Infinite coproduct of rings

Question

I just learned from Wikipedia that coproduct of two (commutative) rings is given by tensor product over integers, and that coproduct of a family of rings is given by a "construction analogous to the free product of groups." Can the tensor product approach be generalized to an arbitrary family of rings? (Infinite tensor product?) I'm a little surprised that coproduct of commutative rings requires noncommutative structure (free group). Does someone have a reference which explicitly constructs the coproduct?

Answer 1

Your question is mildly ambiguous, so let's separate the statements for the categories of commutative rings and rings.

The infinite coproduct of commutative rings in $\text{CRing}$ is not given by an infinite tensor product, the reason being intuitively that you can only multiply finitely many elements of a ring at a time and infinite tensor products deal with attempting to multiply infinitely many things at a time. It is in fact given by a filtered colimit of finite tensor products. More precisely, let $A_i, i \in I$ be an indexed family of commutative rings, and for every finite subset $J$ of $I$ consider the tensor product $A_J = \bigotimes_{i \in J} A_i$. Then the infinite coproduct $\bigsqcup_i A_i$ is the filtered colimit of the $A_J$ equipped with all inclusion maps $A_{J_1} \to A_{J_2}$, where $J_1 \subset J_2$. The inclusion maps look like this:

$$a_{i_1} \otimes ... \otimes a_{i_n} \to a_{i_1} \otimes ... \otimes a_{i_n} \otimes 1_{A_{i_{n+1}}} \otimes ... \otimes 1_{A_{i_m}}$$

where $|J_1| = n$ and $|J_2| = m$.

The infinite coproduct of rings in $\text{Ring}$ is an infinite free product. As an abelian group, it is constructed similarly to the above; for an indexed family $A_i, i \in I$ of rings it consists of a filtered colimit (of abelian groups!) of finite tensor products (as abelian groups!) of the $A_i$ where we can repeat factors (due to the fact that we cannot assume that the images of $A_i$ and $A_j$ commute in general). The multiplication is given by concatenation. I don't know a reference for the construction but it consists more or less in following one's nose. When in doubt, always refer back to the universal property.

In particular, the functor $\text{CRing} \to \text{Ring}$ does not preserve coproducts; the phrase "coproduct of commutative rings" has a different meaning depending on whether you want to take the coproduct in $\text{CRing}$ or in $\text{Ring}$ because the corresponding universal properties are different.

Answer 2

Here is a way to construct the coproduct of a family of commutative rings $R_i$ in the category $\mathsf{CRing}$:

Let $S = \mathbb{Z}[x_{i, r} : i \in I, r \in R_i]$ be the free commutative ring on the indeterminates $x_{i, r}$ for $i \in I$ and $r \in R_i$. Now let $J \subseteq S$ be the ideal generated by expressions of the following form:

• $x_{i, r_1} + x_{i, r_2} - x_{i, r_1 + r_2}$ for $i \in I$ and $r_1, r_2 \in R_i$
• $x_{i, r_1} x_{i, r_2} - x_{i, r_1 r_2}$ for $i \in I$ and $r_1, r_2 \in R_i$
• $x_{i, 1} - 1$ for $i \in I$

Now take $\newcommand\quotient[2]{{^{\Large #1}}/{_{ \Large #2}}} T = \quotient{S}{J}$. Then $T \cong \coprod_{i \in I} R_i$ is the coproduct over all the $R_i$ in $\mathsf{CRing}$. The inclusions $R_i \to T$ are the obvious ones, and the fact that we took the quotient by $J$ ensures that these are actually ring homomorphisms.

---

Here is the completely analogous construction for the case of the coproduct of a family of rings $R_i$ in the category $\mathsf{Ring}$:

Let $S = \mathbb{Z} \langle x_{i, r} : i \in I, r \in R_i \rangle$ be the free noncommutative ring on the indeterminates $x_{i, r}$ for $i \in I$ and $r \in R_i$. Now let $J \subseteq S$ be the ideal generated by expressions of the following form:

• $x_{i, r_1} + x_{i, r_2} - x_{i, r_1 + r_2}$ for $i \in I$ and $r_1, r_2 \in R_i$
• $x_{i, r_1} x_{i, r_2} - x_{i, r_1 r_2}$ for $i \in I$ and $r_1, r_2 \in R_i$
• $x_{i, 1} - 1$ for $i \in I$

Now take $\newcommand\quotient[2]{{^{\Large #1}}/{_{ \Large #2}}} T = \quotient{S}{J}$. Then $T \cong \coprod_{i \in I} R_i$ is the coproduct over all the $R_i$ in $\mathsf{Ring}$. The inclusions $R_i \to T$ are the obvious ones, and the fact that we took the quotient by $J$ ensures that these are actually ring homomorphisms.

---

It is also not at all unexpected that the coproduct of commutative rings in the category $\mathsf{Ring}$ is not commutative. The free product of abelian groups is also not abelian.

---

Here are some related questions:

• Coproduct in the category of (noncommutative) associative algebras
• How to construct the coproduct of two (non-commutative) rings
• About direct sum/product and subdirect product of rings
• How to construct the tensor product of two (non-commutative) algebras?
• Coproduct of non-unital commutative algebras