then Gal(E1,1(Z), k(Y ) =Ap+1 or Sp+1.

Note that if p > 5 and t is an odd prime factor of (p−1)(p−3)4theneither t is an odd prime factor of p−12or t is an odd prime factor of p−32 , and inboth the cases 1 < t < p+12and GCD(p + 1, t) = 1. Also note that, in the casep > 5, there do exist odd prime factors of (p−1)(p−3)4, because otherwise we wouldhave p −1 = 2m with m > 2 and p −3 = 2m′ with m′ > 1, and this would give2m′−1 = p−32= p−1−22= 2m−1 −1 which would be a contradiction since 2m′−1 iseven and 2m−1 −1 is odd.

Thus, without using CT, we have proved the following.98(IV.0*) If p > 5 then t can be chosen so that 1 < t < p+12andGCD(p + 1, t) = 1,and for any such t, upon assuming b =tt−1 and s ≡0(t(p + 1 −t)), we have thatno valuation of k(Y )/k, other than the valuation Y = ∞, is ramified in a splittingfield Es,a(Z) over k(Y ), and Es,a(Z) is irreducible in k(Y )[Z] and has no multipleroots in any overfield of k(Y ). (IV.3*) If p > 5 then t can be chosen so that 1 < t < p+12andGCD(p + 1, t) = 1,and for any such t, upon assuming b =tt−1 and s ≡0(t(p + 1 −t)), we haveGal(Es,a(Z), k(Y )) = Ap+1.The above two results with (Y, Z) changed to (X, Y ), together with the results(IV.0′), (IV.1′), (IV.2′), and (IV.4′) with (n, Y, Z) changed to(n + 1, X, Y ),99 maybe summarized in the followingSummary about the primed roof polynomial.

Let k be an algebraically closedfield of characteristic p ̸= 0, let a, b be nonzero elements in k, let n, s, t be positiveintegers with n + 1 ̸≡0(p) and n > t ̸≡0(p), and consider the monic polynomial ofdegree n in Y with coefficients in k(X) given bybF ′n = h(Y )(Y + b)p −aX−sY twith 0 ̸= b ∈k,where h(Y ) is the monic polynomial of degree n−p in Y with coefficients in k givenbyh(Y ) = (Y + n + 1)n+1−p −Y n+1−p(n + 1)2.Then in the following cases bF ′n is irreducible in k(Y )[Z], has no multiple roots inany overfield of k(Y ), and gives an unramified covering of Lk with the indicatedGalois group bG′n = Gal( bF ′n, k(X)). (IV.1) If n + 1 −p = t > 2 ̸= p and b = t and s ≡0(p −1) and s ≡0(t), thenbG′n = An.

(IV.2) If n + 1 −p = t = 2 and p ̸= 7 and b = t and s ≡0(p −1), then bG′n = An. (IV.3) If n = p + 1 and p > 5, then t can be chosen so that 1 < t < p+12andGCD(p+1, t) = 1, and for any such t, upon assuming b =tt−1 and s ≡0(t(p+1−t)),we have bG′n = An.98This was inspired by discussions with Walter Feit.99Note that with these changes, if n + 1 −p = t ̸≡0(p) then γ(Y ) = (Y +n+1)n+1−p−Y n+1−p(n+1)2.

62S. S. ABHYANKAR(IV.4) If n + 1 −p = t and p = 2 and b = t and s ≡0(t), then bG′n = Sn.Here CT is used only in the p > 5 case of (IV.2).Referring to the summaries about the tilde polynomial and the roof polynomialand the primed roof polynomial, we have established the following four corollariesand we have arranged a proof of the First and the Second Corollaries independentof CT.First Corollary.

For any n ≥p > 2, there exists an unramified covering of theaffine line in characteristic p whose Galois group is An.Second Corollary. For any n ≥p = 2, there exists an unramified covering of theaffine line in characteristic p whose Galois group is Sn.Third Corollary.

Unramified coverings of the affine line in characteristic p witha few more Galois groups have been constructed.Fourth Corollary. Let G be a quasi p-group.

Assume that G has a subgroup Hof index p + 1 such that H does not contain any nonidentity normal subgroup ofG.100 Also assume that p is not a Mersenne prime and p is different from 11 and23. Then there exists an unramified covering of the affine line in characteristic phaving G as Galois group.It only remains to note that, in view of the said three Summaries, the aboveForth Corollary follows from CTT, Special CDT, and the Corollary of the FourthIrreducibility Lemma given in §21.

Moreover, given any n ≥p > 2, as a definitealternative for getting an (An)-covering as asserted in the First Corollary withoutCT : if p+2 < n ̸≡0(p) then use (I.4) with t = n−p; if n = p+2 and p < 7 then use(I.3) with t = 2; if n = p + 2 and p ≥7 then use (IV.1) with t = 3 ands = t(p −1);if n = p + 1 and p < 7 then use (IV.2) with t = 2 and s = p −1; if n = p + 1and p ≥7 then use (IV.3) with t = the smallest odd prime factor of (p−1)(p−3)2andwith s = t(p + 1 −t), [note that in case of n = 8 and p = 7 this gives t = 3 ands = 15]; if n ≡0(p) then use (II.1) with t = 2. Likewise, given any n ≥p = 2, as adefinite alternative for getting an (Sn)-covering as asserted in the Second Corollarywithout CT : if n ̸≡0(p) then use (I.5) with t = n −p; if n ≡0(p) then use (IV.4)with t = n + 1 −p and s = t.The above cited three summaries are transcribed from my e-mail message toSerre dated 28 August 1989.

This was only one out of the nearly a hundred e-mailand s-mail messages which flashed back and forth between him and me in the twoyear period September 1988 to September 1990. Indeed it has been a tremendouspleasure working with him.

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Hamburg 11, 187–220.Appendix by J.-P. Serre∗Harvard, November 15, 1990Dear Abhyankar,Here is my original proof that PSL2(Fq) occurs for the equation Y q+1−XY +1 =0.I use “descending Galois theory,” i.e., I construct a priori the Galois coveringone wants. This is different from your “ascending” method; in particular, I don’tneed any characterization of PGL2(Fq), or PSL2(Fq), as a permutation group onq + 1 letters.Notation.p is a prime; q is a power pe of p.G = PGL2(Fq), i.e., the quotient of GL2(Fq) by its center F∗q.k is an algebraically closed field of characteristic p; all the curves I consider areover k.Preliminary construction.

I start from the obvious fact that G acts in a naturalway (by “fractional linear transformations”) on the projective line P1. In algebraicterms this means that G acts on k(t) by t 7→(at + b)/(ct + d).

The quotient curveP′1 = P1/G is of course (L¨uroth’s theorem!) a projective line.

Equivalently, thefield of invariants of G in k(t) is a purely transcendental field k(x).The first computational problem which arises is to write x explicitly. To do so,let us call (u, v) the homogeneous coordinates on P1, so that t = v/u.

The invarianttheory of k[u, v] with respect to the action of G has been done long ago by Dickson.The basic covariants are the following homogeneous polynomials:A(u, v) = uvq −vuq,B(u, v) = (uvq2 −vuq2)/A(u, v).∗The author expresses his appreciation to J. P. Serre for permission to include the followingletter in this paper.

GALOIS THEORY ON THE LINE IN NONZERO CHARACTERISTIC65They are of degree q + 1 and q2 −q respectively. Hence the ratiox = B(u, v)q+1/A(u, v)q2−qis invariant by G. Its expression in terms of t is easy to find: if we write A(u, v) =uq+1a(t), B(u, v) = uq2−qb(t), we havea(t) = tq −t,b(t) = (tq2 −t)/(tq −t) = a(t)q−1 + 1and(∗)x = b(t)q+1/a(t)q2−q = (a(t)q−1 + 1)q+1/a(t)q2−q.This shows that x is a rational function of t of degree q(q2 −1).

Since it is invariantby G, which has order q(q2 −1), Galois theory shows it generates the field of theG-invariant elements of k(t). Hence we have found our parameter for P′1 = P1/G.Ramification.

It is necessary to study the ramification in the Galois extensionk(t)/k(x) thus constructed. This amounts to looking for the fixed points of theaction of G on the projective line P1.

It is easy to see that these fixed points makeup two orbits. Namely:(a) The Fq-rational points of P1.

This orbit has q + 1 elements. The stabilizerof an element is a triangular subgroup (“Borel subgroup”) of order q(q −1).

Sincethat order is divisible by p, there is wild ramification.The point of P′1 corresponding to this orbit is x = ∞. (b) The “quadratic” points, i.e., the Fq2-rational points of P1 which are notrational over Fq.

There are q2 −q of them. The stabilizer of such a point is a cyclicgroup of order q + 1 (“nonsplit Cartan subgroup”).

Since that order is prime to p,the ramification at such a point is tame.The point of P′1 corresponding to this orbit is x = 0.Hence we see that the covering of P′1 we get in this way is ramified both at 0 and∞, and nowhere else. The next step is thus:Getting rid of the ramification at 0 using Abhyankar’s lemma.

We considerthe cyclic extension k(X) of k(x) defined by the equation Xq+1 = x. By making abase change to that extension (i.e., by considering k(t, X)/k(X)) we get rid of theramification at 0.

Only the ramification at ∞remains. Of course, one has to seewhat the new Galois group is.

There are two cases:(i) p = 2. The extensions k(X)/k(x) and k(t)/k(x) are disjoint.

Hence the newGalois group is equal to the old one, namely G = PGL2(Fq), which happens to beequal to PSL2(Fq). (ii) p ̸= 2.

The extensions k(X)/k(x) and k(t)/k(x) have a quadratic extensionin common, namely k(x1/2). Hence the new Galois group is G′ = PSL2(Fq).In both cases, one thus gets a Galois extension of k(X) with Galois groupPSL2(Fq) which is ramified only at X = ∞.It remains to see that this extension is the same as the one you get by theequation Y q+1 −XY + 1 = 0.

66S. S. ABHYANKARAn equation for the degrees q + 1 extension.

I go back to the k(t)/k(x)extension with Galois group G. Let H be the triangular subgroup of G, of indexq + 1. The fixed field k(t)H is an extension of degree q + 1 of k(x) and we wantto find a generator y for that field (which is also a purely transcendental field, ofcourse).

We may assume that H is the group of transformations t 7→at + b. Thisshows that the polynomialy = a(t)q−1 = (tq −t)q−1is invariant by H. Since its degree is q(q −1) = |H|, the same argument as aboveshows that k(y) is equal to k(t)H.I now write the equation of degree q + 1 relating y to x. This is easy, since byconstruction, we have x = (y + 1)q+1/yq, cf.

(∗) above. We thus get the equation(y + 1)q+1 −xyq = 0.But I want to work on k(X), with x = Xq+1.

We have:(y + 1)q+1 −Xq+1yq = 0.Let me put Y = (y + 1)/yX, i.e., y = 1/(XY −1). The above equation becomesY q+1 −XY + 1 = 0,and we are done.This is the proof I found in 1988 when I started thinking about your problem.The first part is natural enough—and could indeed be applied to other groups, `a laNori.

The second part (the search for the degree q + 1 equation) is not; it looks likea happy coincidence, and I would not have found it if I had not known in advanceyour polynomial Y q+1 −XY + 1.With best regards,YoursJ-P. SerrePSThe determination of the invariants of G in k(t) is not new. I am almost certainto have seen it in print very long ago, as an elementary exercise in Galois theory.

(Indeed: see Lang’s Algebra, 2nd ed., p. 349, exercise 33, and also P. Rivoire, Ann.Inst. Fourier 6 (1955–1956), pp.

121–124. )(Shreeram S. Abhyankar) Mathematics Department, Purdue University, West Lafayette,Indiana 47907

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