Extension of Naturals via Grothendieck Group Construction

Question

So there is a way of extending the set $N$ of natural numbers with 0, equipped with ordinary multiplication, to its Grothendieck group, the group of integers with respect to addition.

This group, $Z$, is the best extension of $N$ in the sense that ...? (question)

So, here are my questions:

1. I know the field of fractions $F$ of a commutative ring $R$ (without a zero divisor) is the best extension of $R$ into a field in the sense that given a natural inclusion map $\iota:R\to F$ (by sending $r\in R$ to $r/1\in F$), and any other field $K$ and an in injective homomorphism $\phi:R\to K$, there exists a (unique?) injective homomorphism $\overline{\phi}:F\to K$ such that the diagram commutes. So in a sense $F$, the field of fraction $F$ is the smallest extension of $R$.

How does this idea then apply to $Z$, the Grothendieck group of $N$? Wikipedia has something like what I said, but slightly different (in that the homomorphisms are not necessarily injective, so one cannot talk about "inclusion relation").

2. Okay, the integers are not only a group with respect to addition, but it is also a ring with respect to addition and multiplication. How do you then extend a commutative monoid (such as $N$) to a commutative ring (such as $Z$)? Do we do so simply by defining one more binary operation on $(Z,+)$? If s0, how do we guarantee that it is indeed a very special structure (in the sense that it is the smallest ring extension of $N$)?

Much appreciated in advance!

Answer 1

Okay, bear with me, because this is going to be rather informal and some details might be off, but I think the general idea should be right:

Concerning 2.: "Free constructions" give rise to functors, for example there is a functor $F : \mathsf{Set}\rightarrow \mathsf{Grp}$ mapping a set $A$ to the free group $F(A)$ on this set and a function $f : A\rightarrow B$ to a group morphism $F(f) : F(A) \rightarrow F(B)$. Of course, you would need to figure out, how the actual construction of a free group works, but to prove, that $F$ is a functor (preserves identities and composition) you can use the UMP (universal mapping property) of the free group on a set:

There is a mapping $i : A \rightarrow U(F(A))$, s.t. for all groups $B$ and mappings $f : A \rightarrow U(B)$ there is a unique group morphism $f^* : F(A) \rightarrow B$, s.t. $f=f^*\circ i$

where $U : \mathsf{Grp} \rightarrow \mathsf{Set}$ is the forgetful functor, mapping groups to their underlying set of elements and group morphisms $f$ to $f$ ("forgetting", that they are group morphisms and not just maps). Strictly speaking, we should also write: "s.t. $f = U(f^*)\circ i$".

The same kind of property holds for all the "free constructions", including the field of fractions, except we have functors $F : \mathsf{CRing} \rightarrow \mathsf{Fld}$ and $U : \mathsf{Fld} \rightarrow \mathsf{CRing}$, a commutative ring $A$ and a ring morphism $i : A \rightarrow U(F(A))$ and so on and so forth.

We observe in all these cases, that there is an adjunction: $F$ is left adjoint to $U$. Observe, that adjoints are unique up to natural isomorphism, so they are in some sense special and $F$ really gives you the most efficient way to turn something into something with more structure or more properties. To prove, that you can not "efficiently" turn a commutative monoid into a commutative ring, you would need to show, that the corresponding forgetful functor $U$ has no left adjoint. I don't know, how to do that though and I only suspect, that this is really true.

Concerning 1.:

I informally determined (that is I think), that you can turn an additively cancellative commutative rig $A$ (that's like a commutative ring with identity, except addition need not be a group operation, but still cancellative) into a commutative ring $F(A)$, by the usual construction, that turns natural numbers into integers (equivalence classes of $\mathbb{N}^2$). Given such a rig $A$, we have a rig monomorphism: $i : A \rightarrow U(F(A)), a \mapsto [(a,0)]$ and of course the UMP:

For all commutative rings $B$ and rig morphisms $f : A \rightarrow U(B)$ there is a unique ring morphism $f^* : F(A) \rightarrow B$, s.t. $f=f^*\circ i$.

This morphism is defined by $f^* : F(A) \rightarrow B, [(a,b)]\mapsto f(a-b)$, I believe. You of course need to check everything carefully (including, that $f^*$ is well-defined).

Note: A rig morphism preserves addition, multiplication as well as $0$ and $1$. (It's a monoid morphism for both addition and multiplication)

Answer 2

As Pavel's comment notes, either you need to take the natural numbers with 0 under addition or the natural numbers beginning at 1 under multiplication.

I will assume for now that you are talking about the natural numbers under addition, but the other case is similar except that the abelian group you should get is the positive rationals under multiplication, also it isn't super important since the construction is similar for all monoids.

The idea specifically for $\mathbb{N}$ and $\mathbb{Z}$ is that if there is a monoid homomorphism $\phi$, from $\mathbb{N}\to A$ for $A$ some abelian group, then there is a unique homomorphism from $\mathbb{Z} \to A$ defined by $\phi'(-n)=-\phi(n)$ for $-n < 0$. It is pretty clear that this will be a homorphism and that it is unique since if such a morphism exists, then $0=\phi'(0)=\phi'(-n)+\phi'(n)=\phi'(-n)+\phi(n)$, so $\phi'(-n)=-\phi(n)$.

The idea is that if you can map $\mathbb{N}$ into a space, then you can also map the integers into it. In particular, if your map is injective, then the map of the integers into the space should also be injective. Otherwise you have Clearly $\phi'(n)\ne \phi'(m)$ for either both positive or both negative integers $n$ and $m$ since $\phi$ is injective. Then if $\phi'(-n)=\phi'(m)$ for $-n<0$, $m\ge 0$ then $-\phi(n)=\phi(m)$, so $\phi(n)+\phi(m)=\phi(n+m)=0$ which is a problem.

I think the best way to answer your second question is yes, we define a second binary operation on $\mathbb{Z}$. It turns out to be a ring, and since $\mathbb{Z}$ is the smallest abelian group containing the natural numbers, it should also be the smallest ring containing them.

Why it turns out to be a ring is perhaps interesting to think about. Maybe not, but it basically boils down to the fact that $n$ added to itself $m$ times is the same as $m$ added to itself $n$ times, so that the natural endomorphism of abelian groups taking an element $a$ to $na$ for $n$ an integer is a symmetric operation.

Every abelian group $A$ is a module with respect to its endomorphism ring $S=\operatorname{End}(A)$, and for every $R$ such that $A$ is a module over $R$, there is a map from $R$ to $S$. In particular if we make $A$ into a ring, there has to be a ring homomorphism from $A$ as a ring into $S$, and in particular there has to be an abelian group homomorphism from $A$ into the additive group of $S$. For the integers this can be done naturally in fact, since we can take the natural endomorphisms mentioned above, and map $n$ to the morphism that takes $a$ to $na$, then identifying $\mathbb{Z}$ with this subring (check closure under composition) of the endomorphism ring of $\mathbb{Z}$ as an abelian group gives a natural ring structure on $\mathbb{Z}$, but I don't think you get such nice morphisms in general.