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 $(R,\cdot)$ (R-\{0\},\cdot)$ and $(S,\cdot)$ (S-\{0\},\cdot)$ are both isomorphic to the direct product of $F-{0}$ 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?
