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*](https://web.math.pmf.unizg.hr/%7Eduje/numbertheorybook.html), 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](https://doi.org/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.



$2$-isogenous to a curve in the Tate normal form
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)*x*y - b*y = x^3 - b*x^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)*x*y - b*y = x^3 - b*x^2 + (-5*b^2 + 5/2*b*c^2 + 5/2*b*c - 5*b - 5/16*c^4 + 5/4*c^3 - 15/8*c^2 + 5/4*c - 5/16)*x + (-3*b^3 + 9/4*b^2*c^2 + 7/2*b^2*c + 10*b^2 - 9/16*b*c^4 + 1/4*b*c^3 + 21/8*b*c^2 - 15/4*b*c + 23/16*b + 3/64*c^6 - 9/32*c^5 + 45/64*c^4 - 15/16*c^3 + 45/64*c^2 - 9/32*c + 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 + 1/4*c^2 - 1/2*c + 1/4)*x^3 + (b^2 - 1/2*b*c^2 + 1/2*b + 1/16*c^4 - 1/4*c^3 + 3/8*c^2 - 1/4*c + 1/16)*x^2 + (-1/2*b^2*c - 3/2*b^2 + 1/8*b*c^3 - 3/8*b*c^2 + 3/8*b*c - 1/8*b)*x + (3/4*b^3 + 1/16*b^2*c^2 - 1/8*b^2*c + 1/16*b^2)) / (x^3 + (-b + 1/4*c^2 - 1/2*c + 1/4)*x^2 + (1/2*b*c - 1/2*b)*x + 1/4*b^2) : (x^6*y + (-2*b + 1/2*c^2 - c + 1/2)*x^5*y + (b^2*c - b^2 - 1/2*b*c^3 + 3/2*b*c - b + 1/16*c^5 - 5/16*c^4 + 5/8*c^3 - 5/8*c^2 + 5/16*c - 1/16)*x^5 + (5/2*b*c - 5/2*b)*x^4*y + (-1/2*b^3*c + b^3 + 3/8*b^2*c^3 - 11/8*b^2*c^2 - 5/2*b^2*c + 7/2*b^2 - 3/32*b*c^5 + 1/2*b*c^4 - 17/16*b*c^3 + 9/8*b*c^2 - 19/32*b*c + 1/8*b + 1/128*c^7 - 7/128*c^6 + 21/128*c^5 - 35/128*c^4 + 35/128*c^3 - 21/128*c^2 + 7/128*c - 1/128)*x^4 + 5*b^2*x^3*y + (1/2*b^3*c^2 + 9/4*b^3*c - 5*b^3 - 1/4*b^2*c^4 + 11/16*b^2*c^3 - 9/16*b^2*c^2 + 1/16*b^2*c + 1/16*b^2 + 1/32*b*c^6 - 3/16*b*c^5 + 15/32*b*c^4 - 5/8*b*c^3 + 15/32*b*c^2 - 3/16*b*c + 1/32*b)*x^3 - 5*b^3*x^2*y + (-5/4*b^4*c + 7/2*b^4 + 1/8*b^3*c^3 - 9/16*b^3*c^2 + 3/4*b^3*c - 5/16*b^3 + 3/64*b^2*c^5 - 15/64*b^2*c^4 + 15/32*b^2*c^3 - 15/32*b^2*c^2 + 15/64*b^2*c - 3/64*b^2)*x^2 + (2*b^4 - 1/2*b^3*c^2 + 1/2*b^3*c)*x*y + (-b^5 + 5/8*b^4*c^2 - 7/8*b^4*c + 1/4*b^4 + 1/32*b^3*c^4 - 1/8*b^3*c^3 + 3/16*b^3*c^2 - 1/8*b^3*c + 1/32*b^3)*x - 1/2*b^4*c*y + (11/32*b^5*c - 1/16*b^5 + 1/128*b^4*c^3 - 3/128*b^4*c^2 + 3/128*b^4*c - 1/128*b^4)) / (x^6 + (-2*b + 1/2*c^2 - c + 1/2)*x^5 + (b^2 - 1/2*b*c^2 + 2*b*c - 3/2*b + 1/16*c^4 - 1/4*c^3 + 3/8*c^2 - 1/4*c + 1/16)*x^4 + (-b^2*c + 3/2*b^2 + 1/4*b*c^3 - 3/4*b*c^2 + 3/4*b*c - 1/4*b)*x^3 + (-1/2*b^3 + 3/8*b^2*c^2 - 3/4*b^2*c + 3/8*b^2)*x^2 + (1/4*b^3*c - 1/4\*b^3)*x + 1/16*b^4) : 1)
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