Galois cohomology group equal the ground field

Question

In our lecture on Algebraic Number Theory, we were given the following task:

Let $L/K$ be a Galois extension with Galois group $G$. Show that $$H^0 (G,L)=K.$$ and $$H^n(G,L)=0 ,$$ for $n>0$.

I already got stuck with the first part. I know that $H^0 (G,L)$ is the set of all maps from $Gal(L/K)$ to $L$. Hence, if I choose $[f] \in H^0 (G,L)$, this means that I have an $f $ which maps a $\sigma \in G=Gal(L/K)$ to some element in $L$. (I assume that the map goes from $G^1$ to $L$ and not from $G^0$ to $L$. The notation in our lecture was unfortunately not consistent - so maybe I'm already mistaken at this point...). However, I'm having some trouble about how to work with the fact that $f$ maps basically another map. I think that I should probably somehow get that all the maps in $H^0 (G,L)$ are constant, but I don't know how.

For my ideas above as well as for the second part, any suggestions/hints/corrections would be greatly appreciated! Thank you!

Answer 1

Actually $H^0(G,L)=L^G$ the fixed points of $G$ acting on $L$. These are $K$ by the Fundamental Theorem of Galois Theory.

By the Normal Basis Theorem $L$ is a free $KG$-module. As $L$ is projective its higher cohomology vanishes: $H^n(G,L)=0$ for $n\ge1$.

Answer 2

This isn't an answer in full generality, but here's an elementary proof when the characteristic of $K$ doesn't divide $[L:K]$, and $[L:K]$ is finite, just for fun.

Consider the functor $-^{G}$ from $K[G]$-modules to $K[G]$-modules sending a module $A$ to its $G$-invariants $A^G$. Then $-^G$ is exact. To see this, let $$0 \to A \overset{f}\to B \overset{p}\to C \to 0$$ be a short exact sequence of $K[G]$ modules and consider the sequence $$0 \to A^G \overset{f^G}\to B^G \overset{p^G} \to C^G \to 0.$$ Clearly $f^G$ is injective, since it's just a restriction of $f$. To see that $p^G$ is surjective, let $x \in C^G$. Since $p$ is surjective, there is some $y \in B$ such that $p(y)=x$. Since the characteristic of $K$ doesn't divide $[L:K]=|G| \lt \infty$, we may let $y'=\frac{1}{|G|}\sum_{g \in G} gy$. Clearly $y' \in B^G$. Then $p(y')=x$ and so $p^G$ is surjective.

Now take an injective resolution $L \to I$. Since group cohomology is the derived functor of $-^G$, applying $-^G$ to $I$ we get that $H^k(G,L)=L^G=K$ when $k=0$ and $H^k(G,L)=0$ otherwise.