Welcome to MathOverflow

Question

2

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 $S$?

edited Jul 29 at 18:05

MTS
5,306●1●9●27

asked Jul 29 at 14:09

zacarias
351●4

	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 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 at 15:10
	thank you Benjamin – zacarias Jul 29 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 at 16:32

Mark Sapir · Accepted Answer · 2012-07-30 15:59:26Z UTC

5

$S$ 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.

answered Jul 30 at 15:59

Mark Sapir
27.5k●2●57●118

	Since S is commutative, Wedderburn seems a bit like overkill, IMHO. – quid Jul 30 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 at 16:36
	@Mark Sapir: somehow I overlooked that final remark. Sorry for the noise and thanks for the reply. – quid Jul 30 at 16:58
	Thank you Mark Sapir – zacarias Jul 30 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 at 1:49

Embedding Semigroups in Rings

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.