Let $L/K$ be a finite Galois extension. I know that $H^1(Gal(L/K), L^{\times}) = 0$ and $H^r(Gal(L/K), (L,+)) = 0$ for all $r>0$.
1) Do we have $H^r(Gal(L/K), L^{\times}) = 0$ for all $r>1$ ?
2) Do we have $H^1(Gal(L/K), (L^{\times})^n) = 0$ for all $n>1$, where I'm using the diagonal action on the product ?
Thank you!
On
We don't in general have $H^2(\textrm{Gal}(L/K),L^\times)=0$. These $H^2$ groups are related to Brauer groups, which classify central division algebras over $K$. In particular $H^2(\textrm{Gal}(L/K),L^\times)$ is the relative Brauer group $\text{Br}(L/K)$, the kernel of the natural map $\text{Br}(K)\to \text{Br}(L)$ between the Brauer groups of $K$ and $L$.
An easy special case is $H^2(\textrm{Gal}(\Bbb C/\Bbb R),\Bbb C^\times) \cong\Bbb Z/2\Bbb Z$.
In the following, write $G=Gal(L/K)$.
1) The main reason behind the nullity of all the cohomology groups $H^r(G, (L,+))$ is the so-called normal basis theorem: there exists $x\in L$ s.t. the system $s(x)$, for $s$ running through $G$, is a basis of the $K$-vector space $L$, in other words, $L$ is a $K[G]$-free module, hence has trivial cohomology.
2) The multiplicative version of Hilbert's thm.90 states that $H^1(G, L^*)=0$, but the higher cohomology groups do not vanish in general. And fortunately so, because what would be the interest of the cohomology theory of fields ? Here is a list of the most interesting cases:
a) If $K$ is a finite field, $G$ is cyclic, generated by the Frobenius automorphism. But in the cyclic case, say with $G=<\phi>$, it is known that $H^2(G, L^*)\cong K^*/NL^*$ and $H^1(G, L^*)\cong KerN/(L^*)^{\phi-1}$, where $N$ denotes the norm map. Since $L^*$ is finite, the kernel and cokernel of $\phi-1$ have the same order, and Hilbert's 90 then immediately implies the nullity of $H^2(G, L^*)$. This last property is known as the surjectivity of the norm in a finite extension of finite fields.
b) In the general case, $H^2(G, L^*)$ is known to be canonically isomorphic to the relative Brauer group $Br(L/K)$, which is a subgroup of the absolute Brauer group $Br(K)$, with exponent dividing the degree $[L:K]$. Recall that $Br(K)$ is a fundamental object which classifies the equivalence classes of central simple $K$-algebras.
c) Even "better", in local CFT, which describes the abelian extensions of a local $p$-adic field, $Br(L/K)$ is cyclic of order $[L:K]$, generated by a so-called fundamental class $u(L/K)$ which encapsulates the main properties of the theory. For instance, the cup-product with $u(L/K)$ defines an isomorphism of the abelian quotient $G^{ab}$ onto $K^*/NL^*$ (the inverse of this isomorphism is usually called the local reciprocity map or the norm residue symbol). More generally, the whole $G$-cohomology of $L^*$ is given by the cup-product with $u(L/K)$, which gives $H^r(G,\mathbf Z)\cong H^{r+2}(G,L^*)$ for all $r\in \mathbf Z$ (modified Tate cohomology groups here).
NB: On the archimedean side, as recalled by @Lord Shark the Unknown, local CFT boils down to an isomorphism $H^2(G(\mathbf C/\mathbf R), \mathbf C^*)\cong \mathbf Z/2\mathbf Z)$, and the fundamental class corresponds to the quaternion algebra.
d) In global CFT (for number fields or, more generally, global fields), one must replace $L^*$ gy the idèle class group $C_L$, and the main key is again a certain fundamental class $u(L/K)\in H^2(G, C_L)$ which has properties similar to those of the local classes. From these one can deduce global-local properties of the cohomology groups $H^r(G,L^*)$. For example, $H^3(G,L^*)$ is cyclic of order $n/n_0$, the global degree divided by the l.c.m. of the local degrees, generated by the Teichmüller class $\delta(u(L/K))$, where $\delta$ is the natural map $H^2(G, C_L) \to H^2(G, L^*)$.
For local and global CFT, I refer to chapters VI and VII of Cassels-Fröhlich.
3) You ask for $H^1(G,(L^*)^n$) with diagonal action of $G$ on the direct product. I wonder why, since this is not a very natural action. Anyway, if $A, B$ are two $G$-modules, then naturally $H^r(G, A\times B)\cong H^r(G,A)\times H^r(G,B)$ ./.