I is a maximal ideal iff R/I is a field

Dependencies:

Let $R$ be a commutative ring with unity. Let $I$ be a proper ideal of $R$ .

$I$ is defined to be a maximal ideal iff no proper ideal of $R$ is a proper superset of $I$ .

$I$ is a maximal ideal iff $R / I$ is a field.

Proof

Since $R$ is a commutative ring with unity, $R / I$ is a commutative ring with unity $1 + I$ .

Proof of 'only-if' part

Let $a + I \in R / I$ be a non-zero element. Thus, $a \notin I$ . Since $0 \in I$ , $a \neq 0$ .

Let $J = {r a + i : r \in R \land i \in I}$ .

$0 = 0 a + 0 \in J$ , so $J$ is non-empty.
Let $r_{1} a + i_{1}, r_{2} a + i_{2} \in J$ . Then $(r_{1} a + i_{1}) - (r_{2} a + i_{2}) = (r_{1} - r_{2}) a + (i_{1} - i_{2}) \in J$ .
Let $r_{1} a + i_{1}, r_{2} a + i_{2} \in J$ . Then $(r_{1} a + i_{1}) (r_{2} a + i_{2}) = (r_{1} a r_{2}) a + (r_{1} a i_{2} + i_{1} r_{2} a + i_{1} i_{2}) \in J$ .
Let $r^{'} \in R \land r a + i \in J$ . Then $r^{'} (r a + i) = (r^{'} r) a + r^{'} i \in J$ and $(r a + i) r^{'} = (r r^{'}) a + i r^{'} \in J$ .

Therefore, $J$ is an ideal of $R$ .

Let $i \in I$ . Then $i = 0 a + i \in J$ . Therefore, $I \subseteq J$ . $a = 1 a + 0 \in J$ , but $a \notin I$ . Therefore, $I \subset J$ and $I \neq J$ .

Let $I$ be a maximal ideal. This means that $J = R$ .

Since $1 \in R, 1 \in J$ , which means $\exists b \exists i, 1 = b a + i$ . $\Rightarrow 1 + I = b a + I = (b + I) (a + I)$ . Therefore, every non-0 element $a + I$ in $R / I$ has an inverse. Therefore, $R / I$ is a field.

Proof of 'if' part

Let $R / I$ be a field. So $0 + I, 1 + I \in R / I$ . Therefore, $I \neq R$ .

Since $I$ is not a maximal ideal, let $J$ be an ideal of $R$ which is a superset of $I$ . Let $a \in J - I$ . Since $a \notin I, a + I \neq 0 + I$ .

Since $R / I$ is a field and $a + I$ is non-0, $a + I$ has an inverse. Therefore, $\exists b, (a + I) (b + I) = a b + I = 1 + I$ . Therefore, $\exists i, a b + i = 1$ .

$b a \in J$ because $a \in J$ and absorption. $b a + i \in J$ because $I \subset J$ and closure of $J$ . Therefore, $1 \in J$ . By absorption, $\forall r \in R, r \in J$ . Therefore, $R = J$ . Therefore, $I$ is maximal.

Dependency for: None

Info:

Depth: 6
Number of transitive dependencies: 14