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).

A few links:

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.

Solution 1:

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 $x$).