ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Sat, 18 Mar 2017 00:39:20 +0100finding a point on the elliptic curve with parameterhttps://ask.sagemath.org/question/36304/finding-a-point-on-the-elliptic-curve-with-parameter/ I wish to know how find a point on the elliptic curve y^2=X^3 A(t)X +B(t) given over Q(t) with coffecients overQ(t)?as an example:y^2=x^3+t^4x+t+t^3. edSun, 15 Jan 2017 23:13:13 +0100https://ask.sagemath.org/question/36304/finding-a-point-on-the-elliptic-curve-with-parameter/Answer by slelievre for <p>I wish to know how find a point on the elliptic curve y^2=X^3 A(t)X +B(t) given over Q(t) with coffecients overQ(t)?as an example:y^2=x^3+t^4x+t+t^3. ed</p>
https://ask.sagemath.org/question/36304/finding-a-point-on-the-elliptic-curve-with-parameter/?answer=36321#post-id-36321Try this:
sage: F = QQ['t'].fraction_field()
sage: t = F.gen()
sage: E = EllipticCurve(F, [t^4, t + t^3])
Tue, 17 Jan 2017 10:29:57 +0100https://ask.sagemath.org/question/36304/finding-a-point-on-the-elliptic-curve-with-parameter/?answer=36321#post-id-36321Comment by AB for <p>Try this:</p>
<pre><code>sage: F = QQ['t'].fraction_field()
sage: t = F.gen()
sage: E = EllipticCurve(F, [t^4, t + t^3])
</code></pre>
https://ask.sagemath.org/question/36304/finding-a-point-on-the-elliptic-curve-with-parameter/?comment=36327#post-id-36327Thank you for your reply. I try the above case, but don't show thingsTue, 17 Jan 2017 15:36:29 +0100https://ask.sagemath.org/question/36304/finding-a-point-on-the-elliptic-curve-with-parameter/?comment=36327#post-id-36327Answer by dan_fulea for <p>I wish to know how find a point on the elliptic curve y^2=X^3 A(t)X +B(t) given over Q(t) with coffecients overQ(t)?as an example:y^2=x^3+t^4x+t+t^3. ed</p>
https://ask.sagemath.org/question/36304/finding-a-point-on-the-elliptic-curve-with-parameter/?answer=36970#post-id-36970The elliptic curve functionality is restricted on general fields, for instance:
sage: R.<t> = QQ[]
sage: F = R.fraction_field()
sage: E = EllipticCurve(F, [1, t + t^3])
sage: E.point( (-t,0) )
(-t : 0 : 1)
sage: P = E.point( (-t,0) )
sage: P.is_finite_order()
---------------------------------------------------------------------------
NotImplementedError Traceback (most recent call last)
<ipython-input-39-27a368cc0c53> in <module>()
----> 1 P.is_finite_order()
::::: many lines
NotImplementedError: Computation of order of a point not implemented over general fields.
I took above a simpler curve with an obvious torsion point.
Even the method `is_finite_order` failed to be executed.
Although in the above case we can at least see that $2P$ is $0$, if we ask for it:
sage: 2*P
(0 : 1 : 0)
sage: 2*P == E(0)
True
So if we get problems with the method `is_finite_order`, asking for the rank and/or for generators...
We can still use sage (and the own devices) to study the posted curve. After a quick pessimistic search of a polynomial $x=x(t)$ (divided by a power of $t$ maybe), such that $x^3 + t^4x + (t^3+t)$ should at least have an even order of $t$, i've got an optimistic point of view, that there *should* be no point on the curve
$$
y^2 = x^3 + t^4x + (t^3+t)\ .
$$
Why? Since if there would be one, say $(x,y) = (\ P_x(t)/Q_x(t)\ ,\ P_y(t)/Q_y(t)\ )$, then using specialization (not everywhere defined) morphisms $\mathbb{Q}(t)\to\mathbb{Q}$, $t\to t_0$ we would get some rational points on the elliptic curves defined over $\mathbb Q$,
$$
y^2 = x^3 + t_0^4x + (t_0^3+t_0)\ .
$$
Unless there is a division by zero in the one or the other denominator, when specializing $t$ to $t_0$. But even so, the denominators would have been already very complicated, since...
# for t0 in [ -29..29 ]:
for t0 in [ -25, -24, -20, -8, -3, -2, 2, 8, 12, 18, 22, 24, 29 ]:
try:
E = EllipticCurve( QQ, ( t0^4, t0^3 + t0 ) )
ETP = E.torsion_points()
if len( ETP ) > 1:
print "... t0 = %3s | rank = ? | torsion points = %s" % ( t0, ETP )
continue
r = E.rank()
if r > 0:
print "... t0 = %3s | rank = %s | torsion points = %s" % ( t0, r, ETP )
continue
print "*** t0 = %3s | rank = %s | torsion points = %s" % ( t0, r, ETP )
except Exception:
pass
And we get:
*** t0 = -25 | rank = 0 | torsion points = [(0 : 1 : 0)]
*** t0 = -24 | rank = 0 | torsion points = [(0 : 1 : 0)]
*** t0 = -20 | rank = 0 | torsion points = [(0 : 1 : 0)]
*** t0 = -8 | rank = 0 | torsion points = [(0 : 1 : 0)]
*** t0 = -3 | rank = 0 | torsion points = [(0 : 1 : 0)]
*** t0 = -2 | rank = 0 | torsion points = [(0 : 1 : 0)]
*** t0 = 2 | rank = 0 | torsion points = [(0 : 1 : 0)]
*** t0 = 8 | rank = 0 | torsion points = [(0 : 1 : 0)]
*** t0 = 12 | rank = 0 | torsion points = [(0 : 1 : 0)]
*** t0 = 18 | rank = 0 | torsion points = [(0 : 1 : 0)]
*** t0 = 22 | rank = 0 | torsion points = [(0 : 1 : 0)]
*** t0 = 24 | rank = 0 | torsion points = [(0 : 1 : 0)]
*** t0 = 29 | rank = 0 | torsion points = [(0 : 1 : 0)]
So an "unexpected" $\mathbb{Q}(t)$-rational point on $E$ is comming - in view of our first (naive) sieve - with denominators containing the factor
$$\small Q=(t+25)(t+24)(t+20)(t+8)(t+3)(t+2)(t-2)(t-8)(t-12)(t-18)(t-22)(t-24)(t-29)\ .$$
(In fact $Q^2$ and $Q^3$ in $x$ and $y$.) This is not much, from an algorithmic point of view, but a good start.
For a family with one generating point $A(x(t),y(t))$, we can still use the specialization to "guess" $A$.Sat, 18 Mar 2017 00:39:20 +0100https://ask.sagemath.org/question/36304/finding-a-point-on-the-elliptic-curve-with-parameter/?answer=36970#post-id-36970