# Abstract Nonsense

## Product of Modules

Point of Post: In this post we discuss the product of modules, including their characterizations via univeral mapping properties.

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Motivation

We now consider the product of modules, which as always, is just endowing the Cartesian product of a set of modules with operations that turn the resulting (set) product into a module. That said, while we have mentioned before that products of things can be characterized via certain universal mapping properties (e.g. for rings and groups) here we shall actually start with thinking of products in terms of these universal mapping properties and then define the “natural product”  only to prove existence of such modules. Why? What precisely is the point of doing? Well, we all have an intuitive idea about what products are, we have seen them as objects for much of our mathematical careers. That said, we understand them intuitively in a concrete “I can see them” sense. What we are now more currently interested is understanding them in a “how do they act” sense–what makes “products” products from the view of mappings. Well, this is precisely what the universal characterization of products tells us. Stated it says that “a product of the set $\left\{M_\alpha\right\}_{\alpha\in\mathcal{A}}$ of left $R$-modules is a left $R$-module $M$ together with a set of maps $f_\alpha:M\to M_\alpha$ with the following property: given any left $R$-module $N$ and maps $g_\alpha:N\to M_\alpha$ there exists a unique map $g:N\to M$ such that $f_\alpha\circ g=g_\alpha$.” Ok, so fine, but what is this big-long-horrible definition really telling us about this $M$? The existence of these $f_\alpha$‘s tell us that $M$ is “put together” in some way via the $M_\alpha$‘s. The fact that any map $N\to M$ is determined by its values on $M_\alpha$ (translated via the “put together maps”, $f_\alpha$) tells us that $M$ is put together in a fairly minimal way–i.e. this $M$ isn’t put together in such a way that there is much “room outside the $M_\alpha$s” for movement. That said, we see that the $M_\alpha$‘s are living inside of $M$ in a fairly faithful manner, in the sense that a functions values on each of the $M_\alpha$s are independent of one another–this tells intuitively that we didn’t squish the $M_\alpha$‘s, or that we didn’t underepresent any of the $M_\alpha$s at any stage of construction. Thus, with this in mind, it’s clear why we’d consider products of modules.

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Products of Modules

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As stated in the motivation we define a (take note of the indefiniteness of the article) product of the set $\left\{M_\alpha\right\}_{\alpha\in\mathcal{A}}$ of left $R$-modules to be a pair $\left(M,\{p_\alpha\}_{\alpha\in\mathcal{A}}\right)$ where $M$ is a left $R$-module and where each $p_\alpha:M\to M_\alpha$, called a projection, is a morphism, such that the set $\left\{p_\alpha\right\}$ satisfies the following universal property: given any left $R$-module $N$ and any set of $R$-morphisms $g_\alpha:N\to M_\alpha$ there exists a unique $R$-morphism $g:N\to M$ with $p_\alpha\circ g=g_\alpha$. The first thing to notice, especially considering that the archetypal example of a product is the usual set product and the $p_\alpha$‘s being the canonical projections, is that we do not require the $p_\alpha$‘s to be epis. That said, it’s not hard to see that they must, in fact, be epis. In particular, taking $N=M_\alpha$ and $g_\beta$ to be $\text{id}_{M_\alpha}$ if $\alpha=\beta$ and $0$ if $\alpha\ne \beta$ we see that there exists a map $g:M_\alpha\to M$ which has, in particular, the property that $p_\alpha\circ g=1_{M_\alpha}$ and so $p_\alpha$ is an epi. Thus, we get the following theorem:

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Theorem: If $\left(M,\left\{p_\alpha:M\to M_\alpha\right\}_{\alpha\in\mathcal{A}}\right)$ is a product then each $p_\alpha$ is an epi.

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The next thing to expect is that while products aren’t literally unique they are surely going to be unique up to isomorphism, and in fact natural isomorphism. To be particular, we have the following fact:

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Theorem: Let $\left\{M_\alpha\right\}_{\alpha\in\mathcal{A}}$ be any set of left $R$-modules and let $\left(M,\{p_\alpha\}\right)$ and $(N,\{q_\alpha\})$ be any two products of the $\left\{M_\alpha\right\}$. Then, there exists a unique isomorphism $f:M\to N$ such that $q_\alpha\circ f=p_\alpha$.

Proof: The proof should be obvious enough. Whenever we maps from a fixed module into each of the factors $M_\alpha$ we are given maps from the fixed module into any product. Well, we’re looking for a map $M\to N$, which by the previous section allows us to only look for maps $M\to M_\alpha$, but, we are already given these for free! Namely, merely by the existence of the maps $p_\alpha:M\to M_\alpha$ we are afforded, since $N$ is a product, a unique map $f:M\to N$ with $q_\alpha\circ f=p_\alpha$. That said, by the symmetry of the situation we know that we are also able to find a map $g:N\to M$ with $p_\alpha\circ g=q_\alpha$. What we’d like, in a perfect world, is that $f=g^{-1}$, but why is this so?  Well, what we really want to prove is that $f\circ g=\text{id}_N$ and $g\circ f=\text{id}_M$, and keeping with our theme that mappings into products are determined by the composition with their projections we are led to examining $q_\alpha\circ(f\circ g)$ and $p_\alpha\circ(g\circ f)$. But, $q_\alpha\circ f\circ g=(q_\alpha\circ f)\circ g=p_\alpha\circ g=q_\alpha$ and similarly $p_\alpha\circ(g\circ f)=p_\alpha$. But, wait a minute! We know that the right hand side of these equations completely determines $f\circ g$ and $g\circ f$, and we already know functions for which $q_\alpha\circ -=q_\alpha$ and $p_\alpha\circ -=p_\alpha$, namely $\text{id}_N$ and $\text{id}_M$. Thus, by uniqueness we know that $f\circ g=\text{id}_N$ and $g\circ f=\text{id}_M$ and so $f$ is an isomorphism as desired. $\blacksquare$

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Thus, we have in essence proved “uniqueness” (up to isomorphism) of products. Now, while slightly silly, we actually have no reason to believe at this point that there actually exist products. Of course, we know there do exist products (or at least there should if there is any justice in the world) in the form of the “usual” product given by $\displaystyle \left(\prod_\alpha M_\alpha,\{\pi_\alpha\}\right)$. To be particular, we define a left $R$-module structure on the set $\displaystyle \prod_{\alpha\in\mathcal{A}}M_\alpha$ (whose elements shall be denoted $(x_\alpha)$) in the usual way: $r(x_\alpha)=(rx_\alpha)$ and $(x_\alpha)+(y_\alpha)=(x_\alpha+y_\alpha)$–it is easily verified that this does define a left $R$-module structure. We then define, as per usual, $\displaystyle \pi_\beta:\prod_{\alpha\in\mathcal{A}}M_\alpha\to M_\beta:(x_\alpha)\mapsto x_\beta$ to be the canonical projection. Of course, as is the case in every other algebraic structure, it’s trivial to verify that with these definitions $\displaystyle \left(\prod_\alpha M_\alpha,\{\pi_\alpha\}\right)$ is a product. Since, as we have pointed out there is, up to one isomorphism one product, we shall often speak of this construct (this particular product) as the product.

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We’d now like to discuss some of the obvious theorems that go along with the product of modules. In particular, letting $\widehat{M_\alpha}$

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Theorem: Let $\left\{M_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a set of left $R$-modules and $\left\{N_\alpha\right\}_{\alpha\in\mathcal{A}}$ a set of submodules. Then, $\displaystyle \prod_{\alpha\in\mathcal{A}}N_\alpha\leqslant\prod_{\alpha\in\mathcal{A}}M_\alpha$ and

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$\displaystyle \left(\prod_{\alpha\in\mathcal{A}}M_\alpha\right)/\left(\prod_{\alpha\in\mathcal{A}}N_\alpha\right)\cong\prod_{\alpha\in\mathcal{A}}\left(M_\alpha/N_\alpha\right)$

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Proof: We merely define a map $\displaystyle \prod_{\alpha\in\mathcal{A}}M_\alpha\to\prod_{\alpha\in\mathcal{A}}(M_\alpha/N_\alpha)$ by sending $(x_\alpha)$ to $\left(x_\alpha+N_\alpha\right)$. This is evidently a $R$-epimorphism and since evidently its kernel is $\displaystyle \prod_{\alpha\in\mathcal{A}}N_\alpha$ we may conclude by the first isomorphism theorem. $\blacksquare$

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Remark: Note that the fact that $\displaystyle \prod_{\alpha\in\mathcal{A}}N_\alpha$ was a kernel implies that its a subspace.

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Note, that when each $M_\alpha$ is secretly $R$, and $\mathcal{A}$ is finite we’ve discussed that a sort of converse to the first part of this theorem where every ideal of the product of finitely many rings is of the prescribed form. While probably fairly obvious, it’s clear that this does not hold here. Namely, while the product of submodules is a submodule, not every submodule of a product is a product of submodules. This has nothing to do with the fact that the ring $R$ may be ugly, it’s also true for vector spaces. For example, $\text{span}_\mathbb{R}((1,1))$ is a subspace of $\mathbb{R}^2$ but is not a product of submodules of $\mathbb{R}$ (indeed, it’s not even a set-theoretic product!).

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The last two theorems we note are to the effect that products are “associative” and commutative in the following sense:

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Theorem: Let $\left\{M_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a set of $R$-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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$\displaystyle \prod_{\alpha\in\mathcal{A}}M_{\sigma(\alpha)}\cong\prod_{\alpha\in\mathcal{A}}M_\alpha\cong\prod_{\beta\in\mathcal{B}}\prod_{t\in P_\beta}M_t$

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this is easily verified since the extreme right and left sides are both products, and so by previous discussion naturally isomorphic to the center module.

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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.

November 11, 2011 -

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