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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?

Mike Shulman
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    Related: http://mathoverflow.net/questions/109/what-do-epimorphisms-of-commutative-rings-look-like , which in particular answers the corresponding question for $\mathbb{Z}$. – Qiaochu Yuan May 22 '14 at 05:13
  • @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". – Mike Shulman May 22 '14 at 05:57
  • That should be straightforward; $\mathbb{N} \to \mathbb{Z}$ is an epimorphism, and epimorphisms compose. – Qiaochu Yuan May 22 '14 at 06:09
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    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 $\sim_{0,n}$. – Ilya Bogdanov May 22 '14 at 07:25
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    @QiaochuYuan but is every epimorphism of rings also an epimorphism of semirings? – Mike Shulman May 24 '14 at 03:26

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