A recent big result proved by $\mathrm{\check{C}}$esnavi$\mathrm{\check{c}}$ius states that

For a regular, integral, noetherian scheme $X$ and an open subset $U \subset X$ whose complement is of codimension at least $2$, the restriction map $\mathrm{Br}(X) \rightarrow \mathrm{Br}(U)$ is an isomorphism.

This is called purity for Brauer groups. I wonder how much of the result can be extended to curves. Say, for example, we have an elliptic curve $$E: Y^2Z = X^3 + aXZ^2 + bZ^3$$ and we remove the point of origin $O$ to obtain the affine model $$C:y^2 = x^3+ax+b$$ whose complement $\{O\}$ is of codimension $1$. How much can be said about the restriction map $\mathrm{Br}(E) \rightarrow \mathrm{Br}(C)$? By a result of Bertuccioni in Brauer groups and cohomology,

Let $X$ be an separated noetherian scheme and $U \subset X$ be a nonempty open subscheme. Assume that $U$ contains every generic point and every singular point of $X$. Then the restriction map $\mathrm{Br}(X) \rightarrow \mathrm{Br}(U)$ is an injective homomorphism.

I would like to know if the example given has any chance of being an isomorphism. Also, what does purity even mean?

  • $\begingroup$ Here $Br(X)=H^2_{et}(X,\mathbb G_m)$? $\endgroup$
    – Kenta S
    Jun 9, 2022 at 5:15
  • 1
    $\begingroup$ Purity in this context is basically "determined precisely by codimension one stuff". The source of the terminology is Zariski & Nagata's theorem on purity of the branch locus. $\endgroup$
    – KReiser
    Jun 9, 2022 at 5:28
  • $\begingroup$ @KentaS Yes, the etale cohomological Brauer group. $\endgroup$
    – oleout
    Jun 9, 2022 at 5:50


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