$\DeclareMathOperator\Sha{Sha}$I calculated the Tate–Shafarevich group $\Sha(E/K)[2]$ of the elliptic curve $E:y^2=x^3+17x$ over $K=\Bbb{Q}(\sqrt{-37})$.
I calculated that by hand and I reached the conclusion that $\Sha(E/K)[2]$ has order $1$ or $4$.
Magma code
K<b>:=QuadraticField(-37);
A:=EllipticCurve([K!0,0,0,17,0]);
Sel2:=TwoSelmerGroup(A); Sel2;
K<b>:=QuadraticField(-37);
A:=EllipticCurve([K!0,0,0,17,0]);
Tor:=TorsionSubgroup(A); Tor;
E := EllipticCurve([0,0,0,17,0]); // Define your elliptic curve E
Q := QuadraticTwist(E, -37); // Compute the quadratic twist of E by -37
rank := Rank(Q); // Calculate the rank of the quadratic twist rank;
outputs
Abelian Group isomorphic to Z/2 + Z/2 + Z/2 Defined on 3 generators in
supergroup:
Sel2.1 = $.1 + $.5
Sel2.2 = $.7
Sel2.3 = $.8 Relations:
2*Sel2.1 = 0
2*Sel2.2 = 0
2*Sel2.3 = 0 Abelian Group isomorphic to Z/2 Defined on 1 generator Relations:
2*Tor.1 = 0 \
2
Thus Magma reads $\Sha(E/K)[2]\cong ((\Bbb{Z}/2\Bbb{Z})^3/((\Bbb{Z} /2\Bbb{Z})\times (\Bbb{Z}/2\Bbb{Z})^2 )\cong 0$ ($E_{tor}$ is isomorphic to $\Bbb{Z}/2\Bbb{Z}$ and last '2' means rank of twist $E_{-37}$ over $\Bbb{Q}$ is $2$. $rank(E/\Bbb{Q})=0$, thus $rank(E/\Bbb{Q}(\sqrt{-37})=rank(E/\Bbb{Q})+rank(E_{-37}/\Bbb{Q})=0+2=2$.
On the other hand, Sage Math's code thanks to Chris Wuthrich's comment on Does $17x^4+y^2=-1$ have solution in $\Bbb{Q}_2(\sqrt{-5})$?:
sage: E = EllipticCurve([17,0])
sage: K.<t> = QuadraticField(-37)
sage: EK = E.base_extend(K)
sage: EK.simon_two_descent(verbose=4)
outputs $\Sha(E/K)[2]$ is nontrivial.
Why does this discrepancy occur and which is the correct Tate–Shafarevich group?
E := … Q
followed by:=
seemed to indicate thatQ
was meant to be at the beginning of the next line; I edited accordingly. I am not sure whether the\` after
2*Tor.1 = 0` was meant to be there. $\endgroup$MordellWeilShaInformation(A: ShaInfo:=true);
$\endgroup$