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.

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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.

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Coproduct of Modules

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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:

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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

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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.

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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.

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

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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

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\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

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

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[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.


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


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