Abstract Nonsense

Crushing one theorem at a time

Inverse Limit of Rings (Pt. III)

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

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Now that we have seen some examples we would now, as we have done at least three times in the recent past, show that we can always find an inverse limit for a given inverse system. The construction is very similar to that of the inverse limit of modules so we shall try to keep things short, just describing the general object. As we have noted before, the reason for the similarity between the construction is due to the fact (as we shall eventually discuss) that inverse limits and direct limits are a more general construct able to be done in (most) categories with the category of rings or of modules over a fixed ring just being one example. Anyways, let’s get on with the construction:

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Theorem: Let \left(\mathcal{A},\leqslant\right) be a preordered set and \left(\{R_\alpha\}_{\alpha\in\mathcal{A}},\{f_{\alpha,\beta}:R_\beta\to R_\alpha\}_{\alpha,\beta\in\mathcal{A}\;\alpha\leqslant\beta}\right) an inverse system of rings over \mathcal{A}. Then, if R denotes the subring of the product  ring over the R_\alpha defined by \{(r_\alpha):f_{\alpha,\beta}(r_\beta)=r_\alpha\text{ for all }\alpha\leqslant\beta\}. Then, R along with the natural projections \pi_\alpha:R\to R_\alpha are an inverse limit of the aforementioned inverse system.

Proof: Let us first verify that R is a (unital if each R_\alpha is) subring of the product ring. To do this we merely have to note that to prove that f_{\alpha,\beta}(r_\beta+s_\beta)=r_\alpha+s_\alpha and f_{\alpha,\beta}(r_\beta s_\beta)=r_\alpha s_\alpha for each \alpha\leqslant\beta, but this follows immediately from the fact that each f_{\alpha,\beta} is a ring homomorphism (moreover, it’s clear that R is unital if each R_\alpha is since, taking each f_{\alpha,\beta} to be unital, we have that f_{\alpha,\beta}(1_\beta)=1_\alpha always). Now, to prove that this ring really is an inverse limit we suppose that we are given a set of homomorphism g_\alpha:S\to R_\alpha for some ring S such that f_{\alpha,\beta}\circ g_\beta=g_\alpha. We then note that by the universal property of products that there exists a unique map j from S into the product ring over the R_\alpha with the (desired) property that \pi_\alpha\circ j=g_\alpha. If we can thus prove that \text{im }j\subseteq R we’ll be done. But, this is simple since, writing j(s)=(\pi_\alpha(j(s)) we have that f_{\alpha,\beta}(\pi_\beta(j(s))=f_{\alpha,\beta}(g_\beta(s))=g_\alpha(s)=\pi_\alpha(j(s)) as desired. The conclusion follows. \blacksquare

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As was the case or the inverse limit of modules we shall almost always refer to the above ‘model’ of the inverse limit of an inverse system of rings as “the” inverse limit, and denote it by \varprojlim R_\alpha.

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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.

[4] Lang, Serge. Algebra. Reading, MA: Addison-Wesley Pub., 1965. Print.


December 26, 2011 - Posted by | Uncategorized | , , , ,

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