The Stacks project

Lemma 28.26.4. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $s \in \Gamma (X, \mathcal{L})$. For any affine $U \subset X$ the intersection $U \cap X_ s$ is affine.

Proof. This translates into the following algebra problem. Let $R$ be a ring. Let $N$ be an invertible $R$-module (i.e., locally free of rank 1). Let $s \in N$ be an element. Then $U = \{ \mathfrak p \mid s \not\in \mathfrak p N\} $ is an affine open subset of $\mathop{\mathrm{Spec}}(R)$. This you can see as follows. Think of $s$ as an $R$-module map $R \to N$. This gives rise to $R$-module maps $N^{\otimes k} \to N^{\otimes k + 1}$. Consider

\[ R' = \mathop{\mathrm{colim}}\nolimits _ n N^{\otimes n} \]

with transition maps as above. Define an $R$-algebra structure on $R'$ by the rule $x \cdot y = x \otimes y \in N^{\otimes n + m}$ if $x \in N^{\otimes n}$ and $y \in N^{\otimes m}$. We claim that $\mathop{\mathrm{Spec}}(R') \to \mathop{\mathrm{Spec}}(R)$ is an open immersion with image $U$.

To prove this is a local question on $\mathop{\mathrm{Spec}}(R)$. Let $\mathfrak p \in \mathop{\mathrm{Spec}}(R)$. Pick $f \in R$, $f \not\in \mathfrak p$ such that $N_ f \cong R_ f$ as a module. Replacing $R$ by $R_ f$, $N$ by $N_ f$ and $R'$ by $R'_ f = \mathop{\mathrm{colim}}\nolimits N_ f^{\otimes n}$ we may assume that $N \cong R$. Say $N = R$. In this case $s$ is an element of $R$ and it is easy to see that $R' \cong R_ s$. Thus the lemma follows. $\square$


Comments (5)

Comment #3508 by Yicheng Zhou on

By using direction (2) to (1) of Lemma 28.11.3, one can give a more schematic proof as follows. By hypothesis we have locally , therefore the open immersion is locally given by a principal open set (in an affine neighborhood). By the lemma cited, is affine, therfore is affine whenever is affine.

Comment #3548 by on

This works, but in the Stacks project we never give forward references to avoid circular arguments. Since this lemma is stated before the lemma about affine morphisms you mention, we cannot use that lemma in its proof.

Comment #4633 by Andy on

Could you include a proof that is commutative? It's not immediately obvious to me.

Comment #4634 by Andy on

Could I also suggest that this statement could be just a consequence of the fact that an affine morphism is local on base, so we can just check on an affine cover where the sheaf is trivial and it reduces to the fact that basic opens in an affine scheme is affine.

Comment #4635 by on

Dear Andy, you can check the commutativity locally so this follows from the local description in the second paragraph of the proof. In comment #3548 above I explain why we can't yet use that affineness of a morphism is local.


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