# Abstract Nonsense

## Inverse Limits of Modules (Pt. II)

Point of Post: This is a continuation of this post.

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General Construction

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As went with the case of direct limits, it is probably prudent to show that any inverse system of modules actually admits an inverse limit. The setup will be similar to the case of direct limits, where we shall “glue” all the modules in the system together, and then mod out by the relation that “identifies” objects in the desired way. More rigorously:

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Theorem: Let $\left(\mathcal{A},\leqslant\right)$ be a preoredred set and $\left(\{M_\alpha\}_{\alpha\in\mathcal{A}},\{f_{\alpha,\beta}M_\beta\to M_\alpha\}_{\alpha,\beta\in\mathcal{A},\; \alpha\leqslant\beta}\right)$ an inverse system of modules. Then, if $P$ denotes the submodule the product of all the modules $M_\alpha$ containing all the elements $(x_\alpha)$ such that $x_\alpha=f_{\alpha,\beta}(x_\beta)$ for all $\beta\geqslant\alpha$, then $P$ along with the natural projections $\pi_\alpha:P\to M_\alpha$ is an inverse limit of our system.

Proof: We need to show first off that $f_{\alpha,\beta}\circ\pi_\beta=\pi_\alpha$ for every $\alpha\leqslant\beta$. But, this is precisely how we defined $P$! Thus, it remains to show that given any set of maps $g_\alpha:N\to M_\alpha$ such that $f_{\alpha,\beta}\circ g_\beta=g_\alpha$ whenever $\alpha\leqslant\beta$, that there exists a unique map $j:N\to P$ such that $\pi_\alpha\circ j=g_\alpha$.  The idea of how to do this is clear, namely by definition of the product we get a unique map $j$ which maps $N$ into the product of the modules, and so it suffices to show that $\text{im }j\subseteq P$. To see this we merely note that $\pi_\alpha(j(n))=g_\alpha(n)=f_{\alpha,\beta}(g_\beta(n))=f_{\alpha,\beta}(\pi_\beta(j(n))$ and so $j(n)\in P$–since $n\in N$ was arbitrary the claim follows. Thus, $j$ is a well-defined morphism $N\to P$, which clearly satisfies $\pi_\alpha\circ j=g_\alpha$. Moreover, it’s clear that this was the only way we could have defined such a $j$ since if $k$ was another such map then $\iota\circ j=\iota\circ k$ (where $\iota$ is the inclusion of $P$ into the full product)  have same values when precomposed with the $\pi_\alpha$‘s and so are equal by the definition of the product, but since $\iota$ is injective we may left cancel to find that $j=k$. $\blacksquare$

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Now, we have seen that all inverse limits of a given inverse system are isomorphic, and so the above construction shall serve as a sort of “model” for the inverse limit. In particular, while there is no “the” inverse limit, we shall make this phrasing make sense by saying that whenever we discuss “the” inverse limit of an inverse system we shall mean the above construction, which we shall denote $\varprojlim M_\alpha$.

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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.

[5] Grillet, Pierre A. Abstract Algebra. New York: Springer, 2007. Print.

December 9, 2011 -

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