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 $V = \{ \mathfrak p \mid s \not\in \mathfrak p N\}$ is an affine open subset of $\mathop{\mathrm{Spec}}(R)$.

Let $A = \bigoplus _{n \geq 0} A_ n$ be the symmetric algebra of $N$ (which is commutative) and view $s$ as an element of $A_1$. Set $B = A/(s - 1)A$. This is an $R$-algebra whose construction commutes with any base change $R \to R'$. Thus $B' = B \otimes _ R R'$ is the zero ring if $s$ maps to zero in $N' = N \otimes _ R R'$. It follows that if $x \in \mathop{\mathrm{Spec}}(R) \setminus V$, then $B \otimes _ R \kappa (x) = 0$. We conclude that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(R)$ factors through $V$ as the fibres over $x \not\in V$ are empty. On the other hand, if $\mathop{\mathrm{Spec}}(R') \subset V$ is an affine open, then $s$ maps to a basis element of $N'$ and we see that $B' = R'[s]/(s - 1) \cong R'$. It follows that $\mathop{\mathrm{Spec}}(B) \to V$ is an isomorphism and $V$ is indeed affine. $\square$

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 $\mathcal{L}\cong\mathcal{O}_X$, therefore the open immersion $\iota: X_s \hookrightarrow X$ is locally given by a principal open set (in an affine neighborhood). By the lemma cited, $\iota$ is affine, therfore $U \cap X_ s = \iota^{-1}(U)$ is affine whenever $U$ 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 $R'$ 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.

Comment #6947 by on

Dear Johan, I think Andy was right. Commutativity indeed holds in the rank-1 case (as you said, just check it locally), but not in general.

I think that for the general construction, one shoud use the symmetrized tensor product. For the case of interest (that of linde bundles), this would make no difference.

Comment #6948 by on

OK, I may have misread Andy's second comment. Anyway, if anybody has a precise suggestion for changing the text of the proof, please let me know. Thanks!

Comment #6963 by Laurent Moret-Bailly on

Here is a suggestion: let $A=\bigoplus_{n\ge0}A_n$ be the symmetric algebra of $N$ (which is commutative) and view $s$ as an element of $A_1$. Then define $R'$ as $A/(s-1)$. This is an $R$-algebra whose construction commutes with any base change $R\to R''$. If $x\in\operatorname{Spec}(R)\setminus U$, then $R'\otimes_R\kappa(x)=\kappa(x)[t]/(0t-1)=0$. So $\operatorname{Spec}(R')\to\operatorname{Spec}(R)$ factors through $U$. And over $U$ we have an isomorphism since $N$ is free with basis $s$.

Comment #6983 by on

Very good! This is indeed better. I have made the changes here. It should be reflected on the website in about 20 minutes.

Comment #8359 by Et on

Is U not supposed to represent the affine specR from the statement of the lemma? Whereas the U in the proof I think refers to the intersection with X_s

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