# Tagged Questions

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### Gerbes on the multiplicative group

Let $k$ be an arbitrary field with absolute Galois group $\Gamma$. The group $\text{Hom}(\Gamma,\mathbb{Q}/\mathbb{Z})$ injects into $H^2(\mathbb{A}^1 \setminus \{ 0 \},\mathbb{G}_m)$, as one can see ...
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### Central division algebras and splitting fields

Let $K$ be a field and $D$ be a central division algebra over $K$ of degree $n$. Suppose that $L\subset D$ is a maximal subfield, so that $[L:K]=n$. Then we know that $L$ is a splitting field, so ...
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### Extension of sheaf of Azumaya algebras and derived equivalence

Suppose there is a smooth variety $X$ and a sheaf of algebra $\mathcal{B}$. Let $Z\subseteq X$ be a closed subvariety, whose codimension is large (say $\geq 2$). If the restriction of $\mathcal{B}$ to ...
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### Can one control the ramification of a Brauer class under birational morphisms?

Assume we are given a Brauer class $\xi\in Br(k(\mathbb{P}^n))$ ramified at some divisor $D\subset \mathbb{P}^n$, here $k=\mathbb{C}$. If $f: \mathbb{P}^n\mathrel{-\,}\rightarrow \mathbb{P}^n$ is a ...
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### Is the Brauer group functor a Zariski sheaf?

Is the functor $$\operatorname{Br} : \operatorname{Sch}^{\operatorname{op}} \to \operatorname{Ab}$$ sending $X \mapsto \operatorname{Br}X$ a sheaf for the Zariski topology on $\operatorname{Sch}$? ...
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### Brauer group of a rational variety

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: ...
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### What is the relation (if any) between Clifford algebras and Azumaya algebras?

Suppose the base field is $\mathbb{C}$ and the Clifford algebra is the classical one (i.e. associated to a quadratic form in $n$ variables). It seems that there are relations between Clifford algebras ...
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### Obstruction and rational points on curves

Is etale-Brauer the only obstruction to the existence of rational points on projective plane curves over number fields?
It is a well-known fact that for an Enriques surface $Y$, if $D^b(Y)\cong D^b(X)$ for some smooth projective variety $X$, then $X\cong Y$. In other words, Enriques surfaces have no non-trivial ...