$\mathbb N$ will denote the set $\{0,1,2,\ldots\}.$ The semigroup $(\mathbb N,+)$ doesn't have a zero element. $\mathbb N^0$ will denote the semigroup $\mathbb N$ with zero adjoined, that is the set $\mathbb N\cup\{\star\},\;\star\not\in\mathbb N,$ with the operation $+_0$ defined by $$a+_0b=\begin{cases}a+b & \text{ for } \{a,b\}\subset\mathbb N\\\star &\text{ for } \{a,b\}\not\subset\mathbb N\end{cases}$$

Is there a ring whose multiplicative structure is isomorphic to $\mathbb N^0?$ I believe there isn't, but I don't see a proof.

I've proved that the analogously defined semigroup $\mathbb Z^0$ isn't isomorphic to the multiplicative structure of any ring, but I could do that because such a ring would have to be a field, which made the task easier. Such a field would have to be of characteristic $2$ because the multiplicative group of any other field contains $\{-1,1\}$ as a finite subgroup isomorphic to $\mathbb Z/2\mathbb Z,$ and $(\mathbb Z,+)$ doesn't have non-trivial finite subgroups. Therefore, the field would have to be an extension of $\mathbb F_2.$ If it were an algebraic extension, its multiplicative group would either be finite or contain a non-trivial finite subgroup, which is again impossible. However, a transcendental extension would have to have a copy of $\mathbb F_2(x)$ in it, whose multiplicative group is not finitely generated. Every subgroup of $\mathbb Z$ is finitely generated.

The characteristic restriction carries over to the question I'm asking. If $1\neq -1$ in a ring, then its group of units has $\mathbb Z/2\mathbb Z$ as a subgroup, which $\mathbb N^0$ doesn't. So any ring whose multiplicative structure is isomorphic to $\mathbb N^0$ must be of characteristic $2.$ Even more, it would have to have a trivial group of units, as there are no invertible elements in $\mathbb N^0$ other than $0.$ It must obviously be a countable commutative ring with unity. There are rings satisfying these conditions: $\mathbb F_2[x]$ does. However, the multiplicative structure of $\mathbb F_2[x]$ isn't isomorphic to $\mathbb N^0.$ (The proof of this I've found seems slightly too complicated to me, so if you have a very simple one, I'd be very glad to see it. Mine involves counting "prime" elements in the semigroups, defined like for rings.) If I could reduce this question to $\mathbb F_2[x],$ similarly to the case of $\mathbb Z$ and $\mathbb F_2(x),$ it would be great but I don't see how to do it.

  • 2
    $\begingroup$ Isn't it fairly clear that there is no $p\in\Bbb F_2[x]$ such that $\{p^n:n\in\Bbb N\}=\Bbb F_2[x]\setminus\{0\}$? The degree of $p$ would have to be $1$, and neither $x$ nor $1+x$ works. $\endgroup$ May 13, 2012 at 1:02
  • $\begingroup$ Wouldn't the field of fractions of any such ring be a ring with multiplicative structure $\mathbb{Z}^0$, which you already proved does not exist? $\endgroup$ May 13, 2012 at 3:47
  • $\begingroup$ @BrianM.Scott Yes, thanks! That's exactly the kind of thing I was afraid I missed. :) $\endgroup$
    – user23211
    May 13, 2012 at 9:12
  • $\begingroup$ @Omar I think this is right: the multiplicative structure of a fraction field of $R$ only depends on the multiplicative structure of $R$ because the definition of multiplication in the fraction ring doesn't use addition anywhere. Is this a correct reasoning? (Perhaps you could add this as an alternative proof in your answer?) $\endgroup$
    – user23211
    May 13, 2012 at 10:05

1 Answer 1


Assume there is such a ring and let $p(a,b)$ be its addition. As ymar mentions in the question, every element $t$ must be its own additive inverse, i.e., $p(t,t) = \star$. This means that the relation $p(a,b)=c$ is completely symmetric in $a$, $b$ and $c$, and furthermore that two of the three are equal iff the third is $\star$. Let $p(0,1) = x$. By what we said above, $x \neq \star, 0, 1$, so $x \ge 2$. Now we get $0 = p(x,1) = p(x-1,0) + 1$ by the distributive law. Since $x-1 \neq 0$, we have that $p(x-1,0) \neq \star$ and so $0 = p(x-1,0) + 1 \ge 1$, a contradiction.

  • $\begingroup$ What a nice proof, thank you very much! $\endgroup$
    – user23211
    May 13, 2012 at 9:38

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