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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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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
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@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
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1 Answer

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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