Universally Closed Affine Morphisms are Integral

I recently read about the following result from Stacks project tag 01WM and tried to prove it:

Theorem. Let \(f: X \to Y\) be morphism of schemes. Then \(f\) is integral iff \(f\) is affine and universally closed.

One direction is obvious, integral morphisms are affine by definition, stable under base change, and are closed maps. In the rest of this post, I will give my proof of the other direction. Our proof will be based on the following basic example of morphisms that are not universally closed (hence not proper):

Lemma 1. Let \(k\) be a field. The morphism \(\DeclareMathOperator{Spec}{Spec} \Spec k[x] \to \Spec k\) is not universally closed.

Proof. By base changing along \(\Spec \overline{k} \to \Spec k\), we may assume \(k\) is algebraically closed. Next consider the base change along \(\Spec k[y] \to \Spec k\). Consider the image of \(V(xy-1)\) under \(\Spec k[x,y] \to \Spec k[y]\). It is easy to see that the image is precisely the affine line minus the origin, hence is not closed. \(\blacksquare\)

Our proof strategy will be a series of reductions. We will show that we can successively make the following assumptions:

1. \(X = \Spec B\) and \(Y = \Spec A\) are affine;

2. \(A \subseteq B\);

3. \(A\) is a local ring;

4. \(B\) is generated by one element over \(A\);

5. \(A = k\) is a field.

The Proof

Now we begin our proof of the theorem. Assume that \(f : X \to Y\) is affine and universally closed.

Since \(f\) is affine, and integrality can be checked affine-locally on the target, we may assume \(X = \Spec B\) and \(Y = \Spec A\). Let \(\phi : A \to B\) be the corresponding map on global sections.

By replacing \(A\) with \(\phi(A)\), we reduce to the case that \(\phi\) is an inclusion:

Lemma 2. The morphism \(\Spec B \to \Spec \phi(A)\) is universally closed.

Proof. Base change \(\Spec B \to \Spec A\) along \(\Spec \phi(A) \to \Spec A\), noting that \(B \otimes_A \phi(A) = B\). \(\blacksquare\)

Thus we shall assume \(A \subseteq B\).

Lemma 3. \(A \subseteq B\) is an integral extension iff for all primes \(\newcommand{\fp}{\mathfrak{p}} \fp\) of \(A\), \(A_\fp \subseteq B_\fp\) is an integral extension.

Here \(B_\fp = (A \setminus \fp)^{-1} B\). The proof of this lemma is the same as Proposition 5.13 in Atiyah-Macdonald.

By base changing \(f\) along \(\Spec A_\fp \to \Spec A\), we shall assume \(A\) is a local ring, say with maximal ideal \(\DeclareMathOperator{\fm}{\mathfrak{m}}\fm\) and residue field \(k = A / \fm\). To show that \(B\) is integral over \(A\), it suffices to show that \(A[b]\) is integral over \(A\) for every \(b \in B\), so we would like to pass to the subring \(A[b]\) of \(B\). In Lemma 7 below we prove that we can do this.

Lemma 4. Assume \(A \subseteq B\). Then the morphism \(\Spec B \to \Spec A\) is dominant.

Proof. If \(g \in A\) is such that \(D(g) \neq \emptyset\), then \(g\) is not nilpotent. The preimage in \(\Spec B\) is \(D(g)\), which is non-empty as \(g\) is not nilpotent in \(B\). \(\blacksquare\)

Corollary 5. Assume \(A \subseteq B\), and that the morphism \(\Spec B \to \Spec A\) is closed. Then the morphism \(\Spec B \to \Spec A\) is surjective.

Lemma 6. Let \(f : X \to Y\) and \(g : Y \to Z\) be morphisms of schemes such that \(f\) is surjective and \(g f\) is (universally) closed. Then \(g\) is (universally) closed.

Proof. Assume \(g f\) is closed. Let \(V\) be a closed subset of \(Y\). Then \(g(V) = gf(f^{-1}(V))\) is closed. In the case \(g f\) is universally closed, note that surjectivity is preserved by base change (see Stacks tag 01S1). \(\blacksquare\)

Lemma 7. Assume \(A \subseteq B\), \(b \in B\), and the morphism \(\Spec B \to \Spec A\) is universally closed. Then

1. \(\Spec B \to \Spec A[b]\) is universally closed; and

2. \(\Spec A[b] \to \Spec A\) is universally closed.

Proof. For the first part, consider the base change along \(\Spec A[b] \to \Spec A\). Let \(I\) be the kernel of \(A[x] \to B\), sending \(x\) to \(b\), so that \(A[b] = A[x]/I\). We have a cartesian square

Thus the left vertical map is universally closed. The homomorphism \(B[x]/I B[x] \to B\) sending \(x\) to \(b\) is surjective, so \(\Spec B \to \Spec B[x] / I B[x]\) is a closed embedding. Composing these two morphisms give the conclusion.

For the next part, by Lemma 6, it suffices to show that \(\Spec B \to \Spec A[b]\) is surjective. This follows by Corollary 5 and part 1. \(\blacksquare\)

By part 2 of the above lemma, we may assume that \(B = A[b]\). We reduce to the case \(A\) is a field:

Lemma 8. Let \((A, \fm, k)\) be a local ring. Let \(A \subseteq B\) and suppose that \(B = A[b]\). Then \(B\) is integral over \(A\) iff \(B/\fm B\) is integral over \(k\).

Proof. Note that \(B / \fm B = k[\overline{b}]\), where \(\overline{b}\) is the image of \(b\). Thus it suffices to prove the above with the word integral replaced by finite. The only if part is trivial, the if part follows from Nakayama’s lemma. \(\blacksquare\)

Base changing along \(\Spec k \to \Spec A\), we find that in the above notation, \(\Spec k[\overline{b}] \to \Spec k\) is universally closed. From Lemma 1, we see that \(\overline{b}\) must be integral over \(k\). This completes the proof of the theorem.