# Are epimorphic endomorphisms of noetherian commutative rings always injective?

[In this post "ring" means "commutative ring with one".]

Let $$A$$ be a noetherian ring, and let $$f:A\to A$$ be an endomorphism which is also an epimorphism.

Is $$f$$ necessarily injective?

Eric Wofsey gave an example of an epimorphic endomorphism of a noetherian ring which is not surjective. (It is well known, and easy to prove, that surjective endomorphisms of noetherian rings are automatically bijective.)

[By definition, a morphism of rings $$f:A\to B$$ is an epimorphism if for all pairs of morphisms $$(g,h):B\rightrightarrows C$$ the equality $$g\circ f=h\circ f$$ implies $$g=h$$. Surjective morphisms are epimorphic, but the converse does not always hold: for instance the inclusion $$\mathbb Z\to\mathbb Q$$ is an epimorphism.]

For more details about epimorphisms see

$$\bullet$$ MathOverflow thread What do epimorphisms of (commutative) rings look like?.

$$\bullet$$ Stacks Project Section Epimorphisms of rings.

$$\bullet$$ Samuel Seminar .

This is false. I will use freely that $$f \colon A \to B$$ is an epimorphism if and only if $$B \otimes_A B \to B$$ is an isomorphism; in particular this is also equivalent to $$\operatorname{Spec} B \to \operatorname{Spec} A$$ being a monomorphism.

Example. Let $$k$$ be a field of characteristic $$0$$, and let $$R = k\left[x,\tfrac{1}{x-1},\tfrac{1}{x-2},\ldots\right].$$ Then $$R$$ has a self-map $$x \mapsto x-1$$, whose image on spectra is the complement $$U$$ of the origin. Note that $$U \amalg \{0\} \to \operatorname{Spec} R$$ is a monomorphism, so the map \begin{align*} g \colon R &\to R \times k\\ x &\mapsto (x-1,0) \end{align*} is an epimorphism. Now let $$A = R \times k$$, and let $$f$$ be the composition $$A \stackrel\pi\twoheadrightarrow R \stackrel g\to A$$, where $$\pi$$ is the natural projection. It is an epimorphism since both $$\pi$$ and $$g$$ are, and it is clearly not injective. $$\square$$

Picture. Here is a picture of the situation:

Remark. However, the result is true if $$A$$ has a unique associated prime $$\mathfrak p$$ (i.e. $$A$$ is irreducible without embedded primes). Indeed, we get an ascending chain $$\mathfrak p \subseteq f^{-1}(\mathfrak p) \subseteq f^{-1}(f^{-1}(\mathfrak p)) \subseteq \ldots,$$ which stabilises by the Noetherian hypothesis. But if $$\operatorname{Spec} A \to \operatorname{Spec} A$$ is a monomorphism, in particular it is injective, so this forces $$\mathfrak p = f^{-1}(\mathfrak p)$$. Then the base change $$A_{\mathfrak p} \to (f_*A)_{\mathfrak p}$$ of $$f \colon A \to f_*A$$ along $$A \to A_{\mathfrak p}$$ is an epimorphism as well, hence so is the composition $$A_{\mathfrak p} \to (f_*A)_{\mathfrak p} \to A_{\mathfrak p}$$ since it is a further localisation. But this is an epimorphism of Artinian rings, hence surjective by this post. Since both rings are Artinian of the same length, we conclude that it's an isomorphism. Since $$\mathfrak p$$ is the unique associated prime, the map $$A \to A_{\mathfrak p}$$ is injective, which gives the result.

I'm not sure what happens if $$A$$ is allowed to have embedded primes. It seems hard to imagine what a counterexample could look like, but I also have no argument to rule it out.

Remark. Note also that there are trivial counterexamples without the Noetherian hypothesis, even if $$A$$ is a domain. For example, $$A = k[x_0,x_1,\ldots]$$ has a surjective endomorphism that is not injective.

• Thanks! I'm confused. You write: "$f:A\to B$ is an epimorphism if and only if $B\otimes_AB\to B$ is an isomorphism; in particular this is also equivalent to $\operatorname{Spec}B\to\operatorname{Spec} A$ being a monomorphism". I don't understand the second phrase: if $A$ and $B$ are fields and $f$ is not surjective, then $f$ is not an epimorphism but $\operatorname{Spec}B\to\operatorname{Spec} A$ is a monomorphism, right? – Pierre-Yves Gaillard Oct 25 '19 at 9:12
• @Pierre-YvesGaillard: sorry, I mean in the category of schemes, not topological spaces. – R. van Dobben de Bruyn Oct 25 '19 at 11:51
• That is, the diagonal $\operatorname{Spec} B \to \operatorname{Spec} B \times_{\operatorname{Spec} A} \operatorname{Spec} B$ is an isomorphism if and only if the underlying ring map $B \otimes_A B \to B$ is. – R. van Dobben de Bruyn Oct 25 '19 at 14:24
• I'm the one who's sorry! I'm most grateful for your awesome answer and your comments! I convinced myself that $R\to R\times k$ is an epi by checking the identity $(p(x),\lambda)\otimes(1,1)=(1,1)\otimes(p(x),\lambda)$ in $(R\times k)\otimes_R(R\times k)$. I hope this argument is correct. Even if it is, I'm sure your approach is more elegant, but I'm not familiar with schemes (to say the least!). I would guess that scheme theory helped you find your example more that it was needed for your proof, but I may be completely wrong... – Pierre-Yves Gaillard Oct 25 '19 at 14:40
• @Pierre-YvesGaillard: indeed everything can be checked on the ring level. But I think the schemes picture is much more enlightening, and I have always been unable to think about epimorphisms of rings. – R. van Dobben de Bruyn Oct 25 '19 at 20:58