Second Zariski cohomology of the multiplicative group

Question

Let $X$ be a scheme, then $H^2_{et}(X, \Bbb G_m)$ is defined as the Brauer group of $X$ and people always say that Zariski topology is so coarse that we need to consider etale cohomology in practice. However, we have $H^1_{et}(X, \Bbb G_m) \cong H^1_{Zar}(X, \Bbb G_m)$ by using the standard spectual sequence from zariski cohomology to etale cohomology, Cech cohomology and f.f descent for line bundles.

Therefore, what is known about $H^2_{Zar}(X, \Bbb G_m)$ for good schemes (i.e a regular proper scheme over $\Bbb Z$ or a field)? Must it be zero in all cases? What about higher zariski cohomologies?

Answer 1

For regular schemes, the answer to your question is hidden in the comparison between Weil-Divisors and Cartier-Divisors. The result there is that for a noetherian integral, locally factorial scheme $X$ there is a short exact sequence $$ 1 \to \mathbb{G}_m \to \mathcal{M} \xrightarrow{div} \bigoplus_{x \in X^{(1)}} (i_x)_* \mathbb{Z} \to 1 $$ of Zariski-sheaves where $\mathcal{M}$ is the constant sheaf with values the function field of $X$, the set $X^{(1)}$ denotes points of codimension 1 and $i_x: \bar{\{x\}} \to X$ is the closed immersion of the associated codimension-1 irreducible reduced subscheme. In other words, the rightmost sheaf is exactly the sheaf of Weil-Divisors, which recieves a map from $\mathcal{M}$.

The long exact cohomology sequence associated to this short exact sequence is usually used to derive the isomorphism $$H_{Zar}^1(X, \mathbb{G}_m) \to Cl(X)$$ but can also be used to see the claimed vanishing of the higher Zariski-cohomology groups of $\mathbb{G}_m$: Indeed, as cohomology is compatible with direct sums and pushforwards along closed immersions, the two rightmost sheaves in the exact sequence have no cohomology away from 0. This forces $$H^i_{Zar}(X, \mathbb{G}_m) = 0 \ \ \mathrm{ for } \ \ i > 1$$ as claimed by Roland.