Flatness of the restriction morphism of an affine scheme to an affine subscheme

Question

Let $X= \operatorname{Spec}(A)$ be an affine scheme, and $Y= \operatorname{Spec}(B)$ be an open affine subscheme of $X$. The inclusion morphism $Y \rightarrow X$ induces a structure of $A$-algebra on $B$. I would like to prove that $B$ is then flat over $A$.

When $Y$ is a principal open subset of $X$, the result is known because $B$ is then a localization of $A$. In the general case, I suspect that $B$ is not always a localization (or is it ?). How could I then prove flatness of $B$ ?

Answer 1

Recall the relevant definitions:

Definition Let $f : X \to Y$ be a morphism of schemes. $f$ is said to be flat if the map of stalks $\mathcal{O}_{Y,f(x)} \to \mathcal{O}_{X,x}$ is a flat morphism of rings for all $x \in X$.

Definition Let $f : X \to Y$ be a morphism of schemes. $f$ is said to be an open immersion if the map on topological spaces $X \to Y$ is open, is a homeomorphism onto its image, and the morphism of sheaves $f^* \mathcal{O}_Y \to \mathcal{O}_X$ is an isomorphism.

Now it should be clear that every open immersion of schemes is flat! No need to assume that the schemes are affine.

Proposition Every open immersion of schemes is flat.

Proof. Let $f : X \to Y$ be an open immersion of schemes. Since $f^* \mathcal{O}_Y \to \mathcal{O}_X$ is an isomorphism, it induces an isomorphism on all stalks, so the map $\mathcal{O}_{Y,f(x)} = (f^* \mathcal{O}_Y)_x \to \mathcal{O}_{X,x}$ is an isomorphism (hence flat) for all $x \in X$. Therefore, $f$ is flat. $\square$

To be extra precise, the relevant open immersion in your question is the inclusion $\operatorname{Spec}(B) \to \operatorname{Spec}(A)$.

In general, an open affine subscheme of an affine scheme need not be a localization; see for example this question.

Answer 2

While diracdeltafunk's answer is great, since it is more general and not much more complicated, here is the similar (but somewhat more elementary) argument for the special case of affine schemes using the following definition of an open immersion, as in Liu's Algebraic Geometry book:

Definition: A morphism $\phi:X\to Y$ between schemes is an open immersion if $\phi$ is a topological open immersion and if for each $x\in X$, $(\phi^\#)_x:\mathcal O_{Y,\phi(x)}\to\mathcal O_{X,x}$ is an isomorphism.

Note: One can show easily that this definition implies the definition of diracdeltafunk, e.g. by using the functor $\phi^*:\operatorname{Sh}(Y)\to \operatorname{Sh}(X)$ as well as the natural isomorphism $\phi^*\phi_*\mathcal O_X\cong\mathcal O_X$ (in the case of $\phi$ being an open immersion). These considerations are not necessary in the affine case.

Proof: Let $\phi:\operatorname{Spec}B\to \operatorname{Spec}A$ be an open immersion, and let $f:A\to B$ be the corresponding ring map, i.e., $\phi=\operatorname{Spec}(f)$. Let $\mathfrak q\in\operatorname{Spec}B$. On the level of stalks, $(\phi^\#)_\mathfrak q$ is just the localisation of $f$: $$ (\phi^\#)_\mathfrak q:A_{f^{-1}\mathfrak q}\to B_\mathfrak q. $$ Since $\phi$ is an open immersion, we know that $(\phi^\#)_\mathfrak q$ is an isomorphism, and hence $B_q$ is flat over $A_{f^{-1}\mathfrak q}$. Since being flat is a local property, it follows that $f:A\to B$ is flat.