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?



Brauer group of projective space
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.


