I asked [this question](https://math.stackexchange.com/q/3376675/660) 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. 



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> Are the rings $A$ and $B$ necessarily isomorphic?
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[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.]




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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?](https://mathoverflow.net/q/109/461).


$\bullet$ Stacks Project Section [Epimorphisms of rings](https://stacks.math.columbia.edu/tag/04VM).


$\bullet$ [Samuel Seminar](http://www.numdam.org/issues/SAC_1967-1968__2_/) . 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](https://mathoverflow.net/q/39460/461). 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](https://math.stackexchange.com/a/3376063/660).



Does the Cantor-Schröder-Bernstein Theorem hold in the category opposite to the category of noetherian commutative rings?
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 $\{0, 1, 2, 3, \ldots\} \cup \{\infty\}$.


