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In SGA 4.5's proof of Hilbert 90, proposition 1.5.2(that the inclusion $V'^G \otimes_k k' \rightarrow V'$ is an isomorphism) is deduced from faithfully flat descent as stated in 1.4.5. The way that I interpret the usage of 1.4.5 is that I am looking at a single cover of $Spec\,k$, $Spec\,k'$ and the quasicoherent-sheaf $\tilde{V'}$.

Question: is the correct way to use it? If so, it doesn't seem that the discussion/definition of semilinear G-action comes up at all - a clarification on this condition will be greatly appreciated!

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The semilinear $G$-action on $V'$ is a concrete way of recording the fact that $\tilde{V}'$ admits descent data from Spec $k$' to Spec $k$.

(Note that $k'\otimes_k k' = \prod_{g \in G} k'$, if $k'/k$ is Galois with group $G$.)

Added in response to comment:

The isomorphism $k'\otimes_k k' \cong \prod_g k'$ of $k'$ algebras is induced by the morphism of $k$-algebras $$k' \to \prod_g k'$$ given by $$a \mapsto (g(a))_{g \in G}.$$ Giving the isomorphism $k'\otimes_k V' \to V' \otimes_k k'$ then amount to giving a $k'$-linear map $V \to \prod_g V,$ where the $k'$-action on the $g$th factor is twisted by $g$.

Now this is the same as giving a collection of morphisms $V \to V$ labelled by $g \in G$ (which you can check will have to be isomorphisms, if the original morphism is supposed to be one) which twist the $k'$-action by the corresponding element $g$. These are the ingredients for the semi-linear action. The fact that they form an action will be forced by the cocycle condition on the descent data.

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  • $\begingroup$ Hi Prof. Emerton - sorry for the late response. So concretely a decent datum is just an isomorphism of $k' \otimes k'$-module between $V' \otimes_k k'$ and $k' \otimes_k V'$. I didn't find it too easy to find the equivalence between this as a semilinear structure: essentially you have to also show that $V' \otimes k \cong \prod_{g \in G} V'$ in two different ways (so I can relate it to the fact you noted above) and then performing relatively lengthy computation. Does it just go like this? Or is there a super simple way to see it. $\endgroup$ – Elden Elmanto Jan 28 '14 at 23:13
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    $\begingroup$ @EldenElmanto: Dear Elden, I've sketched what I think is an efficient way to check this. Basically, the more conceptual you are, the easier this will be. E.g. another approach, maybe better than the one I wrote out, might be to write Spec $k'\otimes_k k' \cong G \times $ Spec $k$. Then the fact that the cocyle condition translates into the conditions of an action should be more-or-less obvious. Regards, $\endgroup$ – Matt E Jan 29 '14 at 2:06
  • $\begingroup$ Hugely appreciated! $\endgroup$ – Elden Elmanto Jan 30 '14 at 1:59

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