TITLE: Is every integral epimorphism of commutative rings surjective?
QUESTION [20 upvotes]: 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.

REPLY [7 votes]: 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"?