Suppose $E/K$ is a Galois extension with group $G=\operatorname{Gal}(E/K)$.

Construct the crossed product algebra $E\rtimes G$, which is the $E$-vector space with basis the elements of $G$ turned into a $K$-algebra in the unique way such that $eg\cdot e'h=eg(e')gh$ for all $e$, $e'\in E$ and all $g$, $h\in G$.

It is easy to check that the $K$-algebra $E\rtimes G$ is simple. Wedderburn's theorem then tells us that all simple $E\rtimes G$-modules are isomorphic. A corollary of this is:

Theorem. $H^1(G,E^\times)=0$.

Indeed, suppose $\phi:G\to E^\times$ is a $1$-cocycle. The $E$-vector space $V=E$ can be endowed in a unique way with an action of $E\rtimes G$ in such a way that $eg\cdot 1=e\phi(g)$ for all $e\in E$ and all $g\in G$. Since $V$ is one-dimensional as an $E$-vector space, it is a simple module over $E\rtimes G$ and it follows that there is an $E\rtimes G$-linear isomorphism $f:E\to V$. If we set $y=f(1_E)$, then the coboundary of $y$ is $\phi$. $\Box$

More generally, from Wedderburn's theorem we get that all $E\rtimes G$-modules are direct sums of copies of $E$, and that implies in pretty much the same way that $H^1(G,GL(n,E))=0$ (and the infinite dimensional version that Serre leaves as an exercise in Corps locaux)

Suppose $E/K$ is a Galois extension with group $G=\operatorname{Gal}(E/K)$.

Construct the crossed product algebra $E\rtimes G$, which is the $E$-vector space with basis the elements of $G$ turned into a $K$-algebra in the unique way such that $eg\cdot e'h=eg(e')gh$ for all $e$, $e'\in E$ and all $g$, $h\in G$.

It is easy to check that the $K$-algebra $E\rtimes G$ is simple. Wedderburn's theorem then tells us that all simple $E\rtimes G$-modules are isomorphic. A corollary of this is:

Theorem. $H^1(G,E^\times)=0$.

Indeed, suppose $\phi:G\to E^\times$ is a $1$-cocycle. The $E$-vector space $V=E$ can be endowed in a unique way with an action of $E\rtimes G$ in such a way that $eg\cdot 1=e\phi(g)$ for all $e\in E$ and all $g\in G$. Since $V$ is one-dimensional as an $E$-vector space, it is a simple module over $E\rtimes G$ and it follows from Wedderburn's theorem that there is an $E\rtimes G$-linear isomorphism $f:E\to V$,for $E$ is also a simple $E\rtimes G$-module. If we set $y=f(1_E)$, then the coboundary of $y$ is $\phi$. $\Box$

More generally, from Wedderburn's theorem we get that all $E\rtimes G$-modules are direct sums of copies of $E$, and that implies in pretty much the same way that $H^1(G,GL(n,E))=0$ (and the infinite dimensional version that Serre leaves as an exercise in Corps locaux)

Suppose $E/K$ is a Galois extension with group $G=\operatorname{Gal}(E/K)$.

Construct the crossed product algebra $E\rtimes G$, which is the $E$-vector space with basis the elements of $G$ turned into a $K$-algebra in the unique way such that $eg\cdot e'h=eg(e')gh$ for all $e$, $e'\in E$ and all $g$, $h\in G$.

It is easy to check that the $K$-algebra $E\rtimes G$ is simple. Wedderburn's theorem then tells us that all simple $E\rtimes G$-modules are isomorphic. A corollary of this is:

Theorem. $H^1(G,E^\times)=0$.

Indeed, suppose $\phi:G\to E^\times$ is a $1$-cocycle. The $E$-vector space $V=E$ can be endowed in a unique way with an action of $E\rtimes G$ in such a way that $eg\cdot 1=e\phi(g)$ for all $e\in E$ and all $g\in G$. Since $V$ is one-dimensional as an $E$-vector space, it is a simple module over $E\rtimes G$ and it follows that there is an $E\rtimes G$-linear isomorphism $f:E\to V$. If we set $y=f(1_E)$, then the coboundary of $y$ is $\phi$. $\Box$

More generally, from Wedderburn's theorem we get that all finite length $E\rtimes G$-modules are direct sums of copies of $E$, and that implies in pretty much the same way that $H^1(G,GL(n,E))=0$.

Suppose $E/K$ is a Galois extension with group $G=\operatorname{Gal}(E/K)$.

Construct the crossed product algebra $E\rtimes G$, which is the $E$-vector space with basis the elements of $G$ turned into a $K$-algebra in the unique way such that $eg\cdot e'h=eg(e')gh$ for all $e$, $e'\in E$ and all $g$, $h\in G$.

It is easy to check that the $K$-algebra $E\rtimes G$ is simple. Wedderburn's theorem then tells us that all simple $E\rtimes G$-modules are isomorphic. A corollary of this is:

Theorem. $H^1(G,E^\times)=0$.

Indeed, suppose $\phi:G\to E^\times$ is a $1$-cocycle. The $E$-vector space $V=E$ can be endowed in a unique way with an action of $E\rtimes G$ in such a way that $eg\cdot 1=e\phi(g)$ for all $e\in E$ and all $g\in G$. Since $V$ is one-dimensional as an $E$-vector space, it is a simple module over $E\rtimes G$ and it follows that there is an $E\rtimes G$-linear isomorphism $f:E\to V$. If we set $y=f(1_E)$, then the coboundary of $y$ is $\phi$. $\Box$

More generally, from Wedderburn's theorem we get that all $E\rtimes G$-modules are direct sums of copies of $E$, and that implies in pretty much the same way that $H^1(G,GL(n,E))=0$ (and the infinite dimensional version that Serre leaves as an exercise in Corps locaux)