Archive for December, 2008

Topological construction of p-adic integers

December 26, 2008

One of the interesting things to observe is what multiplication by p does to p-adic integers:

p(u_1, u_2, u_3, \ldots) = (p u_1, p u_2, p u_3, \ldots).

But what matters is the congruence class of the components modulo p^r, and pu_1 \equiv 0 \pmod p, pu_{r}\equiv pu_{r-1} \pmod{p^r}, so

p(u_1, u_2, u_3, \ldots) = (0, p u_1, p u_2, \ldots).

Iterating gives us

p^n (u_1, u_2, u_3, \ldots) = (\underbrace{0, \ldots, 0}_{n\text{ times}}, p^n u_1, p^n u_2, \ldots).

This is, of course, a lot easier when using the series notation.

So, we can tell that p^n divides u precisely when the first n components of u are zero. As a corollary, we obtain that if u\neq 0, then there exist a largest n such that p^n divides u (i.e., p^{n+1} does not divide u). This n is precisely the number of the initial components of u that are congruent to zero in their respective {\mathbb Z}/p^r{\mathbb Z}.

Definition. The number n as above is called the p-adic valuation of u, and is denoted by \mathrm{ord}_p(u):

\mathrm{ord}_p(u) := \mathrm{max}\{n\in {\mathbb N} | p^n\text{ divides }u\text{ in }{\mathbb Z}_p \}.

We set \mathrm{ord}_p(0) = \infty.

The p-adic valuation function satisfies the following properties:

  • \mathrm{ord}_p(u) = \infty if and only if u=0,
  • \mathrm{ord}_p(uv) = \mathrm{ord}_p(u)\mathrm{ord}_p(v), and
  • \mathrm{ord}_p(u+v) \geq \mathrm{min}(\mathrm{ord}_p(u), \mathrm{ord}_p(v)).

The last property is referred to as the ultrametric property for reasons to be explained right now.

Given any number 0 < \sigma < 1 (customarily, \sigma = 1/p), we can define

|u|_p := \sigma^{\mathrm{ord}_p(u)}.

Then the properties of \mathrm{ord}_p become

  • |u|_p \geq 0 for all u, with |u|_p = 0 if and only if u=0,
  • |uv|_p = |u|_p\, |v|_p, and
  • |u+v|_p \leq \mathrm{max}(|u|_p, |v|_p).

    • In other words, |\bullet|_p: {\mathbb Z}_p \to {\mathbb R} is an ultrametric norm. We can also define a metric

    d_p(u,v) := |u-v|_p,

    which makes {\mathbb Z}_p into an utrametric space. The topology (convergence of sequences) induced by this norm does not change if we choose a different \sigma, so from now on, \sigma = 1/p. Then

    |u|_p = \frac{1}{p^{\mathrm{ord}_p(u)}}.

    In other words, the higher power of p divides you, the smaller you are.

    Note that this norm can be restricted to the rational integers {\mathbb Z} contained in {\mathbb Z}_p. The ring of integers is not complete under this norm, and therefore we can consider its completion, i.e., the smallest complete normed space containing {\mathbb Z}. Then the standard construction of the completion coincides for the p-adic norm with the construction of p-adic integers as sequences. Specifically, the condition that u_r \equiv u_{r+s} \pmod{p^r} translates into the Cauchy criterion d_p(u_r, u_{r+s}) \to 0 as r\to 0 for sequences.

    In light of this norm, the series notation also starts to make sense. Indeed, the truncated series form a Cauchy sequence, since b_r p^r + b_{r+1} p^{r+1} + \cdots + b_{r+s} p^{r+s} has norm no greater than 1/p^r, which goes to zero as r goes off to infinity. Thus the infinite series we work with converge with respect to the ultrametric p-adic norm.

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    The ring of p-adic integers

    December 26, 2008

    In what way can we impose operations of addition, subtraction, and multiplication on the set of p-adic integers? One thing on our wish list is that these operation be compatible with the corresponding operations on the ring of integers embedded in {\mathbb Z}_p. Another, is that they be compatible with the operation on the component considered as elements of {\mathbb Z}/p{\mathbb Z}. It seems that the obvious way to define these operations is componentwise:

    • (u_1, u_2, \ldots) + (v_1, v_2, \ldots) := (u_1 + v_1, u_2 + v_2, \ldots);
    • (u_1, u_2, \ldots) - (v_1, v_2, \ldots) := (u_1 - v_1, u_2 - v_2, \ldots);
    • (u_1, u_2, \ldots) \cdot (v_1, v_2, \ldots) := (u_1 v_1, u_2 v_2, \ldots).

    Another way to define these operations is using the “series” notation. We perform these operations just like base p calculations, only digits extend infinitely to the left (some people write \ldots b_2 b_1 b_0 instead of the series notation b_0 + b_1 p + b_2 p^2 + \cdots). For example, (2 + 1 \cdot 5 + 3 \cdot 5^2 + \cdots)(1 + 2 \cdot 5 + 4 \cdot 5^2 + \cdots) = 2\cdot 1 + (2\cdot 2 + 1\cdot 1) 5 + (2\cdot 4 + 1\cdot 2 + 3\cdot 1) 5^2 + \cdots = 2 + 0\cdot 5 + 4\cdot 5^2 + \cdots. We carry 1 toward 5^2’s and 2 towards 5^3’s. It’s a bit of an exercise to show that these definitions agree. It is not that difficult to check that these operations make {\mathbb Z}_p into a commutative ring.

    Using Abstract Nonsense, the construction of the p-adic numbers as sequences can be reformulated as follows. For each r we have a map

    {\mathbb Z}/p^{r+1}{\mathbb Z} \to {\mathbb Z}/p^r{\mathbb Z}

    of reduction modulo p^r, so rings {\mathbb Z}/p^r{\mathbb Z} form an inverse system. Then

    {\mathbb Z}_p = \displaystyle\mathop{\mathrm{lim}}_{\longleftarrow}\ {\mathbb Z}/p^r{\mathbb Z},

    the inverse limit of this system. Recall that the inverse limit in the category of rings is constructed by taking all elements of the direct product of all rings in the system (a.k.a. sequences) that “agree” with the reduction maps.

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    Introduction to p-adic integers

    December 25, 2008

    I feel like taking a break from elliptic curves and talking about something completely different. For motivation, let’s take a look at solutions of the equation x^2 + x + 3 \equiv 0 \pmod{5^r} for different values of r. For r = 1, we have two solutions: 1, 3 \pmod{5}. For r=2, still two solutions: 8, 16 \pmod{25}. For r=3 we get 33, 91 \pmod{125}. And so on, and so forth.

    Notice something interesting. Reduce 91 modulo 25, and you get 16. Reduce 16 modulo 5, and you get 1. Similarly, reduce 33 modulo 25 to get 8, which reduces to 3 modulo 5. It seems like a solution modulo 5 begets a solution modulo 25, which then yields a solution modulo 125, etc. No surprise there, since all we are doing is reducing the equation x^2 + x + 3 \equiv 0 \pmod{5^r} modulo 5^{r-1}.

    The question is this. Can we go in the opposite direction and keep this string of solutions going ad infimum? Given a solution modulo 125, can we find one modulo 625? (more…)

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