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It seems to be a rather well understood fact that, given commutative rings $R,S$, and a homomorphism $R \to S$ there is a short exact sequence $$\text{Pic}(R) \to \text{Pic}(S) \to F_0 \to \text{Br}(R) \to \text{Br}(S)$$

relating the Picard and Brauer groups. In some sense this is not surprising, as Picard groups are related to first etale cohomology groups, and Brauer groups to the (torsion in the) second. (For example, see here, although I'm unsure what $F_0$ is)

Yet I can't seem to find this result proven in the literature, and it doesn't appear obvious to me how to prove this (clearly first and last maps arise from the fact that the Picard and Brauer groups are both functors $\text{Ring} \to \text{Ab}$, so I am interested in the middle 3 terms).

Is there a 'text-book' level reference for this result? Note that I can find various generalizations of this result using symmetric monoidal categories and fancy category theory, but I'm purely interested in a 'algebraic' proof of the case of rings.

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There doesn't seem to be enough information. What are $S$ and $F_0$? Is $R$ a local domain with fraction field $S$ and residue field $F_0$ or something? –  Matt Jan 16 '13 at 6:13
    
@Matt - thanks for pointing that out, I've updated the question. –  Juan S Jan 16 '13 at 6:17
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1 Answer

Here is one way to do it. Under appropriate conditions on the ring (at the very least Milne's Étale Cohomology has some conditions listed in it, but I won't look them up now) we get that the Azumaya algebra definition of the Brauer group is the same as the étale cohomological one $$Br(R)=H^2(R, \mathbb{G}_m):=H^2_{et}(\operatorname{Spec}(R), \mathbb{G}_m).$$ We always have that $$Pic(R)=H^1(R, \mathbb{G}_m):=H^1_{et}(\operatorname{Spec}(R), \mathbb{G}_m).$$

Now given a ring homomorphism $f:R\to S$ we get a map (we'll just call it the same thing) $f:\operatorname{Spec}(S)\to \operatorname{Spec}(R)$. Consider the Leray spectral sequence for this map and the sheaf $\mathbb{G}_m$, i.e. $$H^p(R, R^qf_*\mathbb{G}_m)\Rightarrow H^{p+q}(S, \mathbb{G}_m)$$ (unfortunately, confusing due to $R$ repetition).

The long exact sequence of low degree terms gives us (using the cohomological interpretation of everything) $$Pic(R)\to Pic(S)\to H^0(R, R^1f_*\mathbb{G}_m)\to Br(R)\to Br(S)\to \cdots$$

This even tells us what the $F_0$ is. It is just the global sections of the first higher direct image of $\mathbb{G}_m$.

I guess you need to know when the cohomological Brauer group is the same as the Brauer group for this to work, and what conditions you need for the Leray spectral sequence to apply.

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Thanks Matt! I'll keep this one open for a few days, as I was hoping to avoid a cohomological argument, but I do appreciate the answer. –  Juan S Jan 16 '13 at 22:29
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Dear @Matt: the Brauer group is the torsion part of $H^2$ (see e.g. this MO thread). And $Pic(R)\to Pic(S)$ is not injective in general (consider the case $R$ is a non principal domain and $S$ its field of fractions) –  user18119 Jan 16 '13 at 22:46
    
@QiL Sentence 2: "Under appropriate conditions on the ring ..." Middle of the page "using the cohomological interpretation of everything ..." End of the post "I guess you need to know when the cohomological Brauer group is the same as the Brauer group ..." I'm not claiming the Brauer group is always $H^2$ as you can see. I pointed to Milne, who for example gives for example that they are the same for a local ring of dimension $\leq 1$. Although that paper references Azumaya algebras, I wouldn't be surprised if in general they only took cohomological version so that it generalizes easier. –  Matt Jan 17 '13 at 2:09
    
@Matt: sorry if my comment seems to be rude. It was just meant to add an information relating $H^2$ and Br. –  user18119 Jan 17 '13 at 8:29
    
To add some more comments, I think it is still unknown in general whether or not the Brauer group is the torsion part of $H^2$ for rings. All that is known is that it injects into the torsion part and in some special cases it is also known to be surjective. –  Matt Jan 17 '13 at 17:14
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