## Coproduct of Modules

**Point of Post: **In this post we shall discuss coproducts (i.e. direct sums) of modules.

*Motivation*

We now discuss the “dual notion” to products–coproducts. Coproducts are often affectionately described as what you get when you take the definition of products and “reverse all the arrows”. In other words, the characterization of coproducts will involve maps from the factors *into* the coproduct. So, what do coproducts represent intuitively? Intuitively coproducts are, just as in the case of products, a way of putting a set of modules together. The difference though is that the way we put them together is not quite as “independent” as it was for products. To be particular, we shall see that while coproducts still don’t have much wiggle room outside of the factors (in the sense that maps are determined by their action on the factors) we shall see that coproducts “squish” the factors more than products–in the sense that the factors aren’t completely independent of one another, at least not always. Coproducts are often more natural in the sense that they are an “internal” object opposed to an “external” one.

*Coproduct of Modules*

Let be a collection of left -modules. We say that an ordered pair is a *coproduct *of the set if is a left -module and each is a -morphism for which satisfies the following universal property: given any left -module and any set of -maps there exists a unique -map such that . The first thing we notice is that just like in the case of products, although we don’t require the ‘s to be any special kind of -map they are forced, by their universal properties, to be something special–in this case monos. For the same reason, basically too, namely fixing some and defining the maps by if and we see we are guaranteed some map such that and so evidently must be a mono. Moreover, it should be clear that there must be only (up to one isomorphism) coproduct of a set of modules. This is the content of the next theorem:

**Theorem: ***Let be a set of left -modules and let and be two coproducts of . Then, there exists an isomorphism with .*

**Proof: **The proof shall flow much the same as it did for products. Namely, we have very little to work with, and what we have to work with all we need. Namely, since we have maps the fact that is a coproduct gives us the existence of a map with . Similarly, we have that induces a map with . Note then that and . But, since and we have by uniqueness that and . The conclusion follows.

Now, just as we did for products, we actually need to prove the existence of coproducts by actually exhibiting one. In particular, given a set of left -modules let, as per usual, the set of all elements of with finite support (i.e. only finitely many nonzero coordinates) be denoted by . We then define the operations on to be such that it is a submodule of . If we then define to be the map which sends to the element where for and , then it’s not hard to show that is a coproduct. We see then that given maps the guaranteed map is given by where this map makes sense precisely because will be zero for all but finitely many .

Just as in the case for products, since is, up to isomorphism, the only coproduct we shall often speak of it as the coproduct.

We end the post for now by noting that similar associative and commutative relations:

**Theorem: ***Let be a set of left -modules, a bijection, and a partition of , then*

**Proof: **All three are easily shown to be products with the obvious maps.

**References:**

[1] Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. *Advanced Modern Algebra*. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. *Module Theory.* Clarendon, 1990. Print.

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