# 02.2.多項式環と単項イデアル整域

## Proposition

### Statement

$F$:Field
$ab = 0 \Rightarrow a = 0$ or $b = 0$

### Proof

if $a \neq 0 \Rightarrow b = 1b = a^{-1}ab = a^{-1}0 = 0$
if $b \neq 0 \Rightarrow a = a1 = abb^{-1} = 0b^{-1} = 0$
QED

## Proposition

### Statement

$R$:Domain $\Rightarrow R[x]$:Domain

### Proof

$f(x),g(x) \in R[x]$
$f(x) = \sum_{i = 0}^n a_i x^i \quad (a_n \neq 0)$
$g(x) = \sum_{i = 0}^m g_i x^i \quad (b_m \neq 0)$
$f(x)g(x) = 0$
Then $f(x)g(x)$ の最高次数の係数は$a_n b_m$
$f(x)g(x) = 0$ より $a_n b_m = 0 \therefore a_n = 0$ or $b_m = 0$
QED

## Proposition

### Statement

$F$:Field
$F[x]$:polynomial over $F$
then $F[x]$ is a Principal Ideal Domain.

### Proof

Let
$I \subset F[x]$
$I$:Ideal
if $I$ is $(0)$, then $I$ is Principal Ideal
if $I$ is not $(0)$, then
$\exists f(x) \in I \subset F[x]$ such that
$f(x) \neq 0$ and $\forall g(x) \in I,\text{deg}f(x) \le \text{deg}g(x)$
$\exists p(x), q(x) \in F[x],g(x) = p(x)f(x) + q(x) \quad (\text{deg} q(x) \lt \text{deg} f(x)) \quad (\because \text{Thm.剰余定理})$
$q(x) = g(x) - p(x)f(x) \in I.$
$\therefore q(x) \in I, \text{deg}q(x) \lt \text{deg}f(x)$ leads to contradiction.
$\therefore q(x) = 0$
$\therefore g(x) = p(x)f(x) \in (f(x))$
QED

## Proposition

### Statement

$D$:Integral Domain
$a,b \in D$
then
$a \sim b \Leftrightarrow a|b, b|a$

### Proof

($\Rightarrow$)
$\exists c \in D,c:\text{単元},a = bc \Rightarrow b|a$
$\Rightarrow ac^{-1} = b \Rightarrow a|b$

($\Leftarrow$)
$a|b \Rightarrow \exists c \in D, ac = b$
$b|a \Rightarrow \exists d \in D, bd = a$
$acd = bd = a$
$0 = acd - a = a(cd - 1)$
$cd = 1 \therefore c,d$:単元
$\Rightarrow a \sim b$
QED.

## Proposition

### Statement

$D$:Integral Domain
$p \in D$
$p \neq 0$
then
$p$:素元 $\Rightarrow$ $p$:既約元

### Proof

$p$:素元
$a,b \in D, p = ab$
$\Rightarrow p|a \text{ or } p|b (\because p$ is 素元)
if $p|a$ then $\exists c \in D, pc = a$
$p = pcb \Rightarrow p(1 - cb) = 0 \Rightarrow cb = 1$
$\therefore b$:単元
if $p|b$ then $a$:単元
$\therefore p$:既約元
QED.

## Proposition

PIDにおける既約元と極大イデアル

### Statement

$D$:PID
$p \in R$
then
$p$:既約元 $\Rightarrow$ $(p)$:極大イデアル

### Proof

Suppose
$I$:Ideal
$(p) \subset I$
then
$\exists a \in I, I = (a) \because D$:PID
$p \in (p) \subset I \therefore \exists c \in D,p = ac \quad (c \in D)$
$p$:既約元より$a$:単元または$c$:単元
if $a$:単元, then $I = R$
$\because \exists d \in D, da = 1 \therefore \forall x \in D, x = xda \in (a) = I$
if $c$:単元, then $I = (p)$
$\because p \sim a$
$\Rightarrow p = ac, a = pc^{-1}$
$\Rightarrow (p) = (ac) \subset (a), (a) = (pc^{-1}) \subset (p)$
$\therefore (p) = (a) = I$
QED.

## Proposition

PIDにおける素元と既約元

### Statement

$D$:PID
$p \in D$
then
$p$:素元 $\Leftrightarrow$ $p$:既約元

### Proof

($\Rightarrow$)
$\text{Prop.素元と既約元}$
($\Leftarrow$)
Suppose
$p$:既約元
$a,b \in D, p|ab$
then
$(p)$:極大イデアル($\because\text{Prop.PIDにおける既約元と極大イデアル}$)
generally $(p) \subset (a,p)$
$\therefore (a,p) = (p)$ or $(a,p) = D$

Suppose $(a,p) = D$ and $(b,p) = D$
$\exists a_1,a_2 \in D, a_1 a + a_2 p = 1 \in D$
$\exists b_1,b_2 \in D, b_1 b + b_2 p = 1 \in D$
$\Rightarrow a_1 b_1 ab + (a_1 a b_2 + b_1 b a_2 + a_2 b_2 p)p = 1$
$a_1 b_1 ab \in (ab) \subset (p)$
($\because p|ab \exists c, pc = ab \therefore (ab) = (pc) \subset (p)$)
$\therefore a_1 b_1 ab + (a_1 a b_2 + b_1 b a_2 + a_2 b_2 p)p = 1 \in (p)$
$\therefore (p) = D$ contradiction
$\therefore (a,p) = (p)$ or $(b,p) = (p)$
$\Rightarrow a \in (p)$ or $b \in (p)$
$\Rightarrow p|a$ or $p|b$
QED

## Proposition

PIDにおける素イデアルと極大イデアル

### Statement

$D$:PID
$I$:素イデアル($I \neq \{0\}$)
then
$I$:極大イデアル

### Proof

$\exists p \in I, I = (p) \quad (p \neq 0) (\because I:\text{PID})$
$p$:素元($\because p$:素元でなければ、
$\exists a,b \in D \text{ s.t.} p|ab \Rightarrow p \nmid a \text{ and } p \nmid b$
すなわち $ab \in (p) \Rightarrow a \notin (p) \text{ and } b \notin (p)$
contradict $I$:素イデアル.)
$p$:既約元($\because \text{Prop.PIDにおける素元と既約元}$)
$(p)$:極大イデアル($\because \text{Prop.PIDにおける既約元と極大イデアル}$)
QED