Maximal Ideal iff Quotient Ring is Field/Proof 1

Proof Outline
The hard part is proving the existence of inverses.

Take an element $x$ of the Ring to invert.

Define a set $K$, that contains the Ideal and is contained by the ring.

It is the set of all members of the Ideal, each added to a multiple of $x$.

Prove that this set is an ideal and contains our original maximal ideal.

$(1) \implies (2)$
Since $J \subset R$, it follows from Quotient Ring of Commutative Ring is Commutative and Quotient Ring of Ring with Unity is 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 $x + J \ne J$, that is: $x \notin J$.

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:
 * $g \circ y = \left({j_1 + r_1 \circ x}\right) \circ y = \left({j_1 \circ y}\right) + \left({r_1 \circ y}\right) \circ x$

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 Test for Ideal can be applied to statements $(1)$ to $(3)$, and it is seen 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.

$(2) \implies (1)$
Let $K$ be a left ideal of $R$ such that $J \subsetneq K \subset R$.

Let $x \in K \setminus J$. Since $x \notin J$ then $x + J \ne J$, the zero in $R / J$.

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

That is:
 * $1_R + J = \paren {s + J} \circ \paren {x + J} = \paren {s \circ x } + J$

By Left Cosets are Equal iff Product with Inverse in Subgroup then:
 * $1_R - s \circ x \in J \subsetneq K$

By the definition of an ideal then:
 * $x \in K$ and $s \in R \implies s \circ x \in K$
 * $1_R - s \circ x \in K$ and $s \circ x \in K \implies \paren {1_R - s \circ x } + \paren {s \circ x } = 1_R \in K$
 * $1_R \in K \implies \forall y \in R, y \circ 1_R = y \in K$

Hence $K = R$

The result follows.