Rings, Ideals, and Modules - Applications

The relationship between maximal ideals and prime ideals reveals fundamental structural properties of commutative rings.

TheoremMaximal Ideals are Prime

Every maximal ideal is prime. More precisely, if $\mathfrak{m}$ is a maximal ideal in $R$ , then $R/\mathfrak{m}$ is a field, hence an integral domain, so $\mathfrak{m}$ is prime.

The converse fails: $(0)$ is prime in $\mathbb{Z}$ but not maximal. However, in zero-dimensional rings, prime ideals coincide with maximal ideals.

TheoremExistence of Maximal Ideals

Every non-zero commutative ring contains at least one maximal ideal. Moreover, every proper ideal is contained in some maximal ideal.

This result requires Zorn's lemma and fails constructively, reflecting the non-constructive nature of maximality in infinite rings.

ExampleMaximal Ideals in Common Rings

In $\mathbb{Z}$ : maximal ideals are $(p)$ for primes $p$
In $k[x]$ ( $k$ a field): maximal ideals are $(f(x))$ for irreducible $f$
In $k[x_1, \ldots, x_n]$ : maximal ideals of form $(x_1 - a_1, \ldots, x_n - a_n)$ over algebraically closed $k$
In $C([0,1])$ (continuous functions): maximal ideals are $\{f : f(x_0) = 0\}$ for fixed $x_0 \in [0,1]$

TheoremNakayama's Lemma (Simple Form)

Let $M$ be a finitely generated $R$ -module and $\mathfrak{m}$ a maximal ideal. If $\mathfrak{m}M = M$ , then $M = 0$ .

Equivalently: if $M = \mathfrak{m}M + Rm_1 + \cdots + Rm_n$ , then $M$ is generated by $m_1, \ldots, m_n$ .

Nakayama's Lemma is one of the most frequently applied results in commutative algebra, often used to "lift" properties from quotients back to the original module.

Remark

The Jacobson radical $J(R) = \bigcap \{\mathfrak{m} : \mathfrak{m} \text{ maximal}\}$ consists of elements that "act like zero" from a certain perspective. An element $a \in J(R)$ if and only if $1 - ra$ is a unit for all $r \in R$ .

Nakayama's Lemma generalizes to: if $M$ is finitely generated and $J(R) \cdot M = M$ , then $M = 0$ .

TheoremKrull's Theorem

The nilradical $\sqrt{(0)}$ equals the intersection of all prime ideals: $\sqrt{(0)} = \bigcap_{\mathfrak{p} \text{ prime}} \mathfrak{p}$

An element $a$ is nilpotent if and only if $a$ belongs to every prime ideal. This characterization links the algebraic notion of nilpotence to the geometric notion of belonging to all irreducible components.

ExampleGeometric Interpretation

In the coordinate ring $k[x,y]/(xy)$ , representing the union of two lines, both $x$ and $y$ are zero-divisors but not nilpotent. However, the ideal $(x) \cap (y) = (xy)$ consists of elements that vanish on both lines, illustrating how prime ideals detect geometric components.

These theorems establish the fundamental role of maximal and prime ideals in ring theory, connecting algebraic properties to geometric and topological structures through the spectrum.