# Does the Cantor-Schröder-Bernstein Theorem hold in the category opposite to the category of noetherian commutative rings?

I asked this question on Mathematics Stackexchange, but got no answer.

Let $$A$$ and $$B$$ be noetherian commutative rings with one, and let $$f:A\to B$$ and $$g:B\to A$$ be epimorphisms.

Are the rings $$A$$ and $$B$$ necessarily isomorphic?

[In this post "ring" means "commutative ring with one", and morphisms are required to map $$1$$ to $$1$$. 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.]

The busy reader is invited to skip the sequel.

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

The answers to the following variants of the above question are known:

(1) If $$f:A\to B$$ and $$g:B\to A$$ are injective morphisms of noetherian rings, are $$A$$ and $$B$$ necessarily isomorphic? The answer is No, as shown by the following example taken from a comment of Sam Lichtenstein to this question. Let $$K$$ be a field, $$x$$ an indeterminate, $$f:K[x^2,x^3]\to K[x]$$ the inclusion, and $$g:K[x]\to K[x^2,x^3]$$ the (clearly injective) morphism defined by $$g(p(x))=p(x^2)$$. Note that $$K[x^2,x^3]$$ is not isomorphic to $$K[x]$$ because the ideal $$(x^2,x^3)$$ of $$K[x^2,x^3]$$ is not principal.

(2) If $$f:A\to B$$ and $$g:B\to A$$ are surjective morphisms of rings, are $$A$$ and $$B$$ necessarily isomorphic? The answer is No, as shown by the following example taken from the same comment of Sam Lichtenstein. Set $$A:=\mathbb Z/(4)\times\mathbb Z/(4)\times\cdots,\quad B:=\mathbb Z/(2)\times A,$$ let $$f:A\to B$$ be defined by $$f(x_1,x_2,\dots)=(h(x_1),x_2,\dots)$$, where $$h$$ is the unique ring morphism from $$\mathbb Z/(4)$$ to $$\mathbb Z/(2)$$, and let $$g:B\to A$$ be defined by $$g(x_1,x_2,\dots)=(x_2,x_3,\dots)$$. The rings $$A$$ and $$B$$ are not isomorphic because the equations $$2x=0$$ and $$x^2=x$$ have no nonzero simultaneous solutions in $$A$$, and one such solution in $$B$$ (namely $$x=(1,0,\dots)$$).

(3) If $$f:A\to B$$ and $$g:B\to A$$ are surjective morphisms of noetherian rings, are $$A$$ and $$B$$ isomorphic? The answer is Yes, because surjective endomorphisms of noetherian rings are isomorphisms. But epimorphic endomorphisms of noetherian rings are not always isomorphisms: see this answer of Eric Wofsey.

• Do you have an example of a non-surjective epimorphism $A\to B$ of noetherian rings with $A,B$ with the Krull dimension of $A$ $\le$ that of $B$? (Say, with finite Krull dimension— otherwise the Krull dimension should be defined as an ordinal, namely as $\mathrm{Kdim}(A)=\sup(\mathrm{Kdim}(A/P)+1)$ for $A\neq 0$, where $P$ ranges over non-minimal prime ideals and $\sup\emptyset=0$.) – YCor Oct 10 '19 at 7:17
• The embedding $\mathbb{Z} \times \mathbb{Z} \subseteq \mathbb{Z} \times \mathbb{Q}$ is such an example. – Angelo Oct 10 '19 at 7:51
• @YCor - I hope you saw Angelo's comment. (At the end of the question I mentioned Eric Wofsey' example of a non-surjective epimorphic endomorphism of a noetherian ring.) – Pierre-Yves Gaillard Oct 10 '19 at 10:42
• @Pierre-YvesGaillard oh thanks. Indeed Eric's example is immediately adaptable to answer the question. – YCor Oct 10 '19 at 14:03

## 1 Answer

No because you can take $$A = \mathbf{Z}[x, 1/(x - n); n\geq 0]$$ and $$B = \mathbf{Z}[x, 1/x, 1/(x - n); n \geq 2]$$ and the maps are $$B \to A$$ is the inclusion and $$A \to B$$ sends $$x$$ to $$x - 2$$. The reason $$A$$ is not isomorphic to $$B$$ is that the gaps between the missing points'' are different for $$A$$ and $$B$$. More precisely, any isomorphism $$A \to B$$ sends $$x$$ to something of the form $$(a x + b)/(cx + d)$$ with $$a, b, c, d \in \mathbf{Q}$$ and there is no such function which sends the set $$\{0, 2, 3, \ldots\} \cup \{\infty\}$$ bijectively to $$\{0, 1, 2, 3, \ldots\} \cup \{\infty\}$$.

• This example (very similar to the one linked at the end of the question) is also an alternative (Noetherian!) answer to this question. What fascinates me is that you don't use any structure inside the rings to conclude they are not isomorphic, but rather write down what an isomorphism (if any) might look like. (Although I guess in principle it might be possible to rephrase this as some complicated statement inside the rings...) – R. van Dobben de Bruyn Oct 10 '19 at 13:29
• @R.vanDobbendeBruyn it seems that $A$ and $B$ are not elementary equivalent: with a little effort one checks, if I'm correct, that $A$ satisfies ($\forall t$, if $t$ and $t-3$ are invertible then so is $t-1$), but $B$ doesn't. [I need $-3$ because in $A$, $u=x(x-1)$ satisfies $u,u-2$ invertible but not $u-1$.] – YCor Oct 10 '19 at 16:07