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I'm trying to understand inverse limits. For this I am looking at the example (mentioned in Atiyah-Macdonald, page 102): We start with the topological abelian group $G = \mathbb Z$ (endowed with the topology induced by $|\cdot|_p$) and observes that we have an inverse system, that is, a sequence of groups $G_n$ and group homomorphisms $f_{ji}: G_i \to G_j$ ($j \leq i$) such that

(i) $f_{ii} = id$

(ii) $f_{kj} \circ f_{ji} = f_{ki}$

given by $G_n = \mathbb Z / p^n \mathbb Z$ and homomorphisms $f_{ji}: \mathbb Z / p^i \mathbb Z \hookrightarrow \mathbb Z / p^j \mathbb Z, \bar{x} \mapsto x \mod p^j$. Then, by definition, the inverse limit is $$\lim_{\longleftarrow_{n \in \mathbb N}} G_n = \{(x_n) \in \prod_{n \in \mathbb N} G_n \mid f_{ji} (x_i) = x_j , \text{ for all } j \leq i \}$$

Now since $f_{ji}$ are the "projections" (they are $\mod p^j$ really) we see that the requirement $f_{ji} (x_i) = x_j$ is nicely fulfilled if $x_{i+1}$ only adds stuff times $p^{i+1}$ since then the added stuff gets deleted when projecting down.

Hence we "see" (we don't really but we already know what it should be) that sequences in this inverse limit look like $p$-adic numbers, since $x_i = \sum_{k=0}^i a_k p^k$.

Question 1: Do I have it right?

Question 2: Why do we need a topology on the group? I don't see where we used the topology.

And question 3: Would someone give me another (very easy!) example of an inverse limit, please?

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Maybe this thread is helpful for question 3. –  t.b. Jul 25 '12 at 21:37
    
@t.b. I thought that if I'm a good boy and do as I'm told it might make it more likely that you will look at my next post and check it : ) No worries of course if you don't. –  Matt N. Aug 6 '12 at 20:36
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2 Answers

up vote 4 down vote accepted
  1. Yes.

  2. The topology happens to mesh well with the inverse limit, but is not needed.

  3. Four easy examples:

    • Consider the system with $G_i = k[x]/(x^i)$ and the $f_{ij}$ again are the projections. Again, since things are surjective, you can only add multiples of $p^i$, etc. so we get the inverse limit is the ring of formal power series $k[\![x]\!]$. Its (unneeded) topology is given by $|\cdot|_x$ which as for $x=p$ measures how divisible a polynomial is by $x$.

    • Consider the system where all $G_i$ are equal, and $f_{ij}$ are identity maps. Then the inverse limit is also $G_i$. I don't believe any topology is implicit.

    • Consider the inverse limit where all the $f_{ij}$ are zero maps. Then the inverse limit is zero.

    • Consider the inverse limit where there aren't any maps. Then the inverse limit is the direct product. The topology is the product topology.

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Don't trust example 3. I've changed my mind a few times. Must be dinner time. Example 4 is more general than A&M allow (they don't allow you to omit functions). The first example is very standard. The second is trivial. –  Jack Schmidt Jul 25 '12 at 21:05
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To elaborate on Jack's point, the topology makes this an inverse limit in the category of topological groups, rather than just groups, so with it and the initial topology on the inverse limit-which is just the product topology-we can show this construction gets us the same structure as the purely topological completion of Z under the p-norm. –  Kevin Carlson Jul 25 '12 at 21:12
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The topology is not, strictly speaking, necessary. But there many reasons why it is desirable.

To begin, many subgroup filtrations on $G$ will induce the same topology. From the topological viewpoint it is clear that the completions of $G$ for these filtrations will be isomorphic. This makes clear the value of the Artin–Rees lemma, for example.

But beyond that there is some psychological value in viewing the higher powers of $p$ and $X$ in $\mathbb Z_p$ and $k[[X]]$, respectively, as being small. Now everything in $\mathbb Z_p$ is truly a limit of something in $\mathbb Z$, and certain infinite sums make sense. You might start to wonder if there is a useful notion of analysis in the $p$-adic world.

We also get a useful universal property which makes no reference to specific filtrations or connecting homomorphisms: if $H$ is complete and separated for such a topology and $f\colon G \to H$ is continuous, then $f$ factors through a unique map $\hat G \to H$.

A good (not so easy) example to look at is a nodal cubic. Let $k$ be a field of characteristic $\neq 2$. Below is a drawing that I ripped off from Silverman‘s book of the curve in the plane corresponding to $A = k[x, y]/(y^2 - x^3 - x^2)$.

enter image description here

The completion of $A$ at the ideal $(x, y)$ is isomorphic to $k[[u, v]]/(uv)$. So locally, at the origin, our curve is just two lines crossing.

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This is a bit ramble-y. Will try to make it better later if I have time. –  Dylan Moreland Jul 25 '12 at 21:47
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