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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?

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Your first sentence uses the word coproduct one more time that you intended, I think. –  Mariano Suárez-Alvarez Sep 17 '12 at 1:22

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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.

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It's worth noting that any category with filtered colimits and finite coproducts has all small coproducts, by the construction you give. –  Zhen Lin Sep 17 '12 at 2:34
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@Zhen: the construction I give in $\text{Ring}$ is not quite a construction using finite coproducts; the tensor products I'm considering are not free products. I'm just trying to describe the free product as an abelian group. I should make this clearer. –  Qiaochu Yuan Sep 17 '12 at 2:47
    
Thanks for your helpful answer. For the filtered colimit approach, do you think one needs to order $I$ or mod out the finite tensor products to isomorphism classes or something, in order to have well-defined maps $A_{J_1}\rightarrow A_{J_2}$ ? –  ashpool Sep 17 '12 at 14:49
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@ashpool: not for $\text{CRing}$. The maps come from inclusions of finite subsets. That definition makes sense without an order. I wrote down the terms in order for simplicity but that's not necessary in $\text{CRing}$ because the factors in a tensor product commute. For the construction in $\text{Ring}$ you need to take all possible orders. –  Qiaochu Yuan Sep 17 '12 at 16:28
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@ashpool: actually the colimit is the same either way; that is, I could either only have $A_1 \otimes A_2$ or I could have both, and as long as the inclusion maps are what they should be it doesn't matter. The colimit still has the same universal property. –  Qiaochu Yuan Sep 17 '12 at 18:07

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