Integral Extensions - Applications

Integral extensions have profound applications in algebraic geometry, number theory, and the structure theory of rings.

TheoremNoether Normalization Lemma

Let $k$ be a field and $R$ a finitely generated $k$ -algebra that is an integral domain. Then there exist algebraically independent elements $y_1, \ldots, y_d \in R$ such that $R$ is integral over $k[y_1, \ldots, y_d]$ .

The number $d$ equals $\dim(R)$ , the Krull dimension. This provides a "coordinate system" making $R$ a finite extension of polynomial algebra.

This theorem reduces studying arbitrary affine varieties to studying finite covers of affine space, a fundamental simplification in algebraic geometry.

ExampleApplication to Curve

For the affine curve $R = k[x,y]/(y^2 - x^3 - x)$ , Noether normalization produces $k[x] \subseteq R$ with $R$ integral over $k[x]$ .

Here $x$ serves as a coordinate, and $y$ satisfies the monic equation $t^2 - x^3 - x = 0$ over $k[x]$ , making $R$ a degree 2 extension.

TheoremNullstellensatz (via Integrality)

Hilbert's Nullstellensatz can be proven using integral extensions. For $k$ algebraically closed and $\mathfrak{m} \subset k[x_1, \ldots, x_n]$ maximal, we have $\mathfrak{m} = (x_1 - a_1, \ldots, x_n - a_n)$ for some $(a_1, \ldots, a_n) \in k^n$ .

The proof uses that $k[x_1, \ldots, x_n]/\mathfrak{m}$ is a finitely generated $k$ -algebra and a field, hence integral over $k$ , and algebraically closed forces it to equal $k$ .

Remark

The weak Nullstellensatz states maximal ideals in $k[x_1, \ldots, x_n]$ correspond bijectively to points in $k^n$ when $k$ is algebraically closed. This is the foundation of classical algebraic geometry.

TheoremZariski's Lemma

If $k \subseteq K$ is a field extension with $K$ finitely generated as a $k$ -algebra, then $K$ is a finite algebraic extension of $k$ (i.e., $[K:k] < \infty$ ).

Equivalently, fields finitely generated as algebras are finitely generated as modules (integral).

ExampleFunction Fields

The function field $k(x,y)$ is not finitely generated as a $k$ -algebra in the ring-theoretic sense, though it is generated by $\{x, y\}$ as a field. Zariski's Lemma explains why: it would require $k(x,y)$ to be finite over $k$ , impossible since $\text{trdeg}_k(k(x,y)) = 2 > 0$ .

TheoremDimension in Integral Extensions

If $R \subseteq S$ is an integral extension of Noetherian rings, then: $\dim(R) = \dim(S)$

Integral extensions preserve dimension, contrasting with arbitrary extensions where dimension can increase. This reflects geometric intuition: finite morphisms don't change dimension.

ExampleRings of Integers

For a number field $K$ of degree $n$ over $\mathbb{Q}$ :

$\dim(\mathbb{Z}) = 1$
$\dim(\mathcal{O}_K) = 1$

The dimension is preserved despite $\mathcal{O}_K$ being much "larger" than $\mathbb{Z}$ in other senses.

TheoremKrull's Principal Ideal Theorem (Variant)

In an integral extension $R \subseteq S$ of Noetherian rings, if $\mathfrak{p} \in \text{Spec}(R)$ has height $h$ , then every prime $\mathfrak{P} \in \text{Spec}(S)$ lying over $\mathfrak{p}$ has height $\geq h$ .

Combined with dimension preservation, this constrains the structure of $\text{Spec}(S) \to \text{Spec}(R)$ significantly.

Remark

In algebraic number theory, class field theory studies abelian extensions of number fields. The integral closure $\mathcal{O}_L$ of $\mathcal{O}_K$ in a finite extension $K \subseteq L$ governs prime factorization, ideal class groups, and Galois theory of primes.

ExampleDedekind Extensions

If $R$ is a Dedekind domain with fraction field $K$ and $L/K$ is a finite separable extension with integral closure $S$ in $L$ , then:

$S$ is also a Dedekind domain
Every prime $\mathfrak{p}$ of $R$ factors as $\mathfrak{p}S = \mathfrak{P}_1^{e_1} \cdots \mathfrak{P}_g^{e_g}$
The exponents $e_i$ (ramification indices) and degrees $f_i = [\kappa(\mathfrak{P}_i):\kappa(\mathfrak{p})]$ satisfy $\sum e_i f_i = [L:K]$

This fundamental formula connects algebra and arithmetic.

TheoremValuative Criterion for Properness

A morphism of schemes is proper if and only if it satisfies the valuative criterion involving lifting diagrams with valuation rings. Integral extensions provide algebraic models for this, with $R \subseteq S$ integral corresponding to finite (proper) morphisms.

These applications demonstrate the central role of integral extensions across algebra and geometry, from dimension theory to arithmetic.