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$\DeclareMathOperator\Sha{Sha}$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?

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    $\begingroup$ Your line E := … Q followed by := seemed to indicate that Q 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. $\endgroup$
    – LSpice
    Commented Jul 13, 2023 at 20:47
  • $\begingroup$ Last '2' means rank of twist $E_{-37}$ over $\Bbb{Q}$ is $2$. $rank(E/\Bbb{Q})=0$, thus $rank(E/\Bbb{Q}(\sqrt{-37})=rank(E/\Bbb{Q})+rank(E_{-37}/\Bbb{Q})=0+2=2$. $\endgroup$
    – Duality
    Commented Jul 14, 2023 at 2:02
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    $\begingroup$ Magma outputs that $\mathrm{Sha}(E/K)[2]$ is trivial: MordellWeilShaInformation(A: ShaInfo:=true); $\endgroup$ Commented Jul 14, 2023 at 8:55
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    $\begingroup$ Observe that this is not the first time sage and magma to disagree about EC: What CASes say about the analytic rank of rank 8 elliptic curve $\endgroup$
    – joro
    Commented Jul 14, 2023 at 16:14

1 Answer 1

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Well spotted. This is a problem. After quite a bit of fiddling I found that the error is in Denis Simon's script used by Sage. In fact, when executed with higher values of the parameters so that the script finds the rational points, it also prints the correct information.

The command

K.<t> = QuadraticField(-37)
E = EllipticCurve([K(17),K(0)])
E.simon_two_descent(lim1=10,lim3=50,limtriv=10, verbose=3)

prints correctly

[E(K):phi'(E'(K))] >= 2
#S^(phi')(E'/K)     = 8
#III(E'/K)[phi']   <= 4

[E'(K):phi(E(K))]   = 8
#S^(phi)(E/K)       = 8
#III(E/K)[phi]      = 1

#III(E/K)[2]       <= 4
#E(K)[2]            = 2
#E(K)/2E(K)        >= 8

2 <= rank          <= 4

When executed with lower parameter it produces the incorrect information

2 <= #III(E/K)[2]       <= 8

I believe the script actually performs the correct calculations, but the part that draws these conclusions about the Tate-Shafarevich group has a bug in it. With the (correct) information that has been calculated before that line, the correct conclusion would be that $Ш(E/K)[2]$ has between $1$ and $2^3$ elements.

Magma is consistent. The following

K<b> := QuadraticField(-37);
A:=EllipticCurve([K!0,0,0,17,0]); 
"Torsion::", TorsionSubgroup(A); 
"Selmer Group::", TwoSelmerGroup(A); 
"Rank::", Rank(A);
"Generators::", Generators(A);
T := A![0,0];
phi := TwoIsogeny(T);
"Phi-Selmer group ::", SelmerGroup(phi);
phihat := DualIsogeny(phi);
"Phi hat =Selmer group ::", SelmerGroup(phihat);

returns the information about the descent by isogenies and for the full $2$-Selmer group, rather than only the ones by isogenies. Let $\varphi \colon E\to E'$ is the isogeny with kernel $\{O,(0,0)\}$. The $\phi$-Selmer group $\operatorname{Sel}^{\phi}(E/K)$ has dimension $3$ and so does the $\hat\phi$-Selmer group $\operatorname{Sel}^{\hat{\phi}}(E'/K)$. The $2$-Selmer group $\operatorname{Sel}^{2}(E/K)$ is also of dimension $3$, which means that the image of $\operatorname{Sel}^{\hat{\phi}}(E'/K) \to Ш(E'/K)/\phi(Ш(E/K))$ has dimension $2$. One finds that $Ш(E'/K)[\hat\varphi]$ is of order $2$, but the Tate-Shafarevich group of $E/K$ has a trivial $2$-part.

This is constistant with the Birch and Swinnerton-Dyer conjecture, I believe.

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    $\begingroup$ ... and I should add that I wish to apologies for my comment on the other question, where I read off the wrong results from the script without questioning it at all. $\endgroup$ Commented Jul 14, 2023 at 15:25
  • $\begingroup$ I really appreciate your help. Since the comment may be longer, I sent an email about the matter on 7/23 at 21:29 from my university's address.If I could receive a response, I would be most grateful. $\endgroup$
    – Duality
    Commented Jul 25, 2023 at 15:25

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