There exists no other proper ideal J of R so that I ⊊ J.

For any ideal J with I ⊆ J, either J = I or J = R.

The quotient ring R/I is a simple ring.

If F is a field, then the only maximal ideal is {0}.

In the ring Z of integers, the maximal ideals are the generated by a prime number.

principal ideals

More generally, all nonzero are maximal in a principal ideal domain.

prime ideals

The ideal ${\displaystyle (2,x)}$ is a maximal ideal in ring ${\displaystyle \mathbb {Z} [x]}$. Generally, the maximal ideals of ${\displaystyle \mathbb {Z} [x]}$ are of the form ${\displaystyle (p,f(x))}$ where ${\displaystyle p}$ is a prime number and ${\displaystyle f(x)}$ is a polynomial in ${\displaystyle \mathbb {Z} [x]}$ which is irreducible modulo ${\displaystyle p}$.

Every prime ideal is a maximal ideal in a Boolean ring, i.e., a ring consisting of only idempotent elements. In fact, every prime ideal is maximal in a commutative ring ${\displaystyle R}$ whenever there exists an integer ${\displaystyle n>1}$ such that ${\displaystyle x^{n}=x}$ for any ${\displaystyle x\in R}$.

The maximal ideals of the ${\displaystyle \mathbb {C} [x]}$ are principal ideals generated by ${\displaystyle x-c}$ for some ${\displaystyle c\in \mathbb {C} }$.

polynomial ring

More generally, the maximal ideals of the polynomial ring K[x₁, ..., x_n] over an K are the ideals of the form (x₁ − a₁, ..., x_n − a_n). This result is known as the weak Nullstellensatz.

algebraically closed field

An important ideal of the ring called the can be defined using maximal right (or maximal left) ideals.

Jacobson radical

If R is a unital commutative ring with an ideal m, then k = R/m is a field if and only if m is a maximal ideal. In that case, R/m is known as the . This fact can fail in non-unital rings. For example, ${\displaystyle 4\mathbb {Z} }$ is a maximal ideal in ${\displaystyle 2\mathbb {Z} }$, but ${\displaystyle 2\mathbb {Z} /4\mathbb {Z} }$ is not a field.

residue field

If L is a maximal left ideal, then R/L is a simple left R-module. Conversely in rings with unity, any simple left R-module arises this way. Incidentally this shows that a collection of representatives of simple left R-modules is actually a set since it can be put into correspondence with part of the set of maximal left ideals of R.

(1929): Every nonzero unital ring has a maximal ideal. The result is also true if "ideal" is replaced with "right ideal" or "left ideal". More generally, it is true that every nonzero finitely generated module has a maximal submodule. Suppose I is an ideal which is not R (respectively, A is a right ideal which is not R). Then R/I is a ring with unity (respectively, R/A is a finitely generated module), and so the above theorems can be applied to the quotient to conclude that there is a maximal ideal (respectively, maximal right ideal) of R containing I (respectively, A).

Krull's theorem

Krull's theorem can fail for rings without unity. A , i.e. a ring in which the Jacobson radical is the entire ring, has no simple modules and hence has no maximal right or left ideals. See regular ideals for possible ways to circumvent this problem.

radical ring

In a commutative ring with unity, every maximal ideal is a . The converse is not always true: for example, in any nonfield integral domain the zero ideal is a prime ideal which is not maximal. Commutative rings in which prime ideals are maximal are known as zero-dimensional rings, where the dimension used is the Krull dimension.

prime ideal

A maximal ideal of a noncommutative ring might not be prime in the commutative sense. For example, let ${\displaystyle M_{n\times n}(\mathbb {Z} )}$ be the ring of all ${\displaystyle n\times n}$ matrices over ${\displaystyle \mathbb {Z} }$. This ring has a maximal ideal ${\displaystyle M_{n\times n}(p\mathbb {Z} )}$ for any prime ${\displaystyle p}$, but this is not a prime ideal since (in the case ${\displaystyle n=2}$)${\displaystyle A={\text{diag}}(1,p)}$ and ${\displaystyle B={\text{diag}}(p,1)}$ are not in ${\displaystyle M_{n\times n}(p\mathbb {Z} )}$, but ${\displaystyle AB=pI_{2}\in M_{n\times n}(p\mathbb {Z} )}$. However, maximal ideals of noncommutative rings are prime in the below.

generalized sense

Prime ideal

Anderson, Frank W.; Fuller, Kent R. (1992), Rings and categories of modules, Graduate Texts in Mathematics, vol. 13 (2 ed.), New York: Springer-Verlag, pp. x+376, :10.1007/978-1-4612-4418-9, ISBN 0-387-97845-3, MR 1245487

doi

Lam, T. Y. (2001), A first course in noncommutative rings, Graduate Texts in Mathematics, vol. 131 (2 ed.), New York: Springer-Verlag, pp. xx+385, :10.1007/978-1-4419-8616-0, ISBN 0-387-95183-0, MR 1838439

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