Timeline for If an abelian variety has an m-torsion point, is the set of all Galois conjugates of the m-torsion all the m-torsion? [closed]
Current License: CC BY-SA 2.5
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| Jan 29, 2015 at 1:03 | history | closed |
user9072 Ian Morris Neil Strickland S. Carnahan♦ |
Needs details or clarity | |
| Jan 28, 2015 at 17:07 | review | Close votes | |||
| Jan 29, 2015 at 1:03 | |||||
| Mar 27, 2011 at 1:43 | comment | added | Pete L. Clark | @James: For the second time: If $A_{/K}$ is an abelian variety (or any algebraic variety), the absolute Galois group of $K$ acts trivially on the $K$-rational points of $A$. So your question is equivalent to: if $A$ has one $K$-rational point of order $m$, must each of the $\overline{K}$-rational points of order $m$ be defined over $K$, the answer to which is clearly "no", as I and several others have explained. If you meant to ask a nontrivial question, now is your third chance... | |
| Mar 27, 2011 at 0:03 | comment | added | shenghao | I don't think this solves Pete's question: "$\sigma$ runs over Gal(K)" is still of no use. | |
| Mar 26, 2011 at 22:39 | comment | added | James D. Taylor | Gal(K) is just the absolute Galois group. I think I know what the problem is. What I mean is all the sigma(p) where sigma runs over Gal(K) AND p runs over all the K-rational m-torsion (p is not fixed!). | |
| Mar 26, 2011 at 22:31 | comment | added | Pete L. Clark | Or by $\operatorname{Gal}(K)$ do you perhaps mean $\operatorname{Aut}(K)$, i.e., the full group of field automorphisms over $K$? (If so, for shame: that automorphism group need not be profinite so has nothing to do with Galois theory in general.) Even so, the simple case where $K = \mathbb{Q}$ and the elliptic curve has exactly one $\Q$-rational $2$-torsion point seems to give a counterexample to what you want. | |
| Mar 26, 2011 at 22:29 | comment | added | Pete L. Clark | Then, as I suspected, I don't understand your question. What do you mean by $\operatorname{Gal}(K)$? The most reasonable thing I can think of is $\operatorname{Gal}(K^{\sep}/K)$ but then notice that if $p \in A(K)$ and $\sigma \in \operatorname{Gal}(K)$, then $\sigma(P) = P$. So this time I read your question as asking "Must all the $m$-torsion be $K$-rational?" and of course the answer is no. So please clarify. | |
| Mar 26, 2011 at 22:19 | comment | added | James D. Taylor | In that case the answer is trivially yes: take sigma=identity. | |
| Mar 26, 2011 at 22:09 | comment | added | Pete L. Clark | The way I am reading the question, the answer is trivially no, e.g. if we have $A(\overline{K})[m] = A(K)[m]$, which can certainly occur. Am I missing something? | |
| Mar 26, 2011 at 22:09 | answer | added | David Lehavi | timeline score: 4 | |
| Mar 26, 2011 at 21:57 | history | edited | James D. Taylor | CC BY-SA 2.5 |
added 130 characters in body; edited title
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| Mar 26, 2011 at 21:46 | history | asked | James D. Taylor | CC BY-SA 2.5 |