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I asked a question similar to the one I am about to ask, and I think I got a satisfactory answer. However, this time I have some more specific question.

Let a semiring $(R,+,\times)$ be an algebraic structure such that $(R,+)$ is a monoid with identity $0$, and $(R,\times)$ is a monoid with identity $1$. Further suppose that the distributive law holds, and $0x=x0=0$ for all $x\in R$. We can then see that $\mathbf{N}$, the natural numbers (with $0$), form a semiring with respect to ordinary addition and multiplication.

My question is as follows.

Suppose now you construct $\mathbf{Z}$, the ring of integers, from $\mathbf{N}$ as follows:

  1. Define an equivalence relation $\sim$ on $\mathbf{N}\times\mathbf{N}$ such that $(a,b)\sim(\alpha,\beta)$ if and only if $a+\beta=b+\alpha$. Define addition $\oplus$ on $\mathbf{N}\times\mathbf{N}/\mathord{\sim}$ as \begin{equation}[(a,b)]\oplus[(\alpha,\beta)] = [(a+\alpha,b+\beta)],\end{equation} and multiplication $\otimes$ as \begin{equation}[(a,b)]\otimes[(\alpha,\beta)] = [(a\alpha+b\beta,a\beta+b\alpha)].\end{equation}

These are indeed well-defined functions on the quotient, and by identifying each element $[(a,b)]\in\mathbf{Z}$ (for $b>a$) as $b-a$ in $\mathbf{N}$, we have $\mathbf{Z}$. Call this map $i:\mathbf{N}\to\mathbf{Z}$. Now for the $real$ question:

Is it true that given any other ring $R$ that contains a homomorphic copy of $\mathbf{N}$, or equivalently, given any injective semiring homomorphism $\phi:\mathbf{N}\to R$, is it true that there exists a unique injective semiring homomorphism $\tilde{\phi}:\mathbf{Z}\to R$ such that $\phi=\tilde{\phi}\circ i$?

Much appreciated in advance!

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  • $\begingroup$ Why do you feel the need to include a construction of the the integers from the natural numbers in your question? $\endgroup$ – Rob Arthan May 6 '15 at 20:11
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It does not even seem to have anything to do with one-to-one morphisms.

$\Bbb N$ is clearly initial in the category of semirings: for a given semiring $S$, there is a unique semiring homomorphism from $\Bbb N \to S$ determined by $\phi(1)=1_S$. (The kernel could even be nonzero.)

Likewise, $\Bbb Z$ is initial in the category of rings, where the unique map from $\Bbb Z\to R$ is given by $\psi(1)=1_R$.

These two maps are fully determined by additivity and preservation of multiplicative identity, and the only difference is their domain. For each ring $R$, $\psi$ is necessarily the only ring homomorphism extending the semiring homomorphism $\phi$.

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  • $\begingroup$ Yes, but the OP didn't put $1$ in the signatures, so it is unclear what notion of homomorphism is relevant: $\mathbb{N}$ is not an initial object in the category of semirings without unit. $\endgroup$ – Rob Arthan May 6 '15 at 20:33
  • $\begingroup$ @robarthan I can't deny that, but my take on the situation is that in the context of category theory and universal properties, one would most likely stick to the categories I describe. It's a judgement call, of course, unless the OP decisively says differently. $\endgroup$ – rschwieb May 7 '15 at 2:28
  • $\begingroup$ perhaps I should have made it clear, but I am not familiar with the language of Category theory (thus the way I phrased the problem)... But thanks anyway! $\endgroup$ – user134070 May 7 '15 at 6:45
  • $\begingroup$ @user134070 "universal properties" are a basic concept of category theory, so that's why I thought you are familiar with at least the basics. $\endgroup$ – rschwieb May 7 '15 at 9:54
  • $\begingroup$ Ah, I see. My math instructor used the concept of universal property without explicitly mentioning of the categories, way of which you can probably see in my question. $\endgroup$ – user134070 May 7 '15 at 11:10

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