Integral Extensions - Key Properties

Integral extensions satisfy remarkable closure and transitivity properties that make them well-behaved for geometric and arithmetic applications.

TheoremCharacterizations of Integrality

For $s \in S$ where $R \subseteq S$ , the following are equivalent:

$s$ is integral over $R$
$R[s]$ is finitely generated as an $R$ -module
$s$ belongs to some subring $T$ with $R \subseteq T \subseteq S$ and $T$ finitely generated as an $R$ -module
There exists a faithful $R[s]$ -module $M$ that is finitely generated as an $R$ -module

These characterizations provide different perspectives—polynomial, module-theoretic, and functorial.

ExampleUsing Module Characterization

To show $\sqrt{2} + \sqrt{3}$ is integral over $\mathbb{Q}$ , observe that $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ is a 4-dimensional $\mathbb{Q}$ -vector space with basis $\{1, \sqrt{2}, \sqrt{3}, \sqrt{6}\}$ .

Since $\sqrt{2} + \sqrt{3} \in \mathbb{Q}[\sqrt{2}, \sqrt{3}]$ and this is finitely generated as a $\mathbb{Q}$ -module, $\sqrt{2} + \sqrt{3}$ is integral (algebraic) over $\mathbb{Q}$ .

TheoremClosure Properties

The integral closure $\overline{R}$ of $R$ in $S$ is a subring of $S$ . Specifically:

If $s, t \in S$ are integral over $R$ , then $s + t$ and $st$ are integral over $R$
The set of integral elements forms a ring
Integrality is transitive: if $S$ is integral over $R$ and $T$ is integral over $S$ , then $T$ is integral over $R$

ProofSketch of Transitivity

If $t \in T$ is integral over $S$ , then $S[t]$ is finitely generated as an $S$ -module, say by $\{m_1, \ldots, m_k\}$ . If $S$ is integral over $R$ generated by $\{s_1, \ldots, s_n\}$ as an $R$ -module, then $S[t]$ is generated as an $R$ -module by products $\{s_i m_j\}$ , which are finite in number. Thus $R[t]$ is finitely generated over $R$ , so $t$ is integral over $R$ .

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ExampleTower of Extensions

Consider $\mathbb{Z} \subseteq \mathbb{Z}[i] \subseteq \mathbb{Z}[i, \sqrt{2}]$ :

$\mathbb{Z}[i]$ is integral over $\mathbb{Z}$ (since $i^2 + 1 = 0$ )
$\mathbb{Z}[i, \sqrt{2}]$ is integral over $\mathbb{Z}[i]$ (since $(\sqrt{2})^2 - 2 = 0$ )
By transitivity, $\mathbb{Z}[i, \sqrt{2}]$ is integral over $\mathbb{Z}$

TheoremFiniteness and Integrality

If $R \subseteq S$ with $S$ finitely generated as an $R$ -module, then $S$ is integral over $R$ .

Conversely, if $S$ is both integral over $R$ and finitely generated as an $R$ -algebra, then $S$ is finitely generated as an $R$ -module.

Remark

This theorem explains why "integral" and "finite" are nearly synonymous for ring extensions in algebraic geometry. A morphism of affine schemes is finite if it corresponds to an integral ring extension where the target is finitely generated as a module.

TheoremIntegral Closure in Localizations

If $\overline{R}$ is the integral closure of $R$ in $S$ and $T$ is a multiplicative set in $R$ , then $T^{-1}\overline{R}$ is the integral closure of $T^{-1}R$ in $T^{-1}S$ .

Integrality behaves well under localization, allowing local-global principles for normality.

ExampleLocalization and Normality

If $R$ is normal (integrally closed in its fraction field), then $R_\mathfrak{p}$ is normal for any prime $\mathfrak{p}$ .

Conversely, $R$ is normal if and only if $R_\mathfrak{m}$ is normal for all maximal ideals $\mathfrak{m}$ (local criterion for normality).

DefinitionConductor Ideal

For $R \subseteq S$ integral with $S$ finitely generated as an $R$ -module, the conductor is: $\mathfrak{f} = \{r \in R : rS \subseteq R\}$

This measures "how far" $S$ is from $R$ . When $R$ is normal, the conductor captures the "ramification locus" in geometric terms.

These properties make integral extensions the natural setting for algebraic geometry over rings, generalizing field-theoretic algebraic geometry while maintaining crucial finiteness and closure properties.