Infinite coproduct of rings

Solution 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.