Prove that the category of Rings is cocomplete

Question

I am trying to prove that the category of Rings is cocomplete. Of course, it suffices to show that it has coproducts and coequalizers. I looked at this page and I think I get the coequalizer, but I am still unsure of how to take the quotients out of the free ring. I would be grateful for your explanation.

Answer 1

Here's a perspective that may be considered. The category $\mathsf{Ring}$ is the category of algebras of a (finitary) algebraic theory. In classical universal algebraic terms this means that a ring $R$ is a set endowed with two binary operations $+, \cdot$, a unary operation $-$, and two $0$-ary operations $0$ and $1$ satisfying a set of equations (e.g. left distributivity: $$ x \cdot (y+z) \equiv x \cdot y + x \cdot z \text{).}$$ Every category of algebras $\mathsf{Alg_\mathsf{T}}$ of an algebraic theory $\mathsf{T}$ is both complete and cocomplete.

To see that $\mathsf{Alg_\mathsf{T}}$ is complete it is not difficult: the forgetful functor $U: \mathsf{Alg_\mathsf{T}} \to \mathsf{Set}$ preserves and reflects limits and $\mathsf{Set}$ is complete (it has products and equalizers).

To see that $\mathsf{Alg_\mathsf{T}}$ is cocomplete is a bit more laborious. The essential ingredients are:

• $\mathsf{Alg_\mathsf{T}}$ is a full subcategory of $[\mathsf{T}, \mathsf{Set}]$, the category of functors from $\mathsf{T}$ to $\mathsf{Set}$. Indeed, $\mathsf{Alg_\mathsf{T}}$ is the category of finite-products-preserving functors from $\mathsf{T}$ to $\mathsf{Set}$.

• $\mathsf{Alg_\mathsf{T}}$ is reflectieve in $[\mathsf{T}, \mathsf{Set}]$, that is the inclusion has a left adjoint. To prove this one relies on the General Adjoint Functor Theorem.

• If $\mathsf{D}$ is a reflective subcategory of $\mathsf{C}$, $\mathsf{D}$ has all colimits that $\mathsf{C}$ has.

• $[\mathsf{T}, \mathsf{Set}]$ is cocomplete since $\mathsf{Set}$ is cocomplete and a functor category has all colimits that the codomain has, constructed componentwise.

Answer 2

$\def\C{\mathcal{C}}$I will explicitly construct the colimit. Since the non-commutative case is very similar to the commutative one, I will be dealing with both cases at the same time. Let $\mathbf{Ring}$ denote the category of (commutative) rings. Let \begin{align*} F:I&\to\mathbf{Ring}\\ i&\mapsto A_i \end{align*} be small diagram. Let $B$ be the free (commutative) ring on the disjoint union of the $A_i$'s, i.e, $B=\mathbb{Z}\left[\coprod_{i\in I}A_i\right]$ (we mean the (non-)commutative polynomial ring in the (non-)commutative case). We will write an element $x\in A_i$ understood as a letter in $B$ as $\langle x\rangle_i$. Let $\mathfrak{b}\subset B$ be the ideal (double-sided in the non-commutative case) generated by $$ S=S_0\cup S_1\cup S_\text{sum}\cup S_\text{mult}\cup S_\text{rel} $$ where \begin{align*} S_n &=\{\langle n\rangle_i-n\mid i\in I\},\quad n=0,1\\ S_\text{sum} &=\{\langle x\rangle_i+\langle y\rangle_i-\langle x+y\rangle_i \mid x,y\in A_i,i\in I\}\\ S_\text{mult} &=\{\langle x\rangle_i\langle y\rangle_i-\langle xy\rangle_i \mid x,y\in A_i,i\in I\}\\ S_\text{rel} &=\{\langle f(x)\rangle_j-\langle x\rangle_i\mid x\in A_i, f=F(\alpha),\alpha:i\to j\in I\}. \end{align*} Set $A=B/\mathfrak{b}$. For each $i\in I$, there is a map \begin{align*} A_i&\to B\\ x&\mapsto\langle x\rangle_i \end{align*} which is just a map of sets; it never is a ring homomorphism. However, the composite $A_i\to B\to A$ is a ring homomorphism. Moreover, the maps $\{A_i\to A\}_{i\in I}$ give a cocone under $F$. It is not difficult to verify that it is the colimit of $F$.

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If one just wanted to show cocompleteness of the category of commutative rings without knowing how the colimit looks like, here's another proof: For a category $\C$, the following properties are equivalent:

1. Cocompleteness,
2. having coproducts and coequalizers,
3. having coproducts and pushouts,
4. having filtered colimits, finite coproducts and pushouts, and
5. having filtered colimits, pushouts and an initial object.

It is clear that condition 1 implies all other ones (in 5 note that an initial object is a colimit of the empty diagram). The converses also hold:

2$\Rightarrow$1. It is known.

3$\Rightarrow$2. Coequalizers may be built using coproducts and pushouts [ref].

4$\Rightarrow$3. If a category has filtered colimits and finite products then it has all coproducts [ref].

5$\Rightarrow$4. A pushout under an initial object is a coproduct, and having binary coproducts implies having finite coproducts.

We shall show condition 5 for $\C=\mathbf{Ring}$ the category of unital commutative rings. The initial object is $\mathbb{Z}$. A filtered colimit in $\mathbf{Ring}$ is preserved by the forgetful functor $\mathbf{Ring}\to\mathbf{Set}$. In $\mathbf{Ring}$, it is computed by endowing the filtered colimit taken in $\mathbf{Set}$ with a natural ring structure [ref]. On the other hand, for any ring morphisms $A\leftarrow R\rightarrow B$ the square $$ \require{AMScd} \begin{CD} R@>>> B\\ @VVV@VVV\\ A@>>> A\otimes_RB \end{CD} $$ is cocartesian. $\square$

(A finite coproduct in $\mathbf{Ring}$ is of the form $A_1\amalg \cdots\amalg A_n=A_1\otimes_\mathbb{Z}\cdots\otimes_\mathbb{Z}A_n$.)