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The natural numbers are the initial commutative semiring. Thus, for any commutative semiring $R$, there is a unique semiring map $\mathbb{N}\to R$.

For which $R$ is this map an epimorphism?

Some examples where it is:

  • Obviously, if $R=\mathbb{N}$.
  • If it is surjective, e.g. $R=\mathbb{Z}/n$.
  • If it adjoins additive inverses, e.g. $R=\mathbb{Z}$.
  • If it adjoins multiplicative inverses, e.g. $R=\mathbb{Q}_{\ge 0}$ (or smaller localizations of $\mathbb{N}$).
  • If it does both, e.g. $R=\mathbb{Q}$ (or smaller localizations of $\mathbb{Z}$).

Are these the only possibilities?

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    $\begingroup$ Related: mathoverflow.net/questions/109/… , which in particular answers the corresponding question for $\mathbb{Z}$. $\endgroup$ – Qiaochu Yuan May 22 '14 at 5:13
  • $\begingroup$ @QiaochuYuan thanks; I should go read the Bousfield-Kan paper. A first question to ask would then be whether every solid ring is also a "solid semiring". $\endgroup$ – Mike Shulman May 22 '14 at 5:57
  • $\begingroup$ That should be straightforward; $\mathbb{N} \to \mathbb{Z}$ is an epimorphism, and epimorphisms compose. $\endgroup$ – Qiaochu Yuan May 22 '14 at 6:09
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    $\begingroup$ Recall that there are other homomorphic images of $\mathbb N$, namely the factors by a congruence $\sim_{a,b}$ defined as $x\sim_{a,b}y\iff x,y\geq a \;\wedge\; b\,\mid\,x-y$. A ring $\mathbb Z/n$ corresponds to $\sim_{0,n}$. $\endgroup$ – Ilya Bogdanov May 22 '14 at 7:25
  • $\begingroup$ @QiaochuYuan but is every epimorphism of rings also an epimorphism of semirings? $\endgroup$ – Mike Shulman May 24 '14 at 3:26

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