p-adic numbers

Formal power series

Let $R$ be a commutative ring with $1$. Formal power series in the indeterminate $x$ with coefficients from ring $R$ are all formal infinite sums:

$\sum_{n=0}^{\infty} a_nx^n = a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots$

The addition and multiplication are defined as if the power series were polynomials of infinite degree. The ring of formal power series with coefficients from ring $R$ is denoted by $R[[x]]$. $R[[x]]$ is a commutative ring with $1$.

Proposition:

$\sum_{n=0}^{\infty} a_nx^n$ is a unit in $R[[x]]$ if and only if $a_0$ is a unit in $R$.

Let’s first observe:

$\frac{1}{1-x} = 1 + x + x^2 + \cdots$

And:

$\frac{1}{1+x} = 1 - x + x^2 - \cdots$

$\sum_{n=0}^{\infty} a_nx^n$ can be written as:

$a_0 \cdot (1 + \sum_{n=1}^{\infty} \frac{a_n}{a_0}x^n)$

Let’s denote $f = \sum_{n=1}^{\infty} \frac{a_n}{a_0}x^n$. Note that we need the inverse of $a_0$ here.

We have:

$\frac{1}{1+f} = 1 - f + f^2 - \cdots$

Thus, the inverse of $\sum_{n=0}^{\infty} a_nx^n$ is:

$a_0^{-1}(1 - f + f^2 - \cdots)$

Definition:

Let $R$ be a ring. Ring $R((x))$ of formal Laurent series with coefficients from $R$ is defined by: $R((x)) = \{ \sum_{n\geq N}^{\infty} a_nx^n \mid a_n \in R$ and $N \in \mathbb{Z} \}$.

A formal Laurent series is a generalization of a formal power series in which finitely many negative exponents are permitted.

Proposition:

If $F$ is a field, then $F((x))$ is a field.

To get a bit of a feeling why this is true, let’s observe whether $\frac{a}{x^2} + \frac{b}{x}$ has an inverse:

$\frac{a}{x^2} + \frac{b}{x} = \frac{a + bx}{x^2}$

The question is whether we can write $\frac{x^2}{a + bx}$ as a formal Laurent series. That’s the same as asking whether $\frac{1}{a + bx}$ can be written as a formal Laurent series. And we can, since:

$\frac{1}{a + bx} = \frac{1}{a} \cdot \frac{1}{1 + a^{-1}bx}$

We can develop $\frac{1}{1 + a^{-1}bx}$ into a formal Laurent series as we did above for $\frac{1}{1 + x}.$

Thus, in $F((x))$, we can find the inverse of $\frac{a}{x^2} + \frac{b}{x}$.

Proposition:

If $F$ is a field, the field of fractions of $F[[x]]$ is the ring $F((x))$.

The proposition states that $a_0 + a_1x + a_2x^2 + \cdots$ has an inverse in $F((x))$. That’s the same as saying that $\frac{1}{a_0 + a_1x + a_2x^2 + \cdots}$ can be written as a formal Laurent series. Indeed, since $a_0$ has an inverse, we can use the same trick as above for $\frac{1}{1 + x}$.

Note that a formal power series is a generalization of a polynomial, where the number of terms is allowed to be infinite and there is no requirement of convergence.

This last thing about convergence is crucial. Note that the equation below is true only for $|x| < 1$:

$\frac{1}{1+x} = 1 - x + x^2 - \cdots$

For $|x| > 1$, we do:

$\frac{1}{1+x} = \frac{1}{x} \cdot \frac{1}{1 + \frac{1}{x}}$

Now, in the expression $\frac{1}{1 + \frac{1}{x}}$, it holds $|\frac{1}{x}| < 1$ and we can use the same trick again. But note that the two series we get for $|x| < 1$ and $|x| > 1$ are different.

But the theory of formal power series ignores the question of convergence by assuming that the variable $x$ does not denote any numerical value.

p-adic

A p-adic integer is an infinite formal sum of the form:

$\sum_{n\geq 0}^{\infty} a_np^n$

Having an infinite formal sum, it means:

$\frac{1}{1-p} = 1 + p + p^2 + p^3 + ...$

That means we ignore the question of convergence as above.

For example:

$\frac{1}{-6} = \frac{1}{1-7} = 1 + 7 + 7^2 + 7^3 + ...$

7-adic number $\frac{-1}{6}$ is thus the infinite series given above.

What about $\frac{1}{2}$? We need to transform the denominator to the form $1 - 7^{i}$.

$\frac{1}{2} = \frac{-24}{1-7^2} = \frac{-(3+3 \cdot 7)}{1 - 7^2} = -(3 + 3 \cdot 7)(1 + 7^2 + 7^4 + ...) = -(3 + 3 \cdot 7 + 3 \cdot 7^2 + 3 \cdot 7^3 ...)$

A wonderful introduction to p-adic numbers is p-adic numbers and Diophantine equations (opens new window). There is an example of computing 7-adic square of $2$.

We first compute $x^2 = 2 \; mod \; 7$. We get $x = 3$ or $x = -3$. Let’s take the first solution and denote it by $x_0$.

Then we compute $x^2 = 2 \; mod \; {7^2}$. We get $x_1 = 3 + 1 \cdot 7 = 10$

Then we compute $x^2 = 2 \; mod \; {7^3}$. We get $x_2 = 3 + 1 \cdot 7 + 2 \cdot 7^2 = 108$.

The sequence $(x_0, x_1, x_2, ...)$ is a 7-adic approximation of $\sqrt 2$.

Note that $x_{k-1} = x_k \; mod \; {7^k}$ for $k = 1, 2, 3, ...$

In fact, this is how p-adic number is defined.

Definition: Let $p$ be a prime number. We say that a sequence of integers $(x_0, x_1, x_2, ...)$ gives a p-adic integer if

$x_k = x_{k-1} \; mod \; {x^k} \; for \; k = 1,2,3,...$

$(x_0, x_1, x_2, ...)$ is called a canonical sequence. The p-adic expansion in the above case is $(3, 1, 2, ...).$