Maximal Ideal iff Quotient Ring is Field/Proof 1/Maximal Ideal implies Quotient Ring is Field

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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 $R$.


Let $J$ be a maximal ideal.

Then the quotient ring $R / J$ is a field.


Proof

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.

$\blacksquare$


Sources

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