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Let $E$ be an elliptic curve with a 3-torsion point $P$ and $G = \operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. Let $X = \{O, P, -P\}$ where $O$ is the point at infinity and $X$ is a $G$-module. Why does $H^{1}(G, X)$ inject into $\operatorname{Sel}_{3}(E/\mathbb{Q})$?

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It's not generally the case that $E[3]$ has $G_{\mathbb{Q}}$ invariant subspaces. For example the mod-$3$ representation associated to the curve $Y^2 = X^3 - 27X - 42$ has full image in $GL_2(\mathbb{F}_3).$ So in general $G$ won't act on $X$ and taking cohomology is therefore out of the question. – jspecter Jan 6 '13 at 19:22
Wait, ok I see your point. But what would happen if I did happen to suppose $X$ indeed was a $G_{\mathbb{Q}}$ invariant subspace? – user24764 Jan 6 '13 at 19:32
up vote 4 down vote accepted

It doesn't (always)!

Nor would you expect it to. Assume $X$ is $G_{\mathbb{Q}}$ stable. Then in general, the group $H^1(\mathbb{Q},X)$ will be infinite whereas the selmer group $Sel_{3}(E/\mathbb{Q})$ finite.

Even more it may not even be the case $H^1(\mathbb{Q},X)$ injects into $H^1(\mathbb{Q},E[3]).$ Here's an exmple:

Consider the curve $E$ given by $y^2 +xy + y = x^3 -171x -874.$ One checks that $E[3]$ has three $\mathbb{Q}(\mu_3)$-points and one $\mathbb{Q}$-point. Therefore, if you let $X$ be the subgroup of $E[3]$ consisting of these $\mathbb{Q}(\mu_3)$-points, then $X$ is Galois stable and $G_\mathbb{Q}$ acts on $X$ through the nontrivial character of $Gal(\mathbb{Q(\mu_3)}/\mathbb{Q}).$ It follows that $X \cong \mu_3$ as a Galois module.

Because $E[3]$ is reducible with a stable one dimensional subspace $X,$ we have $E[3]$ is an extension of a character $\chi$ by $X \cong \mu_3.$ By the weil pairing $\chi \otimes X = \chi \otimes \mu_3 = \mu_3$ and hence $\chi$ is the trivial character.

From the short exact sequence $$0\rightarrow X \rightarrow E[3] \rightarrow \chi\rightarrow 0,$$ one obtains the exact sequence

$$E[3](\mathbb{Q}) \rightarrow \mathbb{F}_3 \rightarrow H^1(\mathbb{Q}, X) \rightarrow H^1(\mathbb{Q}, E[3]).$$

And since $E[3](\mathbb{Q}) = 0,$ it follows that the map $H^1(\mathbb{Q}, X) \rightarrow H^1(\mathbb{Q}, E[3])$ is not injective.

(Conversely, observe that if $H^1(\mathbb{Q}, X) \rightarrow H^1(\mathbb{Q}, E[3])$ is not injective then $E[3]$ must be a nonsplit extension of the trivial character by $\mu_3.)$

The problem is the incongruity in your question. You have assumed lots of local conditions on $Sel_3(E/\mathbb{Q})$ but none on $H^1(\mathbb{Q},X).$ What are the local conditions one might consider for cocylces in $H^1(\mathbb{Q},X)?$

Well, recall that we can attach a selmer group to any $\mathbb{Q}$-isogeny between elliptic curves. For example $Sel_3(E/\mathbb{Q})$ is the selmer group attached to the multiplication by $3$-isogeny. There is a natural isogeny attached to $X,$ namely the projection map $\psi_X: E \rightarrow E/X,$ and because $X$ is Galois stable $\psi_X$ is defined over $\mathbb{Q}.$ We define the Selmer group attached to $\psi := \psi_X$ as the kernel

$$Sel^{\psi}(E/\mathbb{Q}) := \ker(H^1(\mathbb{Q},X) \rightarrow \prod_p H^1( \mathbb{Q}_p,E)).$$

Because multiplication by $3$ factors through $X,$ (or rather because $X \subset E[3]$), we have a map $ Sel^{\psi}(E/\mathbb{Q}) \rightarrow Sel_3(E/\mathbb{Q})$ induced by the map $X \rightarrow E[3].$ One easily checks that the map

$$H^1(\mathbb{Q},X) \rightarrow \prod_p H^1( \mathbb{Q}_p,E)$$

factors through

$$H^1(\mathbb{Q},E[3]) \rightarrow \prod_p H^1( \mathbb{Q}_p,E)$$

and hence $ Sel^{\psi}(E/\mathbb{Q}) \rightarrow Sel_3(E/\mathbb{Q})$ is injective whenever $H^1(\mathbb{Q}, X) \rightarrow H^1(\mathbb{Q}, E[3])$ is injective.

What about when $H^1(\mathbb{Q}, X) \rightarrow H^1(\mathbb{Q}, E[3])$ isn't injective i.e. when $X \cong \mu_3$ and $E[3]$ is a nonsplit extension of the trivial representation by $X?$ Here one sees using the argument above that the map between selmer groups still has a 1-dimensional kernel. Oh well.

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