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This is a follow-up question to this question. There and here $X$ is a normal projective rational surface over $\mathbb{C}$ with finitely generated divisor class group $\text{Cl}(X)$. My question is:

Is the Brauer group $\text{Br}(X)$ of the rational surface $X$ always trivial?

This might be true and well-known in a more general setting but I can't find a good reference. Let me give some context and show how the question arose. There is an exact sequence $$0 \rightarrow \text{Pic}(X) \rightarrow \text{Cl}(X) \rightarrow \bigoplus_{x \in X^{\text{sing}}} \text{Cl}(\mathcal{O}_{X,x}) \rightarrow H^{2}(X,\mathcal{O}_{X}^{\ast}) \rightarrow 0$$ of abelian groups. If $X$ has at most ADE singularities one obtains $$\text{Br}(X) = H^{2}(X,\mathcal{O}_{X}^{\ast})_{\text{tor}} = H^{2}(X,\mathcal{O}_{X}^{\ast})$$ where the first equality follows from $X$ being a normal surface (see here) and the second one from $\bigoplus_{x \in X^{\text{sing}}} \text{Cl}(\mathcal{O}_{X,x})$ being a torsion group (because $X$ is ADE).

If it was true that the Brauer group is trivial in the rational case I could make more precise statements about $\text{Cl}(X)$ and $\text{Pic}(X)$.

Is there a reference for this? I'm new to this and I'm not even entirely sure if the above is right. So please correct me if I'm wrong. Thanks in advance.

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    $\begingroup$ The Brauer group of a rational surface with non-rational singularities will in general not be trivial. For example, this is the case for the surface obtained by blowing up $\mathbb{P}^2$ at (suitably chosen) 10 points on an elliptic curve $E$ and then blowing down the strict transform of $E$. $\endgroup$
    – naf
    Commented Sep 24, 2015 at 12:10
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    $\begingroup$ My answer to this question gives an explicit example of an element in the Brauer group of Cayley's cubic surface, which has four A1 singularities. Working out your exact sequence for that surface is a nice exercise. $\endgroup$
    – Martin Bright
    Commented Sep 24, 2015 at 12:22
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    $\begingroup$ Could you please stop using the notation $H^{2}(X,\mathcal{O}_{X}^{\ast})$? This is very confusing. What you mean (I suppose) is $H^{2}_{\rm ét}(X,\mathbb{G}_m)$. $\endgroup$
    – abx
    Commented Sep 24, 2015 at 13:53
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    $\begingroup$ @abx Sorry, I've adopted the notation from the answer of the question I have linked in the beginning. It is also used in some articles, for example in this one. And yes, it is meant étale cohomology here. (Isn't $\mathbb{G}_{m}$ just the same as $\mathcal{O}_{X}^{\ast}$?) $\endgroup$
    – user269218
    Commented Sep 24, 2015 at 14:13
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    $\begingroup$ @user269218: perhaps abx's concern is that $H^2(X, \mathcal{O}_X^{\ast})$ looks like it is supposed to depict the cohomology of, say, a quasicoherent sheaf, but $\mathcal{O}_X^{\ast}$ isn't such a thing... $\endgroup$
    – Qiaochu Yuan
    Commented Sep 24, 2015 at 19:50

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