# 02.4.素イデアルと極大イデアル

## Proposition

### Statement

$D$:Integral Domain
$p \in D$
Then
$p$:Prime Element $\Leftrightarrow$ $(p)$:Prime Ideal.

## Proof

$a,b \in D$
then
$p$:Prime Element
$\Leftrightarrow p|ab \Rightarrow p|a \text{ or } p|b$
$\Leftrightarrow ab \in (p) \Rightarrow a \in (p) \text{ or } b \in (p)$
$(p)$:Prime Ideal
QED

## Proposition

### Statement

$R$:Ring
$I \varsubsetneqq R$:Ideal
Then
$I$:Prime Ideal $\Leftrightarrow R/I$:Integral Domain

### Proof

($\Rightarrow$)
$\bar a (= a + I), \bar b (= b + I) \in R/I$
then
$\bar a \bar b = 0$
$\Rightarrow \bar{ab} = 0$
$\Rightarrow ab \in I$
$\Rightarrow a \in I$ or $b \in I \quad$($\because I$:素イデアル)
$\Rightarrow \bar a = 0$ or $\bar b = 0$
($\Leftarrow$)
$ab \in I$
$\Rightarrow \bar{ab} = 0$
$\Rightarrow \bar a \bar b = 0$
$\Rightarrow \bar a = 0$ or $\bar b = 0 \quad$($\because R/I$:整域)
$\Rightarrow a \in I$ or $b \in I$
QED.

## Proposition

### Statement

$R$:Ring
$I \varsubsetneqq R$:Ideal
Then
$I$:Maximal Ideal $\Leftrightarrow R/I$:Field

### Proof

($\Rightarrow$)
$(R/I)^\times = (R/I)-\{I\}$
$a + I \in (R/I)^\times$
$a \notin I$
$I + (a)$:Ideal
$I \varsubsetneqq I + (a)$
$\therefore I + (a) = R$
$\therefore \exists (b \in I, c \in (a)), b + ac = 1 \in R$
$\therefore (a + I)(c + I) = ac + I = 1 - b + I = 1 + I$
$\therefore c + I = (a + I)^{-1}$
($\Leftarrow$)
$I \varsubsetneqq J \subset R$
$\exists a, a \notin I, a \in J$
$a + I \in (R/I)^\times \therefore \exists a',(a + I)(a' + I) = aa' + I = 1 + I$
$\therefore \exists c \in I, aa' + c = 1$
$a \in J, c \in I \subset J, \therefore aa' + c = 1 \in J$
$\therefore J = R$

## Proposition

### Statement

$I$:Maximal Ideal $\Rightarrow$ $I$:Prime Ideal

### Proof

$I$:Maximal Ideal $\Leftrightarrow R/I$:Field
$\Rightarrow R/I$:Domain
$\Leftrightarrow I$:Prime Ideal

## Statement

$R$:Ring
$I$:Ideal ($I\subset R$)
then
$\exists M \supset I \text{ s.t. Maximal Ideal}$

## Proof

Let
$A = \{J \subset R|I \subset J, J:\text{Ideal}\}$
then
$(A, \subset)$ is a 半順序集合

$I \in A \therefore A \neq \emptyset$
Let
$\forall B = \{J_\lambda\}_{\lambda \in \Lambda}$:全順序部分集合 of $A$
then
$U = \cup_{\lambda \in \Lambda} J_\lambda \in A \quad(\because (I \subset U), (\forall r \in R, \forall u \in U \Rightarrow \exists J_\lambda, u \in J_\lambda, \therefore ru \in J_\lambda \subset U))$
and $U$ is upper bound of $B$
$\therefore (A,\subset)$ is 帰納的順序集合
$\exists M \supset I \text{ s.t. Maximal Ideal} \quad (\because \text{ツォルンの補題})$