📂Abstract Algebra

Maximal Ideal

Definition ¹

An ideal $M \ne R$ of a ring $R$ is called a maximal ideal of $R$ if it is not contained within any other ideal $N \ne R$ of $R$, other than $R$ itself. In other words, $M \subsetneq R$ being a maximal ideal means the following. $$ \nexists N : M \subsetneq N \subsetneq R $$

Explanation

What is referred to as ‘maximal’ in algebra is almost the same as maximal in set theory.

Of course, this definition alone does not ensure uniqueness. For example, in the ring of integers $\mathbb{Z}$, both $2 \mathbb{Z}$ and $3 \mathbb{Z}$ do not have a superideal outside of $\mathbb{Z}$, thus they are both maximal ideals of $\mathbb{Z}$. Similarly, for any prime number $p$, $p \mathbb{Z}$ becomes a maximal ideal of $\mathbb{Z}$.

Meanwhile, maximal ideals have the following property in relation to fields, which is similar to the relationship between prime ideals and integral domains.

Theorem

Let the commutative ring $R$ have a unity $1_{R}$.

• $M$ is a maximal ideal of $R$ $\iff$ $R / M$ is a field

Proof

$( \implies )$

$R / M$ is a commutative ring with unity $( 1_{R} + M )$. For any non-unity $a \ne 1_{R}$, let’s say $( a + M ) \in R / M$, for all $r \in R$ there’s $a r \in R$, and since $M$ is an ideal of $R$, then $M + a R = R$ applies. Hence, $$ 1_{R} = m + ar $$ there must exist $m \in M$, $r \in R$ satisfying this, which means in $R / M$, $$ 1_{R} + M = ar + M $$ If considered separately, $$ 1_{R} + M = (a + M)(r + M) $$ thus, for all $(a + M)$ there exists an inverse $(a + M)^{-1} = (r + M)$. Therefore, $R / M$ becomes a field.

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$( \impliedby )$

Let’s assume there’s an ideal $N \triangleleft R$ that satisfies $M \subsetneq N \subsetneq R$, thus making $M$ not maximal.

Then, there must exist some element $n \in N$ that belongs to $N$ but not to $M$. Since $R / M$ was assumed a field, $$ (n + M) (s + M ) = ns + M = 1_{R} + M $$ an $s \in R$ satisfying this must exist. Now let’s assume $n ' := ( 1 - ns ) \in M \subsetneq N$, $$ 1_{R} = ( n' + ns ) \in N $$ But since the ideal $N$ has a unity $1_{R}$, it leads to $N = R$, which is a contradiction to our assumption.

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