Is every integral epimorphism of commutative rings surjective?

Question

That's the question. Recall that a morphism $f\colon A\to B$ of commutative rings is integral if every element in $B$ is the root of a monic polynomial with coefficients in the image of $A$ and that $f$ is an epimorphism if and only if the multiplication map $$B\otimes_A B\to B$$ is an isomorphism.

If we make the additional assumption that $B$ is finitely generated as an $A$-algebra, then it is true. This can be proven by Nakayama's lemma, for example.

This came up not so long ago when I was trying to show that the Witt vector functor (of finite length) preserves separatedness of algebraic spaces. In this application I was able to reduce things to the finitely generated case and could therefore use the weaker statement above, but I still wonder about the general case.

Answer 1

If I'm not mistaken, there is a counter-example. Have a look at Lazard's second counter-example in: "Deux mechants contre-exemples" in Séminaire Samuel, Algèbre commutative, 2, 1967-1968.

For any field $k$, Lazard provides a non-surjective epimorphism of local $k$-algebras $C\to D$, both of Krull dimension zero, and both of residue field equal to $k$. It is then easy to show that $D$ is also integral over $C$, which is what we need here. Indeed, every $d\in D$ can be written as $d=a+b$ with $a\in k$ and $b$ in the maximal ideal (and unique prime) of $D$, which is therefore nilpotent $b^n=0$, hence trivially integral. Since $a\in k$ is also in $C$, our $d$ is the sum of two integral elements. (Or simply, $D$ is a $k$-algebra, hence a $C$-algebra, generated by nilpotent, hence integral, elements.)

In cash, for those who don't want to click, the rings are constructed as follows: Consider the local ring in countably many pairs of variables $S=(k[X_i,Y_i]_{i\geq 0})_M$ localized at $M=\langle X_i,Y_i\rangle_{i\geq0}$. For every $i\geq 0$ choose an integer $p(i) > 2^{i-1}$. Define $J=\langle Y_i-X_{i+1} Y_{i+1}^2 \ ,\ X_i^{p(i)}\rangle_{i\geq0}\subset S$ and define $D=S/J$. Note immediately that $D$ is a local $k$-algebra, say with maximal ideal $m$ and with residue field $D/m\cong S/M\cong k$. Finally, he defines $C$ to be the localization (at $C_0\cap m$) of the subalgebra $C_0:=k[x_i,x_iy_i]_{i\geq 0}\subset D$ where the $x_i$ are the classes of the $X_i$ in $D$ and I let you guess what the $y_i$ are. By construction, the residue field of $C$ is an extension of $k$ which is also a subfield of $D/m=k$, so the residue field of $C$ must be $k$ and we are in the announced situation.

Answer 2

If $A$ is noetherian, then every integral epimorphism $f \colon A \to B$ is surjective. This has been proven by Ferrand: Prop. 3.8 in Monomorphismes de schémas noethérien", Exp. 7 in Séminaire Samuel, Algèbre commutative, 2, 1967-1968. For schemes, this is phrased as:

Theorem: Let $f \colon X \to Y$ be a morphism of schemes such that $Y$ is locally noetherian. Then $f$ is a closed immersion if and only if $f$ is a universally closed monomorphism.

(It can be seen any universally closed injective morphism of schemes is affine and integral [EGA IV, 18.12.10], so the scheme version is equivalent to the affine version.)

In Ferrand's theorem, one can also replace $X$ and $Y$ with algebraic spaces: In fact, the question is local on $Y$ so we can assume that $Y$ is a scheme. Since $f$ has affine fibers and is universally closed, it follows that $f$ is affine by arXiv:0904.0227 Thm 8.5.

PS. How come that you needed this question for non-finite morphisms for your application to Witt vectors? If the diagonal of an algebraic space is a universally closed monomorphism then (since the diagonal of an algebraic space always is locally of finite type), it is a proper monomorphism, hence a closed immersion ([EGA IV 18.12.6]). In particular, it follows that if $f \colon X \to Y$ is universally closed and surjective and $X$ is separated, then so is $Y$. Perhaps this was what you meant by "reduce things to the finitely generated case"?

Answer 3

First assume $A \subset B$, since the notions of surjectivity, epimorphism and integral morphism all depend only on the image of $A$ in $B$.

Then we can assume that $A$ and $B$ are local and that the inclusion is a local homomorphism. Indeed the inclusion will be surjective if and only if $A_P \subset B_P$ is surjective for all primes $P \subset A$. We can find a prime $Q$ of $B$ such that $Q \cap A = P$ since the extension is integral; so if we show that for all such pairs $A_P \subset B_Q$ is surjective we are done. The inclusion $A_P \subset B_Q$ is easily seen to be still and epimorphism.

Noww assume $A$ and $B$ are local and the inclusion is a local homomorphism; assume $A$ is not all of $B$.

EDIT: The following is wrong (see Brian's comment).

Let $K$ and $L$ be the residue field, $M \supset L$ any field having an automorphism $\sigma$ over $K$ which is nontrivial over $L$. Let $g \colon B \to M$ the composition $B \to L \to M$.

Then $\sigma \circ g$ and $g$ are two different homomorphisms $B \to M$ which agree on $A$, contradicting the hypothesis that $A \subset B$ is an epimorphism.