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The projective curve $3x^3+4y^3+5z^3=0$ is often cited as an example (given by Selmer) of a failure of the Hasse Principle: the equation has solutions in any completion of the rationals $\mathbb Q$, but not in $\mathbb Q$ itself.

I don't think I've ever seen a proof of the latter claim — is someone able to provide an outline? What are the necessary tools?

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5 Answers 5

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My friend has written an introduction to algebraic number theory before, which contains a short proof of this statement, but I didn't check its validity.

Edit: updated the link of the document, http://www.2shared.com/document/2d6M7kNU/Introduction_to_Algebraic_Numb.html p. 41 of the document, or p. 45 of the PDF.

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  • $\begingroup$ This is certainly a lot more elementary than the other methods mentioned, although it (therefore) does not present a clear example of what the general obstruction is (for the failure of Hasse). Still, thanks a lot - at least this is a proof I can easily follow :-) $\endgroup$
    – Alon Amit
    Commented Oct 27, 2009 at 7:26
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    $\begingroup$ For what it's worth, I once worked out a completely elementary proof that the equation has p-adic solutions for all p and put it on an UG example sheet here: www2.imperial.ac.uk/~buzzard/maths/teaching/04Lent/M4P32/… $\endgroup$ Commented Nov 20, 2009 at 23:27
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    $\begingroup$ The link in the answer is now broken. $\endgroup$
    – KConrad
    Commented Apr 16, 2011 at 16:32
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    $\begingroup$ Hi, the link is updated. $\endgroup$
    – user709
    Commented Apr 19, 2011 at 23:44
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    $\begingroup$ I think the link is broken again :( $\endgroup$ Commented Dec 10, 2020 at 6:40
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This problem is in Cassels' book "Local Fields" and I wrote up a solution once along those lines, for an algebraic number theory class. See this paper, but I should advise that it comes out seeming pretty tedious. Solutions that involve elliptic curves are more conceptual. Others have already provided pointers to references for that approach.

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  • $\begingroup$ Ten years too late, but I thought I'd point it out: the section headings in this paper are flipped! You wrote "No local solutions" and then "Global solutions". Regardless, thank you for this, and your numerous other awesome papers. $\endgroup$
    – Alon Amit
    Commented Apr 13, 2020 at 20:26
  • $\begingroup$ @AlonAmit fixed. You can email me directly the next time you find a typo. $\endgroup$
    – KConrad
    Commented Apr 14, 2020 at 2:39
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I think I saw a proof of that in Cassel's "Diophantine Equations with special reference to elliptic curves" and in some surveys by Mazur in the Bull. AMS (perhaps this, but I have in the moment no time to look).

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  • $\begingroup$ Interesting. The obstruction, as presented by Mazur, is that the curve 60x^3+y^3+z^3=0 does have a rational point, and is a "companion" (Q-twist of) the Selmer curve. I'll need to dig a lot more to understand this. Thanks! $\endgroup$
    – Alon Amit
    Commented Oct 27, 2009 at 7:22
  • $\begingroup$ "This" is now dead, but it appears to have been a link to Mazur - On the passage from local to global in number theory, with the relevant result beginning on p. 22. $\endgroup$
    – LSpice
    Commented May 14, 2020 at 22:26
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The "standard" technique for killing the Hasse priniciple for elliptic curves is to show that the Tate-Shafarevich group has a copy of (Z/mZ)^2 for some m - see chapter X in Silverman's the arithmetic of Eliptic curves, both for the theory and examples. All the examples which Silverman presents ar with m = 2. Selmers example requires m = 3, which requires (much) more computations. Poonen has an example on his web page of a family of elliptic curves violating the Hasse principle, and containing Selmers example, but you'd have to dive through a labirinth of references.

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  • $\begingroup$ (Hello there :-) ) That's pretty heavy machinery for this humble reader - but an interesting view of the obstruction. Thanks a lot! $\endgroup$
    – Alon Amit
    Commented Oct 27, 2009 at 7:28
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    $\begingroup$ (Hello indeed :) ) The advantage this approach has is certainly not simplicity, it is rather that it can be - and is - mechanised. Google up Hasse Tate Shafarevich and Magma. $\endgroup$ Commented Oct 27, 2009 at 8:26
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There's a proof in Cassels' little blue book on elliptic curves which the OP might find more to his taste than some others mentioned here.

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