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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?

edited Jan 15 2010 at 19:25

Ilya Nikokoshev
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asked Oct 27 2009 at 5:40

Alon Amit
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Ho Chung Siu · Accepted Answer · 2011-04-19 23:44:33Z UTC

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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: Update 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.

edited Apr 19 2011 at 23:44

answered Oct 27 2009 at 7:04

Ho Chung Siu
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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 :-) – Alon Amit Oct 27 2009 at 7:26

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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/… – Kevin Buzzard Nov 20 2009 at 23:27

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The link in the answer is now broken. – KConrad Apr 16 2011 at 16:32

@Ho Chung Siu, do you happen to have a copy of that paper anywhere accessible? – Alon Amit Apr 18 2011 at 23:26

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Hi, the link is updated. – Ho Chung Siu Apr 19 2011 at 23:44

Thomas Riepe · Answer 2 · 2009-10-27 06:37:39Z UTC

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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).

answered Oct 27 2009 at 6:37

Thomas Riepe
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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! – Alon Amit Oct 27 2009 at 7:22

David Lehavi · Answer 3 · 2009-10-27 07:01:23Z UTC

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.

Kevin Buzzard · Answer 4 · 2009-11-20 18:59:13Z UTC

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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.

answered Nov 20 2009 at 18:59

Kevin Buzzard
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KConrad · Answer 5 · 2010-01-15 14:39:06Z UTC

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

http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/selmerexample.pdf,

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.

Proof of no rational point on Selmer’s Curve 3x^3+4y^3+5z^3=0

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