# Is an epimorphic endomorphism of a noetherian commutative ring necessarily an isomorphism?

Let $$A$$ be noetherian commutative ring with one, and let $$f:A\to A$$ be an epimorphic endomorphism of $$A$$.

Is $$f$$ necessarily an isomorphism?

("An epimorphic endomorphism" means of course "an endomorphism which is an epimorphism". In this post "ring" means "commutative ring with one", and morphisms are required to map $$1$$ to $$1$$.)

A few reminders:

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 hold: for instance the inclusion $$\mathbb Z\to\mathbb Q$$ is an epimorphism.

If $$A$$ is noetherian and $$f:A\to A$$ is a surjective endomorphism, then $$f$$ is an isomorphism, because if $$f$$ were not injective, then the kernels of the iterated endomorphisms $$f^n$$ would form an ascending chain of ideals of $$A$$.

If $$A$$ is a nonzero ring, then the map $$(a_1,a_2,\dots)\to(a_2,a_3,\dots)$$ is a surjective endomorphism of $$B:=A\times A\times\cdots$$ which is not an isomorphism (but of course $$B$$ is not noetherian).

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

Stacks Project Section Epimorphisms of rings.

Samuel Seminar . See in particular Section 2 of Exposé Number 7 by Daniel Ferrand.

No. For instance, let $$A=\mathbb{Z}[x,x^{-1},(x+1)^{-1},(x+2)^{-1},\dots]$$. Then $$A$$ is Noetherian, since it is a localization of $$\mathbb{Z}[x]$$. Now consider the homomorphism $$f:A\to A$$ which maps $$x$$ to $$x+1$$. The image of $$f$$ is $$B=\mathbb{Z}[x,(x+1)^{-1},(x+2)^{-1},\dots]$$ and in particular $$f$$ is not surjective since the image does not contain $$x^{-1}$$. But $$f$$ is epimorphic, since $$A$$ is a localization of $$B$$ (just invert $$x$$).