Timeline for Over which fields does the Mordell-Weil theorem hold?
Current License: CC BY-SA 3.0
10 events
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| Nov 6, 2012 at 12:47 | comment | added | François Brunault | @Pete : Thank you very much for your answer. Probably one can even prove that $E(K) \otimes \mathbf{Q}$ is infinite dimensional as a $\mathbf{Q}$-vector space. This is clear if $K$ has characteristic zero and probably also holds in general. | |
| Nov 1, 2012 at 1:31 | comment | added | Pete L. Clark | @Laurent: okay, that works. Thanks! (My intuition was right, at least...) | |
| Oct 30, 2012 at 8:39 | comment | added | Laurent Moret-Bailly | @Pete: right, then take an elliptic curve $E$ of rank one (say) over $\mathbb{Q}$. If $K$ is a nontrivial ultrapower of $\mathbb{Q}$, then $E(K)$ contains the corresponding ultrapower of $\mathbb{Z}$, which is not finitely generated. | |
| Oct 30, 2012 at 7:58 | comment | added | Pete L. Clark | Sorry: I've made a terminology mistake (and not for the first time). What I meant to ask was: is the class of fields for which the Mordell-Weil theorem holds for all elliptic curves (or all abelian varieties...) closed under elementary equivalence? | |
| Oct 30, 2012 at 7:26 | comment | added | Pete L. Clark | @Laurent: but a (non)principal ultraproduct of finite fields is not elementarily equivalent to a finite field. Indeed, such a field is PAC, whereas finite fields are not. Or am I just not understanding you properly? | |
| Oct 30, 2012 at 7:23 | comment | added | Laurent Moret-Bailly | @Pete: the answer is indeed no, consider (ultraproducts of) finite fields! | |
| Oct 30, 2012 at 3:18 | history | edited | Pete L. Clark | CC BY-SA 3.0 |
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| Oct 30, 2012 at 3:17 | comment | added | Pete L. Clark | @François: yes, you are understanding correctly: at least, this is a useful way to view what I am talking about. And indeed I was thinking along these lines myself. A more explicitly model-theoretic question along these lines is: is the class of fields for which the Mordell-Weil Theorem holds for all elliptic curves elementary? (But I suspect that the answer to this is "no"...) | |
| Oct 30, 2012 at 1:45 | comment | added | François G. Dorais | Pete, if I'm understanding correctly, this would be an(other) example of a theorem proved using second-order axioms (completeness) but actually follows from first-order consequences (Henselianness). Is this at least on the right track? | |
| Oct 29, 2012 at 19:37 | history | answered | Pete L. Clark | CC BY-SA 3.0 |