24
$\begingroup$

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?

$\endgroup$
0

5 Answers 5

8
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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
    Oct 27, 2009 at 7:26
  • 3
    $\begingroup$ 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/… $\endgroup$ Nov 20, 2009 at 23:27
  • 2
    $\begingroup$ The link in the answer is now broken. $\endgroup$
    – KConrad
    Apr 16, 2011 at 16:32
  • 2
    $\begingroup$ Hi, the link is updated. $\endgroup$
    – user709
    Apr 19, 2011 at 23:44
  • 1
    $\begingroup$ I think the link is broken again :( $\endgroup$ Dec 10, 2020 at 6:40
10
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
    Apr 13, 2020 at 20:26
  • $\begingroup$ @AlonAmit fixed. You can email me directly the next time you find a typo. $\endgroup$
    – KConrad
    Apr 14, 2020 at 2:39
6
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ 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
    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
    May 14, 2020 at 22:26
6
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ (Hello there :-) ) That's pretty heavy machinery for this humble reader - but an interesting view of the obstruction. Thanks a lot! $\endgroup$
    – Alon Amit
    Oct 27, 2009 at 7:28
  • 2
    $\begingroup$ (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. $\endgroup$ Oct 27, 2009 at 8:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.