Question

About a year ago, a colleague asked me the following question:

Suppose $(R,+,\cdot)$ and $(S,\oplus,\odot)$ are two rings such that $(R,+)$ is isomorphic, as an abelian group, to $(S,\oplus)$, and $(R,\cdot)$ is isomorphic (as a semigroup/monoid) to $(S,\odot)$. Does it follow that $R$ and $S$ are isomorphic as rings?

I gave him the following counterexample: take your favorite field $F$, and let $R=F[x]$ and $S=F[x,y]$, the rings of polynomials in one and two (commuting) variables. They are not isomorphic as rings, yet $(R,+)$ and $(S,+)$ are both isomorphic to the direct sum of countably many copies of $F$, and $(R-\{0\},\cdot)$ and $(S-\{0\},\cdot)$ are both isomorphic to the direct product of $F-\{0\}$ and a direct sum of $\aleph_0|F|$ copies of the free monoid in one letter (and we can add a zero to both and maintain the isomorphism).

He mentioned this example in a colloquium yesterday, which got me to thinking:

Question. Is there a counterexample with $R$ and $S$ finite?

Answer 1

Here are some initial thoughts. Put $$ X(R)=\{e\in R: e^2=e \text{ and } er=re \text{ for all } r\in R\} $$ We can partially order this by declaring that $e\leq f$ iff $ef=e$. We then put $$ Y(R)=\{e\in X(R): 0\lt e \text{ and there is no } f\in X(R) \text{ with } 0 \lt f \lt e \} $$ One can check that $X(R)$ is a finite Boolean algebra under this order (with meet operation $e\wedge f=ef$ and join $e\vee f=e+f-ef$) so it is isomorphic to the lattice of subsets of its set of atoms, which is $Y(R)$. In particular, if $|Y(R)|=n$ then $|X(R)|=2^n$. For $e\in Y(R)$ we put $$ R[e] = Re = \{ x\in R : ex=xe=x\} $$ We can then define $p:R\to\prod_{e\in Y(R)}R[e]$ by $p(x)_e=ex$. It is standard that this is an isomorphism of rings.

Next, by hypothesis we have a bijection $f:R\to S$ that preserves multiplication. It follows that $f$ gives an isomorphism $X(R)\to X(S)$ of posets, and thus a bijection $Y(R)\to Y(S)$. As the sets $R[e]$ and the maps $p_e$ are defined using only the multiplicative structure, we see that $f$ gives an isomorphism $R[e]\to S[f(e)]$ of multiplicative monoids for each $e\in Y(R)$. However, we do not obviously have an additive isomorphism from $R[e]$ to $S[f(e)]$, so this does not succeed in reducing the problem to the indecomposable case.

Nonetheless, it is worth thinking about the ring structure of $R[e]$. The quotient by the Jacobson radical is a finite simple ring and so is a matrix algebra over a finite division ring, but finite division rings are fields by a theorem of Wedderburn, so this quotient is quite tractable.