# Brauer group of $k$-rational scheme

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?

• 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/…) – Alex Youcis Dec 2 '19 at 2:57
• @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 $Cokernel(d) = Ker(r)=Br(\mathbb{P}^n)[r]$? – user705174 Dec 5 '19 at 22:57

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 $$Br(k)\cong Br(\mathbb{P}_k^n)$$.
• 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 $H^2(G_k,\mu_l)\cong \text{coker}(\text{Pic}(\mathbb{P}_k^n)\rightarrow H^2(\mathbb{P}_k^n,\mu_l))[l]\cong Br(\mathbb{P}_k^n)[l]$? – user705174 Dec 5 '19 at 20:42
• 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? – user705174 Dec 7 '19 at 16:17