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Let $A$ be an artinian ring and $B$ be an $A$-algebra such that $B \otimes_A B \to B$ is an isomorphism, i.e. that $A \to B$ is an epimorphism in the category of commutative rings, see the Seminar Les épimorphismes d'anneaux for a detailed account. In Exposé 3, page 10, it is claimed that in this situation $A \to B$ is surjective. But the proof is only sketched and I don't understand it. Can someone fill in the details, or even better give a different proof? Remark that we may assume that $A$ is local artinian.

Martin Brandenburg
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2 Answers2

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I won't be surprised if you already know this, but here is a proof for $A$ noetherian and/or $B$ finite over $A$:

0) We can replace $A$ with its image in $B$ and assume $A\rightarrow B$ is injective.

1) The result is clear if $A$ is a field.

2) The isomorphism $B\otimes_AB\rightarrow B$ descends to an isomorphism $\overline{B}\otimes_{\overline{A}}\overline{B}$, where the overbar means reduction mod the maximal ideal. From this and 1) we have surjectivity of $\overline{A}\rightarrow\overline{B}$.

3) Now if $A$ is noetherian we are done: By 2), $B=A+MB$ where $M$ is the maximal ideal in $A$; from this we have $B=A+M^nB$ for all $n$; but if $A$ is noetherian then $M$ is nilpotent so $B=A$.

4) Alternatively, if $B$ is finite as an $A$-module, then the cokernel of $A\rightarrow B$ vanishes mod $M$ and hence vanishes by Nakayama.

Steven Landsburg
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  • "if $A$ is noetherian then $M$ is nilpotent" here you use that $A$ is artinian, right? – Martin Brandenburg Jul 07 '12 at 01:33
  • Martin: Yes, I used that $A$ is both artinian and noetherian. – Steven Landsburg Jul 07 '12 at 02:11
  • Martin: But of course I should have remembered that artinian implies noetherian, so "if $A$ is noetherian" is unnecessary here. Which means, I think, that this is a complete proof. – Steven Landsburg Jul 07 '12 at 02:26
  • @Steven: I understand your proof if $R$ is local Artinian, but I think it's false/incomplete if $R$ isn't local (and neither the question nor your answer makes this assumption). The problem is that you suppose $M$ is nilpotent, but this isn't true in general: Take a direct product of fields $R=F\times F$. Then $F\times 0$ is a maximal ideal of $R$ that isn't nilpotent! – tj_ Mar 18 '13 at 19:58
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    TJ: My answer does implicitly assume that $A$ is local when it mentions the maximal ideal. I should have stated this upfront. But I think this is harmless, because to check that $A\rightarrow B$ is surjective, it's enough to check locally on $A$, and localizing preserves the hypotheses. – Steven Landsburg Mar 18 '13 at 22:24
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For completeness and so that I find something useful the next time I google for this:

This is not true if the algebras are not commutative.

A simple example is the inclusion $T\to M_2(k)$ of the algebra $T$ of upper triangular $2\times 2$ matrices into the full matrix algebra — that it is epic can be proven by hand.

A general nonsense reason for this map to be epic is that it is a Cohn localization. Indeed, and more generally, if $Q$ is an acyclic quiver, the natural map from the path algebra $kQ$ to the Leavitt path algebra $L(Q)$ on $Q$ is a Cohn localization (one forces certain maps between certain projectives of $kQ$ to become isomorphisms) If $Q$ is of type $A_2$, then $kQ$ is $T$ and $L(Q)$ is $M_2(k)$.

Mariano Suárez-Álvarez
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