$\mathbb{Z}_p$ , $\mathbb{Q}_p$ , and the ring of Witt vectors

No.05:

In this lecture, rings are assumed to be unital, associative and commutative unless otherwise specified.

DEFINITION 05.1   A (unital commutative) ring $A$ is said to be a local ring if it has only one maximal ideal.

LEMMA 05.2   Let $A$ be a ring. Then the following conditions are equivalent:

1. $A$ is a local ring.
2. $A\setminus A^\times$ forms an ideal of $A$ .

PROPOSITION 05.3   $\mathbb{Z}_p$ is a local ring. Its maximal ideal is equal to $p \mathbb{Z}_p$ .

We may do some ``analysis'' such as Newton's method to obtain some solution to algebraic equations.

Newton's method for approximating a solution of algebraic equation.

Let us solve an equation

$$x^2=2$$

in $\mathbb{Z}_7$ . We first note that

$$3^2\equiv 2\quad (7)$$

hold. So let us put $x_0=3=[0.3]_7$ as the first approximation of the solution. The Newton method tells us that for an approximation $x$ of the equation $x^2=2$ , a number $x'$ calculated as

$$x'=\frac{1}{2}(x+ \frac{2}{x})$$

gives a better approximation.

$$x_0'=\frac{1}{2}([0.3]_7+[0.3\dot{2}]_7=[0.3\dot{1}]_7$$

So $[0.3\dot{1}]_7$ is a better approximation of the solution. In order to make the calculation easier, let us choose $x_1=[0.31]_7$ (insted of $x_0'$ ) as a second approximation.

$$x_1' =\frac{1}{2}([0.31]_7+2/[0.31]_7) =\frac{1}{2}([0.31]_7+[0.3\dot{1}45\dot{2}]_7)\fallingdotseq [0.312]_7$$

We choose $x_2=[0.312]_7$ as a second approximation.

$$x_2' =\frac{1}{2}([0.312]_7+2/[0.3\dot{1}2534066\dot{2}]_7)\fallingdotseq [0.31261]_7$$

We choose $x_3=[0.31261]_7$ as a third approximation.

$$x_3' = =\frac{1}{2}([0.31261]_7+[0.3126142465066\dots]_7)\fallingdotseq [0.312612124...]_7$$

We choose $x_4=[0.312612124]_7$ as a third approximation.

$$x_4' =\frac{1}{2}([0.312612124]_7+[0.312612124565220422662213135351\dots]_7)$$

$$\fallingdotseq [0.3126121246621102]_7$$

EXERCISE 05.1   Compute $[0.5]_7/[0.11]_7$

EXERCISE 05.2   Find a solution to

$$x^3\equiv 5 \pmod {11^5}$$

such that $x \equiv 3 \pmod {11}$ .