Let $A$ be a complete discrete valuation ring with fraction field $K$ and perfect residue field $\kappa$.


Let $K\_{nr}$ be the maximal unramified extension of $K$ and let $A\_{nr}$ be its ring of integers.


Is there an easy way to see that $\text{Br}(A)$ is isomorphic to $H^2(\text{Gal}(K\_{nr}/K),A\_{nr}^\star)$?



Brauer group of complete DVR
This is more trivial (in the sense that we have hidden all the non-trivial parts among general preliminaries...) than the identification of $\text{Br}(K)$ that Alex is talking about. We have that $\text{Spec } A\_{nr}\to \text{Spec } A$ is an algebraic universal covering map and $\text{Spec } A\_{nr}$ is acyclic so that $H^\ast(\text{Spec} A,\mathbb G\_m)=H^\ast(\pi,A^\ast\_{nr})$, where $\pi$ is the Galois group of $\text{Spec } A\_{nr}\to \text{Spec } A$ (which is also the Galois group of $K\_{nr}/K$). Hence $H^2(\text{Gal}(K\_{nr}/K),A^\ast\_{nr})$ is the cohomological Brauer group of $A$ which in this case is easily seen to be equal to the Brauer group.


If one looks at what actually goes into this proof one has the following:


* The Brauer group of $A\_{nr}$ is trivial. This comes from the fact that the special fibre of an Azumaya algebra is trivial as the residue field is separably closed and as $A\_{nr}$ is Henselian idempotents lift.

* This immediately gives an embedding $\text{Br}(A)\subseteq H^2(\pi,A^\ast\_{nr})$. The surjectivity is done in the same way as for fields, one explicitly constructs the Azumaya algebra associated to a $2$-cocycle.


