Let $f \colon R \to S$ be an epimorphism of commutative rings, where $R$ and $S$ are integral domains. Suppose that $\mathfrak{p} \subset S$ is a prime such that $f^{-1}(\mathfrak{p}) = 0$. Does it follow that $\mathfrak{p} = 0$?

If answer is "yes", then it follows that for any epimorphism of commutative rings $R \to S$, strictly increasing chains of prime ideals in $S$ lift to strictly increasing chains in $R$; hence, Krull dimension is non-increasing along ring epimorphisms.

All of the above hold for quotients and localizations, which are the only examples of ring epimorphisms that come readily to my mind.

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    $\begingroup$ @JC Ottern: You can check it directly, or it follows easily from the universal property of localization. @OP: I think you can assume $R$ is a field and $S$ is local, by first tensoring $f$ with the fraction field $K$ of $R$, then passing to the localization of $S\underset{R}{\otimes} K$ at $\mathfrak{p}$. $\endgroup$ Commented Jan 16, 2011 at 1:02
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    $\begingroup$ @Daniel: and, unless I am mistaken, an epimorphism from a field to an integral domain is surjective. So the answer is yes. $\endgroup$ Commented Jan 16, 2011 at 1:18
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    $\begingroup$ Thanks! (To elaborate on George Lowther's point: If $k \to S$ is an epimorphism, then $S \to S \otimes_k S$ is an isomorphism. But the theory of tensor products of vector spaces tells us that this is not so unless $k = S$.) $\endgroup$ Commented Jan 16, 2011 at 1:58
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    $\begingroup$ See this MO question mathoverflow.net/questions/109/… for some examples and theorems about epimorphisms of rings. In particular, not everyone is a composition of quotients and localizations. $\endgroup$ Commented Jan 16, 2011 at 8:00
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    $\begingroup$ By the way, perhaps it should be mentioned in the question for the readers who are used to the standard terminology of "epimorphism" in (commutative) algebra that this does not mean "surjective homomorphism" here, but rather a categorical epimorphism. $\endgroup$ Commented Jan 16, 2011 at 8:04

3 Answers 3


Yes. Letting $k$ be the field of fractions of $R$, we have the following commutative diagram. $$ \begin{array}{ccc} R&\stackrel{f}{\rightarrow}&S\\\\ \downarrow\scriptstyle{}&&\downarrow\scriptstyle{}\\\\ k&\stackrel{g}{\rightarrow}&S_{\mathfrak{p}} \end{array} $$ However, $f$ and the localization $S\to S_{\mathfrak{p}}$ are epimorphisms, so $g$ is an epimorphism with domain a field. This means that it is surjective, so $S_{\mathfrak{p}}$ is a field, and $\mathfrak{p}=0$.

To see that an epimorphism $g\colon k\to A$ of commutative rings with domain a field $k$ is surjective, consider the morphisms $u,v\colon A\to A\otimes_k A$ given by $u(a)=a\otimes1$ and $v(a)=1\otimes a$. Then, $u\circ g=v\circ g$ and, from the definition of epimorphism, $u=v$, in which case $g(k)=A$.

  • $\begingroup$ I was a bit slow here, and see that with Daniel's answer and Charles' comment mentioning the tensor product argument, this answer is a bit redundant now. But its a nice simple argument, and I quite like it, so think its worth leaving up. $\endgroup$ Commented Jan 16, 2011 at 2:17
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    $\begingroup$ +1: I like this argument a lot. $\endgroup$ Commented Jan 16, 2011 at 2:19
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    $\begingroup$ Dear George, I am amazed at the elegance of this argument. Especially since, according to your blog, you seem to be a specialist in stochastic processes! Would it be very rude on my part to beg you to explain this prowess? $\endgroup$ Commented Jan 16, 2011 at 22:46
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    $\begingroup$ @Georges: Many thanks for that! Actually, I'm not in academia at all - I have a PhD, in which I specialized in stochastic processes. Otherwise, I like to read up on and think about a wide variety of mathematical topics. For fun. $\endgroup$ Commented Jan 16, 2011 at 23:11
  • $\begingroup$ Dear George, thank you for satisfying my shameful curiosity. I am more happy that you might think at this answer. It always delights me that people outside academia do mathematics: our field is there not only for professional mathematicians but for all earthlings (and even maybe ...) Anyway, once more: kudos! $\endgroup$ Commented Jan 17, 2011 at 0:06

Let's make several reductions. First, the condition implies $f$ is injective, so letting $K=\operatorname{Frac}(R)$, we have that $f: K\to S'$ is an epimorphism, letting $S'$ be the localization of $S$ at the zero ideal of $R$; it is easy to check that this is an epimorphism via the universal property of localization.

But as George Lowther points out, an epimorphism from a field to an integral domain must be surjective. To see this, assume to the contrary that $K\to S'$ is not surjective; thus $K'=\operatorname{Frac}(S')$ is not equal to $K$. But $S'\to K'$ is an epimorphism, so composing, we've reduced to the case of a map between two fields $f: K\to K'$. But $K'$ admits many embeddings into (say) its algebraic closure which agree on $K$, contradicting that $f$ was an epimorphism.

EDIT: As the commenters point out, the algebraic closure doesn't quite work in the case $K'/K$ is purely inseparable, but this case is not difficult; see George Lowther's answer for an easy general argument.

  • $\begingroup$ One minor point: If $f\colon K\to K^\prime$ is a purely inseparable field extension, then no two distinct morphisms into the algebraic closure (or any reduced ring) can agree on $K$. So, we need to look at non-reduced rings to show that $f$ is surjective. I say a minor point, because its easy to see that $\mathfrak{p}=0$ in this case anyway. $\endgroup$ Commented Jan 16, 2011 at 2:31
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    $\begingroup$ More simply: a homomorphism $K\to R$ from a field to a ring must be surjective if it is an epimorphism, because if the dimension of $R$ as a $K$-vector space is more than one then the two evident ring homomorphisms $R\to R\otimes_K R$ are distinct. $\endgroup$ Commented Jan 16, 2011 at 2:47
  • $\begingroup$ @Tom: Yes, that was my argument (and Charles' in the comments). Actually I had a similar method to Daniel's in mind when I commented that epimorphisms from fields are surjective - but thinking about the inseparable case led to the nicer tensor product argument. $\endgroup$ Commented Jan 16, 2011 at 2:55
  • $\begingroup$ I've added a reference to your answer, George; I don't have time now but I'll come back at some point and add an argument of this type for the inseparable case. $\endgroup$ Commented Jan 16, 2011 at 3:55

First, pick a maximal chain of prime ideals in $S$ and mod out by the minimal one. Now $S$ is an integral domain of the same dimension. Similarly, you might as well assume $f$ is injective, since that can only decrease the Krull dimension of $R$.

So, now, we have a map, which must induce an isomorphism on fraction fields, and both algebras inject into their fraction fields. Now, take an ideal $I\subset S$ such that $I\cap R=0$, and let $s\neq 0$ be an element of $I$. Then $s=r'/r''$ for $r',r''\in R$. Thus $sr''\in R\cap I$, and we have arrived at a contradiction.

  • $\begingroup$ This is an interesting answer; I think Daniel Litt's is a little bit better explained because this one omits to explain why the map must induce an isomorphism of fraction fields. (Although this is not hard to see from the comments or Daniel Litt's answer.) $\endgroup$ Commented Jan 16, 2011 at 2:10

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