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ON THE APPROXIMATE ROOTS OF POLYNOMIALS

arXiv:alg-geom/9305009v1 19 May 1993ON THE APPROXIMATE ROOTS OF POLYNOMIALSby Janusz Gwo´zdziewicz and Arkadiusz P loskiAbstract. We give a simplified approach to the Abhyankar–Moh theory of approxi-mate roots.

Our considerations are based on properties of the intersection multiplicityof local curves.1. The main resultsFor any power series f, g ∈C[x, y]we define the intersection number (f, g)0 by(f, g)0 = dimC C[x, y]/(f, g).

Suppose that f = f(x, y) is an irreducible power se-ries and let n = (f, x)0 = ord f(0, y) < +∞. Then there exists a power series y(t) ∈C[t], ord y(t) > 0 such that f(tn, y(t)) = 0.

We have (f, g)0 = ord g(tn, y(t)) forany g = g(x, y) ∈C[x, y]. The mapping g 7→(f, g)0 induces a valuation vf of thering C[x, y]/(f).

Let Γ(f) be the semigroup of vf i.e. Γ(f) = { (f, g)0 ∈N : g ̸≡0mod (f) }.According to the well known structure theorem for the semigroup Γ(f) ([1], [2],[10] and Sect.

3 of this paper) there is a unique sequence of positive integers ¯b0, ¯b1,. .

. , ¯bh such that(i) ¯b0 = (f, x)0,(ii) ¯bk = min(Γ(f) \ (N¯b0 + · · · + N¯bk−1))for k = 1, .

. .

, h.(iii) Γ(f) = N¯b0 + · · · + N¯bhi.e. Γ(f) is generated by ¯b0,¯b1, .

. .

,¯bhIf the conditions (i), (ii), (iii) are satisfied, we write Γ(f) = ⟨¯b0, . .

. ,¯bh⟩.

Alsodefine Bk = gcd(¯b0, . .

. ,¯bk) for k = 0, 1, .

. .

, h and nk = Bk−1/Bk for k = 1, . .

. , h.We have(iv) nk > 1 and the sequence Bk−1¯bk is strictly increasing for k ≥1.Let O be an integral domain of characteristic zero.

Let g ∈O[y] be a monicpolynomial and let d be a positive integer such that d divides deg g. Accordingto Abhyankar and Moh [2] the approximate dth root of g denotedd√g is definedto be the unique monic polynomial satisfying deg(g −( d√g)d) < deg g −degd√g.Obviously degd√g = deg g/d.Let 1 ≤k ≤h.Theorem 1.1. Let g = g(x, y) ∈C[x][y] be a monic polynomial, degy g = n/Bk.Suppose that (f, g)0 > nk¯bk.

Then (f,nk√g)0 = ¯bk.The proof of (1.1) is given in Sect. 4 of this paper.

We shall follow the methodsdeveloped by Abhyankar and Moh in the fundamental paper [2] and simplified byAbhyankar in [1].Let 1 ≤k ≤h + 1.Typeset by AMS-TEX1

2BY JANUSZ GWO´ZDZIEWICZ AND ARKADIUSZ P LOSKITheorem 1.2. Let φ = φ(x, y) ∈C[x, y]be such that (φ, x)0 = n/Bk−1.

Then(f, φ)0 ≤¯bk (we put ¯bh+1 = +∞). If additionaly, (f, φ)0 > nk−1¯bk−1 then φ isirreducible and Γ(φ) = ⟨¯b0Bk−1 , .

. .

,¯bk−1Bk−1 ⟩The proof of (1.2) is given in Sect. 5 of this paper.

Note that the second part of(1.2) is a generalization of Abhyankar’s irreducibility criterion [1, Chapter V].Lemma 1.3. Let g ∈O[y] be a monic polynomial of degree n > 0 and let d, e > 0be integers such that de divides n. Thenepd√g =ed√g.Proof of (1.3).

Let h =epd√g, so deg h =nde.We haved√g = ( epd√g)e + R,deg R

This shows that h =ed√g.Now, we can prove the Abhyankar–Moh theorem.Theorem 1.4 [2]. Let f = f(x, y) ∈C[x][y] be an irreducible distinguished poly-nomial of degree n > 1 with Γ(f) = ⟨¯b0,¯b1, .

. .

,¯bh⟩and ¯b0 = (f, x)0 = n. Let1 ≤k ≤h + 1. Then we have(1.4.1) (f,Bk−1√f)0 = ¯bk(1.4.2)Bk−1√f is an irreducible distinguished polynomial of degreenBk−1 such that Γ( Bk−1√f) =⟨¯b0Bk−1 , .

. .

,¯bk−1Bk−1 ⟩.Proof of (1.4). First we check (1.4.1) using (1.1) and induction on k. If k = h + 1then Bk−1 = Bh, ¯bk = ¯bh+1 = +∞and (1.4.1) is obvious.

Let k ≤h and suppose(f,Bk√f)0 = ¯bk+1 holds true.The polynomialBk√f is of degree n/Bk and (f,Bk−1√f)0 > nk¯bk(¯bk+1 > nk¯bkby (iv)) so we can apply (1.1) to g =Bk√f.By (1.1) we get (f,nk√g)0 = ¯bkand by (1.3) we havenk√g =nkpBk√f =Bk−1√f, consequently (f,Bk−1√f)0 = ¯bkand (1.4.1) is proved.In order to prove (1.4.2) put φ =Bk−1√f. Thus φ is a distinguished polynomialof degree n/Bk−1, hence (φ, x)0 = n/Bk−1.

On the other hand (f, φ)0 = ¯bk >nk−1¯bk−1 by (1.4.1).According to theorem 1.2 φ =Bk−1√f is irreducible andΓ(φ) = ⟨¯b0Bk−1 , . .

. ,¯bk−1Bk−1 ⟩.Note.

Theorems (1.1) and (1.2) could be formulated and proved for meromorphiccurves. However we shall see in the next section that the algebroid case consideredby us is sufficient to get important applications of approximate roots.2.

The Abhyankar–Moh inequalityWe give here a geometrical version of the Abhyankar–Moh inequality which is thebasic tool for proving the Embedding Theorem [3]. Let C ⊂P2 be an irreducibleprojective plane curve of degree n > 1 and let O ∈C be its singular point.

Weassume that C is analytically irreducible at O i.e. the analytic germ (C, O) isirreducible, and let L be the unique tangent to C at O.Let Γ(C, O) be thesemigroup of the branch of C passing through O and let Γ(C, O) = ⟨¯b0,¯b1, .

. .

,¯bh⟩with ¯b0 = (C L)O.

ON THE APPROXIMATE ROOTS OF POLYNOMIALS3Theorem 2.1 [3]. Suppose that C∩L = {O}, i.e.

¯b0 = n. Then we have Bh−1¯bh

It is easyto see that f is a distinguished, irreducible in C[x, y]polynomial and Γ(f) =⟨¯b0,¯b1, . .

. ,¯bh⟩.We have deg f = n and consequently degBh−1√f =nBh−1 , thus (1.4.1) and Be-zout’s theorem imply ¯bh = (f,Bh−1√f)0 ≤nnBh−1 that is Bh−1¯bh ≤n2.

In factBh−1¯bh < n2 because the equality Bh−1¯bh = n2 implies ¯bh = nnBh−1 which contra-dicts the relation ¯bh ̸≡0 mod Bh−1.The inequality (2.1) has an application to the polar curves [4].Theorem 2.2 (with the assumptions as above). Let (D, O) be an irreduciblecomponent of the local polar of C with respect to L.Then (C D)O < (C L)O(D L)O.Proof.

In the coordinates x, y introduced in the proof of (2.1) the local polar isgiven by equation ∂f∂y = 0 and its irreducible component is given by g = 0 where gis irreducible (in C[x, y]) divisor of ∂f∂y. By the Merle formula for polar invariants[6], [4], [5] and theorem 2.1 we get(C D)O(D L)O= (f, g)0(g, x)0= Bk−1¯b0¯bk ≤Bh−1¯b0¯bh < n = (f, x)0 = (C L)Oand the theorem follows.3.

Characteristic, semigroup of ananalytic curve and the Noether formulaIn this section we recall some well-known notions of the theory of analytic curves.Our main reference is [10]. Let f = f(x, y) be an irreducible power series y–regularof order n = ord f(0, y) > 1.

There exists a power series y(t) ∈C[t], ord y(t) > 0such that f(tn, y(t)) = 0. Moreover every solution of the equation f(tn, y) = 0is of the form y(ǫt) for some ǫ such that ǫn = 1.Let y(t) = P ajtj.We putS(f) = { j ∈N : aj ̸= 0 }.

Note that S(f) depends only on f.The characteristic b0, b1, . .

. , bh of f is the unique sequence of positive integerssatisfying the conditions:(i) b0 = n(ii) bk+1 = min{ j ∈S(f) : gcd(b0, .

. .

, bk, j) < gcd(b0, . .

. , bk) }(iii) gcd(b0, .

. .

, bh) = 1In the sequel we put Bk = gcd(b0, . .

. , bk)for k = 0, 1, .

. .

, hand ¯bk = bk +1Bk−1Pk−1i=1 (Bi−1 −Bi)bifor k = 1, . .

. , h.We assume that the sum of an empty family is equal to zero.

Thus we have¯b1 = b1. We put ¯b0 = b0.

One checks easily that gcd(¯b0, . .

. ,¯bk) = Bk for k = 0,. .

. , h and bk = ¯bk −Pk−1i=1 ( Bi−1Bi−1)¯bifor k = 0, 1, .

. .

, h. Therefore thesequence ¯b0, . .

. , ¯bk determines sequences B0, .

. .

, Bk and b0, . .

. , bk for any

4BY JANUSZ GWO´ZDZIEWICZ AND ARKADIUSZ P LOSKIk = 0, . .

. ,h. We have ¯bk+1 −Bk−1Bk ¯bk = bk+1 −bk for k = 1, .

. .

, h −1, whichshows that the sequence Bk−1¯bk is increasing.For any k, 1 ≤k ≤h we set Sk = { j ∈S(f) : j < bk+1 } We put bh+1 =¯bh+1 = +∞, so Sh = S(f). Let yk(t) =Pj∈Skajtj.

There exists an irreducible,monic polynomial fk = fk(x, y) ∈C[x][y] such that fk(tn, yk(t)) = 0.Lemma 3.1 [10]. degy fk(x, y) = n/Bk,(f, fk)0 = ¯bk+1Proof.

[10, p. 15]Note that fh+1 is the distinguished polynomial associated with f.Proposition 3.2 [10]. If ψ(x, y) ∈C[x][y], degy ψ(x, y) < n/Bk and ψ ̸≡0mod (f), then (f, ψ)0 ∈N¯b0 + · · · + N¯bkProof.

[10, p. 16]Note that from (3.1) and (3.2) we get Γ(f) = N¯b0+· · ·+N¯bh. Now, let g = g(x, y)be an irreducible power series y–regular of order p = ord f(0, y) < +∞.

Supposethat f, g are coprime. Let z(t) ∈C[t], ord z(t) > 0 be such that g(tp, z(t)) = 0.We putof(g) = max{ ord(y(ǫx1/n) −z(νx1/p)) : ǫn = 1, νp = 1 }.It is easy to check, thatof(g) = max{ ord(y(x1/n) −z(νx1/p)) : νp = 1 }= max{ ord(y(ǫx1/n) −z(x1/p)) : ǫn = 1 }In particular of(g) = og(f).The classical computation leads to the following formula due to Max Noether:Proposition 3.3 [6], [5].

Suppose that f, g are irreducible, y–regular power series,f of characteristic (b0, b1, . .

. , bh) and let k be the smallest strictly positive integersuch that of(g) ≤bkb0 ( bh+1b0= +∞).

Then(f, g)0(g, x)0=k−1Xi=1(Bi−1 −Bi) bib0+ Bk−1of(g)Remark [9]. The Noether formula is really symetric.

Let (c0, c1, . .

. , cm) be thecharacteristic of g. Then k ≤m,cic0 = bib0 for i = 1, .

. .

, k −1 and of(g) ≤ckc0 . IfCi = gcd(c0, .

. .

, ci) then the formula can be rewritten in the following form(f, g)0(f, x)0=k−1Xi=1(Ci−1 −Ci) cic0+ Ck−1og(f).Using (3.3) we check easilyLemma 3.4. Let l > 0 be an integer.Then of(g) ≤blb0 iff(f,g)0(g,x)0 ≤Bl−1¯bl¯b0 .Moreover of(g) = blb0 is equivalent to (f,g)0(g,x)0 = Bl−1¯bl¯b0 .To end with, let us note

ON THE APPROXIMATE ROOTS OF POLYNOMIALS5Corollary 3.5 to theorem 1.2. If f = f(x, y) ∈C[x][y] is an irreducibledistinguished polynomial, then of( Bk−1√f) = bkb0for k = 1, .

. .

, h.Proof. We have ( Bk−1√f, x)0 = degyBk−1√f =nBk−1 and (f,Bk−1√f)0 = ¯bk by(1.4.1).Hence (f,Bk−1√f)0( Bk−1√f, x)0= Bk−1¯bk¯b0.The power seriesBk−1√f is irreducibleby (1.4.2), thus we get by (3.4)of( Bk−1√f) = bkb0 .4.

Proof of theorem 1.1Let g ∈O[y] be a monic polynomial with coefficients in the integral domain Oof charecteristic zero and let d be a positive divisor of deg g. Given any monicpolynomial h ∈O[y] of degree deg g/d we have h–adic expansion of g, namelyg = hd + a1hd−1 + · · · + ad,ai ∈O[y],deg ai < deg hThe polynomials ai are uniquely determined by g, h. The Tschirnhausen operatorτg(h) = h + 1da1 changes h to τg(h) which is again monic of degree deg g/d.Lemma 4.1 [1].d√g = τg(τg . .

. (τg(h)) .

. . ) with τg repeated degg/d times.Proof [1, p. 16].To prove (1.1) it suffices to check the following(*) if h(x, y) ∈C[x][y] is a monic polynomial of degree n/Bk−1 such that (f, h)0 =¯bk, then (f, τg(h))0 = ¯bk.Indeed, to get the relation (f,nk√g)0 = ¯bk we take h = fk−1 (cf.

lemma 3.1) andapply Tschirnhausen operator τg to hdeg g/nk = n/Bk−1 times.To prove (*) fix a monic polynomial h(x, y) ∈C[x][y] such that deg h = n/Bk−1and (f, h)0 = ¯bk and let us consider the h–adic expansion of g:(1)g = hnk + a1hnk−1 + · · · + ank,degy ai < degy h = n/Bk−1Let I be the set of all i ∈{ 1, . .

. , nk } such that ai ̸= 0.

Therefore (f, ai)0 < +∞fori ∈I and by Proposition 3.2 we have (f, ai)0 ∈N¯b0 +· · ·+N¯bk−1 hence (f, ai)0 ≡0mod Bk−1 for i ∈I.We have(2)(f, aihnk−i)0 ̸= (f, ajhnk−j)0for i ̸= j ∈ISuppose that (2) is not true, So there exist i, j ∈I such that i < j and (f, aihnk−i)0 =(f, ajhnk−j)0. Therefore (f, ai)0 + (nk −i)¯bk = (f, aj)0 + (nk −j)¯bk and (j −i)¯bk =(f, aj)0 −(f, ai)0 ≡0 mod Bk−1.

The last relation implies (j −i)¯bkBk ≡0 mod nkand consequently j −i ≡0 mod nk because ¯bk/Bk and nk are coprime. We get acontradiction because 0 < j −i < nk.From (1) and (2) we have(3)(f, g −hnk)0 = minnki=1(f, aihnk−i)0By assumption (f, g)0 > nk¯bk = (f, hnk)0, so (f, g −hnk)0 = nk¯bk and (3) impliesnk¯bk ≤(f, aihnk−i)0 = (f, ai)0 + (nk −i)¯bk for i = 1, .

. .

, nk. Therefore we get(4)(f, ai)0 ≥i¯bkfor i = 1, .

. .

, nk

6BY JANUSZ GWO´ZDZIEWICZ AND ARKADIUSZ P LOSKIMoreover we have(5)if(f, ai)0 = i¯bk,1 ≤i ≤nktheni = nk.Indeed, from (f, ai)0 = i¯bk it follows that i¯bk ≡0 mod Bk−1 and i¯bkBk ≡0 mod nk,so i ≡0 mod nk because ¯bk/Bk, nk are coprime. Hence we get i = nk.By (5) we get (because nk > 1)(6)(f, a1)0 > bkTherefore (f, τg(h))0 = (f, h +1nk a1)0 = (f, h)0 = ¯bk.5.

Proof of theorem 1.2The proof of (1.2) is based on the followingLemma 5.1. Let g = g(x, y) be an irreducible power series, p = (g, x)0 < +∞andlet 1 < k ≤h+1.

If (f,g)0(g,x)0 > Bk−2¯bk−1b0, then (g, x)0 ≡0 modb0Bk−1 . If, additionally(g, x)0 =b0Bk−1 , then Γ(f) = ⟨¯b1Bk−1 , .

. .

,¯bk−1Bk−1 ⟩.Proof of 5.1. Let (c0, c1, .

. .

, cm), c0 = p be the characteristic of g. By lemma 3.4we have of(g) > bk−1b0 , so there exist Puiseux expansions determined by f(x, y) = 0and g(x, y) = 0 respectively which coincide up to the ‘monomials’ of degree bk−1n .Therefore k −1 ≤m andb1n =c1p , . .

. ,bk−1n=ck−1p .

There exist integers a0,. .

. , ak−1 such that Bk−1 = a0b0 + a1b1 + · · · + ak−1bk−1,consequently we getpBk−1 = (a0p)n + a1(nc1) + · · · + ak−1(nck−1) ≡0 mod n and p ≡0 modnBk−1which proves the first part of (5.1).Suppose now that p =nBk−1 .

We have ci = pnbi =biBk−1 for i = 1, . .

. , k −1,hence Γ(f) = ⟨¯b1Bk−1 , .

. .

,¯bk−1Bk−1 ⟩.Now, we can pass to the proof of (1.2).Let φ = φ(x, y) ∈C[x, y]be y–regular, (φ, x)0 =nBk−1 . We shall check that(f, φ)0 ≤¯bk.

If k = h + 1 it is obvious (¯bh+1 = +∞), so we assume k ≤h. Writeφ = g1 · .

. .

· gs,gj ∈C[x, y]irreducibleWe have(1)(f, gj)0gj, x)0≤Bk−1¯bknfor all j = 1, . .

. , sIndeed, if we had (f,gj)0(gj,x)0 > Bk−1¯bknfor some j then, by lemma 5.1 we would get(gj, x)0 ≡0 modnBk and consequently (gj, x)0 ≥nBk .

It is impossible, because(gj, x)0 ≤(φ, x)0 =nBk−1 .Now, from (1) we get(f, φ)0 =Xj(f, gj)0 ≤XjBk−1¯bkn(gj, x)0 = Bk−1¯bkn(φ, x)0 = ¯bk

ON THE APPROXIMATE ROOTS OF POLYNOMIALS7Having proved the first part of (1.2) let us assume that (f, φ)0 > nk−1¯bk−1. Weclaim that there exists a j ∈{ 1, .

. ., s } such that(2)(f, gj)0(gj, x)0> Bk−2¯bk−1nSuppose, contrary to our claim, that (f,gj)0(gj,x)0 ≤Bk−2¯bk−1nfor all j = 1, .

. .

, s. Thuswe would have(f, φ)0 =Xj(f, gj)0 ≤Bk−2¯bk−1nXj(gj, x)0 = Bk−2¯bk−1n(φ, x)0 = nk−1¯bk−1which contradicts our assumption.From (2) it follows, by lemma 5.1, that (gj, x)0 = qnBk−1 for some integer q ≥1.On the other hand (gj, x)0 ≤(φ, x)0 =nBk−1 . Therefore we get q = 1 and (gj, x)0 =(φ, x)0.

Recall that gj divides φ, gj is irreducible and ord gj(0, y) = ord φ(0, y),thus gj is associated to φ which proves irreducibility of φ.References[1]S. S. Abhhyankar, Expansion Techniques in Algebraic Geometry, Tata Inst. Found.

Research,Bombay, 1977.[2]S. S. Abhhyankar, T. Moh, Newton-Puiseux expansion and generalized Tschirnhausen trans-formation, J. reine.

angew. Math.

260 (1973), 47–83; 261 (1973), 29–54. [3], Embeddings of the line in the plane, ibid.

276 (1975), 148–166.[4]R. Ephraim, Special Polares and Curves with one Place at Infinity, Proceedings of Symposiain Pure Mathematics 40 (1985), 353–359, Part I.[5]J.

Gwo´zdziewicz and A. P loski, On the Merle formula for polar invariants, Bull. Soc.

Sci.Letters XLI, 7 (1991), L´od´z, 61–67.[6]M. Merle, Invariants polaires des courbes planes, Invent.

Math. 41 (1977), 103–111.[7]T.

T. Moh, On the concept of approximate roots for algebra, J. of Algebra 65 (1980), 347–360. [8], On two fundamental theorems for the concept of approximate roots, J.

Math. Soc.Japan 34, no.

4 (1982).[9]A. P loski, Bezout’s theorem for affine curves with one branch at infinity, Univ.

Iaq. ActaMath.

Math. Fasc 28 (1991), 77–80.

[10] O. Zariski, Le problem des modules pour les branches planes, Centre de Mathematiques del’Ecole Polytechnique, 1973.Department of Mathematics,Technical University,Al. 1000 LPP 7, 25–314 Kielce,PolandE-mail address: mat-jg@srv1.tu.kielce.pl

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