TITLE: Why are central simple algebras classified by cohomology?
QUESTION [5 upvotes]: In their article on the Brauer group Wikipedia writes:

Since all central simple algebras over a field $K$ become isomorphic to the matrix algebra over a separable closure of $K$, the set of isomorphism classes of central simple algebras of degree $n$ over $K$ can be identified with the Galois cohomology set $H^1(K, \mathrm{PGL}(n))$.

I understand all the words here, but the reasoning goes too quick for me to follow it.  Could someone explain why this is true?
I believe $H^1(K, \mathrm{PGL}(n))$ is really the group cohomology $H^1(G, \mathrm{PGL}(n,K))$ where $G$ is the Galois group of the separable closure of the field $K$.  The group cohomology $H^1(G,M)$ is explained here.  Perhaps it's not explained in sufficient generality, since they seem to define $H^1(G,M)$ only where $M$ is an abelian group acted on by a group $G$.   But I think the same thing should work for any set acted on by $G$, and I know how $PGL(n,K)$ is acted on by the Galois group $G$.
I guess I need to see how a

central simple algebra $A$ over a field $K$ that becomes isomorphic to an $n \times n$ matrix algebra when tensored with the separable closure of $A$ 

gives rise to a

1-cocycle $c_A \colon G \to M$

and why isomorphic algebras of this sort give cocycles that differ by a coboundary.  (Also how to go back.)

REPLY [2 votes]: I think the best reference is the book [G-S] by Gille & Szamuely, "Central Simple Algebras and Galois Cohomology", Cambridge Univ. Press, 2017.
Over a field $k$, recall that a (finite dimensional) $k$-algebra $A$ is called simple if it has no (two-sided) ideal other than $0$ and $A$. Moreover it is called $central$ if its centre equals $k$. A finite-dimensional $k$-algebra $A$ is central simple iff there exists an integer $n>0$ and a finite Galois extension $K/k$ s.t. $A\otimes_k K$  is isomorphic to the matrix ring $M_n(K)$. Such a field extension $K/k$ is called a splitting field for $k$. Let us denote by $CSA_K(n)$ the pointed set of $k$-isomorphisms classes of central simple $k$-algebras of degree $n$ split by $K$, the base point being the class of the matrix algebra $M_n(k)$.
A first cohomological description of $CSA_K(n)$ is as follows. Recall that for a group $G$ acting on another group (not necessarily abelian) group $A$, a 1-cocycle of $G$ with values in $A$ is a map $s\in G \to a_s\in A$ satisfying $a_{st}=a_s.s(a_t)$, and two 1-cocycles $a_s$ and $b_s$ are equivalent if there exists $c\in A$ s.t. $a_s=c^{-1}b_s s(c)$ (I repeat the definitions because here $A$ is not commutative). This is an equivalence relation between 1-cocycles, and the first cohomology $set$ $H^1(G,A)$ is the  set of the equivalence classes of 1-cocycles, pointed by the trival cocycle $s\to Id_A$. The main result is the existence of a base point preserving bijection between $CSA_K(n)$ and $H^1(G, PGL_n(K))$, where $G=Gal(K/k))$. See [G-S], thms. 2.3.3 and 2.4.3. This should answer your questions, but unfortunately, the proof of thm. 2.4.3 relies on thm. 2.3.3, which is rather elaborate and cannot be summarized here.
For $H^i(G,A)$ to be a group (with the usual natural definition) you need $A$ to be abelian. In the non abelian case, the tensor product induces natural associative and commutative products  $CSA_K(m)\times CSA_K(n)$ $\to CSA_K(mn)$, which can be naturally "transported" to the other side to define analogous products  $H^1(G, PGL_m(K))\otimes H^1(G, PGL_n(K))\to H^1(G, PGL_{mn}(K))$. Fixing a separable closure $k_s$ of $k$ and writing $G_k=Gal(k_s/k)$, one can then take inductive limits to get an isomorphism of abelian groups $Br(k)$ (by definition) $\cong H^1(G_k, PGL_{\infty})$. But this is rather cumbersome. The second definition of $Br(k)$ via Brauer-equivalence classes of central simple $k$-algebras is more supple and yields natural isomorphisms of abelian groups $Br(K/k)\cong H^2(G, K^*)$ and $Br(k)\cong H^2(G_k,{k_s}^*)$. Anyway, these are the versions of Brauer groups which are used in class-field theory.