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I believe this is the case, but I couldn't come up with a proof off the top of my head, so I want to make sure.

If $A$ is an abelian variety over some field $K$ (I'm in fact interested only in Jacobians, but I don't think this should matter), if $A$ has some $m$-torsion point, is the set of all $\sigma(p)$ for $\sigma \in Gal(K)$ and $p$ a $K$-rational $m$-torsion point of $A$, all of the $m$-torsion of $A \times_K \overline{K}$?

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closed as unclear what you're asking by quid, Ian Morris, Neil Strickland, S. Carnahan Jan 29 '15 at 1:03

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

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? – Pete L. Clark Mar 26 '11 at 22:09
In that case the answer is trivially yes: take sigma=identity. – James D. Taylor Mar 26 '11 at 22:19
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. – Pete L. Clark Mar 26 '11 at 22:29
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. – Pete L. Clark Mar 26 '11 at 22:31
@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... – Pete L. Clark Mar 27 '11 at 1:43

No. Take $E=Z(zy^2 = x(x^2+z^2))$, identifying zero with $(0:1:0)$. Then $(-1:0:1)$ is the only real non trivial 2-torsion point of $E(\mathbb{R})$.

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