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.