Schroeder-Bernstein for Rings 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!
 A: Here is another counterexample for fields: If $K=\overline{\mathbb{Q}(x_1,x_2,...)}$, then there are monomorphisms $K(x_0) \to K \to K(x_0)$, but no isomorphism since $K(x_0)$ is not algebraically closed.
A: There are many pairs of as-nice-as-possible compact topological spaces $X,Y$ with continuous surjections $X \to Y$ and $Y \to X$ but no homeomorphism.  For example, let $X$ be a closed interval and $Y$ a circle.  Then you get injections on algebras of functions: $\mathcal C(X) \hookrightarrow \mathcal C(Y)$ and $\mathcal C(Y) \hookrightarrow \mathcal C(X)$.  For sufficiently nice spaces, Gelfand-Naimark, for example, says that the functor $\mathcal C$ that takes a space to its $*$-algebra of continuous $\mathbb C$-valued functions is a full and faithful contravariant functor to commutative algebras, and in particular a complete invariant, so in particular the two rings are not isomorphic.

Edit:  There are complaints in the comments, and I didn't think very carefully before writing down all this.  This has something to do with the fact that I tend to conflate the words "algebra" and "ring".
So let me switch meanings, and denote by $\mathcal C(X)$ the continuous $\mathbb R$-valued functions on $X$.  Suppose that $X$ is Hausdorff and compact (and if that's not good enough, let's just go all the way to being a manifold with corners, where then everything absolutely works).  Since $\mathbb R$ has no ring homomorphisms, the points of $X$ are precisely the same as ring homomorphism $\mathcal C(X) \to \mathbb R$.  Actually, this is true for $X$ not compact provided it is regular and not too large: it suffices for there to be a function $f \in \mathcal C(X)$ so that every level set is finite.  Anyway, then any ring map $\mathcal C(X) \to \mathcal C(Y)$ automatically induces a set map $Y \to X$.  But also the closed sets are precisely the vanishing sets of functions, i.e. a subset $S\subseteq \operatorname{Hom}(\mathcal C(X),\mathbb R) = X$ is closed iff there is $f\in \mathcal C(X)$ so that $s\in S$ iff $s(f) = 0$.  Anyway, the point is, pick a closed subset of $\mathcal C(X)$, pick a function $f$ determining it, look at the image of $f$ under the map, and its vanishing set in $Y$ is precisely the preimage of the closed subset under the map.  So every ring homomorphism determines a continuous map.  Since a continuous map is determined pointwise, we have the full-and-faithful functor that I wanted.
Note that for manifolds (with corners if you want) you can play the same game with $\mathcal C$ meaning "smooth real-valued functions".
A: This is not even true for fields. Let $E_1$ and $E_2$ be isogenous but not
isomorphic elliptic curves over $K=\mathbb{Q}$ or $k=\mathbb{F}_p$
for some prime $p$. Then the isogeny $E_1\to E_2$ and its dual isogeny
$E_2\to E_1$ induce field injections $k(E_2)\to k(E_1)$ and
$k(E_1)\to k(E_2)$. But $k(E_1)$ and $k(E_2)$ are not isomorphic;
a putative isomorphism must extend the identity on $k$ and it
would induce an isomorphism between the elliptic curves $E_1$ and $E_2$.
A: Hey Damien, I think the following should work $\mathbb{C}$ and $\mathbb{C}(x)$. There is only one uncountable algebraically closed field of each cardinality in characteristic 0 and the algebraic closure of the right hand guy should have cardinality the continuum it should be isomorphic to $\mathbb{C}$. Probably, this assumes the axiom of choice though.
A: 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. 
