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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    $\begingroup$ 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. $\endgroup$
    – user6976
    Sep 2, 2011 at 23:45
  • $\begingroup$ @Mark: Thank you for the suggestion; I'll give it some thought! $\endgroup$ Sep 3, 2011 at 4:17
  • 5
    $\begingroup$ 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. $\endgroup$ Oct 13, 2011 at 19:36

2 Answers 2


There do exist pairs of finite unital rings whose additive structures are isomorphic and whose multiplicative structures are isomorphic, yet the rings themselves are not isomorphic.

To see this, let $\mathbb F$ be a field and let $X = \{x_1,\ldots, x_n\}$ be a set of variables. The polynomial ring $\mathbb F[X]$ is graded by degree $$ \mathbb F[X] = H_0\oplus H_1\oplus H_2\oplus\cdots. $$ Let $Q(x_1,\ldots,x_n)$ be a quadratic form over $\mathbb F$. Let $$I = \mathbb F\cdot Q(X)\oplus H_3\oplus H_4\oplus\cdots$$ be the ideal generated by $Q(X)$ and the homogeneous components of degree at least $3$. Let $S_{\mathbb F,Q}$ denote the $\mathbb F$-algebra $\mathbb F[X]/I$. It is a commutative, local ring, which encodes properties of the quadratic form $Q$.

Two quadratic forms $Q_1$ and $Q_2$ are equivalent if they differ by an invertible linear change of variables.

Claim. Let $\mathbb F$ be a finite field of odd characteristic $p$. Let $Q_1(x_1,\ldots,x_n)$ and $Q_2(x_1,\ldots,x_n)$ be nonzero quadratic forms over $\mathbb F$.

  1. $S_{\mathbb F,Q_1}$ and $S_{\mathbb F,Q_2}$ have isomorphic $\mathbb F$-space structures.

  2. If $n>4$ and $Q_1$ and $Q_2$ are nondegenerate, then $S_{\mathbb F,Q_1}$ and $S_{\mathbb F,Q_2}$ have isomorphic multiplicative monoids.

  3. $S_{\mathbb F,Q_1}\not\cong S_{\mathbb F,Q_2}$ as $\mathbb F$-algebras, unless $Q_1$ is equivalent to a nonzero scalar multiple of $Q_2$.

Proof. Exercise! \\

So let $\mathbb F = \mathbb F_3$ be the $3$-element field. It is known that over a finite field of odd characteristic the quadratic forms are classified by the dimension and by the determinant of the form modulo squares. The determinant of $$ Q(x_1,\ldots,x_n)=a_1x_1^2+a_2x_2^2+\cdots+a_nx_n^2 $$ is $a_1\cdots a_n$. If $\alpha\in \mathbb F_3^{\times}=\{\pm 1\}$, then $\alpha\cdot Q$ has determinant $\alpha^n a_1\cdots a_n=(\pm 1)^n a_1\cdots a_n$. If $n$ is even, then the determinants of $Q$ and $\alpha\cdot Q$ will be equal, so $Q$ will be equivalent to $\alpha\cdot Q$ for every $\alpha\in \mathbb F_3^{\times}$. This implies that, when working over $\mathbb F_3$ in an even dimension, if $Q_1$ is not equivalent to $Q_2$, $Q_1$ will also not be equivalent to any nonzero scalar multiple of $Q_2$. In particular, no scalar multiple of $$ Q_1 = x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2, $$ is equivalent to $$ Q_2 = x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 - x_6^2 $$ over $\mathbb F_3$. For these forms we have that $S_{\mathbb F,Q_1}$ and $S_{\mathbb F,Q_2}$ are nonisomorphic finite unital rings with isomorphic additive and multiplicative structures. (These rings have size $3^{27}$.)

Minor side comment 1: If you allow nonunital rings, there is a pair of nonisomorphic $8$-element rings whose additive and multiplicative structures are isomorphic.

Minor side comment 2: The solution to the exercise above (that is, the proof of the Claim) can be found here.

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    $\begingroup$ You are just going around answering old long-dormant questions... Thank you for this. Looks great! $\endgroup$ Aug 22, 2015 at 20:12
  • 1
    $\begingroup$ Could you give the operation tables of this pair of rings of order $8$? $\endgroup$ Jan 9, 2016 at 5:00
  • $\begingroup$ @David Zhang: Define two multiplications on $A:=\mathbb Z_2\times \mathbb Z_4$. $A$ has a Klein subgroup $K$ generated by $a:=(1,0)$ and $b:=(0,2)$. $A$ is the union of cosets $K$ and $K':=(0,1)+K$. In both multiplication tables, set all products equal to $0$ except products of elements of $K'$ with elements of $K'$: In the first multiplication table the product of any element of $K'$ with any element of $K'$ is $a$. In the second multiplication table the product of any element of $K'$ with any element of $K'$ is $b$. $\exists x\exists y(2x=y^2\neq 0)$ is true in 2nd ring and not the 1st. $\endgroup$ Feb 3, 2016 at 6:22

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.

  • $\begingroup$ 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? $\endgroup$
    – Qfwfq
    Sep 2, 2011 at 23:24
  • $\begingroup$ @Neil: Interesting; I'll think about this as well. Thanks! $\endgroup$ Sep 3, 2011 at 4:19

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