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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 $I$?

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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 $R.$

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Don't you assume $-1\neq 1$ in $R$ here? –  Stefan Walter Aug 1 '12 at 17:28
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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 $\mathbb Z$.

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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

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