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 13:52 | comment | added | Laurent Moret-Bailly | @François: infinite algebraic extensions of finite fields are counterexamples ($A(K)\otimes\mathbb{Q}$ is zero). But they are a bit too trivial. Whether they are the only ones is a good question. | |
| Nov 6, 2012 at 12:50 | comment | added | François Brunault | By the way, if $K$ is ferile, will $A(K) \otimes \mathbf{Q}$ be necessarily be infinite-dimensional? | |
| Nov 6, 2012 at 12:40 | comment | added | François Brunault | Thank you very much for your answer and this interesting result that fertile fields satisfy Q2. Looking for references on fertile fields, I found out the recent article "Large implies henselian" by Pop, who proves that fraction fields of henselian rings are fertile. So for example, fields like $K((x_1,\ldots,x_n))$ or more generally fraction fields of $I$-adically complete rings will satisfy Q2. | |
| Nov 4, 2012 at 6:31 | comment | added | Pete L. Clark | @Laurent: Thanks again. That one at least I should have seen for myself. | |
| Nov 2, 2012 at 11:05 | history | edited | Laurent Moret-Bailly | CC BY-SA 3.0 |
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| Nov 1, 2012 at 9:21 | history | edited | Laurent Moret-Bailly | CC BY-SA 3.0 |
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| Nov 1, 2012 at 8:08 | comment | added | Laurent Moret-Bailly | @Pete: enough to prove $A(K_1(x))=A(K_1)$. An element of $A(K_1(x))$ is the same thing as a $K_1$-rational map from the affine line to $A$. These are all constant. | |
| Nov 1, 2012 at 1:37 | comment | added | Pete L. Clark | I was thinking of the example (1) as well, but I couldn't decide whether $A(K) = A(K_1)$: why is this? | |
| Oct 30, 2012 at 11:22 | history | edited | Laurent Moret-Bailly | CC BY-SA 3.0 |
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| Oct 30, 2012 at 10:28 | history | answered | Laurent Moret-Bailly | CC BY-SA 3.0 |