We know that the tensor product is the coproduct in the category of R-algebras for any ring R.

What about the infinite tensor product on an index set I, defined as colimit of the finite tensor products with indices in I? I think it can be proved that the infinite tensor product is actually the infinite coproduct in the above category (true? However, I'm still working on it).

And, apart from that, can we find any hypothesis on the category under which the colimit of finite coproducts is the infinite coproduct? (Of course we must be able to define a colimit in the first place.)

If the question seems too vague, just consider this one: is there any easy counter-example to the property "finite coproducts exist, but infinite coproducts don't, or they exist but cannot be obtained from the finite ones in a natural way"?

In this very detailed post by Martin Brandenburg


you can find another definition and some useful properties.

  • 2
    $\begingroup$ The infinite tensor product Martin Brandenburg describes is not the one that's the coproduct in $R$-algebras. In order to define this colimit over finite tensor products the fact that you can insert identities is crucial. Also, it is always true that infinite coproducts are (filtered) colimits of finite coproducts. $\endgroup$ – Qiaochu Yuan Dec 11 '16 at 17:04

I assume you mean to assume your algebras are commutative (otherwise tensor products are not coproducts!). As Qiaochu commented, the correct definition of an infinite tensor product of algebras (if you want it to be the coproduct) is not the thing that classifies "infinite-fold multilinear maps". Rather, it is the subset consisting of "infinite tensors" where all but finitely many of the coordinates must be $1$.

To be precise, given a collection $(A_i)$ of $R$-algebras, their tensor product is defined as follows. Let $S\subseteq\prod A_i$ be the set of elements of the product which have only finitely many coordinates which are not $1$. We think of a function $i\mapsto x_i$ as an infinite "tensor" $\bigotimes x_i$. We then take the free $R$-module on $S$, and then take the quotient by the usual bilinearity relations, saying that the operation $\bigotimes x_i$ is $R$-linear on each coordinate when all the other coordinates are held fixed. This quotient is the tensor product $\bigotimes A_i$, and has a natural algebra structure by taking products "coordinatewise" on elementary tensors.

Letting $I$ be the index set of this tensor product, the collection of finite subsets of $I$ forms a directed set $J$. For each $F\in J$, we can consider the ordinary finite tensor product $A_F=\bigotimes_{i\in F} A_i$. These algebras form a directed system indexed by $J$: when $F\subseteq G$, we have a canonical map $A_F\to A_G$, obtained by "tensoring with $1$" on all the new coordinates. It is then clear that our infinite tensor product $\bigotimes_{i\in I}A_i$ is the colimit of this directed system. Indeed, every element of $\bigotimes_{i\in I}A_i$ can be thought of as coming from some $A_F$, since every tensor must be $1$ on all but finitely many coordinates. The relations imposed are then just exactly the relations imposed on all the $A_F$.

If all the algebras $A_i$ are commutative, the infinite tensor product $\bigotimes_{i\in I}A_i$ is then their coproduct in the category of commutative $R$-algebras. You can prove this directly from the definition much the same way as for finite tensor products. You can also deduce from a general statement: if $(A_i)$ is a collection of objects in any category, then their coproduct is the colimit of the coproducts of finite subcollections (assuming these coproducts exist). The proof is straightforward: you just check that the universal properties are equivalent. To give a map out of the colimit of the finite coproducts, you must give compatible maps out of each finite coproduct. Compatibility ends up just meaning that these maps are all determined by the maps on the $1$-fold coproducts, i.e. the maps out of each $A_i$. So this is equivalent to having a map out of each $A_i$, which is the same thing as a map out of the coproduct of all of them.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.