Abstract Nonsense

Crushing one theorem at a time

Coproduct of Modules


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

\text{ }

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 M_\alpha\to M 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.

\text{ }

Coproduct of Modules

\text{ }

Let \left\{M_\alpha\right\}_{\alpha\in\mathcal{A}} be a collection of left R-modules. We say that an ordered pair \left(M,\{i_\alpha\}\right) is a coproduct of the set \left\{M_\alpha\right\}_{\alpha\in\mathcal{A}} if M is a left R-module and each i_\alpha:M_\alpha\to M is a R-morphism for which \{i_\alpha\} satisfies the following universal property: given any left R-module N and any set of R-maps g_\alpha:M_\alpha\to N there exists a unique R-map g:M\to N such that g\circ i_\alpha=g_\alpha. The first thing we notice is that just like in the case of products, although we don’t require the i_\alpha‘s to be any special kind of R-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 \alpha_0\in\mathcal{A} and defining the maps g_\alpha:M_\alpha\to M_{\alpha_0} by g_\alpha=0 if \alpha\ne \alpha_0 and g_{\alpha_0}=\text{id}_\alpha we see we are guaranteed some map g:M\to M_\alpha such that g\circ i_\alpha=\text{id}_\alpha and so evidently i_\alpha 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:

\text{ }

Theorem: Let \left\{M_\alpha\right\}_{\alpha\in\mathcal{A}} be a set of left R-modules and let \left(M,\{i_\alpha\}\right) and \left(N,\{j_\alpha\}\right) be two coproducts of \left\{M_\alpha\right\}_{\alpha\in\mathcal{A}}. Then, there exists an isomorphism f:M\to N with f\circ i_\alpha=j_\alpha.

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 i_\alpha:M_\alpha\to M the fact that N is a coproduct gives us the existence of a map g:N\to M with g\circ j_\alpha=i_\alpha. Similarly, we have that j_\alpha:M_\alpha\to N induces a map f:M\to N with f\circ i_\alpha=j_\alpha. Note then that f\circ g\circ j_\alpha=f\circ i_\alpha=j_\alpha and g\circ f\circ i_\alpha=g\circ j_\alpha=i_\alpha. But, since \text{id}_M\circ i_\alpha=i_\alpha and \text{id}_N\circ j_\alpha=j_\alpha we have by uniqueness that f\circ g=\text{id}_M and g\circ f=\text{id}_{N}. The conclusion follows. \blacksquare

\text{ }

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 \left\{M_\alpha\right\}_{\alpha\in\mathcal{A}} of left R-modules let, as per usual, the set of all elements of \displaystyle \prod_{\alpha\in\mathcal{A}}M_\alpha with finite support (i.e. only finitely many nonzero coordinates) be denoted by \displaystyle \bigoplus_{\alpha\in\mathcal{A}}M_\alpha. We then define the operations on \displaystyle \bigoplus_{\alpha\in\mathcal{A}}M_\alpha to be such that it is a submodule of \displaystyle \prod_{\alpha\in\mathcal{A}}. If we then define \displaystyle \iota_{\alpha_0}:M_{\alpha_0}\to\bigoplus_{\alpha\in\mathcal{A}}M_\alpha to be the map which sends x\in M_{\alpha_0} to the element (x_\alpha) where x_\alpha=0 for \alpha\ne \alpha_0 and x_{\alpha_0}=x, then it’s not hard to show that \displaystyle \left(\bigoplus_{\alpha\in\mathcal{A}}M_\alpha,\left\{\iota_\alpha\right\}_{\alpha\in\mathcal{A}}\right) is a coproduct. We see then that given maps g_\alpha:M_\alpha\to N the guaranteed map \displaystyle g:\bigoplus_{\alpha\in\mathcal{A}}M_\alpha\to N is given by \displaystyle g((x_\alpha))=\sum_{\alpha\in\mathcal{A}}g_\alpha(x_\alpha) where this map makes sense precisely because g(x_\alpha) will be zero for all but finitely many \alpha.

\text{ }

Just as in the case for products, since \displaystyle \left(\bigoplus_{\alpha\in\mathcal{A}}M_\alpha,\{\iota_\alpha\}\right) is, up to isomorphism, the only coproduct we shall often speak of it as the coproduct.

\text{ }

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

\text{ }

Theorem: Let \left\{M_\alpha\right\}_{\alpha\in\mathcal{A}} be a set of left R-modules, \sigma:\mathcal{A}\to\mathcal{A} a bijection, and \left\{P_\beta:\beta\in\mathcal{B}\right\} a partition of \mathcal{A}, then

\text{ }

\displaystyle \bigoplus_{\alpha\in\mathcal{A}}M_{\sigma(\alpha)}\cong\bigoplus_{\alpha\in\mathcal{A}}M_\alpha\cong\bigoplus_{\beta\in\mathcal{A}}\bigoplus_{t\in P_\beta}M_t

\text{ }

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

\text{ }

\text{ }

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.

Advertisements

November 12, 2011 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , , ,

12 Comments »

  1. […] claim is that is bijective. Indeed, if , then  one has by assumption that (where denotes the canonical injection) and so is trivial. To see that is surjective let be arbitrary, we claim that . Indeed, by pure […]

    Pingback by Dual Modules « Abstract Nonsense | November 12, 2011 | Reply

  2. […] We shall in the future have many occasions to deal with homomorphism groups (modules). In particular, we shall often deal with one of the Hom functors, and surprisingly often we shall have that in the free variable there is a product or coproduct. Consequently, it would be nice if we had some way of simplifying such Hom’s in terms of nicer groups. That is precisely the content of this post. It shall be good practice for us applying our notions of product and coproduct. […]

    Pingback by Homomorphism Groups of Products and Coproducts (Pt. I) « Abstract Nonsense | November 12, 2011 | Reply

  3. […] first natural thing to ask is whether, as in the case of products and coproducts whether this map has to have any special properties that, while not present in the definition, […]

    Pingback by Free Modules (Pt. I) « Abstract Nonsense | November 16, 2011 | Reply

  4. […] have seen that the category of modules has notions of coproducts, namely given a set of modules, we can form the module which is the submodule of the product  of […]

    Pingback by Internal Direct Sum of Modules « Abstract Nonsense | November 20, 2011 | Reply

  5. […] we first prove is the following obvious theorem which says, in essence, that the arbitrary coproduct of free modules are free, and moreover that what we want to be a basis is a basis. In […]

    Pingback by Coproduct of Free Modules are Free, but not Arbitrary Products « Abstract Nonsense | November 23, 2011 | Reply

  6. […] up seeing that this limit is a fairly faithful representation of the individual in the sense that coproducts are good representation of the factor modules–this shouldn’t be surprising since […]

    Pingback by Direct Limit of Modules « Abstract Nonsense | November 29, 2011 | Reply

  7. […] given a set of morphisms where we have defined the trivial directed system then the coproduct is a direct limit. Indeed, if one notes that since the only map defined is for each we see that […]

    Pingback by Direct Limit of Modules (Pt. II) « Abstract Nonsense | November 30, 2011 | Reply

  8. […] some ‘degenerate’ cases which are very useful. First and foremost in my mind is the coproduct of a set of modules which is obtained by defining the trivial preorder on . Thus, we […]

    Pingback by Direct Limits of Directed Sets « Abstract Nonsense | December 1, 2011 | Reply

  9. […] limits and reversing all the arrows. We have already seen this kind of duality between products and coproducts, and in fact this shall serve as the main kind of duality between direct and inverse limits […]

    Pingback by Inverse Limits of Modules (Pt. I) « Abstract Nonsense | December 9, 2011 | Reply

  10. […] functor to get an -map . Note though that by previous discussion  and are nothing more than the coproducts  and respectively. What we now claim is that is the map often denoted which sends to . To see […]

    Pingback by Category of Directed/Inverse Systems and the Direct/Inverse Limit Functor (Pt. II) « Abstract Nonsense | December 28, 2011 | Reply

  11. […] actually tell us for some particular direct limits? In particular, the most important direct limit, coproducts. To figure out what the above says about coproducts let’s simplify things and suppose we are […]

    Pingback by Tensor Products Naturally Commute with Direct Limits « Abstract Nonsense | January 19, 2012 | Reply

  12. […] the objects that one gets if we reverse all the arrows. These generalize the common notion of coproducts of modules, but we shall see that general coproducts are not always as […]

    Pingback by Categorical Coproducts « Abstract Nonsense | February 7, 2012 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: