MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $S$ be a finite commutative semigroup with identity. Under what conditions (on the semigroup $S$) it is possible to find a ring $R$ such that the multiplicative structure of $R - \{0\}$ is isomorphic to $S$?

share|cite|improve this question
R-0 Is rarely a semigroup. You want to look at which commutative semigroups with zero are multiplicative semigroups of rings. The question of which semigroups can be the multiplicative semigroup of a ring is classical. I suggest you google this subject. I am not sure how complete the answers are in general but for finite commutative maybe the answer is known. – Benjamin Steinberg Jul 29 '12 at 14:46
Googling would seem to indicate that this is in general an open question but the answer is known if each element is idempotent (since we are in the Boolean ring case here). – Benjamin Steinberg Jul 29 '12 at 15:10
thank you Benjamin – zacarias Jul 29 '12 at 16:04
Note that his conditions would have R be a ring with no zero divisors, so the situation gets close to asking about the structure of multiplication of skew fields and ordinary fields. Gerhard "Ask Me About System Design" Paseman, 2012.07.29 – Gerhard Paseman Jul 29 '12 at 16:32
up vote 6 down vote accepted

$S$ must be a cyclic group of order $p^n-1$ for some prime $p$ and natural $n$. Indeed, since $R\setminus \{0\}$ is a semigroup under multiplication, $R$ does not have zero divisors. Hence $R$ is a division ring. Since $S$ is finite, $R$ is a finite division ring, hence, by Wedderburn, a finite field. Therefore $S$ must be the multiplicative group of a finite field, hence a cyclic group of order $p^n-1$. Note that you do not need to assume that $S$ is commutative.

share|cite|improve this answer
Since S is commutative, Wedderburn seems a bit like overkill, IMHO. – user9072 Jul 30 '12 at 16:09
@quid: Of course, that is why I wrote that one does not need to assume that $S$ is commutative, just finite. – Mark Sapir Jul 30 '12 at 16:36
@Mark Sapir: somehow I overlooked that final remark. Sorry for the noise and thanks for the reply. – user9072 Jul 30 '12 at 16:58
Thank you Mark Sapir – zacarias Jul 30 '12 at 22:33
The interesting question is when S is a semigroup with zero when is it the semigroup of a ring. – Benjamin Steinberg Jul 31 '12 at 1:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.