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Context: In many usual categories, surjective morphisms and epimorphisms are one and the same. Clearly in Set, but also in Ab, in Grp and FinGrp (although it is not obvious), in A-Mod, in CHaus, etc. But this isn't true in the category of rings since the inclusion map $\mathbb{Z}\to\mathbb{Q}$ is an epimorphism without being surjective. More generally, localization maps $A\to S^{-1}A$ are always epimorphisms but may fail to be surjective if $S$ has a non-unit that is not a zero divisor. For a while epimorphisms of rings seemed intangible to me but there's one result, which we discuss below, that makes it seem very concrete to me.

Let $f:A\to B$ be a morphism of rings. It is true that $f$ is surjective if and only if it is an epimorphism and is finite. A proof can be found in the tag 04VT of the Stacks Project.

There, Johan de Jong says that this result "seems to have been reproved many times in the literature, and has many different proofs". Nevertheless, searching the usual commutative algebra books I didn't found any proofs of this result. I would like to know where I can find other proofs of this result.

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    $\begingroup$ Actually, $A \to S^{-1} A$ fails to be surjective as soon as $S$ has a non-unit that is not a zero divisor, which is the case that people usually care about. $\endgroup$
    – Zhen Lin
    Oct 1 '20 at 15:04
  • $\begingroup$ @ZhenLin fixed. $\endgroup$
    – Gabriel
    Oct 1 '20 at 15:33
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I can't remember where to look in the literature but here's a proof that comes to mind. As in the proof on stacks, the problem boils down to showing that $S/R$, a finite $R$-module, vanishes. For me that has Nakayama's lemma written all over it, so let's go that route.

(1) Use the fact that being finite, being surjective, and being epic are all local properties to reduce to the case that $R$ is local with maximal ideal $\mathfrak{m}$.

(2) If $R \rightarrow S$ is epic and finite, then factoring $R \rightarrow S$ as $R \twoheadrightarrow R' \subseteq S$ yields $R' \subseteq S$ epic and finite, so further reduce to the case that $R \subseteq S$.

(3) Because a finite morphism has lying over, we know $\mathfrak{m}S \not= S$. Deduce $\mathfrak{m}S \cap R = \mathfrak{m}$. Thus we get an epimorphism $R/\mathfrak{m} \subseteq S/\mathfrak{m}S$, which is a surjection because epis of fields are surjective (*). This implies $R/\mathfrak{m} \cong S/\mathfrak{m}S$, and further $R/\mathfrak{m} \otimes S/R = 0$.

(4) Tensor $0 \rightarrow \mathfrak{m} \rightarrow R \rightarrow R/\mathfrak{m} \rightarrow 0$ by $S/R$ to get that $\mathfrak{m} \otimes S/R \rightarrow S/R \rightarrow 0$ is exact, hence $\mathfrak{m} (S/R) = (S/R)$.

(5) Use that $S/R$ is finite and apply Nakayama's lemma to get that $S/R = 0$.

(*) This is Stacks 04VV. It can be argued in many ways. One way I like is using the zig-zag characerization of dominions. If $k$ is a field and $k \subseteq R$ is epic, then for any $r \in R$ we get a zig-zag representation $r = XMY$ where $X,Y$ are vectors with entries in $R$, $M, XM, MY$ have entries in $k$. Picking $P, Q$ invertible matrices such that $PMQ$ is diagonal, replace $X$ by $XP^{-1}$, $M$ by $PMQ$, $Y$ by $Q^{-1}Y$. Thus we can assume $r = \sum x_i m_i y_i$ where $x_i m_i \in k, m_iy_i \in k$. But since $m_i$ is invertible in $k$, we see that $x_i , y_i \in k$, hence $r \in k$.

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  • $\begingroup$ Beautiful proof! I surely agree that this has Nakayama's lemma written all over it. I'll leave the question open for a while because I would love to see other proofs but I'll accept yours soon enough! $\endgroup$
    – Gabriel
    Oct 3 '20 at 9:10
  • $\begingroup$ @Gabriel sorry you didn't get any other proofs. I think I remember seeing a cute category-theoretic proof once. maybe if you want to see it done from a particular angle you could reask and tag the question more specifically $\endgroup$ Oct 5 '20 at 14:20
  • $\begingroup$ I would love to see it. Do you recall where it was? $\endgroup$
    – Gabriel
    Oct 7 '20 at 8:34

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