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.

  • $\begingroup$ Fixed LaTeX $\endgroup$ Commented Mar 25, 2010 at 6:36
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    $\begingroup$ As I guess you know, James, it's equivalent to prove it in the case that f is the inclusion of A as a subring of B. For suppose we know this case. Take an arbitrary integral epi f: A --> B. Factorize f as a surjection s: A --> C followed by an inclusion i: C --> B. Then i is integral epi, hence surjective, so f = si is surjective. $\endgroup$ Commented Mar 25, 2010 at 12:38
  • $\begingroup$ Sorry, I misunderstood the question and make a previous wrong comment. $\endgroup$ Commented Mar 25, 2010 at 12:49
  • $\begingroup$ By the way, is this the usual notion of epimorphism (from a categorical point of view)? If so, you may want to answer the question mathoverflow.net/questions/109/… $\endgroup$ Commented Mar 25, 2010 at 12:54
  • $\begingroup$ @Tom: Yes, indeed I did know that, but I should have mentioned it! It's also enough to assume that $A$ is a local ring. $\endgroup$
    – JBorger
    Commented Mar 25, 2010 at 12:59

3 Answers 3


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.


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"?

  • $\begingroup$ Thanks for the remark that it's true with a noetherian base. That's good to know. Regarding your postscript, assuming it is correct (I haven't checked it is correct, but have no reason to doubt it), then yes, it appears you can apply it to the case where $f$ is the ghost component map to get what I wanted. I can't remember what I was thinking when I wrote that part of the paper (BGWV II, on my web page, prop 16.8). It's clear I was using the surjectivity and integrality of the ghost map to descend the separatedness. $\endgroup$
    – JBorger
    Commented Feb 10, 2011 at 10:59
  • $\begingroup$ I can't remember, though, whether I considered the general statement that you gave and failed to find an argument, or whether I just used the first argument that came to mind. It is a rather unfortunate argument, in that it uses the Conrad-Lieblich-Olsson result that any qcqs algebraic space is affine over a separated algebraic space of finite type. $\endgroup$
    – JBorger
    Commented Feb 10, 2011 at 11:00
  • $\begingroup$ Yes, your prop 16.8 is way overkill. $b\circ a=d\circ c$ is integral and in particular universally closed. As $c$ is surjective, we have that $d$ is universally closed. A monomorphism that is locally of finite type and universally closed (such as $d$) is a closed immersion. In fact, it follows (by purely topological arguments) that it is of finite type and thus proper and a proper monomorphism is a closed immersion by EGA IV 18.12.6. (I realize that I more or less repeated my argument. Sorry!) Another argument: $d$ is affine and universally closed, hence integral, hence a closed immersion. $\endgroup$
    – David Rydh
    Commented Mar 7, 2011 at 20:49

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.

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    $\begingroup$ The hypotheses force $Q$ to be unique over $P$ with the same residue field (i.e., if $A$ is a field then one can conclude by using fpqc descent theory), so passing to the local case doesn't ruin integrality. But in such cases $L = K$. So the proposed argument doesn't work. $\endgroup$
    – BCnrd
    Commented Mar 25, 2010 at 14:27

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