Maximal Ideal iff Quotient Ring is Field/Proof 2

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
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 Preserves Rings 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.