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let $X$ be a smooth, projective and geometrically integral $k$-scheme. the Brauer group of $X$ is defined by $Br(X)=H^2_{ét}(X, \mathbb{G}_m)$.

I'm searching for a proof of this Theorem: assume that $X$ as above and $X$ is $k$-rational, ie birational equivalent to some $\mathbb{P}^n_k$. then $Br(X)=Br(k)$.

obviously the problem can be splitted in two statements:

1) $X,Y$ smooth, projective and geometrically integral $k$-schemes which are birationally equivalent to each other. then $Br(X)=Br(Y)$.

2) $Br(\mathbb{P}^n_k)=Br(k)$

could anybody sketch these proofs or give a reference? additionally: do we need for 1) and 2) really every of the smooth, projective and geometrically integral conditions or can it be weakened?

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  • $\begingroup$ Do you need this for char p k? For char 0 1) is proven by Grothendieck ine one of the Theorie de Brauer (by explicitly using resolution of singularities iirc). The second case is well-known (e.g. see this mathoverflow.net/questions/75774/…) $\endgroup$ – Alex Youcis Dec 2 '19 at 2:57
  • $\begingroup$ @AlexYoucis:I took a glance at ulrich's answer in linked discussion. do you see why surjectivity of the map $\mathbb{Z} = Pic(\mathbb{P}^n) \to H^2(\mathbb{P}^n, \mu_r) \to H^2(\mathbb{P}_{\bar{k}}^n, \mu_r) = \mathbb{Z}/r$ imply that $H^2(Gal(\bar{k}/k),\mu_r)$ is isomorphic to $Cokernel(d) = Ker(r)=Br(\mathbb{P}^n)[r]$? $\endgroup$ – user705174 Dec 5 '19 at 22:57
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Let me spell out how the computation for (2). We first note that we have the Kummer short exact sequence, which is $$ 0\rightarrow \mu_l \rightarrow \mathbb{G}_m\rightarrow \mathbb{G}_m\rightarrow 0.$$ The induced long exact sequence on etale cohomology is then $$ H^1(\mathbb{P}_k^n,\mathbb{G}_m) \rightarrow H^2(\mathbb{P}_k^n, \mu_l) \rightarrow H^2(\mathbb{P}_k^n,\mathbb{G}_m)\rightarrow H^2(\mathbb{P}_k^n,\mathbb{G}_m).$$ Note that $H^1(\mathbb{P}_k^n,\mathbb{G}_m)=\text{Pic}(\mathbb{P}^n_k)=\mathbb{Z}$. Then we want to compute $H^2(\mathbb{P}_k^n,\mu_l)$. For this we use the Hochschild-Serre spectral sequence. This then gives us $H^i(G_k, H^j(\mathbb{P}^n_{\bar{k}} ,\mu_l))$ $\Rightarrow$ $H^{i+j}(\mathbb{P}_{\bar{k}}^n, \mu_l)$. Hence we want to compute $H^0(\mathbb{P}_{\bar{k}}^n,\mu_l)$, $H^1(\mathbb{P}_{\bar{k}}^n,\mu_l)$ and $H^2(\mathbb{P}_{\bar{k}}^n,\mu_l)$, which are $\mu_l,0$ and $\mathbb{Z}/l$ respectively. Thus we have $$0 \rightarrow H^2(G_k,\mu_l)\rightarrow H^2(\mathbb{P}_k^n,\mu_l) \rightarrow H^0(G_k,\mathbb{Z}/l)\rightarrow 0.$$ Thus $H^2(G_k,\mu_l)\cong \text{coker}(\text{Pic}(\mathbb{P}_k^n)\rightarrow H^2(\mathbb{P}_k^n,\mu_l))[l]\cong Br(\mathbb{P}_k^n)[l].$ Then we note that $H^2(G_k,\mu_l)\cong Br(k)[l]$. Since this is true for all $l$, we see that in fact $Br(k)\cong Br(\mathbb{P}_k^n)$.

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  • $\begingroup$ why the exactness of $0 \rightarrow H^2(G_k,\mu_l)\rightarrow H^2(\mathbb{P}_k^n,\mu_l) \rightarrow H^0(G_k,\mathbb{Z}/l)\rightarrow 0$ imply $H^2(G_k,\mu_l)\cong \text{coker}(\text{Pic}(\mathbb{P}_k^n)\rightarrow H^2(\mathbb{P}_k^n,\mu_l))[l]\cong Br(\mathbb{P}_k^n)[l]$? $\endgroup$ – user705174 Dec 5 '19 at 20:42
  • $\begingroup$ I opened a separate thread on the aspect how we conclude from the bunch of exact sequences above that $H^2(G_k,\mu_l)\cong \text{coker}(\text{Pic}(\mathbb{P}_k^n)$. could you loose a few words on how you combine these sequences in order to obtain the desired isomorphism? $\endgroup$ – user705174 Dec 7 '19 at 16:17

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