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Just for completeness: The "correct" way to understand the Brauer group of $X$ using its codimension $1$ points is via residue maps. Specifically: Let $X$ be a regular integral noetherian scheme. The …

An explicit simple counter-example is the following: just take the quaternion algebra $(x,y)$ over $k(x,y)$, where $k$ is a field with $\mathrm{char}(k) \neq 2$. This is non-zero on any open subset of …

Let $k$ be a field and $\mathrm{Br}(k)$ the Brauer group of $k$. Let $k \subset L$ be a field extension. Let $b \in \mathrm{Br}(k)$ and denote by $b \otimes L \in \mathrm{Br}(L)$ the base-change of $b …

I consider only smooth curves for simplicity. In which case this is expected to be true, but certainly not known in general. In fact, it is even expected that the Brauer-Manin obstruction is already e …

I think the following should work, but I have not checked all the details. Let $k$ be an algebraically closed field and let $\ell$ be coprime to the characteristic of $k$. Then the Grothendieck purit …

Let $X$ be a non-singular projective variety over a field $k$ (perhaps not of characteristic $2$), and let $\pi:Y\to X$ be a conic bundle over $X$ i.e. a proper morphism all of whose fibres are isomor …