# Quotients of the initial semiring

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?

• Related: mathoverflow.net/questions/109/… , which in particular answers the corresponding question for $\mathbb{Z}$. – Qiaochu Yuan May 22 '14 at 5:13
• @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 5:57
• That should be straightforward; $\mathbb{N} \to \mathbb{Z}$ is an epimorphism, and epimorphisms compose. – Qiaochu Yuan May 22 '14 at 6:09
• 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 7:25
• @QiaochuYuan but is every epimorphism of rings also an epimorphism of semirings? – Mike Shulman May 24 '14 at 3:26