TITLE: Schroeder-Bernstein for Rings
QUESTION [8 upvotes]: Suppose $f: A \to B$ and $g: B \to A$ are injections of rings
(commutative with identity).  Must $A$ and $B$ be isomorphic as
rings?
According to this
question, this answer should be "no", but can someone give an example?
Thanks!

REPLY [2 votes]: Sam Lichtenstein poses the dual question in comments:  

What's a counterexample to "dual Schroeder-Bernstein" for rings? (That is, same question but with surjections rather than injections.) Is there one with A,B finite type over a field? 

That is,

Do there exist finite type $k$-algebras $A, B$ not isomorphic to each other, and surjections $A\to B, B\to A$? (*)

He gives an example in the non-Noetherian case; I claim the "dual Schroeder-Bernstein theorem" is true if $A$ and $B$ are Noetherian.  And in general, if two Noetherian schemes $X, Y$ admit maps $i: X\to Y, j: Y\to X$ exhibiting each as a closed subscheme of the other, then $i, j$ are isomorphisms.   So the answer to (*) is "no".
I'll prove the more general claim.  Assume to the contrary that one of $i,j$ is not an isomorphism.  Then $j\circ i: X\to X$ exhibits $X$ as a proper closed subscheme of itself, say $X_1$.  But then $X_1$ is isomorphic to some proper closed subscheme of itself, say $X_2$; continuing in this manner, we may construct a sequence $X_n$ where each $X_i$ is a proper closed subscheme of $X_{i-1}$.  Let $\mathcal{I}_n$ be the ideal sheaf of $X_i$ in $\mathcal{O}_X$.  By Noetherianness we must have that $\mathcal{I}_1\subset \mathcal{I}_2\subset \mathcal{I}_3\subset\cdots$ stabilizes, however, which contradicts the claim that each $X_i\subset X_{i-1}$ is a proper inclusion.  Here's a more formal write-up of the affine case.
This provides an example of a "surjunctive" category in the sense of John Goodnick's answer to this question.