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 1

 * 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 not the zero element of $R / J$.

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

Now $0_R \in K$ as $0_R \in J$ and $0_R \in R$, giving $0_R + 0_R \circ x = 0_R$.

So:
 * $(1): \quad K \ne \varnothing$.

Now let $g, h \in K$.

That is, $g = j_1 + r_1 \circ x, h = j_2 + r_2 \circ x$.

Then:
 * $-h = -j_2 + \left({-r_2}\right) \circ x$

But $j_1 - j_2 \in J$ from Test for Ideal.

Similarly $-r_2 \in R$.

So $-h \in K$ and we have:
 * $(2) \quad g + \left({-h}\right) = \left({j_1 - j_2}\right) + \left({r_1 - r_2}\right) \circ x $

Now consider $g \in J, y \in R$ Then:
 * $g \circ r = \left({j_1 + r_1 \circ x}\right) \circ y = \left({j_1 \circ y}\right) + \left({r_1 \circ y}\right) \circ x$

which is valid by the fact that $R$ is commutative.

But as $J$ is an ideal, $\left({j_1 \circ y}\right) \in J$, while $r_1 \circ y \in R$.

Thus:
 * $(3) \quad g \circ y \in K$

and similarly:
 * $(3) \quad y \circ g \in K$

So we can apply Test for Ideal on statements $(1)$ to $(3)$, and we see that $K$ is an ideal of $R$.

Now:

... 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.

Proof 2
Let $\mathbb L_J$ be the set of all ideals of $R$ which contain $J$.

Let the poset $\left({\mathbb L \left({R / J}\right), \subseteq}\right)$ be the set of all ideals of $R / J$.

Let the mapping $\Phi_J: \left({\mathbb L_J, \subseteq}\right) \to \left({\mathbb L \left({R / J}\right), \subseteq}\right)$ be defined as:


 * $\forall a \in \mathbb L_J: \Phi_J \left({a}\right) = q_J \left({a}\right)$

where $q_J: a \to a / J$ is the canonical epimorphism from $a$ to $a / J$ from the definition of quotient ring.

Then from Ideals Containing Ideal Isomorphic to Quotient Ring, $\Phi_J$ is an isomorphism.

Now from Quotient Ring Defined by Ring Itself is Null Ideal, $q_J \left({J}\right)$ is the null ideal of $R / J$.

At the same time, $q_J \left({R}\right)$ is the entire ring $R / J$.

If $R / J$ is not the Null Ring then $R / J$ is a commutative ring with unity by Epimorphism from Ring and Epimorphism Preserves Commutativity.

By definition, $J$ is a maximal ideal of $R$ iff $\mathbb L_J = \left\{{J, R}\right\}$ and $J$ is a proper ideal of $R$.

By Ideals of a Field, $R / J$ is a field iff $\mathbb L \left({R / J}\right) = \left\{{q_J \left({J}\right), q_J \left({R}\right)}\right\}$ and the null ideal $q_J \left({J}\right)$ is a proper ideal of $R / J$.

As $\Phi_J: \mathbb L_J \to \mathbb L \left({R / J}\right)$ is an isomorphism, $J$ is a maximal ideal iff $J$ is a field.