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?

$\begingroup$ 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. $\endgroup$– WandererSep 18 '11 at 18:32

$\begingroup$ 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. $\endgroup$– WandererSep 18 '11 at 18:38

$\begingroup$ 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. $\endgroup$– user19475Sep 18 '11 at 18:50

$\begingroup$ Dix exposés, page 50, says that $Br(X) = H^3(X, \mathbf{Z})$, which vanishes for $X = \mathbf{P}^n$. $\endgroup$– user19475Sep 18 '11 at 19:05

1$\begingroup$ Hi, i don't know if this is what you are looking for: there is the concept of the unramified Brauer group. For a projective regular variety $X$ over $k$ with function field $K$ the Brauer group of $X$ equals the unramified Brauer group of $K$. One can further show that if $K/k$ is rational, then the unramified Brauer group of $K$ equals the Brauer group of $k$. This can be found in Chapter 5 of arXiv:math/0507154 by ColliotThelene and Sansuc $\endgroup$– TonySSep 18 '11 at 21:37
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 HochschildSerre 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.

$\begingroup$ why does surjectivity of $\mathbb{Z} = Pic(\mathbb{P}^n) \to H^2(\mathbb{P}^n, \mu_r) \to H^2(\mathbb{P}_{\bar{k}}^n, \mu_r) = \mathbb{Z}/r$ imply that $H^2(Gal(\bar{k}/k),\mu_r)$ is isomorphic to $Cokernel(d)$? $\endgroup$ Dec 6 '19 at 0:05


$\begingroup$ first of all by composing $H^2(Gal(\bar{k}/k),\mu_r) \to H^2(\mathbb{P}_k^n, \mu_r)$ with $H^2(X, \mu_r) \to Ker(Br(X) \stackrel{r}{\to} Br(X))=Br(X)[r]$ we obtain a promising candidate. I not fully understand how do you 'combine' surjective $ Pic(\mathbb{P}^n) \to \to H^2(\mathbb{P}_{\bar{k}}^n, \mu_r) $ with other sequences to obtain the desired isomorphy. by functoriality of $H$ we can 'add' to exact $H^1(X,\mathbb{G}_m) = Pic(X) \stackrel{d}{\to} H^2(X, \mu_r) \to Br(X)[r] \to 0$ below corresponding row with $\overline{X}:=X\times_k \overline{k}= \mathbb{P}^n_{\overline{k}}$. $\endgroup$ Dec 7 '19 at 16:13

$\begingroup$ what do we know about induced $Br(X) \to Br(\overline{X})$? is it the key? $\endgroup$ Dec 7 '19 at 16:13

$\begingroup$ The first sequence tells us that $H^2(X, \mu_r)$ surjects onto $Br(X)[r]$ and the second tells us the structure of this group. Combining this with the third sequence, one gets a direct sum decomposition of $H^2(\mathbb{P}^n, \mu_r)$ as $H^2(Gal(\bar{k}/k), \mu_r) \oplus \mathbb{Z}/r$. The second summand maps to $0$ in the Brauer group. $\endgroup$– nafDec 8 '19 at 13:53
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 HochschildSerre 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.

$\begingroup$ The Brauer group of a smooth projective variety $X/k$ equals $H^2(X,\mathbf{G}_m)$, which is torsion. So it suffices to show that $\mathrm{Br}(\mathbf{P}^N_{k^s})[\ell^n] = 0$ for all $\ell,n$. The Kummer sequence gives $0 \to \mathrm{Pic}(\mathbf{P}^N)/\ell^n \to H^2(\mathbf{P}^N_{k^s},\mathbf{Z}/\ell^n(1)) \to \mathrm{Br}(\mathbf{P}^N_{k^s})[\ell^n] \to 0$ for $\ell$ invertible in $k$. Now the Picard group of projective space for $N > 0$ is $\mathbf{Z}$ and the étale cohomology of projective space over separably closed fields is known. (This follows ulrich's lines; ... $\endgroup$– user19475Jan 22 '18 at 16:02

$\begingroup$ ... for $\ell = p$, one could use flat cohomology.) $\endgroup$– user19475Jan 22 '18 at 16:06

$\begingroup$ One would be finished if $H^2_{fppf}(\mathbf{P}^N_{k^s},\mu_{p^n}) = \mathbf{Z}/p^n$ for $N > 0$. One also has $\mathrm{cd}_p(k) \leq 1$ for $\mathrm{char}(k) = p > 0$. $\endgroup$– user19475Jan 22 '18 at 16:20
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 [AuslanderGoldman, "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.
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 BrauerSeveri scheme to the Brauer group of the base scheme. Theorem 2 in page 193 says the following:
Let $\pi : X \to S$ be a BrauerSeveri 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.)
Here is a proof that works in all characteristics. The idea comes from ColliotThélène ["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 $\pi \colon X \to \mathbb{P}^{n1}$, where $X$ is the blowup of $\mathbb{P}^n$ at $P$. By induction we have $\mathrm{Br}\,\mathbb{P}^{n1} = \mathrm{Br}\,k$. 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 $\pi^* \colon \mathrm{Br}\,\mathbb{P}^{n1} \to \mathrm{Br}\,X$ is surjective.
Let $\eta\colon \mathrm{Spec}\,K \to \mathbb{P}^{n1}$ 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 $\sigma \colon \mathbb{P}^{n1} \to X$ 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 $\mathrm{Br}\,\mathbb{P}^{n1}$, that is, $\mathrm{Br}\,\mathbb{P}^{n1} \to \mathrm{Br}\,X$ is surjective.
I've not been able to find this proof given explicitly in the literature.

$\begingroup$ The proof is found in The BrauerGroethendieck Group by JeanLouis ColliotThélène and Alexei N. Skorobogatov. This is theorem 6.1.3. The book was written after your comment though. $\endgroup$ Sep 19 '21 at 19:38