Brauer group of projective space

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 Schröer's paper "Brauer groups for quiver moduli", http://arxiv.org/abs/1410.0466.

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EDIT: Perhaps this is what Wanderer is referring to when he says "not so easy in the general case": Gabber in his thesis proved a more general claim relating the Brauer group of a Brauer-Severi scheme to the Brauer group of the base scheme. Theorem 2 in page 193 says the following:

Let $\pi : X \to S$ be a Brauer-Severi scheme. Then $$ \pi^{\ast} : \mathrm{H}^{2}(S,\mathbb{G}_{m})_{\mathrm{tors}} \to \mathrm{H}^{2}(X,\mathbb{G}_{m})_{\mathrm{tors}} $$ is surjective and $\ker \pi^{\ast} = \mathrm{H}^{0}(S,\mathbb{Z}) \cdot \delta(\mathrm{cl}(X))$.

In his notation, $\mathrm{cl}(X) \in \mathrm{H}^{1}(S,\mathrm{PGL}_{n+1})$ is the class corresponding to $X$ and $\delta : \mathrm{H}^{1}(S,\mathrm{PGL}_{n+1}) \to \mathrm{H}^{2}(S,\mathbb{G}_{m})$ is the boundary map. Thus in particular if $X$ is the projectivization of $\mathbb{P}(\mathcal{E})$ of a finite locally free $\mathcal{O}_{S}$-module $\mathcal{E}$ of rank $n+1$, then $\mathrm{cl}(X)$ is in the image of $\mathrm{H}^{1}(S,\mathrm{GL}_{n+1}) \to \mathrm{H}^{1}(S,\mathrm{PGL}_{n+1})$ so $\delta(\mathrm{cl}(X)) = 0$. (Gabber actually reduces the proof of Theorem 2 to this case, see the top of page 195.)

Answer 4

Here is a proof that works in all characteristics. The idea comes from Colliot-Thélène ["Formes quadratiques multiplicatives et variétés algébriques: deux compléments", Bull. Soc. Math. France, 108(2), 213–227].

We'll use several times that the Brauer group of a smooth variety injects into the Brauer group of its function field.

First we prove $\mathrm{Br}\,\mathbb{P}^1 = \mathrm{Br}\,k$. By Tsen's theorem, this is true if $k$ is algebraically closed. By Grothendieck ["Le Groupe de Brauer III", Corollaire 5.8], it is true if $k$ is separably closed, and so for general $k$ we have an isomorphism $\mathrm{Br}\, \mathbb{P}^1 = \mathrm{Br}_1\,\mathbb{P}^1$. But $\mathrm{Br}_1\,\mathbb{P}^1$ is contained in $\mathrm{Br}_1\,\mathbb{A}^1$, and Auslander–Goldman ["The Brauer group of a commutative ring", Theorem 7.5] have proved $\mathrm{Br}_1\,\mathbb{A}^1 = \mathrm{Br}\,k$.

Now we prove $\mathrm{Br}\,\mathbb{P}^n = \mathrm{Br}\, k$ by induction. Pick any point $P \in \mathbb{P}^n(k)$. Resolving the projection away from $P$ gives a morphism $\pi \colon X \to \mathbb{P}^{n-1}$, where $X$ is the blow-up of $\mathbb{P}^n$ at $P$. By induction we have $\mathrm{Br}\,\mathbb{P}^{n-1} = \mathrm{Br}\,k$. Since the injective map $\mathrm{Br}\,\mathbb{P}^n \to \mathrm{Br}\,k(\mathbb{P}^n)$ factors through $\mathrm{Br}\,X$, it is enough to prove that $\pi^* \colon \mathrm{Br}\,\mathbb{P}^{n-1} \to \mathrm{Br}\,X$ is surjective.

Let $\eta\colon \mathrm{Spec}\,K \to \mathbb{P}^{n-1}$ be the generic point. We have $X_\eta \cong \mathbb{P}^1_K$, so the natural map $\pi^* \colon \mathrm{Br}\,K \to \mathrm{Br}\,X_\eta$ is an isomorphism.

Now let $\sigma \colon \mathbb{P}^{n-1} \to X$ be a section (just take any hyperplane in $\mathbb{P}^n$ not containing $P$). Let $\alpha$ lie in $\mathrm{Br}\,X \subset \mathrm{Br}\,X_\eta$. We've just shown that $\alpha$ lies in the image of $\mathrm{Br}\,K$, that is, there exists $\beta \in \mathrm{Br}\,K$ such that $\alpha = \pi^* \beta$. Applying $\sigma^*$ shows that $\beta = \sigma^* \pi^* \beta = \sigma^* \alpha$ lies in $\mathrm{Br}\,\mathbb{P}^{n-1}$, that is, $\mathrm{Br}\,\mathbb{P}^{n-1} \to \mathrm{Br}\,X$ is surjective.

I've not been able to find this proof given explicitly in the literature.