# Inverse of open affine subscheme is affine

This seems ridiculously simple, but it's eluding me.

Suppose $f:X\rightarrow Y$ is a morphism of affine schemes. Let $V$ be an open affine subscheme of $Y$. Why is $f^{-1}(V)$ affine?

I noted that $V$ is quasi-compact and wrote it as a finite union of principal open sets. Because the pullbacks of principal open sets are principal open sets, we can write $f^{-1}(V)$ as the union of such sets. But I'm not sure how to show this union is affine. I don't think this is the right way to go, because such unions are not affine in general.

I'm particularly perplexed because this occurs as an exercise in the chapter on separated morphisms and base change in Liu. Of course, all affine schemes are separated, but I don't see the relevance of that here.

• I think the idea is to identify $f^{-1}(V)$ with $V\times_Y X$, and use the fact that the fiber product of affine schemes is affine. – Jeff Jun 23 '13 at 2:07
• Dear @Potato, It is formal. You can verify directly that $f^{-1}(V)$, together with its canonical maps to $Y$ (the restriction of $f$) and $X$ (the inclusion) satisfies the universal property of the fiber product. But so does the spectrum of a certain tensor product, so $f^{-1}(V)$ must be affine by the uniqueness of fiber products up to canonical isomorphism. – Keenan Kidwell Jun 23 '13 at 2:47
• Dear @Potato, It is implicit in the construction of fiber products by gluing. – Keenan Kidwell Jun 23 '13 at 2:48
• Woops, sorry, I'm an idiot. It looked like what I wanted but it isn't. I'll just write something. – TTS Jun 23 '13 at 3:03
• Dear @Potato, It is in the second paragraph of his proof, when he says "It suffices for this to take $U\times_SY:=p^{-1}(U)$..." – Keenan Kidwell Jun 23 '13 at 3:16

## 2 Answers

The slogan I remember is, "open immersions are stable under base change". Suppose you have a diagram $\begin{matrix} & & X' \\ & & \downarrow {\scriptstyle f} \\ U & \xrightarrow{i} & X \end{matrix}$ and $i$ is an open immersion, in which case we might as well identify $U$ with an open subset of $X$. The fibered product exists and one realization of it is the open subscheme $f^{-1}(U)$ of $X'$. To show this you would just verify that it has the universal property. [You don't need to know that fibered products exist in general to do this.]

In the setting of the problem $X, X'$, and $U$ are affine, so you have another realization of $U \times_X X'$ as the spectrum of a ring. Hence there is an isomorphism between it and $f^{-1}(U)$.

I don't think this is trivial by any means. It is a consequence of the following result:

If $X$ is a scheme with $f_1,\ldots,f_n\in\mathscr{O}_X(X)$ which generate the unit ideal and each $X_{f_i}$ is affine, then $X$ is affine.

Here, for a scheme $X$ and a global section $f\in\mathscr{O}_X(X)$, $X_f=\{x\in X:f_x\notin\mathfrak{m}_x\}$ is the open set where $f$ doesn't vanish." If $X=\mathrm{Spec}(B)$ is an affine scheme, then $X_f=D(f)$ is the standard open associated to $f$.

First I'll assume the result and explain why it implies the result you're interested in. Say $X=\mathrm{Spec}(B)$ and $Y=\mathrm{Spec}(A)$, and let $\varphi:X\rightarrow Y$ be the morphism corresponding to the ring map $\varphi^\sharp:A\rightarrow B$. Let $V$ be an affine open of $\mathrm{Spec}(A)$. Write $V=\bigcup_{i=1}^n D(f_i)$ for $f_i\in A$. Then $\varphi^{-1}(V)=\bigcup_{i=1}^n \varphi^{-1}(D(f_i))=\bigcup_{i=1}^n D(\varphi^\sharp(f_i))$ is an affine open cover of $\varphi^{-1}(V)$. The global sections $f_i\vert_V\in\mathscr{O}_Y(V)$ generate the unit ideal (because $V$ is covered by the $D(f_i)=D(f_i\vert_V)$), so their images under $\varphi^\sharp$ generate the unit ideal in $\mathscr{O}_X(\varphi^{-1}(V))$. Since $D(\varphi^\sharp(f_i))=\varphi^{-1}(V)_{\varphi^\sharp(f_i)\vert_{\varphi^{-1}(V)}}$ is affine for each $i$, the result above implies that $\varphi^{-1}(V)$ is affine.

Now I'll sketch the proof of the result. There is a unique morphism $\varphi:X\rightarrow\mathrm{Spec}(\mathscr{O}_X(X))$ which induces the identity on global sections (for any scheme $X$ and even any locally ringed space). This morphism is an isomorphism if and only if $X$ is an affine scheme. The inverse image of $D(f)$ for any $f\in\mathscr{O}_X(X)$ under this morphism is $X_f\subseteq X$. The hypothesis on $X$ implies that it is quasi-compact and separated, meaning it is covered by finitely many affine opens (the $X_{f_i}$) whose intersections are also affine ($X_{f_i}\cap X_{f_j}=X_{f_if_j}$ is a standard open of each of $X_{f_i}$, $X_{f_j}$). This implies (Exercise 2.16 in Hartshorne, I think also proved in Liu) that the natural map $\mathscr{O}_X(X)_{f_i}\rightarrow\mathscr{O}_X(X_{f_i})$ is an isomorphism of rings. Now the assumption that $X_{f_i}$ is affine implies that the restriction of the natural map $\varphi$ to $X_{f_i}$ induces an isomorphism $X_{f_i}\cong D(f_i)$. This is because the map on global sections of the morphism $X_{f_i}\rightarrow D(f_i)$ is, by construction of the map $\varphi$, the natural map $\mathscr{O}_{\mathrm{Spec}(\mathscr{O}_X(X))}(D(f_i))=\mathscr{O}_X(X)_{f_i}\rightarrow\mathscr{O}_X(X_{f_i})$, which I've said is an isomorphism, and a morphism of affine schemes which induces an isomorphism on global sections is an isomorphism. Since the $D(f_i)$ cover $\mathrm{Spec}(\mathscr{O}_X(X))$, as follows from the assumption that the $f_i$ generate the unit ideal, this implies that $\varphi$ is an isomorphism, so $X$ is affine.

EDIT: Jeff Tolliver's comment shows that my argument is overkill for this particular statement; however, my argument can be used to show that for a morphism of schemes $f:X\rightarrow Y$, if there is an affine open cover $Y=\bigcup_i V_i$ such that $f^{-1}(V_i)$ is affine for all $i$, then $f^{-1}(V)$ is affine for any affine open $V$ of $Y$.

• If you do not assume $X$ and $Y$ to be affine, the proof would be exactly the same right? (I do not know if V as a open affine subset of a (not necesary affine scheme) can be expressed as a union of principal affine open subsets...) If it would be same, I have a problem to see where is used that the morphism is affine... – Patricio Apr 9 '16 at 14:19