Maximal Ideal iff Quotient Ring is Field

Theorem
Let $$\left({R, +, \circ}\right)$$ be a commutative ring with unity whose zero is $$0_R$$ and whose unity is $$1_R$$.

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

Then $$J$$ is a maximal ideal iff the Quotient Ring $$R / J$$ is a field.

Proof

 * Since $$J \subset R$$, it follows from Commutative Quotient Ring and Quotient Ring with Unity that $$R / J$$ is a commutative ring with unity.


 * We now need to prove that every non-zero element of $$\left({R / J, +, \circ}\right)$$ has an inverse for $$\circ$$ in $$R / J$$.

Let $$x \in R$$ such that $$x + J \ne J$$, i.e. $$x \notin J$$.

Thus $$x + J \in R / J$$ is non-zero.

Take $$K \subseteq R$$ such that $$K = \left\{{j + r \circ x: j \in J, r \in R}\right\}$$, that is, the set of all elements of $$R$$ which can be expressed as a sum of an element of $$J$$ and a product in $$R$$ of $$x$$.

From Test for Ideal it can be shown that $$K$$ is an ideal of $$R$$.

$$ $$ $$ $$

... and since $$x = 0_R + 1_R \circ x$$ (remember $$0_R \in J$$), then $$x \in K$$ too.

So, since $$x \notin J$$, $$K$$ is an ideal such that $$J \subset K \subseteq R$$.

Since $$J$$ is a maximal ideal, then $$K = R$$.

Thus $$1_R \in K$$ and thus $$\exists j_0 \in J, s \in R: 1_R = j_0 + s \circ x$$.

So $$1_R + \left({- s \circ x}\right) = j_0 \in J$$.

Hence $$1_R + J = s \circ x + J = \left({s + J}\right) \circ \left({x + J}\right)$$.

So in the commutative ring $$\left({R / J, +, \circ}\right)$$, the inverse of $$x + J$$ is $$s + J$$.

The result follows.