TITLE: Brauer group of projective space
QUESTION [6 upvotes]: 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?

REPLY [4 votes]: 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.