# Thread: Universal arrow from an abelian group to the forgetful functor U: Rng --> Ab

1. ## Universal arrow from an abelian group to the forgetful functor U: Rng --> Ab

This is one of the exercises in Maclane's book.

Find, for any given abelian group $\displaystyle G$, a universal arrow from $\displaystyle G$ to the forgetful functor $\displaystyle U:\mbox{Rng}\to\mbox{Ab}$ which forgets the multiplicative structure.
I can describe the following construction, which I think fits the bill.

My first guess is that $\displaystyle R_G$ should be a quotient of a suitably defined free object on the elements of $\displaystyle G$. Take the free monoid $\displaystyle M$ on the elements of $\displaystyle G$, and then take the monoid ring $\displaystyle \mathbb{Z}[M]=F$ on the elements of $\displaystyle M$. Then, let $\displaystyle I$ be the ideal of $\displaystyle F$ generated by elements of the form $\displaystyle \tilde a+ \tilde b- \tilde c$, where $\displaystyle \tilde a, \tilde b, \tilde c$ are the canonical images of $\displaystyle a, b, c \in G$ via the obvious "inclusion" $\displaystyle G \to F$, with $\displaystyle a+b-c=0$ in $\displaystyle G$. Then, if I am not mistaken, $\displaystyle R_G=F/I$ is the desired "universal ring" on the abelian group $\displaystyle G$.

Does this seem correct? If we have an morphism of abelian groups $\displaystyle h:G\to U(S)$, we can extend $\displaystyle h$ to a morphism of monoids $\displaystyle M \to S$ by defining $\displaystyle h(a_1\dots a_k)=h(a_1)\dots h(a_k)$ for any word $\displaystyle a_1\dots a_k$ in M (and by definition of the empty product, the identity of $\displaystyle M$ is mapped to the identity of $\displaystyle S$). Then we can extend this map to a morphism of rings $\displaystyle \mathbb{Z}[M]\to S$ by linearity. But there is no guarantee at this point that the additive structure imposed by $\displaystyle G$ will be respected, so we have to quotient out by the ideal described above. The resulting map $\displaystyle \tilde h$ seems to do the job.

In Maclane's book, the category $\displaystyle \mbox{Rng}$ is meant to represent the category of rings, with arrows the ring morphisms preserving identity elements. However, I can read on Wikipedia that this category is now called $\displaystyle \mbox{Ring}$, and that $\displaystyle \mbox{Rng}$ denotes the category of pseudo-rings. This distinction seems reasonable. If this is the case, we should replace $\displaystyle \mbox{Rng}$ by $\displaystyle \mbox{Ring}$ above. I think that the construction of a "universal" pseudo-ring on $\displaystyle G$ would be the same as above, except that we would take the free semigroup on the elements of $\displaystyle G$ instead of the free monoid.

If you have any comments, or insight, or if you can suggest a simplification of the above construction, I would be glad to hear it.

2. Originally Posted by Bruno J.
This is one of the exercises in Maclane's book.

I can describe the following construction, which I think fits the bill.

My first guess is that $\displaystyle R_G$ should be a quotient of a suitably defined free object on the elements of $\displaystyle G$. Take the free monoid $\displaystyle M$ on the elements of $\displaystyle G$, and then take the monoid ring $\displaystyle \mathbb{Z}[M]=F$ on the elements of $\displaystyle M$. Then, let $\displaystyle I$ be the ideal of $\displaystyle F$ generated by elements of the form $\displaystyle \tilde a+ \tilde b- \tilde c$, where $\displaystyle \tilde a, \tilde b, \tilde c$ are the canonical images of $\displaystyle a, b, c \in G$ via the obvious "inclusion" $\displaystyle G \to F$, with $\displaystyle a+b-c=0$ in $\displaystyle G$. Then, if I am not mistaken, $\displaystyle R_G=F/I$ is the desired "universal ring" on the abelian group $\displaystyle G$.

Does this seem correct? If we have an morphism of abelian groups $\displaystyle h:G\to U(S)$, we can extend $\displaystyle h$ to a morphism of monoids $\displaystyle M \to S$ by defining $\displaystyle h(a_1\dots a_k)=h(a_1)\dots h(a_k)$ for any word $\displaystyle a_1\dots a_k$ in M (and by definition of the empty product, the identity of $\displaystyle M$ is mapped to the identity of $\displaystyle S$). Then we can extend this map to a morphism of rings $\displaystyle \mathbb{Z}[M]\to S$ by linearity. But there is no guarantee at this point that the additive structure imposed by $\displaystyle G$ will be respected, so we have to quotient out by the ideal described above. The resulting map $\displaystyle \tilde h$ seems to do the job.

In Maclane's book, the category $\displaystyle \mbox{Rng}$ is meant to represent the category of rings, with arrows the ring morphisms preserving identity elements. However, I can read on Wikipedia that this category is now called $\displaystyle \mbox{Ring}$, and that $\displaystyle \mbox{Rng}$ denotes the category of pseudo-rings. This distinction seems reasonable. If this is the case, we should replace $\displaystyle \mbox{Rng}$ by $\displaystyle \mbox{Ring}$ above. I think that the construction of a "universal" pseudo-ring on $\displaystyle G$ would be the same as above, except that we would take the free semigroup on the elements of $\displaystyle G$ instead of the free monoid.

If you have any comments, or insight, or if you can suggest a simplification of the above construction, I would be glad to hear it.
That looks good to me. I mean, you kind of just did what you "should" do. That said, I myself have only recently started studying $\displaystyle \textbf{Cat}$-theory myself.

P.S. Does Mclane define a forgetful functor to be anything that "forgets"? I am using Adamek et. al and they reserve it merely for the obvious functors $\displaystyle F:\textbf{C}\to\textbf{Set}$

3. Yes, the notion of "forgetful functor" is wider than, say, the notion of a faithful set-valued functor. You'll find a good description on Wikipedia.

Thank for your input. It seems good to me also; I did just what I "should" do, perhaps, but it still took me all day to come up with!