📂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

In algebra, ‘maximal’ is almost the same as maximal in set theory.

Obviously, the uniqueness cannot be guaranteed by definition alone. For example, in the integer ring $\mathbb{Z}$, both $2 \mathbb{Z}$ and $3 \mathbb{Z}$ have no superideal besides $\mathbb{Z}$, making them the maximal ideals of $\mathbb{Z}$. Similarly, for any prime number $p$, $p \mathbb{Z}$ are all maximal ideals of $\mathbb{Z}$.

On the other hand, maximal ideals have the following properties in relation to fields. This is similar to the relationship between prime ideals and integral domains.

Theorem

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

• $M$ is a maximal ideal of $R$ if and only if $\iff$ $R / M$ is a field.

Proof

$( \implies )$

$R / M$ is a commutative ring with unity $( 1_{R} + M )$. For any non-identity element $a \notin M$ of $R$, let it be $( a + M ) \in R / M$, then for any $r \in R$, $a r \in R$ holds, and since $M$ is a maximal ideal of $R$, $$R = M + a R$$ holds. That is, there exist $m \in M$ and $r \in R$ satisfying $$1_{R} = m + ar$$, and since $R / M$ has unity $( 1_{R} + M )$ as mentioned before, $$1_{R} + M = ar + M$$ can be expressed as follows. Factoring out gives $$1_{R} + M = (a + M)(r + M)$$. Therefore, there exists an inverse $(a + M)^{-1} = (r + M)$ for every non-identity element $(a + M)$ of $M$. Thus, $R / M$ becomes a field.

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$( \impliedby )$

Assume that there exists an ideal $N \triangleleft R$ which satisfies $M \subsetneq N \subsetneq R$, making $M$ not maximal.

There will exist an element $n \in N$ which does not belong to $M$ but belongs to $N$. Since $R / M$ is a field, $$(n + M) (s + M ) = ns + M = 1_{R} + M$$ implies the existence of $s \in R$. Now, let it be $n ' := ( 1 - ns ) \in M \subsetneq N$, we have $$1_{R} = ( n' + ns ) \in N$$. However, ideal $N$ having element $1_{R}$ implies $N = R$, which contradicts our assumption.

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