Maximal Ideal iff Quotient Ring is Field/Proof 3

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Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $J$ be an ideal of $R$.

The following are equivalent:

$(1): \quad$ $J$ is a maximal ideal.
$(2): \quad$ The quotient ring $R / J$ is a field.


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:

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


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

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 $a \notin J$.


$J + \ideal a \subsetneq K$

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 $r + J$:

$\paren {r + J} \paren {a + J} = 1_R + J$


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

That is:

$r a + \paren {-1_R} \in J$


$j + r a = 1_R$

and from Ideal of Unit is Whole Ring: Corollary this implies:

$J + \ideal a = R$


$K = R$

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