Maximal Ideal iff Quotient Ring is Field

Theorem

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

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

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

Proof 1

Since $J \subset R$, it follows from Commutative Quotient Ring and Quotient 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 $x + J \ne J$, i.e. $x \notin J$.

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:

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

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 we can apply Test for Ideal on statements $(1)$ to $(3)$, and we see that $K$ is an ideal of $R$.

Now:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle j\)	\(\in\)	\(\displaystyle J\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle j + 0_R \circ x\)	\(\in\)	\(\displaystyle K\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle j\)	\(\in\)	\(\displaystyle K\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle J\)	\(\subseteq\)	\(\displaystyle K\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

... 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.

$\blacksquare$

Proof 2

Let $\mathbb L_J$ be the set of all ideals of $R$ which contain $J$.

Let the poset $\left({\mathbb L \left({R / J}\right), \subseteq}\right)$ be the set of all ideals of $R / J$.

Let the mapping $\Phi_J: \left({\mathbb L_J, \subseteq}\right) \to \left({\mathbb L \left({R / J}\right), \subseteq}\right)$ be defined as:

$\forall a \in \mathbb L_J: \Phi_J \left({a}\right) = q_J \left({a}\right)$

where $q_J: a \to a / J$ is the canonical epimorphism from $a$ to $a / J$ from the definition of quotient ring.

Then from Ideals Containing Ideal Isomorphic to Quotient Ring, $\Phi_J$ is an isomorphism.

Now from Quotient Ring Defined by Ring Itself is Null Ideal, $q_J \left({J}\right)$ is the null ideal of $R / J$.

At the same time, $q_J \left({R}\right)$ is the entire ring $R / J$.

If $R / J$ is not the Null Ring then $R / J$ is a commutative ring with unity by Epimorphism Preserves Rings and Epimorphism Preserves Commutativity.

By definition, $J$ is a maximal ideal of $R$ iff $\mathbb L_J = \left\{{J, R}\right\}$ and $J$ is a proper ideal of $R$.

By Ideals of a Field, $R / J$ is a field iff $\mathbb L \left({R / J}\right) = \left\{{q_J \left({J}\right), q_J \left({R}\right)}\right\}$ and the null ideal $q_J \left({J}\right)$ is a proper ideal of $R / J$.

As $\Phi_J: \mathbb L_J \to \mathbb L \left({R / J}\right)$ is an isomorphism, $J$ is a maximal ideal iff $J$ is a field.

$\blacksquare$

Sources

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 5.23$: Theorem $43$