-1

  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?)

rutruttt
  • 111
  • 5
  • 1
    Does this answer question 2? As for 1, this holds whenever $F$ is a left adjoint, in particular when $F$ is a free functor (which I take to mean "left adjoint to a forgetful functor"). – Ben Steffan Mar 30 '24 at 15:44
  • 1
    Left adjoints respect all colimits, right adjoints respect all limits. In particular, the coproduct, which is a colimit, is respected by any "free" functor, which by definition is the left adjoint of the underlying set functor. This holds for semigroups, monoids, abelian groups, groups, commutative rings, rings, posets, topological spaces, etc: the free object on the disjoint union is the coproduct of the free objects. See here – Arturo Magidin Mar 30 '24 at 15:52
  • 3
    This is a duplicate of two separate questions: Coproduct of free monoids. has a categorical argument that shows that left adjoints respect coproducts, and the question linked to by Ben Steffan shows the tensor product over $\mathbb{Z}$ is the coproduct for commutative rings with unity. – Arturo Magidin Mar 30 '24 at 15:52

1 Answers1

0

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.

diracdeltafunk
  • 14,911