1

In the answer of this question, it says that the complement of an effective Cartier divisor in an irreducible affine scheme is affine. However, I can't think of a proof (actually I'm not sure how to use the property of being locally principal, as affineness is a global thing).

(solved) One more question is that, again in the linked answer, an example is given where the complement of Weil divisor is not affine. This is confusing me as any Cartier divisor is a Weil divisor almost by definition - at least under the assumption that $X$ is regular in codimension one (context in Hartshorne's text). But it looks to me that the mentioned $X$ is regular in codimension one, only singular at the vertex.

I would greatly appreciate any proof for the first question and clarification on the second one.

KReiser
  • 74,746
okabe rintarou
  • 945

1 Answers1

2

Certainly the claim is true locally: an effective Cartier divisor can be specified by an open cover $\{U_i\}_{i\in I}$ of your scheme $X$ and an non-zero-divisor $f_i\in \mathcal{O}_X(U_i)$ for each $i\in I$ subject to a compatibility condition on the overlaps - by subdividing, we may assume that each $U_i$ is actually affine, and so the complement of the Cartier divisor inside $U_i$ is $D(f_i)\subset U_i$, an affine scheme.

Now a key fact about affine morphisms swoops in and saves the day.

Lemma. (Stacks 01S8) The following conditions on a morphism $f:X\to S$ of schemes are equivalent, and such a morphism is called affine:

  1. The inverse image of any affine open of $S$ is an affine open of $X$,
  2. There exists an open affine cover $\{S_i\}_{i\in I}$ of $S$ so that $f^{-1}(S_i)$ is affine,
  3. There exists a quasi-coherent sheaf of $\mathcal{O}_S$-algebras $\mathcal{A}$ and an isomorphism $X\cong \underline{\operatorname{Spec}}_S \mathcal{A}$ of schemes over $S$.

(See also here on MSE for another take on the issue.)

Applying the implication $2\Rightarrow 1$ to the inclusion $X\setminus D \to X$ lets us use the first paragraph to conclude that $X\setminus D$ is actually affine since $X$ is.

KReiser
  • 74,746