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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?

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There might be a counterexample of two 3-nilpotent rings. More specifically, take two non-isomorphic nilpotent of class 3 semigroups S,T of the same size, and consider algebras ${\mathbb F}_2S$ and ${\mathbb F}_2T$. Then these algebras are isomorphic as Abelian (additive) groups but might be non-isomorphic and might have isomorphic multiplicative semigroups. I do not have concrete examples, but I would search in this direction. –  Mark Sapir Sep 2 '11 at 23:45
@Mark: Thank you for the suggestion; I'll give it some thought! –  Arturo Magidin Sep 3 '11 at 4:17
If you assume that your finite rings are algebras over a finite field $F$, then their additive groups are isomorphic as soon as they have the same cardinality since they are then vector spaces of the same dimension. Thus in this case one can ask is it true that two finite dimensional algebras over a finite field are isomorphic iff their multiplicative monoids are isomorphic. Seems hard to believe. –  Benjamin Steinberg Oct 13 '11 at 19:36

1 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.

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Is it correct to say that elements of the set $X(R)$, i.e. the central idempotents, correspond to subsets of the set of connected components of $\mathrm{Spec}(R)$ and elements of $Y(R)$ (positive minimal central idempotents) correspond to individual connected components? –  Qfwfq Sep 2 '11 at 23:24
@Neil: Interesting; I'll think about this as well. Thanks! –  Arturo Magidin Sep 3 '11 at 4:19

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