# 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!

• Oops, I posted and deleted an answer because I misread the question. Missed "commutative"! – Jonas Meyer Sep 21 '10 at 3:39
• Consider $k[x^2,x^3]\subset k[x]$. There is an injection $k[x]\hookrightarrow k[x^2,x^3]$ given by $x\mapsto x^2$, for example. These rings are not isomorphic because one is normal and the other isn't -- indeed, $k[x]$ is the normalization of $k[x^2,x^3]$. – Sam Lichtenstein Sep 21 '10 at 3:44
• In the same direction as Sam Lichtenstein's answer, see mathoverflow.net/questions/23478/#25231 – Theo Johnson-Freyd Sep 21 '10 at 4:18
• There are some examples with fields discussed at the following blog post: sbseminar.wordpress.com/2007/10/30/… – S. Carnahan Sep 21 '10 at 8:57
• Well for irreducible affine varieties, if two contain one another as closed subvarieties then they coincide for dimension reasons. I'm wondering whether in the presence of non-reducedness weird things can happen. For an infinite example, take $\mathbf{Z}/2\times \prod_{i=1}^\infty \mathbf{Z}/4$ and $\prod_{i=1}^\infty \mathbf{Z}/4$. There are obvious surjections both ways. (Project away the $\mathbf{Z}/2$ from left to right, mod out by $2$ in the first factor from right to left.) – Sam Lichtenstein Sep 21 '10 at 20:30

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.

• In your second sentence, you should replace "cardinality" with "transcendence degree" to cover the countable cases. Alternatively, you may choose to restrict consideration to uncountable fields. – S. Carnahan Sep 21 '10 at 8:54
• Good point, i was thinking about uncountable fields. – Daniel Pomerleano Sep 21 '10 at 17:30

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.

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 $E_2$.

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".

• Isn't it possible that the rings are isomorphic, but not $*$-isomorphic? – Martin Brandenburg Sep 21 '10 at 8:11
• For example, every $*$-homomorphism between $C^*$-algebras is norm-decreasing and thus continuous, but this is not true for ring homomorphisms. Thus $*$ is a rather strong condition. – Martin Brandenburg Sep 21 '10 at 8:21
• Martin, what's the example of a discontinuous ring HM? At least for X and Y compact Hff, I think that C-algebra isomorphism of C(X) and C(Y) implies that X and Y are homeomorphic (and in fact the homeomorphism induces the original given isomorphism of C-algebras). – Yemon Choi Sep 21 '10 at 9:13
• ... My reasoning is that a unit-preserving C-algebra HM from C(X) to C(Y) cannot increase spectral radius, hence is continuous (because norm=spectral radius in a uniform algebra), hence (Gelfand duality) induces a continuous map from Y to X; so by general functoriality, any C-algebra isomorphism between C(X) and C(Y) is going to induce homeomorphism between Y and X. But I could have made a stupid error somewhere during this argument... – Yemon Choi Sep 21 '10 at 9:15
• @Martin: Don't apologize for nagging --- that's what keeps MO honest (and in particular, I was thinking about $\mathbb R$ when I wrote the post, but figured "I'll just cite a big theorem rather than explain what I'm thinking", and got it wrong). In any case, I think one of the biggest arguments for participating in MO is that just the act of writing up questions and answers helps users (certainly helps me) to clarify our thoughts, and receiving feedback helps even more. – Theo Johnson-Freyd Sep 22 '10 at 19:11

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.