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.