Abstract Nonsense

Crushing one theorem at a time

Inverse Limit of Rings (Pt. II)

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

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Examples of Inverse Limits

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Now that we have defined inverse limits of inverse systems of rings, let’s see if we can explicitly find inverse limits for each of the inverse systems we submitted as examples previously in this post.

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For our first example consider the module P=\left\{(a,b)\in M_\alpha\oplus M_\beta:f_{\gamma,\alpha}(a)=f_{\gamma,\beta}(b)\right\}. It’s easy to see that P with the natural projections \pi_\alpha:P\to M_\alpha,\pi_\beta:P\to M_\beta, and the map \pi_\gamma:P\to M_\gamma:(a,b)\mapsto f_{\gamma,\alpha}(a) is an inverse limit for this inverse system. This is called a pull-back of the diagram associated to the inverse system. These, and their direct limit analogues, will have their own post soon enough–they are of the utmost importance.

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Now, for our second example let’s now take a specific example where we have a chain of the form \{1,p,p^2,p^3,\cdots\} where p is some prime. What we are then looking at is the set of rings \left\{\mathbb{Z}_{p^n}\right\}_{n\in\mathbb{N}} and the set of morphisms \{f_{n,m}:\mathbb{Z}/(p^m)\to\mathbb{Z}/(p^n)\}_{n,m\in\mathbb{N},\; n\leqslant m}. We call the inverse limit of this inverse system (which, as we shall shortly show, always exists) the p-adic integers and denote it by \mathbf{Z}_p. There is a natural way to realize \mathbf{Z}_p, namely we can view \mathbf{Z}_p as the subring of \displaystyle \prod_{n\in\mathbb{N}}\mathbb{Z}_{p^n}  consisting of those tuples (a_n) such that a_{n+1}\equiv a_n\text{ mod }p^{n+1} (it’s a slightly annoying, but totally straightforward exercise, to check that this defines a unital subring) . So, what are the natural maps \pi_n:\mathbf{Z}_p\to\mathbb{Z}_{p^n}? Why just the projections onto the n^{\text{th}} coordinate of course! Ok, to prove this we must verify that f_{n,m}\circ\pi_m=f_n for each n\leqslant m\in\mathbb{N} and that the \pi_n satisfy the universal property of inverse limits. For this first we must merely note that f_{n,m}(\pi_m((a_k)))=f_{n,m}(a_m)=a_m\text{ mod }p^n but, by assumption (on the way the coordinates interact) this is equal to a_n\text{ mod }p^n which is equal to \pi_n((a_k)). To see that the \pi‘s satisfy the universal mapping properties we suppose that \{g_n:S\to\mathbb{Z}_{p^n}\} is a set of ring homomorphisms with S some unital ring such that f_{n,m}\circ g_m=g_n for each n\leqslant m\in\mathbb{N}. We must now define a map j:S\to\mathbf{Z}_p, but clearly it suffices to define a map \displaystyle j:S\to\prod_{n\in\mathbb{N}}\mathbb{Z}_{p^n} and merely prove that \text{im }j\subseteq\mathbf{Z}_p. But, by the universal characterization of products to do this we must merely define what we want \pi_n\circ j to be for each n. But, we merely define \pi_n\circ j=g_n. Now, all we have to do (as already stated) is prove that \text{im }j\subseteq\mathbf{Z}_p, but this is equivalent to showing that \pi_m(j(s))\equiv \pi_n(j(s))\text{ mod }p^n whenever n\leqslant m but note that this is equivalent to g_m(s)\equiv g_n(s)\text{ mod }p^n or that f_{n,m}(g_m(s))=g_n(s) which is true by assumption. Thus, our map j is clearly a well-defined and satisfies \pi_n\circ j=g_n. We must now check that this was the only feasible way to define j. More concretely, suppose that k:S\to\mathbf{Z}_p is another map such that \pi_n\circ k=g_n. We note that, if we include \mathbf{Z}_p into the product by the inclusion mapping \iota we have by assumption that \pi_n(\iota\circ j)=\pi_n(\iota\circ k) for all n\in\mathbb{N}, but by the definition of product this tells us that \iota\circ j=\iota\circ k, and since \iota is injective this implies that j=k.

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To get an example of the third type of inverse system we consider the special case where we look at the polynomial ring k[x] for some field and consider the chain of ideals (x)\supseteq (x)^2=(x^2)\supseteq (x)^3=(x^3)\supseteq\cdots. We then have the naturals maps f_{n,m}:k[x]/(x)^m\to k[x]/(x)^n which are just the modulo maps. This gives us a nice, well-defined, inverse system of rings. What we claim is that an inverse limit of this system is the ring k[[x]] of formal power series over k. To see this we define the obvious maps \pi_n:k[[x]]\to k[x]/(x)^n by \displaystyle \sum_{k\geqslant 0}a_k x^k\mapsto \sum_{k=0}^{n-1}a_kx^k+(x^n). This clearly satisfies

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\displaystyle \begin{aligned}f_{n,m}\left(\pi_m\left(\sum_{k\geqslant0}a_kx^k\right)\right) &=f_{n,m}\left(\sum_{k=0}^m a_kx^k+(x^m)\right)\\ &=\sum_{k=1}^{n}a_kx^k+(x^n)\\ &=\pi_n\left(\sum_{k\geqslant0}a_kx^k\right)\end{aligned}

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It remains now to show that given any set of maps g_n:R\to k[x]/(x^n) such that f_{n,m}\circ g_m=g_n then there exists a unique j:M\to k[[x]] such that \pi_n\circ j=g_n. To do this let r\in R be arbitrary. We note then that to define j(r)\in k[[x]] it suffices to define numbers a_k for each k\geqslant 0. To do this we merely note that there exists a unique a_{1,n+1},\cdots,a_{n,n+1}\in R such that \displaystyle g_{n+1}(r)=\sum_{k=1}^{n}a_{k,n+1}x^k+(x^{n+1}). We then define a_k=a_{k,k+1}. The fact that f_{n,m}\circ g_m=g_n guarantees that j is a ring homomorphism, and trivially we see that \pi_n\circ j=g_n. Thus, k[[x]] is the inverse limit of the k[x]/(x^n).

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What we claim is that if we take the trivial inverse system on a set \{R_\alpha\}_{\alpha\in\mathcal{A}} of rings we get the product ring as a result. To see this we note that we have natural maps \displaystyle \pi_\alpha:P\to R_\alpha (where P is the product ring) given by the natural projections. These trivially satisfy the compatibility relations since we must only check that \pi_\alpha=f_{\alpha,\alpha}\circ\pi_\alpha=\text{id}_{R_\alpha}\circ\pi_\alpha. Moreover, the universal property of the inverse limits then clearly just translates to the usual universal characterization of products.

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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 | Algebra, Ring Theory, Uncategorized | , , , , , ,

1 Comment »

  1. […] Inverse Limit of Rings (Pt. III) Point of Post: This is a continuation of this post. […]

    Pingback by Inverse Limit of Rings (Pt. III) « Abstract Nonsense | December 26, 2011 | Reply

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