# Infinite coproduct of rings

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?

#### Solutions Collecting From Web of "Infinite coproduct of rings"

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.