Elon University Logo UNC Logo

REU Computational Research on Local Fields and Galois Groups

Home   People   Results  

A Shallow Introduction to $p$-adic Numbers

Let $p$ be a prime number. We give a quick introduction to $p$-adic numbers and some of their fun properties.

The $p$-adic numbers $\mathbb{Q}_p$ are infinite series of the following form for some integer $k$: \[ \sum_{i=k} a_i\cdot p^i \qquad\qquad a_i\in\{0,1\} \]

Examples. Here are some examples for $p=2$. The first column is in decimal representation:

$36$ $2^2+2^5$ $(k=2)$
$\frac{165}{2}$ $\frac{1}{2}+2+2^4+2^6$ $(k=-1)$
$-1$ $1+2+2^2+2^3+2^4+2^5+\cdots+2^i+\cdots$ $(k=0$)

The last one is my favorite $2$-adic number. But why does this sum converge?

The $p$-adic valuation

Let $x=\sum_{i=k} a_i\cdot p^i$ and suppose $a_k\ne 0$. The $p$-adic valuation of $x$ is the integer $k$; we set $v_p(0)=\infty$. We write: $v_p(x) = k$.

Notice that $v_p(x)$ measures how divisible $x$ is by $p$.

Examples. Here are some examples for $p=2$. The first column is in decimal representation:

36 $2^2+2^5$ $(k=2)$
$\frac{165}{2}$ $\frac{1}{2}+2+2^4+2^6$ $(k=-1)$
$-1$ $1+2+2^2+2^3+2^4+2^5+\cdots+2^i+\cdots$ $(k=0$)

I still haven't explained why the bottom $2$-adic numbers is $-1$. But I will soon.

The $p$-adic absolute value

Again let $x=\sum_{i=k} a_i\cdot p^i$ where $a_k\ne 0$.

Definition. Define the $p$-adic absolute value of $x$ to be the reciprocal of ${p^{v_p(x)}}$ and set $|0|_p=0$. We write: $|x|_p = p^{-v_p(x)}$. Notice that $|x|_p$ is small when $v_p(x)$ is large.

Examples. Here are some examples for $p=2$. The first column is in decimal representation:

$x$ $\mathbf{v_2(x)}$ $\mathbf{|x|_2}$
$36$ $2$ $0.25$
$\frac{165}{2}$ $-1$ $2$
$-1$ $0$ $1$

So 36 is ``smaller'' than $-1$ in the 2-adic world.

Exercise. Can you prove the following properties of $v_p$ and $|\cdot|_p$ ? Let $x$ and $y$ be $p$-adic numbers.

  1. $v_p(xy) = v_p(x) + v_p(y)$
  2. $v_p(x+y) \geq \min\{v_p(x), v_p(y)\}$
  3. $|xy|_p = |x|_p|y|_p$
  4. $|x+y|_p \leq \max\{|x|_p, |y|_p\}$

Example. I can now explain why $\sum_{i=0} 2^i = -1$ in the 2-adic world. It's because this series is a geometric series with first term 1 and common ratio 2. The (2-adic) absolute value of the common ratio is less than 1. Thus we can use the formula for the sum of a geometric series: \[ 1+2+2^2+2^3+\cdots = \frac{\text{first term}}{1-\text{common ratio}} = \frac{1}{1-2} = -1.\]

More properties of $p$-adic numbers

The $p$-adic numbers have lots of properties. Here are a few:
  1. $0,1\in\mathbb{Q}_p$ and $0\cdot x=x\cdot 0 =0$ and $1\cdot x=x\cdot 1=x$ for all $x\in\mathbb{Q}_p$
  2. For each $x\in\mathbb{Q}_p$, $-x\in\mathbb{Q}_p$
  3. For each $0\neq x\in\mathbb{Q}_p$, $\frac{1}{x}\in\mathbb{Q}_p$
  4. For all $x,y\in\mathbb{Q}_p$, $x+y=y+x$ and $x\cdot y=yx$
  5. For all $x,y,z\in\mathbb{Q}_2$, $(x+y)+z=x+(y+z)$ and $(x\cdot y)\cdot z=x\cdot(y\cdot z)$ and $(x+y)\cdot z = x\cdot z+y\cdot z$
  6. Every Cauchy sequence in $\mathbb{Q}_p$ converges
The first 5 properties show that $\mathbb{Q}_2$ is field. The last property shows that $\mathbb{Q}_2$ is complete; so we can do calculus with the 2-adic numbers. But I want to do algebra with the 2-adic numbers.

Field Extensions

Let $f(x)$ be a polynomial of degree $n$ with $p$-adic coefficients. Suppose that $f$ is irreducible (it can't be factored into polynomials of strictly smaller degree). Let $r$ be a root of $f$. (Technically, we should fix an algebraic closure of $\mathbb{Q}_p$).

Consider the following set: \[ K = \{ b_0+b_1\cdot r+\cdots+b_{n-1}\cdot r^{n-1} : b_i\in\mathbb{Q}_p \} \]

It turns out that $K$ is a vector space over $\mathbb{Q}_p$ and it's a field. Since $K$ contains $\mathbb{Q}_p$ plus some extra elements, we call $K$ an extension of $\mathbb{Q}_p$. As a vector space over $\mathbb{Q}_p$, $K$ has dimension $n$. So we say $K$ is a degree n extension of the $p$-adic numbers.

Masses of Extensions

Suppose $K$ is an extension of $\mathbb{Q}_p$ obtained by adjoining a root $r$ of the irreducible polynomial $f(x)$ of degree $n$.

How many roots of $f$ are contained in $K$? Let $d$ denote this number.

We know $1\leq d\leq n$, since $r\in K$.

In general, $d$ must divide $n$.

Define the mass of $K$ to be $n/d$. If $\text{mass}(K)=1$, then $K$ is called a Galois extension. This really is the ``best'' case.

Galois Groups

Suppose $K$ is a Galois extension of $\mathbb{Q}_2$ of degree $n$ defined by a root $r_1$ of the irreducible polynomial $f$.

Let $r_2, \ldots, r_n$ be the other roots. So $r_i\in K$ for all $i$.

Define functions $\sigma_i: K \to K$ by $\sigma_i(r_1) = r_i$ and $\sigma_i(a)=a$ for all $a\in \mathbb{Q}_2$.

Then each $\sigma_i$ has the following properties:

  1. $\sigma_i(x+y)=\sigma_i(x)+\sigma_i(y)$
  2. $\sigma_i(xy)=\sigma_i(x)\cdot\sigma_i(y)$
  3. $\sigma_i$ is 1-1 and onto
  4. $\sigma_i$ permutes the roots of $f$
  5. The collection of $\sigma_i$ forms a group $G$ under function composition

The group $G$ is called the Galois group of $f$.

Example. Let $f(x)=x^4+4x^2+2$, $r$ a root of $f$, and $K = \mathbb{Q}_2(r)$.

If we factor $f$ over $K$, the roots of the linear factors correspond to the roots of $f$ that are contained in $K$. Using Magma, I found the roots to be: $\textbf{1}=r$, $\textbf{2}=-r$, $\textbf{3}=r^3+3r$, and $\textbf{4}=-r^3-3r$.

We can define four functions, $\sigma_1, \ldots, \sigma_4$, where $\sigma_i$ sends root $\mathbf{1}$ to root $\mathbf{i}$. We can also determine how these functions permute the roots. Here is a summary:

$\sigma_1(r) = r$ ()
$\sigma_2(r) = -r$ (1,2)(3,4)
$\sigma_3(r) = r^3+3r$ (1,3,2,4)
$\sigma_4(r) = -r^3-3r$ (1,4,2,3)

So $G$ is the cyclic group of order 4.


Chad Awtrey, Department of Mathematics and Statistics, Elon University
Sebastian Pauli, Number Theory Group Department of Mathematics and Statistics, UNCG