Algebraic Geometry – 2-Isogenous to a Curve in the Tate Normal Form

ag.algebraic-geometryelliptic-curvesmagmant.number-theoryrational-points

It is well-known that an elliptic curve $E$ that has a point of order $2$ and is represented as $E=[0,a,0,b,0]$ has a $2$-isogenous curve $E^\prime=[0,-2a,0,a^2-4b,0]$, see e.g. p. 507 in

A. Dujella, Number Theory, University of Zagreb, Školska knjiga, Zagreb, 2021, ISBN: 978-953-0-30897-8, 621 pp.

Question: Does a similar simple formula for $E^\prime$ exist for a curve with $2$-torsion expressed in the Tate normal form $E=[1-c,-b,-b,0,0]$? We may assume that all torsion points on $E$ are known.

Rationale: I am working on $\mathbb{Z}/16\mathbb{Z}$ curves over cubic fields. The curves are generated by the formulas on p. 584 in

D. Jeon, C. H. Kim, and Y. Lee, Families of elliptic curves over cubic number fields with prescribed torsion subgroups, Mathematics of Computation, Volume 80, Number 273, January 2011, pp. 579–591, doi:10.1090/S0025-5718-10-02369-0.

For $t=\frac{4}{7}$, Magma struggles to calculate the last generator.

The DescentInformation is very limited over number fields in Magma ($2$-descent only). IsogenousCurves and IsIsogenous are not implemented at all.

Sometimes, feeding a $2$-isogenous curve or a different model of the curve helps.

Best Answer

John Cremona has some explicit code for calculating the 2-torsion points of curves in the general Weierstrass [a1,a2,a3,a4,a6] format, and part of that formula is finding rational roots to the cubic equation

(1) $ P(x,[W]) = 4(x^3+e_{a2}x^2+e_{a4}x+e_{a6})+(e_{a1}x+e_{a3})^2)$

where [a1,a2,a3,a4,a6] are from the Weierstrass form W.

If you stick in a curve of the form [0,a,0,b,0] then (1) becomes

(2) $4x^3 + 4ax^2 + 4bx = 0$ (to find the 2-torsion points)

and it is easy to see that $x=0$ is one torsion point. The other 2 are the quadratic roots of $x^2 + 4ax + b$ which need to be rational, if three 2-torsion points are to be found.

However if we use the Tate form above of $[1-c,-b,-b,0,0]$ and stick it into (1) we derive

(2) $P(x,b,c) = 4x^3 + (c^2 - 2c + (-4b + 1))x^2 + (2bc - 2b)x + b^2$

and we want rational roots of this cubic for x.

Maxima tells us that the 3 roots of this cubic in [b,c] is

$x_1=\left({{-1}\over{2}}-{{\sqrt{3}\,i}\over{2}}\right)\,\left({{b\, \sqrt{-b\,\left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+\left(-20\,b-1 \right)\,c+16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}}}}}+{{-{{3\,b^ 2}\over{4}}-{{\left(4\,b-c^2+2\,c-1\right)\,\left(\left(c-1\right)\, b\right)}\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c-1\right)^3}\over{ 1728}}\right)^{{{1}\over{3}}}-{{\left({{\sqrt{3}\,i}\over{2}}+{{-1 }\over{2}}\right)\,\left({{\left(c-1\right)\,b}\over{6}}-{{\left(4\, b-c^2+2\,c-1\right)^2}\over{144}}\right)}\over{\left({{b\,\sqrt{-b\, \left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+\left(-20\,b-1\right)\,c+ 16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}}}}}+{{-{{3\,b^2}\over{4}} -{{\left(4\,b-c^2+2\,c-1\right)\,\left(\left(c-1\right)\,b\right) }\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c-1\right)^3}\over{1728}} \right)^{{{1}\over{3}}}}}+{{4\,b-c^2+2\,c-1}\over{12}}$

$x_2 = \left({{\sqrt{3}\,i}\over{2}}+{{-1}\over{2}}\right)\,\left({{b\, \sqrt{-b\,\left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+\left(-20\,b-1 \right)\,c+16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}}}}}+{{-{{3\,b^ 2}\over{4}}-{{\left(4\,b-c^2+2\,c-1\right)\,\left(\left(c-1\right)\, b\right)}\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c-1\right)^3}\over{ 1728}}\right)^{{{1}\over{3}}}-{{\left({{-1}\over{2}}-{{\sqrt{3}\,i }\over{2}}\right)\,\left({{\left(c-1\right)\,b}\over{6}}-{{\left(4\, b-c^2+2\,c-1\right)^2}\over{144}}\right)}\over{\left({{b\,\sqrt{-b\, \left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+\left(-20\,b-1\right)\,c+ 16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}}}}}+{{-{{3\,b^2}\over{4}} -{{\left(4\,b-c^2+2\,c-1\right)\,\left(\left(c-1\right)\,b\right) }\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c-1\right)^3}\over{1728}} \right)^{{{1}\over{3}}}}}+{{4\,b-c^2+2\,c-1}\over{12}}$

and

$ x_3 = \left({{b\,\sqrt{-b\,\left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+ \left(-20\,b-1\right)\,c+16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}} }}}+{{-{{3\,b^2}\over{4}}-{{\left(4\,b-c^2+2\,c-1\right)\,\left( \left(c-1\right)\,b\right)}\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c -1\right)^3}\over{1728}}\right)^{{{1}\over{3}}}-{{{{\left(c-1\right) \,b}\over{6}}-{{\left(4\,b-c^2+2\,c-1\right)^2}\over{144}}}\over{ \left({{b\,\sqrt{-b\,\left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+ \left(-20\,b-1\right)\,c+16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}} }}}+{{-{{3\,b^2}\over{4}}-{{\left(4\,b-c^2+2\,c-1\right)\,\left( \left(c-1\right)\,b\right)}\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c -1\right)^3}\over{1728}}\right)^{{{1}\over{3}}}}}+{{4\,b-c^2+2\,c-1 }\over{12}}$

I really don't see a simple form here for rational roots. unless b and c take very specific values so that one of $x_1$, $x_2$ or $x_3$ become rational.

I did also consider the elliptic curve invariants $c_4$ and $c_6$ equivalency, but it means moving from a sextic to a quartic equation for the two variables (b,c --> a,b)

(continuing) after reading up a bit more on 2-isogenies and Magma defining fields, the following Magma code and results is your answer, no, there is no simple elliptic curve:

F<a,b,c>:=FunctionField(Rationals(),3);

E:=EllipticCurve([1-c,-b,-b,0,0]);

E;

Elliptic Curve defined by y^2 + (-c + 1)xy - by = x^3 - bx^2 over Multivariate rational function field of rank 3 over Rational Field

E1, f := IsogenyFromKernel(E, DivisionPolynomial(E, 2));

E1;

Elliptic Curve defined by y^2 + (-c + 1)xy - by = x^3 - bx^2 + (-5b^2 + 5/2bc^2 + 5/2bc - 5b - 5/16c^4 + 5/4c^3 - 15/8c^2 + 5/4c - 5/16)x + (-3b^3 + 9/4b^2c^2 + 7/2b^2c + 10b^2 - 9/16bc^4 + 1/4bc^3 + 21/8bc^2 - 15/4bc + 23/16b + 3/64c^6 - 9/32c^5 + 45/64c^4 - 15/16c^3 + 45/64c^2 - 9/32c + 3/64) over Multivariate rational function field of rank 3 over Rational Field

f;

Elliptic curve isogeny from: CrvEll: E to CrvEll: E1

taking (x : y : 1) to ((x^4 + (-b + 1/4c^2 - 1/2c + 1/4)x^3 + (b^2 - 1/2bc^2 + 1/2b + 1/16c^4 - 1/4c^3 + 3/8c^2 - 1/4c + 1/16)x^2 + (-1/2b^2c - 3/2b^2 + 1/8bc^3 - 3/8bc^2 + 3/8bc - 1/8b)x + (3/4b^3 + 1/16b^2c^2 - 1/8b^2c + 1/16b^2)) / (x^3 + (-b + 1/4c^2 - 1/2c + 1/4)x^2 + (1/2bc - 1/2b)x + 1/4b^2) : (x^6y + (-2b + 1/2c^2 - c + 1/2)x^5y + (b^2c - b^2 - 1/2bc^3 + 3/2bc - b + 1/16c^5 - 5/16c^4 + 5/8c^3 - 5/8c^2 + 5/16c - 1/16)x^5 + (5/2bc - 5/2b)x^4y + (-1/2b^3c + b^3 + 3/8b^2c^3 - 11/8b^2c^2 - 5/2b^2c + 7/2b^2 - 3/32bc^5 + 1/2bc^4 - 17/16bc^3 + 9/8bc^2 - 19/32bc + 1/8b + 1/128c^7 - 7/128c^6 + 21/128c^5 - 35/128c^4 + 35/128c^3 - 21/128c^2 + 7/128c - 1/128)x^4 + 5b^2x^3y + (1/2b^3c^2 + 9/4b^3c - 5b^3 - 1/4b^2c^4 + 11/16b^2c^3 - 9/16b^2c^2 + 1/16b^2c + 1/16b^2 + 1/32bc^6 - 3/16bc^5 + 15/32bc^4 - 5/8bc^3 + 15/32bc^2 - 3/16bc + 1/32b)x^3 - 5b^3x^2y + (-5/4b^4c + 7/2b^4 + 1/8b^3c^3 - 9/16b^3c^2 + 3/4b^3c - 5/16b^3 + 3/64b^2c^5 - 15/64b^2c^4 + 15/32b^2c^3 - 15/32b^2c^2 + 15/64b^2c - 3/64b^2)x^2 + (2b^4 - 1/2b^3c^2 + 1/2b^3c)xy + (-b^5 + 5/8b^4c^2 - 7/8b^4c + 1/4b^4 + 1/32b^3c^4 - 1/8b^3c^3 + 3/16b^3c^2 - 1/8b^3c + 1/32b^3)x - 1/2b^4cy + (11/32b^5c - 1/16b^5 + 1/128b^4c^3 - 3/128b^4c^2 + 3/128b^4c - 1/128b^4)) / (x^6 + (-2b + 1/2c^2 - c + 1/2)x^5 + (b^2 - 1/2bc^2 + 2bc - 3/2b + 1/16c^4 - 1/4c^3 + 3/8c^2 - 1/4c + 1/16)x^4 + (-b^2c + 3/2b^2 + 1/4bc^3 - 3/4bc^2 + 3/4bc - 1/4b)x^3 + (-1/2b^3 + 3/8b^2c^2 - 3/4b^2c + 3/8b^2)x^2 + (1/4b^3c - 1/4*b^3)x + 1/16b^4) : 1)

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Related Solutions

[Math] Counting isomorphism classes of elliptic curves with specific torsion

$\newcommand\F{\mathbf{F}}$ $\newcommand\Z{\mathbf{Z}}$ $\newcommand\SL{\mathrm{SL}}$

Because of the existence of the Weil paring, elliptic curves with such a subgroup only exist when $p \equiv 1 \mod \ N$.

Let $S_N$ denote the set of elliptic curves over $\F_p$ such that $E[N]$ is defined over $\F_p$. It will be slightly easier to assume that $N \ge 3$. In this case, $Y(N)$ is a fine moduli space, and an $\F_p$-point on $Y(N)$ corresponds to a pair $(E,\alpha:E[N] \simeq \Z/N\Z \times \Z/N \Z)$ defined over $\F_p$. Given an elliptic curve $E \in S_N$, how many points does it contribute to $Y(N)$? For a curve $E$ whose automorphism group is $\Z/2\Z$, We see that out of the $|\SL_2(\Z/N\Z)|$ possible choices of $\alpha$ (technical remark, we have fixed a Weil pairing so that $Y(N)$ is connected), $(E,\alpha) \simeq (E,\alpha')$ only if $\alpha' = \alpha$ or $\alpha' = [-1] \alpha$. Thus $E$ contributes $|\SL_2(\Z/N\Z)|/2$ points to $Y(N)(\F_p)$. In general, $E$ may have slightly more automorphisms, and we deduce that (for $N \ge 3$): $$|Y(N)(\F_p)| = |\SL_2(\Z/N\Z)| \sum_{E \in S_N} \frac{1}{|\mathrm{Aut}(E)|}.$$ Note that the quantity on the right is very close to $|\SL_2(\Z/N\Z)| \cdot |S_N|/2$, one only has to worry about the elliptic curves with $j = 0$ or $j = 1728$, and this can be done by hand if one wants to cross all the i's and dot all the t's.

Suppose that $X(N)$ has $c_N$ cusps and genus $g_N$ (there are some explicit slightly unpleasant formulas for these numbers, which can be found (for example) in Shimura's book. All the cusps are defined over $\F_p$ (with $p \equiv 1 \mod N$) so by the Riemann hypothesis for finite fields, $$|Y(N)(\F_p) - (1+p) + c_N| = |X(N)(\F_p) - (1+p)| \le 2 g_N \sqrt{p}.$$ If $g_N = 0$ (which only happens if $N \le 5$), this leads to an exact formula for $|S_N|$. In general, at least for large $p$, we see that $$|S_N| \sim \frac{2p}{|\SL_2(\Z/N \Z)|}.$$

To make this all completely explicit for $N = 3$ (for example), one gets, presuming I have not made a horrible computational error which is quite possible: $$S_3 = \begin{cases} (p+11)/12, & p \equiv 1 \mod \ 12, \\\ (p+5)/12, & p \equiv 7 \mod \ 12. \end{cases}$$ (note that $p \equiv 1 \mod 3$):

Of course, "exact formulas" will only exist for $N \le 5$. Some related and slightly more difficult counting problems are also nicely explained by Lenstra here (See 1.10):

https://openaccess.leidenuniv.nl/bitstream/1887/3826/1/346_086.pdf

Degree 17 Number Fields Ramified Only at 2

Thank you for calling this problem to my attention. I computed $K$ en route to AWS (though this year's topics are a rather different flavor of number theory...). After some simplification (gp's $\rm polredabs$), it turns out that the field $K$ is generated by a root of $$ f(x) = x^{17} - 2x^{16} + 8x^{13} + 16x^{12} - 16x^{11} + 64x^9 - 32x^8 - 80x^7 $$ $$ \qquad\qquad\qquad {} + 32x^6 + 40x^5 + 80x^4 + 16x^3 - 128x^2 - 2x + 68. $$ gp quickly confirms:

p = Pol([1,-2,0,0,8,16,-16,0,64,-32,-80,32,40,80,16,-128,-2,68])
poldegree(p)
F = nfinit(p);
factor(F.disc)

returns

[2 79]

So the field has discriminant $2^{79}$.

Let $F = {\bf Q}(\zeta_{64}^{\phantom{0}} - \zeta_{64}^{-1})$, with $F/\bf Q$ cyclic of degree $16$ of $\bf Q$. It was known that $K$ is contained in an unramified cyclic extension $L/F$ of degree $17$. Hence $L(\zeta_{17})$ is a Kummer extension of $F_{17} := F(\zeta_{17})$. Now $F_{17}$ is a $({\bf Z} / 16 {\bf Z})^2$ extension of $\bf Q$. Fortunately it was not necessary to work in the unit group of this degree $256$ field: the Kummer extension is $F_{17}(u^{1/17})$ with $u \in F_{17}^* / (F_{17}^*)^{17}$ in an eigenspace for the action of ${\rm Gal}(F_{17}/{\bf Q})$, which makes it fixed by some index-$16$ subgroup of this Galois group. Thus $u$ could be found in one of the cyclic degree-$16$ extension of $\bf Q$ intermediate between $\bf Q$ and $F_{17}$, and eventually a generator of $K$ turned up whose minimal polynomial, though large, was tractable enough for gp's number-field routines to do the rest.

Best Answer

Related Solutions

[Math] Counting isomorphism classes of elliptic curves with specific torsion

Degree 17 Number Fields Ramified Only at 2

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