Maximal Ideal iff Quotient Ring is Field/Proof 3

Proof
Let $J$ be a maximal ideal.

We have by definition of quotient ring that $J$ is the zero element of $R / J$.

Let $A \in R / J$ be a non-zero element of $R / J$.

Let $x \in A$.

Since $A \ne J$, we have that $x \notin J$.

Let the ideal $K = J + A$ of $R$ be formed.

This contains all the elements of the form $j + r a$, with $j \in J$ and $r \in R$.

As $J$ is maximal and $J \subsetneq K$, it follows that:
 * $K = R$

and so:
 * $1_R \in K$

That is:
 * $\exists j \in J, r \in R: j + r a = 1_R$

Thus:
 * $\paren {r + J} \paren {a + J} = \paren {1 - u} J = 1_R + J$

and so $\paren {r + J}$ is the product inverse of $\paren {a + J}$.

So every non-zero element of $R / J$ has a product inverse.

That is, $R / J$ is a field.

Let $R / J$ be a field.

Let $K$ be an ideal of $R$ such that:
 * $J \subsetneq K \subseteq R$

Let $a \in K$ such that $a \notin J$.

Then:
 * $J + \ideal a \subsetneq K$

But as $a \notin J$, we have that $a + J$ is a non-zero element of $R / J$.

Thus as $R / J$ is a field, $a + J$ has a product inverse $r + J$:
 * $\paren {r + J} \paren {a + J} = 1_R + J$

So:
 * $\exists r \in R, j \in J: r a + \paren {-1_R} = j$

That is:
 * $r a + \paren {-1_R} \in J$

So:
 * $j + r a = 1_R$

and from Ideal of Unit is Whole Ring: Corollary this implies:
 * $J + \ideal a = R$

So:
 * $K = R$

and it follows by definition that $J$ is maximal.