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!


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.