Brauer group of complete DVR - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T22:13:09Z http://mathoverflow.net/feeds/question/78508 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78508/brauer-group-of-complete-dvr Brauer group of complete DVR Wanderer 2011-10-18T23:40:50Z 2011-10-19T14:53:44Z <p>Let $A$ be a complete discrete valuation ring with fraction field $K$ and perfect residue field $\kappa$.</p> <p>Let $K_{nr}$ be the maximal unramified extension of $K$ and let $A_{nr}$ be its ring of integers.</p> <p>Is there an easy way to see that $\text{Br}(A)$ is isomorphic to $H^2(\text{Gal}(K_{nr}/K),A_{nr}^\star)$?</p> http://mathoverflow.net/questions/78508/brauer-group-of-complete-dvr/78516#78516 Answer by Alex Bartel for Brauer group of complete DVR Alex Bartel 2011-10-19T01:29:25Z 2011-10-19T01:29:25Z <p>The proof of this is spelled out very clearly and nicely by Serre in Cassels and Fröhlich, Algebraic Number Theory. He does it for the fields, rather than their rings of integers, i.e. $\text{Br}(K)\cong H^2(K_{nr}/K,K^\times)$. But quickly browsing through it, the proof should adapt almost verbatim to the rings of integers.</p> <p>The basic idea is as follows. By definition, $H^2(K_{nr}/K,K^\times) \subseteq H^2(K_{s}/K,K^\times) = \text{Br}(K)$, where $K_s$ is the separable closure of $K$. $H^2(K_{s}/K,K^\times)$ in turn is the direct limit of $H^2(L/K,L^\times)$ over the finite Galois extensions of $L/K$, so it is enough to show that each $H^2(L/K,L^\times)$ is canonically a subgroup of $H^2(K_{nr}/K,K^\times)$. There are several steps to this, but the heart of the argument is a comparison between maps $$inv_K:H^2(K_{nr}/K,K^\times)\rightarrow \mathbb{Q}/\mathbb{Z} \;\;\;\text{and}\;\;\;inv_L:H^2(L_{nr}/L,L^\times)\rightarrow \mathbb{Q}/\mathbb{Z}$$ for $L/K$ a finite extension. I won't do it any better than Serre does, so I refer you to him for the remaining details.</p> http://mathoverflow.net/questions/78508/brauer-group-of-complete-dvr/78531#78531 Answer by Torsten Ekedahl for Brauer group of complete DVR Torsten Ekedahl 2011-10-19T04:35:09Z 2011-10-19T04:35:09Z <p>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.</p> <p>If one looks at what actually goes into this proof one has the following:</p> <ul> <li>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. <li>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. </ul> http://mathoverflow.net/questions/78508/brauer-group-of-complete-dvr/78565#78565 Answer by Filippo Alberto Edoardo for Brauer group of complete DVR Filippo Alberto Edoardo 2011-10-19T10:57:39Z 2011-10-19T14:53:44Z <p>I think an even easier answer (although it should more or less boil down to Ekedahl's) is to apply snake lemma to the following diagram - noting that the middle vertical arrow is an isomorphism $$0\to H^2(K_{nr}/K,A_{nr}^\times)\to H^2(K_{nr}/K,K_{nr}^\times)\to H^2(K_{nr}/K,Q)\to 0$$ $$0\to H^2(\bar{K}/K,A_\bar{K}^\times)\to H^2(\bar{K}/K,\bar{K}^\times)\to H^2(\bar{K}/K,Q)$$ where I apologize for not putting the vertical arrows but I do not how to write diagrams: they are inflation maps (they go down). These sequences are the cohomology sequences coming from the exact sequence $$1\rightarrow A_\bar{K}^\times \rightarrow \bar{K}^\times\rightarrow Q\rightarrow 1$$ given by taking $p$-adic valuation. I write $Q$ instead of $\mathbb{Q}$ because the latter is not discrete, so profinite Galois cohomology must be taken with a grain of salt: simply define $Q=\lim \frac{1}{e}\mathbb{Z}$ and filter $\mathrm{Gal}(\bar{K}/K)$ by subgroups fixing an extension of $K$ with absolute ramification index $e$. Then you can apply Proposition 8 in Serre's "Cohomologie Galoisienne". The point is that the inductive limit topology makes any $\frac{1}{e}\mathbb{Z}\subset\mathbb{Q}$ open (and the limit is indeed discrete), but this is not the case with the usual euclidean topology of $\mathbb{Q}$.</p> <p>Of course, in order to apply Snake Lemma, one needs to check some facts:</p> <p>1) The three zero showing up are really $0$</p> <p>2) The right vertical arrow is injective.</p> <p>For what concerns the zero on the right of the first sequence, it is just cohomological triviality of units in unramified extension (going to the limit): you find it in Serre's paper in Cassels and Fröhlich, Proposition 1 (it is also referred to in Ekedahl's answer). For what concerns the two zeros on the right AND injectivity of the right vertical arrow, just apply inf-res sequence to $$1\to\mathrm{Gal}(\bar{K}/K_{nr})\to \mathrm{Gal}(\bar{K}/K)\to \mathrm{Gal}(K_{nr}/K)\to 1$$ for the module $Q$ showing up above.</p>