The setup for my question is as follows: $k$ is a field, $K$ a Galois extension of $k$ with group $G$, $k^\prime$ an arbitrary extension of $k$, and $K^\prime/k^\prime$ another Galois extension of fields with group $G^\prime$ (the notation, as well as the setup, is essentially verbatim from Chapter X, Section 4 of Serre's Local Fields). Given a $k$-monomorphism $f:K\rightarrow K^\prime$, we get a homomorphism $\overline{f}:G^\prime\rightarrow G$ (sending $s^\prime\in G^\prime$ to the unique $s\in G$ with $f\circ s=s^\prime\circ f$), and if we have another $k$-monomorphism $g:K\rightarrow K^\prime$, then there is some unique $s\in G$ with $f\circ s=f^\prime$ (both these facts are consequences of the normality of $K$ over $k$, which implies that every $k$-monomorphism of $K$ into $K^\prime$ has the same image), so the homomorphism $\overline{g}:G^\prime\rightarrow G$ induced by $g$ is just that induced by $f$ followed by a conjugation of $G$.

Now, if $A$ is a commutative group scheme over $k$, then such an $f$ (as above) gives a map $A(K)\rightarrow A(K^\prime)$ that is compatible (in the sense of group cohomology) with $\overline{f}:G\rightarrow G^\prime$, so we get maps $H^n(G,A(K))\rightarrow H^n(G^\prime,A(K^\prime))$. The key fact which comes from the considerations in the previous paragraph is that if we use a different $k$-monomorphism $g:K\rightarrow K^\prime$, then the induced maps on cohomology differ by the map $H^n(G,A(K))\rightarrow H^n(G,A(K))$ induced by the pair $x\mapsto sxs^{-1}$ and $p\mapsto s^{-1}p$, which is the identity.

My question regards a particular case of this situation but with (what appears to me) to be a twist. $k$ is a global field, $k^\prime=k_v$ is the completion of $k$ at some place $v$, $K=k^s$ is a separable closure of $k$, and $K^\prime=k_v^s$ is a separable closure of $k$. Any two embeddings of $k^s$ into $k_v^s$ over $k\rightarrow k_v$ differ by an element of the absolute Galois group $G_k=Gal(k^s/k)$ of $k$ (these are just two choices of places of $k^s$ extending $v$). As before, with such an embedding, I get a map $G_{k_v}=Gal(k_v^s/k_v)\rightarrow G_k$ which is (by Krasner's lemma) an injection with image the decomposition group in $G_k$ of the corresponding place. Any two such injections differ by conjugation in $G_k$. So, if I have, for instance, an abelian variety $A$ over $k$, I get canonical "localization" maps $H^n(k,A(k^s))\rightarrow H^n(k_v,A(k_v^s))$, i.e., they are independent of the embedding of $k^s$ into $k_v^s$. But now if I start with an arbitrary $G_k$-module $M$, the homomorphism $G_{k_v}\rightarrow G_{k}$ makes $M$ into a $G_{k_v}$-module and is compatible with the identity $M\rightarrow M$ (where the target is regarded as a module over the local Galois group), so I get a map on cohomology $H^n(k,M)\rightarrow H^n(k_v,M)$. My question is how can this map be independent of the embedding $k^s\rightarrow k_v^s$? It's true that any two embeddings give rise to conjugate injections of the local Galois group into the global one, but the action of the local group on $M$ is defined in terms of the map $G_{k_v}\rightarrow G_k$. There is no canonical choice of $G_{k_v}$-action on $M$ as there is in the case of a commutative $k$-group scheme. As far as I can tell, if I use a different embedding, I get a conjugate homomorphism $G_{k_v}\rightarrow G_k$, and this gives a DIFFERENT action of $G_{k_v}$ on $M$. So, strictly speaking, the map on cohomology induced by a second embedding shouldn't even have the same target as the one induced by the first. Even in dimension zero, shouldn't the Galois invariants of $G_{k_v}$ with respect to the different actions differ by an element of $G_k$ (i.e. if $a$ is fixed by $G_{k_v}$ with respect to the first action then $sa$ is fixed by $G_{k_v}$ with respect to the second action, where $s\in G_k$). I would greatly appreciate if somebody could clarify what I'm not getting right here. I know these maps have to be canonical, and everywhere this is stated it's attributed to the fact that conjugation induces the identity on cohomology, but when I've tried to apply this, it doesn't seem to work out if my Galois module isn't already endowed with an action of $G_{k_v}$.

I apologize for this post being so long. I suspect there's something simple I've overlooked that will warrant maybe a sentence or two, but I felt like I needed to try to motivate this question somewhat.

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    $\begingroup$ When you learn etale cohomology and its relation with Galois cohomology you'll see how to express all of this stuff in terms which entirely avoid any mention of separable closures at all (apart from the back-of-envelope calculation that the pullback maps which have you worried here really are computed by pullback maps in etale cohomology). Then you never have to get caught up in these kinds of contortions ever again. It's like working with homology instead of the fundamental group: all of that confusion about base points just fizzles away. $\endgroup$ – BCnrd May 5 '10 at 3:57
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    $\begingroup$ +1 BCnrd. If it helps, I remember being ridiculously confused about this sort of thing once too. Here's the hint: be clear that H^n(G,M) means what you think it means for M a G-module, but H^n(k,M) is more delicate. It means that for any extension K of k, M(K) makes sense and is functorial in K, and if K is a sep closure of k then H^n(k,M) is isomorphic to H^n(Gal(K/k),M(K)), and if K1 and K2 are two sep closures then don't try and identify H^n(Gal(K1/k),M(K1)) and H^n(Gal(K2/k),M(K2)) until you have chosen an iso K1=K2, and once you chose it, use it to identify both the groups and the mods. $\endgroup$ – Kevin Buzzard May 5 '10 at 9:44
  • $\begingroup$ Thanks for the comments guys. It does help to know that I'm not the only person that's ever been perplexed by this issue before. I look forward to less contortions. $\endgroup$ – Keenan Kidwell May 5 '10 at 10:57

As you note, if we choose two different embeddings $k^s \to k_v^s$, say $\imath_1$ and $\imath_2$, then we get two different $G_{k_v}$-module structures on $M$, call them $M_1$ and $M_2$, and two different restriction maps $r_1:H^n(k,M) \to H^n(k_v,M_1)$ and $r_2:H^n(k,M) \to H^n(k_v,M_2)$.
The point is that there will also be a canonical isomorphism $i:H^n(k_v,M_1) \cong H^n(k_v,M_2)$ such that $i\circ r_1 = r_2,$ given by a formula analogous to the one you gave in the abelian variety context.

Namely, if $\imath_2 = \imath_1\circ g,$ then the isomorphism $i$ will be induced by $m \mapsto g\cdot m$ (if I have things straight; you can easily check if this is correct, of if I should have $g^{-1}$ instead). The fact that $i\circ r_1 = r_2$ will then depend on the fact that the automorphism of $H^n(k,M)$ induced by conjugation by $g^{-1}$ and the map $m\mapsto g\cdot m$ is the identity. So one does not have a literal independence of the embedding, but rather, the restriction is defined up to a canonical isomorphism independent of the embedding, and this latter fact does depend upon conjugation inducing the identity on cohomology (which is why people often summarize it in that way).

Note also that your abelian variety example is actually a special case of this, because the $M$ is $A(k^s)$, and the $G_{k_v}$-action on $M$ does depend on the embedding of $k^s$ in $k_v^s$. But the natural map $A(k^s) \to A(k_v^s)$ also depends on this embedding, in such a way that, when you compose the restriction from $G_k$ to $G_{k_v}$ (with coefficients in $A(k^s)$) with the map on $G_{k_v}$-cohomology induced by the embedding $A(k^s) \hookrightarrow A(k_v^s)$, you do obtain a map on cohomology that is independent of the embedding.

But it is not that in this case $M$ has a well-defined action of $G_{k_v}$ independent of the choice of embedding $k^s \hookrightarrow k_v^s$. It is rather that $M$ embeds into a bigger module $M_v$ (in a way that also depends on the embedding) so that the composite $H^n(k,M)\to H^n(k_v,M) \to H^n(k_v,M_v)$ is independent of the embedding. It is the embeddings $M \hookrightarrow M_v$ that are missing in the more general context (i.e. when $M$ is not of the form $A(k^s)$).

  • $\begingroup$ That makes things vastly more clear. Thank you for the detailed explanation. $\endgroup$ – Keenan Kidwell May 5 '10 at 10:51

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