Coproduct of free objects

Question

1. In the category of groups, the coproduct is free product, and for any $A,B$ sets: the coproduct of $F(A),F(B)$ is $F(A\sqcup B)$. Does this property hold in any other categories?

2. What is the coproduct in the category of commutative rings? Or if the definition of it is too complicated, then just what's the coproduct of free objects (I know that the free commutative ring over a set $A$ is $\mathbb{Z}[A]$, does the property above hold here too?)

Answer 1

1.

Let $\mathsf{C}$ be a category of algebraic objects: sets, groups, abelian groups, rings, $\mathbb{R}$-vector spaces, etc. Of course I am not being precise about what exactly counts as a "category of algebraic objects" -- we really just need the following two assumptions:

($\star$) There is a "forgetful functor" $U : \mathsf{C} \to \mathsf{Set}$ which sends objects to their underlying sets.

($\star$) There is also a "free functor" $F : \mathsf{Set} \to \mathsf{C}$, defined by the condition that $F$ is left adjoint to $U$.

Now left adjoints preserve colimits, so in particular $F(A \amalg B)$ is the coproduct of $F(A)$ and $F(B)$ for all sets $A$ and $B$. QED

---

2.

We can also understand the coproduct in commutative rings via this adjunction. For what follows, let $\mathsf{C}$ denote the category of commutative rings.

The counit of the adjunction $F(U(X)) \to X$ is surjective for all commutative rings $X$. By the first isomorphism theorem, this means $X \cong \mathbb{Z}[U(X)]/I$ for some ideal $I$ of $\mathbb{Z}[U(X)]$.

Now take two commutative rings $X$ and $Y$, and let $I, J$ be ideals such that $\mathbb{Z}[U(X)]/I \cong X$ and $\mathbb{Z}[U(Y)]/J \cong Y$. We can view $\mathbb{Z}[U(X)]$ and $\mathbb{Z}[U(Y)]$ as subrings of $\mathbb{Z}[U(X) \amalg U(Y)]$. Let $K$ denote the ideal of $\mathbb{Z}[U(X) \amalg U(Y)]$ generated by $I \cup J$. Then the coproduct of $X$ and $Y$ is given by $\mathbb{Z}[U(X) \amalg U(Y)]/K$.

Try to prove it! It's really not so hard to do.

This turns out to be the same as the tensor product over $\mathbb{Z}$ -- i.e. the coproduct of $X$ and $Y$ is given by $X \otimes_{\mathbb{Z}} Y$. If you're familiar with tensor products of commutative rings this is also easy to show.