Maximal Ideal iff Quotient Ring is Field/Proof 1/Quotient Ring is Field implies Ideal is Maximal

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Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $J$ be an ideal of $R$.

Let the quotient ring $R / J$ be a field.

Then $J$ is a maximal ideal.


Let $R / J$ be a field.

Let $K$ be a left ideal of $R$ such that $J \subsetneq K \subseteq R$.

We have that $J$ is the zero of $R / J$.

Let $x \in K \setminus J$.

Because $x \notin J$ then $x + J \ne J$.

Because $R / J$ is a field then $x + J \in R / J$ has a product inverse, say $s + J$.


$1_R + J = \paren {s + J} \circ \paren {x + J} = \paren {s \circ x } + J$

By Left Cosets are Equal iff Product with Inverse in Subgroup:

$1_R - s \circ x \in J \subsetneq K$

By the definition of an ideal:

$x \in K, s \in R \implies s \circ x \in K$
$1_R - s \circ x \in K, s \circ x \in K \implies \paren {1_R - s \circ x} + \paren {s \circ x} = 1_R \in K$
$1_R \in K \implies \forall y \in R: y \circ 1_R = y \in K$

Hence $K = R$.

The result follows.