📝 Original Info
- Title: A point of order 8
- ArXiv ID: 1110.0357
- Date: 2011-10-30
- Authors: Semjon Adlaj
📝 Abstract
A formula expressing a point of order 8 on an elliptic curve, in terms of the roots of the associated cubic polynomial, is given. Doubling such a point yields a point of order 4 distinct from the well-known points of order 4 given in standard references such as "A course of Modern Analysis" by Whittaker and Watson.
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Deep Dive into A point of order 8.
A formula expressing a point of order 8 on an elliptic curve, in terms of the roots of the associated cubic polynomial, is given. Doubling such a point yields a point of order 4 distinct from the well-known points of order 4 given in standard references such as “A course of Modern Analysis” by Whittaker and Watson.
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arXiv:1110.0357v1 [math.GM] 3 Oct 2011
A POINT OF ORDER 8
S. ADLAJ
Abstract. A formula expressing a point of order 8 on an elliptic
curve, in terms of the roots of the associated cubic polynomial, is
given. Doubling such a point yields a point of order 4 distinct from
the well-known points of order 4 given in standard references such
as “A course of Modern Analysis” by Whittaker and Watson.
Let
(1)
y2 = 4(x −e1)(x −e2)(x −e3), ei ∈C, i = 1, 2, 3,
be the defining equation for an elliptic curve E over the complex field
C, and let
β :=
re1 −e3
e1 −e2
, γ :=
p
(e1 −e3)(e1 −e2), i :=
√
−1.
The roots of the cubic on the right hand side of the defining equation
(1) need not sum to zero but assume that β > 1, and introduce the
values
β1 :=
s
β + 1
β −1 +
r
2
β −1 > β2 :=
r
2
β + 1 + 1
β .
The point P = (x, y) on E, where
x = e1 −γ −γ
r
β + 1
2
−1
!
1 −1
β +
r
1 + 1
β
p
β1 + β2 + i
p
β1 −β2 +
r
1 −1
β
,
is then a point of order 8.
Note that doubly doubling the afore-indicated point P cannot possibly
yield either (e1, 0) nor (e3, 0), and so must, if P is indeed of order 8,
yield the point (e2, 0).
Date: May 30, 2011.
1991 Mathematics Subject Classification. 11G05.
Key words and phrases. Torsion point on an elliptic curve over the complex field,
roots of cubic, Weierstrass normal form, doubling formula.
2
S. ADLAJ
For an example, consider an elliptic curve E given in Weierstrass normal
form via equation (1), where
e1 = i, e2 = 0, e3 = −i.
Then
γ = i
√
2, β =
√
2, β1 = 1+
√
2+
r
2
√
2 + 1
, β2 = 1
√
2+
r
2
√
2 −1
,
and the x–coordinate of P is calculated to be
x(P) =
√
2 −1 −i
r
2
√
2 −1
.
One might employ the well-known doubling formula, found in stan-
dard sources such as [1], for successively calculating the points 2P =
(1, ±2
√
2) and 4P = (0, 0), the latter evidently being a point of order
2 on E.
Incidentally, the formulas given in [1, §20.33, p. 444] apply here yielding
two pairs of points of order 4, whose x-coordinates are e1 ± γ = i(1 ±
√
2), so all four points differ from either point of the already computed
pair 2P, as they match when doubled the points (±i, 0), which are,
aside from the point 4P, the two remaining points of order 2 on E.
References
[1] Whittaker E. T. Watson G. N. A Course of Modern Analysis. Cambridge
University Press; 4th edition (January 2, 1927).
CC RAS, Vavilov st. 40, Moscow, Russia, 119333
E-mail address: SemjonAdlaj@gmail.com
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