# Maximal Ideal iff Quotient Ring is Field/Proof 1

## Theorem

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.

## 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.

## Proof

### Maximal Ideal implies Quotient Ring is Field

Let $J$ be a maximal ideal.

Because $J \subset R$, 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 $\struct {R / J, +, \circ}$ has a product inverse 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:

$K = \set {j + r \circ x: j \in J, r \in R}$

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:

$(1): \quad K \ne \O$

Now let $g, h \in K$.

That is:

$g = j_1 + r_1 \circ x, h = j_2 + r_2 \circ x$

Then:

$-h = -j_2 + \paren {-r_2} \circ x$

But $j_1 - j_2 \in J$ from Test for Ideal.

Similarly $-r_2 \in R$.

So $-h \in K$ and we have:

$(2) \quad g + \paren {-h} = \paren {j_1 - j_2} + \paren {r_1 - r_2} \circ x$

Now consider $g \in J, y \in R$.

Then:

$g \circ y = \paren {j_1 + r_1 \circ x} \circ y = \paren {j_1 \circ y} + \paren {r_1 \circ y} \circ x$

which is valid by the fact that $R$ is a commutative ring.

But as $J$ is an ideal:

$\paren {j_1 \circ y} \in J$

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:

 $\ds j$ $\in$ $\ds J$ $\ds \leadsto \ \$ $\ds j + 0_R \circ x$ $\in$ $\ds K$ $\ds \leadsto \ \$ $\ds j$ $\in$ $\ds K$ $\ds \leadsto \ \$ $\ds J$ $\subseteq$ $\ds K$

and because $x = 0_R + 1_R \circ x$ (remember $0_R \in J$), then $x \in K$ too.

So, because $x \notin J$, $K$ is an ideal such that $J \subset K \subseteq R$.

Because $J$ 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:

$1_R + \paren {-s \circ x} = j_0 \in J$

Hence:

$1_R + J = s \circ x + J = \paren {s + J} \circ \paren {x + J}$

So in the commutative ring $\struct {R / J, +, \circ}$, the product inverse of $x + J$ is $s + J$.

The result follows.

$\Box$

### Quotient Ring is Field implies Ideal is Maximal

Let $R / J$ be a field.

Let $K$ be a left ideal of $R$ such that $J \subsetneq K \subseteq R$.

We have that $J$ is the zero of $R / J$.

Let $x \in K \setminus J$.

Because $x \notin J$ then $x + J \ne J$.

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

Hence:

$1_R + J = \paren {s + J} \circ \paren {x + J} = \paren {s \circ x } + J$
$1_R - s \circ x \in J \subsetneq K$

By the definition of an ideal:

$x \in K, s \in R \implies s \circ x \in K$
$1_R - s \circ x \in K, s \circ x \in K \implies \paren {1_R - s \circ x} + \paren {s \circ x} = 1_R \in K$
$1_R \in K \implies \forall y \in R: y \circ 1_R = y \in K$

Hence $K = R$.

The result follows.

$\blacksquare$