Question

In a commutative ring without unity a maximal ideal
(a) need to be prime
(b) need not to be prime
(c) both (a) and (b)
(d) none of these

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO CHOOSE THE CORRECT OPTION

In a commutative ring without unity a maximal ideal

(a) need to be prime

(b) need not to be prime

(c) both (a) and (b)

(d) none of these

ANSWER

In a commutative ring without unity a maximal ideal

(b) need not to be prime

EXPLANATION

$\sf{Consider \: the \: ideal \: 4 \mathbb{Z} \: in \: the \: ring \: 2 \mathbb{Z}}$

$\sf{Since \: \: 2 \in 2 \mathbb{Z} \: \: but \: \: 2 \notin \: 4 \mathbb{Z} \: }$

$\sf{So \: 4 \mathbb{Z} \: is \: a \: proper \: subset \: of \: 2 \mathbb{Z}}$

$\sf{Let \: U \: be \: an \: ideal \: of \: the \: ring \: \: 2 \mathbb{Z}}$

$\sf{Which \: properly \: contains \: \: 4 \mathbb{Z}}$

$\sf{Then \: there \: exists \: an \: element \: \: p \in \: U}$

$\sf{Such \: that \: \: p \notin \: 4 \mathbb{Z}}$

$\sf{Then \: \: p = 4m+2 \: \: where \: m \: is \: an \: integer}$

$\sf{4m \in 4 \mathbb{Z} \: \: therefore \: \: 4m \in U}$

$\sf{Since \: U \: is \: an \: Ideal \: , \: p \in U \: and \: 4m \in U}$

$\implies \sf{ (p-4m) \in U}$

$\sf{So \: \: 2 \in U}$

$\sf{And \: this \: again \: implies \: that \: U = 2 \mathbb{Z}}$

$\sf{Thus \: 4 \mathbb{Z} \subset \: U \: \subset \: 2 \mathbb{Z}}$

$\sf{ \implies \: \: U= 2 \mathbb{Z }}$

$\sf{This \: proves \: that \: 4 \mathbb{Z} \: is \: a \: maximal \: Ideal \: in \: 2 \mathbb{Z}}$

$\sf{Since \: 2.2 \in \: 4 \mathbb{Z} \: \: but \: 2 \notin 4 \mathbb{Z}}$

$\sf{The \: ideal \: 4 \mathbb{Z }\: in \: the \: ring \: 2 \mathbb{Z } \: is \: not \: prime}$

$\sf{Therefore \: 4 \mathbb{Z} \: \: is \: a \: maximal \: Ideal}$

$\sf{but \: not \: prime \: ideal \: in \: the \: ring \: \: 2 \mathbb{Z}}$

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