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

No.05: \fbox{$\mathbb {Z}_p$ as a local ring.}

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

$\displaystyle x^2=2
$

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

% latex2html id marker 604
$\displaystyle 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

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

gives a better approximation.

$\displaystyle 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.

% latex2html id marker 624
$\displaystyle 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.

% latex2html id marker 628
$\displaystyle 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.

% latex2html id marker 632
$\displaystyle 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.

    $\displaystyle x_4'$ $\displaystyle =\frac{1}{2}([0.312612124]_7+[0.312612124565220422662213135351\dots]_7)$
      % latex2html id marker 637
$\displaystyle \fallingdotseq [0.3126121246621102]_7$

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

EXERCISE 05.2   Find a solution to

% latex2html id marker 651
$\displaystyle x^3\equiv 5 \pmod {11^5}
$

such that % latex2html id marker 653
$ x \equiv 3 \pmod {11}$ .


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2008-06-10