# Isomorphism of Brauer groups

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?

• Here $Br(X)=H^2_{et}(X,\mathbb G_m)$? Jun 9, 2022 at 5:15
• 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. Jun 9, 2022 at 5:28
• @KentaS Yes, the etale cohomological Brauer group. Jun 9, 2022 at 5:50