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!

  • $\begingroup$ 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

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 $\phi$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.