It's well-known that quadratic forms over the rational numbers $\mathbf{Q}$ satisfy the Hasse-Minkowski theorem. This is to say that they are isotropic over $\mathbf{Q}$ if and only if they are isotropic over $\mathbf{R}$ and over $\mathbf{Q}_p$ for all prime numbers $p$.

Of course the same statement does not hold for cubic forms, as demonstrated by Selmer, who showed that the ternary cubic form $3x^3 + 4y^3 + 5z^3$ has nontrivial zeros over $\mathbf{R}$ and over $\mathbf{Q}_p$ for all $p$. We say then that Selmer's example violates the Hasse Principle.

Now I saw a claim recently, which seems a little too good to be true, that if you take a ternary cubic form which violates the Hasse Principle then you can show that it must be diagonal.

Now I don't want to talk about approaches to this or potential proofs or anything of the sort. Merely, when I saw this I thought there ought to be some easy counterexample, but I couldn't immediately come up with one. If you're like me, when you think of Hasse Principle counterexamples, you first think of Selmer's example or of a hyperelliptic example, or maybe something that's just harder to write down. No obvious non-diagonal cubic forms lie in that bunch. As such, I pose the following question.

Is there a non-diagonal ternary cubic form which violates the Hasse Principle? Can you write it down explicitly?

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    $\begingroup$ How about $3(x+y)^3 + 4y^3 + 5z^3$? $\endgroup$ Oct 2, 2013 at 14:13
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    $\begingroup$ The jacobian of a diagonal cubic has $j$-invariant zero. Just find an elliptic curve with $j\ne0$ and with $Sha[3] \ne 0$ in Cremona's table. $\endgroup$ Oct 2, 2013 at 14:21
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    $\begingroup$ Ah. Yes I should have thought of the Jacobian. 182b3 should work just fine. $\endgroup$
    – stankewicz
    Oct 2, 2013 at 14:26

1 Answer 1


Continuing with Martin Bright's comment: if $F(X,Y,Z)$ is a ternary cubic form, say with integer coefficients and $M\in GL_3(\mathbf{Z})$ then $M$ acts on the variables $X,Y,$ and $Z$ in an obvious way and with this action, $F(MX, MY,MZ)$ is another ternary cubic form and any reasonable invariants of $F$ will also be true of this new cubic form. This of course includes anything about rational points and also the Jacobian as Prof. Voloch pointed out.

Of course this was not really the sort of example I was looking for. I was hoping to find something quantitatively different (say, they should not be equivalent to a diagonal under the action of $GL_3(\mathbf{Z})$ at least). And what I mean in my comment above about 182b3 is that using the search of Cremona's tables provided by LMFDB, we can find an example with $j$-invariant $-\frac{424962187484640625}{182}$.

Below are 4 ternary cubic forms, all of which are inequivalent under the action of $GL_3(\mathbf{Z})$ and all of which violate the Hasse Principle.

[ -x^3 + 17*x*y*z + 2*y^3 + 91*z^3,

2*x^3 + 17*x*y*z - 7*y^3 + 13*z^3,

-x^3 + 17*x*y*z - 7*y^3 - 26*z^3,

-x^3 - 17*x*y*z - 13*y^3 + 14*z^3 ]

These were generated with the magma commands

E := EllipticCurve("182b3");


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    $\begingroup$ Instead of "diagonal" you should have written "diagonalizable". In your examples, stylistically, it would look nicer to make an overall sign change if needed so that $x^3$ has a positive coefficient rather than a negative coefficient. $\endgroup$
    – KConrad
    Nov 25, 2013 at 15:54

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