Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $I=\{0, 1, \ldots \}$ be the multiplicative semigroup of non-negative integers. It is possible to find a ring $R$ such that the multiplicative semigroup of $R$ is isomorphic (as a semigroup) to $I$?

share|cite|improve this question
up vote 2 down vote accepted

Suppose that there is such an isomorphism, $\phi:R\to\mathbb N.$ In particular, such an $R$ would have to be commutative, since $\mathbb N$ is a commutative semigroup, and unital, since $1\in\mathbb N$ is an identity. Then, $-1\in R$ maps to some $\phi(-1)\neq 1\in\mathbb N.$ But $\phi(-1)^2=\phi((-1)^2)=\phi(1)=1,$ which is clearly impossible. Thus there cannot exist such a ring $R.$

share|cite|improve this answer
Don't you assume $-1\neq 1$ in $R$ here? – Stefan Aug 1 '12 at 17:28
Does this proof work for $R = \mathbb F_2[x]$? – Erick Wong Aug 1 '12 at 17:28
Ah, good point, I may need a different argument for characteristic $2$... – Andrew Aug 1 '12 at 17:34

Let $F$ be any field and take $R = F[x_1,\ldots,x_n]$ for any finite $n\ge 1$. Every non-zero polynomial in $R$ factors uniquely into monic irreducibles times a constant in $F^\times$.

When $F = \mathbb F_2$, the factorization into monic irreducibles is exactly unique. This gives $R \setminus \{0\}$ the structure of a free semigroup on countably many generators (all non-constant irreducible polynomials).

The structure of $I \setminus \{0\}$ is also the free semigroup on countably many generators (the primes).

It also seems like taking $F = \mathbb F_3$ should give the multiplicative semigroup of $\mathbb Z$.

share|cite|improve this answer
This is a nice example. It seems to show that $\mathbb F_2[x]\not\cong I,$ since for example we have $0\cdot m=0\cdot n,\forall m,n\in I$ so $I$ cannot be free. – Andrew Aug 1 '12 at 21:03

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.