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I've read that $\text{Br} \mathbb{P}^n_k$ (here $\text{Br}$ is the cohomological Brauer group, i.e. $H^2_{ét}(-,\mathbb{G}_m)$) is just isomorphic to $\text{Br} k$. As proof of this fact seems to be not so easy in the general case, but there should be a simple and conceptual proof when $k$ has characteristic zero. Does anyone know this simple proof?

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I don't know it, but have you looked into Dix exposés, Le groupe de Brauer or Milne? – Timo Keller Sep 18 '11 at 18:23
See also here:… – Timo Keller Sep 18 '11 at 18:31
Thanks for the reactions. I've looked through the "Dix exposés", but didn't find any simple proof for the characteristic zero case - which doesn't mean it isn't there, it might be well hidden somewhere, as is sometimes the case with Grothendieck. – Wanderer Sep 18 '11 at 18:32
Don't find it in Milne either, at least not in the paragraph on the Brauer group. He focuses on comparing the cohomological Brauer group and the Brauer group in terms of Azumaya algebras. – Wanderer Sep 18 '11 at 18:38
Hm, at least it gives us an injection $Br(\mathbf{P}^n) \hookrightarrow Br(k(X_1, \ldots, X_n))$, and the Brauer group of the function field of a curve over an algebraically closed field is trivial (Tsen). I don't know if this helps. – Timo Keller Sep 18 '11 at 18:50
up vote 12 down vote accepted

I don't think the assumption of characteristic zero simplifies the proof a great deal. However, it does allow us to avoid having to give a more involved proof for the $p$-power torsion (where $p = char (k)$) so I will assume that below.

Firstly, by Proposition 1.4 of Grothendieck's "Groupe de Brauer II", $H^2(X, \mathbb{G}_m)$ is torsion for a smooth variety $X$ so we may use the Kummer sequence $$ 1 \to \mu_r \to \mathbb{G}_m \stackrel{r}{\to} \mathbb{G}_m \to 1 $$ of etale sheaves on $X$ to compute the $r$-torsion for all $r$ and hence compute all of $Br(X)$.

The long exact sequence of etale cohomology on $X$ gives an exact sequence $$ H^1(X,\mathbb{G}_m) = Pic(X) \stackrel{d}{\to} H^2(X, \mu_r) \to Br(X) \stackrel{r}{\to} Br(X) $$ so we need to compute the cokernel of $d$.

Since $H^0(\mathbb{P}_{\bar{k}}^n,\mu_r) = \mu_r$, $H^1(\mathbb{P}_{\bar{k}}^n, \mu_r) = 0$, and $H^2(\mathbb{P}_{\bar{k}}^n, \mu_r)) = \mathbb{Z}/r$, the Hochschild-Serre spectral sequence gives an exact sequence $$ 0 \to H^2(Gal(\bar{k}/k),\mu_r) \to H^2(\mathbb{P}_k^n, \mu_r) \to H^0(Gal(\bar{k}/k), \mathbb{Z}/r) \to 0 .$$

The map $\mathbb{Z} = Pic(\mathbb{P}^n) \to H^2(\mathbb{P}^n, \mu_r) \to H^2(\mathbb{P}_{\bar{k}}^n, \mu_r) = \mathbb{Z}/r$ is surjective so it follows that $H^2(Gal(\bar{k}/k),\mu_r) = Br(k)[r]$ maps isomorphically onto $Cokernel(d) = Br(\mathbb{P}^n)[r]$. Since this is true for any integer $r$ it follows that the map $Br(k) \to Br(\mathbb{P}^n)$ is an isomorphism.

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Let $X$ be an $n$-dimensional projective space over a field $k$. Let $k_s$ be a separable closure of $k$, and $X_s$ the base change of $X$ to $k_s$. The algebraic part $\textrm{Br}_1(X)$ of the Brauer group of $X$ is defined as $\textrm{Br}_1(X) = \ker(\textrm{Br}(X) \rightarrow \textrm{Br}(X_s))$ and sits inside a short exact sequence $$0 \rightarrow \textrm{Br}(k) \rightarrow \textrm{Br}_1(X) \rightarrow H^1(G_k,\textrm{Pic}(X_s)) \rightarrow 0$$ given by the Hochschild-Serre spectral sequence. Since $\textrm{Pic}(X_s) = \mathbf{Z}$, it follows that we have a canonical isomorphism between $\textrm{Br}_1(X)$ and $\textrm{Br}(k)$.

So now we are reduced to showing that $\textrm{Br}(X_s)$ is trivial. I do not how to do this at the moment.

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Hey Rene, I added some Latex to your answer. Hope you don't mind. – Ariyan Javanpeykar Sep 19 '11 at 12:55

Consider the commutative diagram

\begin{array}{ccc} \operatorname{Br}k & \xrightarrow{f_1} & \operatorname{Br} \mathbb{P}_{k}^{n} \\ \scriptsize{f_2}\ \downarrow & \swarrow \scriptsize{f_3}& \downarrow \scriptsize{f_4}\\ \operatorname{Br}\mathbb{A}_{k}^{n} & \xrightarrow[f_5]{} & \operatorname{Br}k(x_{1},\dotsc,x_{n}) \end{array}

induced by pullback.

By [Milne, Etale Cohomology, Corollary IV.2.6], $f_{4}$ is injective; hence $f_{3}$ is injective. If $k$ has characteristic zero, $f_{2}$ is an isomorphism by [Auslander-Goldman, "The Brauer group of a commutative ring", Proposition 7.7]; thus $f_{3}$ is surjective.

EDIT: See also Proposition 1.5 of Reineke and Schroer's paper "Brauer groups for quiver moduli",

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