Maximal ideal
Let $A$ be a commutative ring with unit.
Definition. A proper ideal $M\subset A$ is called maximial if any of the following equivalent conditions hold:
- $A/M$ is a field
- $M$ is not contained in any larger proper ideal.
Corollary. The set of invertible elements is contained in the complement of the union of all maximal ideals.
Local rings
A ring with exactly one maximal ideal (e.g. a field) is called a local ring.
Existence
Theorem. Every nonzero (commutative, with identity) ring has at least one maximal idea.
Pf. Zorn’s lemma
Corollary. If an ideal $I\subset A$ is nontrivial (i.e. not $(1)$), then there exists a maximal ideal containing $I$.
Corollary. Suppose there exists $a\in A$ such that $a$ is not invertible. Then there exists a maximal ideal $M\subset A$ such that $a\in M$.
Properties
- Every maximal ideal is prime.
- If $f:A\to B$ is a ring homomorphism and $\mathfrak{b}$ is a maximial ideal in $B$, then $f^{-1}(\mathfrak{b})$ is a prime ideal. (Not necessarily maximal.)
Proofs
1
Let $I$ be the maximal ideal. By definition $A/I$ is a field, hence in particular an integral domain, hence $I$ is a prime ideal.
2
In the Prime ideal we prove the result for prime ideals, and we have just showed every maximal ideal is a prime ideal. To give an example of where $f^{-1}(\mathfrak{b})$ is not maximal, consider the natural map $\mathbb{Z}\to\mathbb{Q}$. Take $\mathfrak{b}=(0)$. This is maximal because $\mathbb{Q}\setminus{0}$ is a field. But $\mathbb{Z}\setminus{0}$ is not a field.