Integral Extensions and Krull Dimension

Introduction

An element of a ring extension is integral over the base when it satisfies a monic polynomial equation with coefficients in the base. The condition is the ring-theoretic generalisation of being algebraic over a field, and the set of integral elements of an extension forms a subring, the integral closure. Rings equal to their integral closure in their fraction field are integrally closed, and this is the hypothesis that makes the arithmetic of a Dedekind domain work.

Integrality is a finiteness condition: an element is integral exactly when the subring it generates is finitely generated as a base-submodule. This is why integral extensions preserve so much: they preserve the prime ideal lattice in a controlled way, and they preserve Krull dimension, the supremum of the lengths of chains of prime ideals. The dimension theory that results — Noether normalisation, the dimension of a polynomial ring, the principal ideal theorem of Krull — is the tool that measures how far a ring is from a field, and it is what makes primary decomposition geometrically meaningful.

This article develops integral dependence and the integral closure, states and proves the Cohen–Seidenberg theorems on primes in integral extensions, defines Krull dimension and height, proves that integral extensions preserve dimension, proves Noether normalisation and computes the dimension of $k[x_1, \ldots, x_n]$, and proves Krull's principal ideal theorem and its general form. Throughout, $R \subseteq S$ is an extension of commutative rings with $1 \neq 0$; integral domains, fraction fields and divisibility are from Integral Domains and Unique Factorisation Domains, localisation from Localization and the Fraction Field, chain conditions and primary decomposition from Noetherian and Artinian Rings and Primary Decomposition. The ring of integers of a number field is treated; here it appears only as an example.

Integral Elements

Definition and Elementary Properties

Definition. Let $R \subseteq S$ be an extension of commutative rings. An element $\alpha \in S$ is integral over $R$ if it is a root of a monic polynomial with coefficients in $R$:

$$ \alpha^n + a_{n-1}\alpha^{n-1} + \cdots + a_1 \alpha + a_0 = 0, \qquad a_i \in R. $$

If every element of $S$ is integral over $R$, then $S$ is an integral extension of $R$.

The definition is deliberately restricted to monic polynomials; without monicity it defines algebraic dependence over the fraction field and is far weaker (see the remark below).

Proposition. Let $R \subseteq S$, let $\alpha \in S$, and let $R[\alpha]$ be the subring generated by $R$ and $\alpha$. Then $\alpha$ is integral over $R$ if and only if $R[\alpha]$ is a finitely generated $R$-submodule of $S$, that is,

$$ R[\alpha] = R + R\alpha + \cdots + R\alpha^{m} $$

for some $m$.

Proof. If $\alpha$ is integral, say $\alpha^n = -(a_{n-1}\alpha^{n-1} + \cdots + a_0)$, then multiplication by $\alpha$ preserves the $R$-span of $1, \alpha, \ldots, \alpha^{n-1}$, so every power of $\alpha$ lies in that span and $R[\alpha]$ equals it, with $m = n-1$.

Conversely, if $R[\alpha] = R + R\alpha + \cdots + R\alpha^m$, then $\alpha^{m+1}$ lies in this span, say $\alpha^{m+1} = r_0 + r_1 \alpha + \cdots + r_m \alpha^m$ with $r_i \in R$. Then $\alpha$ is a root of the monic polynomial $x^{m+1} - r_m x^m - \cdots - r_0$, so $\alpha$ is integral.

Theorem. Let $R \subseteq S$ be an extension. The set of elements of $S$ integral over $R$ is a subring of $S$ containing $R$.

Proof. Let $\alpha, \beta \in S$ be integral, of degrees $m$ and $n$ over $R$ in the sense above. The subring $R[\alpha, \beta]$ is spanned over $R$ by the finitely many products $\alpha^i \beta^j$ with $0 \leq i < m$ and $0 \leq j < n$: indeed this span is closed under multiplication by $\alpha$ and by $\beta$, hence under multiplication by any element of $R[\alpha,\beta]$, and it contains $1$. It follows that $R[\alpha + \beta]$ and $R[\alpha\beta]$ are contained in a finitely generated $R$-submodule of $S$; the same determinant argument as in the proposition, applied to the submodule generated by $1, \alpha+\beta, \ldots, (\alpha+\beta)^N$ or by $1, \alpha\beta, \ldots$, shows that $\alpha + \beta$ and $\alpha\beta$ are integral. More explicitly, one uses that if $M$ is a finitely generated $R$-submodule of $S$ closed under multiplication by $\gamma$, then $\gamma$ satisfies a monic polynomial equation obtained by expanding $\det(\gamma I - A) = 0$, where $A$ is a matrix representing multiplication by $\gamma$ on a generating set. Thus the integral elements are closed under addition, subtraction and multiplication and contain $R$, so they form a subring.

Definition. The subring of integral elements is the integral closure of $R$ in $S$. If $R$ is an integral domain and $R$ equals its integral closure in its fraction field $\operatorname{Frac}(R)$, then $R$ is integrally closed, or normal. If $R \subseteq S$ and $S$ is equal to the integral closure of $R$ in $S$, then $S$ is the integral closure and $S/R$ is integral. An integral domain that is integrally closed in its fraction field is a normal domain.

Proposition. Integrality is transitive: if $R \subseteq S \subseteq T$, if $S$ is integral over $R$ and $T$ is integral over $S$, then $T$ is integral over $R$.

Proof. Let $\gamma \in T$ satisfy $\gamma^n + s_{n-1}\gamma^{n-1} + \cdots + s_0 = 0$ with $s_i \in S$. Each $s_i$ is integral over $R$, so the subring $S_0 = R[s_0, \ldots, s_{n-1}]$ is a finitely generated $R$-submodule of $S$, and $S_0[\gamma]$ is a finitely generated $S_0$-submodule. Hence $S_0[\gamma]$ is a finitely generated $R$-submodule of $T$ containing $\gamma$ and closed under multiplication by $\gamma$, so $\gamma$ is integral over $R$.

Proposition. Let $R \subseteq S$ be integral and let $T \subseteq R$ be a subring. Then $R$ is integral over $T$ if and only if $S$ is integral over $T$.

Proof. If $S$ is integral over $R$ and $R$ is integral over $T$, transitivity gives $S$ integral over $T$. Conversely if $S$ is integral over $T$ then $R \subseteq S$ consists of elements integral over $T$, since $R \subseteq S$.

Examples

Example (algebraic integers). Let $\overline{\mathbb{Q}}$ be the field of algebraic numbers and let $R = \mathbb{Z} \subseteq S = \overline{\mathbb{Q}}$. The integral elements over $\mathbb{Z}$ are the algebraic integers: roots of monic polynomials with integer coefficients. The element $\sqrt{2}$ is an algebraic integer; the element $1/2$ is not, although it is algebraic, being a root of $2x - 1$. The integral closure of $\mathbb{Z}$ in $\mathbb{Q}$ is $\mathbb{Z}$ itself, which is therefore integrally closed: this is the classical statement that a rational algebraic integer is an integer, and it is the reason the arithmetic of $\mathbb{Z}$ extends to rings of integers.

Example (integer roots). More generally, if $R$ is a unique factorisation domain with fraction field $K$ and $\alpha \in K$ is integral over $R$, then $\alpha \in R$: writing $\alpha = a/b$ in lowest terms and clearing denominators in a monic integral equation gives $b \mid a^n$ and hence $b$ is a unit. Thus every unique factorisation domain is integrally closed. In particular $\mathbb{Z}$ and $k[x_1, \ldots, x_n]$ are integrally closed.

Example. The domain $R = k[t^2, t^3] \subseteq k[t]$ is not integrally closed: the element $t$ of the fraction field of $R$ is integral over $R$, since it satisfies $x^2 - t^2 = 0$ with $t^2 \in R$, but $t \notin R$. The integral closure of $R$ in $k(t)$ is $k[t]$.

Example. The ring $\mathbb{Z}[\sqrt{5}]$ is not integrally closed: the golden ratio $\varphi = (1+\sqrt5)/2$ lies in the fraction field $\mathbb{Q}(\sqrt5)$, is integral over $\mathbb{Z}$ as a root of $x^2 - x - 1$, and does not lie in $\mathbb{Z}[\sqrt5]$. Its integral closure is $\mathbb{Z}[(1+\sqrt5)/2]$.

Remark. Integral is strictly stronger than algebraic. If $R = \mathbb{Z}$ and $S = \mathbb{Q}$, every element of $S$ is algebraic over the fraction field $\mathbb{Q}$ of $R$, but only the elements of $\mathbb{Z}$ are integral over $R$. The two notions coincide when the base is a field: for a field $K$, an element of an extension is integral over $K$ exactly when it is algebraic over $K$, since a monic polynomial over a field can be divided by its leading coefficient for free.

The Cohen–Seidenberg Theorems

Throughout this section $R \subseteq S$ is an integral extension of commutative rings and, where primes are discussed, both are integral domains. For a prime $\mathrm{Q}$ of $S$, its contraction $\mathrm{Q} \cap R$ is prime in $R$, and $\mathrm{Q}$ is said to lie over $\mathrm{Q} \cap R$.

Lying Over and Incomparability

Theorem (lying over). Let $R \subseteq S$ be integral and let $\mathrm{P}$ be a prime ideal of $R$. Then there is a prime ideal $\mathrm{Q}$ of $S$ with $\mathrm{Q} \cap R = \mathrm{P}$.

Proof sketch. Localize to assume $R$ local with maximal ideal $\mathrm{P}$. The ideal $\mathrm{P}S$ is proper: if $1 = \sum c_i s_i$ with $c_i \in \mathrm{P} \subseteq R$ and $s_i \in S$, the subring generated over $R$ by the $s_i$ is a finitely generated $R$-submodule, and the determinant argument of the first section exhibits $1$ as a root of a monic polynomial with lower coefficients in $\mathrm{P}$, forcing $1 \in \mathrm{P}$, which is impossible. So $\mathrm{P}S$ is contained in a maximal ideal $\mathrm{Q}$ of $S$. Then $\mathrm{Q} \cap R$ is a prime containing $\mathrm{P}$, and in a local ring the only prime contained in the maximal ideal $\mathrm{P}$ is $\mathrm{P}$ itself, so $\mathrm{Q} \cap R = \mathrm{P}$.

Theorem (incomparability). Let $\mathrm{Q}_1 \subseteq \mathrm{Q}_2$ be primes of $S$ with $\mathrm{Q}_1 \cap R = \mathrm{Q}_2 \cap R$. Then $\mathrm{Q}_1 = \mathrm{Q}_2$. Equivalently, distinct primes of $S$ lying over the same prime of $R$ are incomparable.

Proof sketch. Localise at $\mathrm{Q}_2$ and replace $R$ by its localisation at $\mathrm{Q}_2 \cap R$, so that the common contraction is the unique maximal ideal $\mathrm{M}$ of $R$ and $\mathrm{Q}_2$ is the maximal ideal of $S$. If $\mathrm{Q}_1 \subsetneq \mathrm{Q}_2$, then $S/\mathrm{Q}_1$ is a domain integral over the field $R/(\mathrm{Q}_1 \cap R)$; by the corollary below it is a field, so $\mathrm{Q}_1$ is maximal, and since $\mathrm{Q}_1 \subseteq \mathrm{Q}_2$ with $\mathrm{Q}_2$ maximal we get $\mathrm{Q}_1 = \mathrm{Q}_2$, a contradiction.

Corollary. In an integral extension $S/R$, two primes of $S$ with the same contraction are equal: the map $\mathrm{Q} \mapsto \mathrm{Q} \cap R$ is injective on the set of primes of $S$ lying over a fixed prime of $R$. In particular an integral extension induces a surjection from the primes of $S$ onto the primes of $R$.

Corollary. If $R \subseteq S$ is integral and $S$ is a domain, then $R$ is a field if and only if $S$ is a field.

Proof. If $S$ is a field and $0 \neq x \in R$, then $x$ has an inverse $x^{-1} \in S$; since $x^{-1}$ is integral over $R$, the subring $R[x^{-1}]$ is a finitely generated $R$-submodule of $S$ contained in $R[x^{-1}]$, and a determinant argument shows $x$ is a unit in $R$. Conversely if $R$ is a field and $S$ is a domain integral over $R$, then $S$ is algebraic over $R$; an integral domain algebraic over a field is a field, since for $0 \neq y \in S$ the subring $R[y]$ is a finite-dimensional $R$-algebra and the multiplication map $R[y] \to R[y]$, $z \mapsto yz$, is injective hence bijective, so $1 = yz$ for some $z$.

Going Up and Going Down

Theorem (going up). Let $R \subseteq S$ be integral, let

$$ \mathrm{P}_1 \subseteq \mathrm{P}_2 $$

be primes of $R$, and let $\mathrm{Q}_1$ be a prime of $S$ lying over $\mathrm{P}_1$. Then there is a prime $\mathrm{Q}_2 \supseteq \mathrm{Q}_1$ of $S$ lying over $\mathrm{P}_2$.

Proof sketch. Working in $\bar S = S/\mathrm{Q}_1$ and $\bar R = R/\mathrm{P}_1$, the extension is integral and $\mathrm{P}_2$ maps to a prime $\bar{\mathrm{P}}_2$, which by lying over is the contraction of some prime of $\bar S$; its preimage in $S$ is the required $\mathrm{Q}_2$.

Theorem (going down). Let $R \subseteq S$ be an integral extension with $R$ and $S$ integral domains and $R$ integrally closed. Let $\mathrm{P}_1 \subseteq \mathrm{P}_2$ be primes of $R$ and let $\mathrm{Q}_2$ be a prime of $S$ lying over $\mathrm{P}_2$. Then there is a prime $\mathrm{Q}_1 \subseteq \mathrm{Q}_2$ of $S$ lying over $\mathrm{P}_1$.

Proof sketch. This is the one Cohen–Seidenberg theorem that does not hold for an arbitrary integral extension; the requirement that $R$ be integrally closed is essential. One reduces to a localised situation, writes the multiplicative set of the elements of $S$ that lie over $R \setminus \mathrm{P}_1$, and shows that a minimal prime of the localised ideal is the required $\mathrm{Q}_1$, using that an element of $S$ integral over the integrally closed domain $R$ and lying in $\mathrm{P}_1 S_{\mathrm{P}_2}$ satisfies a monic equation whose constant coefficient lies in $\mathrm{P}_1$.

Summary of the theorems. Lying over and going up hold for every integral extension; incomparability holds for every integral extension; going down requires $R$ integrally closed. Together the four go under the name of the Cohen–Seidenberg theorems of 1946.

Krull Dimension

Definition and Elementary Properties

Definition. Let $R$ be a commutative ring. A chain of prime ideals is a strictly increasing sequence

$$ \mathrm{P}_0 \subsetneq \mathrm{P}_1 \subsetneq \cdots \subsetneq \mathrm{P}_n . $$

The length of the chain is $n$, the number of strict inclusions. The Krull dimension $\dim R$ is the supremum of the lengths of all chains of prime ideals of $R$, and $\dim R = \infty$ if the lengths are unbounded. The height of a prime ideal $\mathrm{P}$, written $\operatorname{ht}(\mathrm{P})$, is the supremum of the lengths of chains of primes below $\mathrm{P}$, that is, the dimension of the local ring $R_\mathrm{P}$.

Equivalently, $\operatorname{ht}(\mathrm{P}) = \dim R_\mathrm{P}$, and

$$ \dim R = \sup_{\mathrm{P} \in \operatorname{Spec}(R)} \operatorname{ht}(\mathrm{P}) = \sup_{\mathrm{M} \text{ maximal}} \operatorname{ht}(\mathrm{M}). $$

Examples.

(a) A field has dimension $0$: it has only the prime ideal $(0)$.

(b) A principal ideal domain has dimension $1$: the primes are $(0)$ and the maximal ideals $(p)$, and every chain has the form $(0) \subsetneq (p)$. Thus $\dim \mathbb{Z} = 1$ and $\dim k[x] = 1$.

(c) $\mathbb{Z}/n\mathbb{Z}$ has dimension $0$ for $n > 1$, since every prime is maximal; more generally a commutative Artinian ring has dimension $0$ by the structure theory of Noetherian and Artinian Rings.

(d) A local ring has dimension equal to the length of its longest chain of prime ideals; the localisation $k[x]_{(x)}$ has dimension $1$.

(e) The ring $k[x_1, \ldots, x_n]$ has dimension $n$, proved below.

(f) The ring $k[x_1, x_2, x_3, \ldots]$ of polynomials in infinitely many variables has infinite dimension: the chain $(x_1) \subsetneq (x_1, x_2) \subsetneq \cdots$ has unbounded length. This ring is not Noetherian.

Proposition. Let $I$ be an ideal of a Noetherian ring $R$. Then

$$ \dim R = \max_{\mathrm{P} \text{ minimal over } I} \bigl(\dim R/\mathrm{P} + \operatorname{ht}(\mathrm{P})\bigr), $$

and for an integral domain $R$ and a prime $\mathrm{P}$ one has $\operatorname{ht}(\mathrm{P}) + \dim R/\mathrm{P} \leq \dim R$ with equality when $R$ is a finitely generated algebra over a field.

Proof sketch. Every maximal ideal contains a minimal prime over $I$, and a saturated chain from a minimal prime to a maximal ideal has length $\dim R/\mathrm{P} + \operatorname{ht}(\mathrm{P})$; the supremum over all chains gives the formula. Equality in the finitely generated case follows from the dimension theorem for finitely generated algebras over a field.

Integral Extensions Preserve Dimension

Theorem. Let $R \subseteq S$ be an integral extension of commutative rings. Then

$$ \dim R = \dim S . $$

Proof. Let $\mathrm{P}_0 \subsetneq \mathrm{P}_1 \subsetneq \cdots \subsetneq \mathrm{P}_n$ be a chain of primes in $R$. By lying over, $\mathrm{P}_0$ is the contraction of a prime $\mathrm{Q}_0$ of $S$; by going up applied repeatedly, the chain extends to primes $\mathrm{Q}_0 \subseteq \mathrm{Q}_1 \subseteq \cdots \subseteq \mathrm{Q}_n$ of $S$ with $\mathrm{Q}_i \cap R = \mathrm{P}_i$. By incomparability the inclusions $\mathrm{Q}_i \subseteq \mathrm{Q}_{i+1}$ are strict, since their contractions are. Hence $\dim S \geq n$, and since $n$ was arbitrary, $\dim S \geq \dim R$.

Conversely, let $\mathrm{Q}_0 \subsetneq \mathrm{Q}_1 \subsetneq \cdots \subsetneq \mathrm{Q}_m$ be a chain of primes in $S$. Their contractions form a chain in $R$, and by incomparability the contractions are strictly increasing: if $\mathrm{Q}_i \cap R = \mathrm{Q}_{i+1} \cap R$ then the two primes of $S$ lie over the same prime of $R$ and incomparability gives $\mathrm{Q}_i = \mathrm{Q}_{i+1}$, contrary to strictness. Hence $\dim R \geq m$ and $\dim R \geq \dim S$.

Corollary. If $R \subseteq S$ is integral and $S$ is a field, then $R$ is a field; if $R$ is a field and $S$ is a domain, then $S$ is a field. In particular the dimension of $S$ is $0$ exactly when the dimension of $R$ is $0$.

Noether Normalisation and the Dimension of a Polynomial Ring

Noether Normalisation

Theorem (Noether normalisation). Let $k$ be a field and let $A$ be a finitely generated commutative $k$-algebra. Then there exist elements $y_1, \ldots, y_d \in A$, algebraically independent over $k$, such that $A$ is integral over the polynomial subalgebra $k[y_1, \ldots, y_d]$.

Proof sketch. Write $A = k[x_1, \ldots, x_n]/I$ and argue by induction on $n$. If $x_1, \ldots, x_n$ are algebraically independent, take $d = n$ and $y_i = x_i$. Otherwise there is a nontrivial polynomial relation $f(x_1, \ldots, x_n) = 0$. Introduce a change of variables $x_i = y_i + y_n^{m_i}$ for $i < n$, chosen so that the leading term of $f(y_1 + y_n^{m_1}, \ldots, y_n)$ is a monic power of $y_n$: this is possible by choosing the exponents $m_i$ to be strictly increasing and large, so that the monomials of $f$ have distinct $y_n$-degrees. Then the relation exhibits $y_n$ as integral over $k[y_1, \ldots, y_n]$, and one of the variables has been eliminated; induction finishes.

Theorem (Zariski's lemma). Let $k$ be a field and let $A$ be a field that is finitely generated as a $k$-algebra. Then $A$ is a finite algebraic extension of $k$.

Proof. By Noether normalisation, $A$ is integral over a polynomial subalgebra $k[y_1, \ldots, y_d]$. Since $A$ is a field and the extension is integral, the corollary above forces $k[y_1, \ldots, y_d]$ to be a field; a polynomial ring over $k$ in $d$ variables is a field only when $d = 0$, since for $d \geq 1$ the element $y_1$ is not invertible. Hence $d = 0$ and $A$ is integral, equivalently algebraic and finite, over $k$.

Zariski's lemma is the input from integral extension theory that proves the weak Nullstellensatz in Primary Decomposition.

Dimension of a Polynomial Ring

Theorem. Let $k$ be a field. Then

$$ \dim k[x_1, \ldots, x_n] = n . $$

Proof. The chain

$$ (0) \subsetneq (x_1) \subsetneq (x_1, x_2) \subsetneq \cdots \subsetneq (x_1, \ldots, x_n) $$

consists of prime ideals, since each quotient $k[x_1, \ldots, x_n]/(x_1, \ldots, x_i) \cong k[x_{i+1}, \ldots, x_n]$ is a polynomial ring over a field and hence a domain. It has length $n$, so $\dim \geq n$.

For the reverse inequality, let $A = k[x_1, \ldots, x_n]/I$ be an arbitrary quotient by a prime ideal, so that $A$ is a domain; by Noether normalisation there is an integral extension $k[y_1, \ldots, y_d] \subseteq A$. By the dimension theorem for integral extensions, $\dim A = \dim k[y_1, \ldots, y_d]$. Every chain of primes in a polynomial ring can be analysed by induction on the number of variables using the principal ideal theorem below, which gives $\dim k[y_1, \ldots, y_d] \leq d$. Applying this to $I = \mathrm{P}$ a prime and using that $d \leq$ the number of variables gives the bound $\dim k[x_1, \ldots, x_n] \leq n$.

Corollary (dimension of an affine algebra). If $A$ is a finitely generated $k$-algebra and a domain, then $\dim A$ equals the transcendence degree of $\operatorname{Frac}(A)$ over $k$, namely the integer $d$ of the Noether normalisation. Chains of primes of $A$ have length at most $d$, and the dimension is attained.

Krull's Principal Ideal Theorem

The Theorem

Theorem (Krull's principal ideal theorem). Let $R$ be a Noetherian commutative ring and let $a \in R$ be a non-unit. Then every prime ideal minimal among those containing the principal ideal $(a)$ has height at most $1$. Equivalently, if $\mathrm{P}$ is a prime of $R$ minimal over $(a)$, then there are no primes strictly between a minimal prime of $R$ and $\mathrm{P}$.

Proof sketch. Let $\mathrm{P}$ be minimal over $(a)$. Localising at $\mathrm{P}$ reduces to the case where $R$ is a Noetherian local ring with maximal ideal $\mathrm{P}$ and $\operatorname{rad}((a)) = \mathrm{P}$. If $\operatorname{ht}(\mathrm{P}) \geq 2$ there is a prime $\mathrm{Q}_1 \subsetneq \mathrm{Q}_2 \subsetneq \mathrm{P}$. The minimal primes of $(a)$ are the associated primes of the primary component, and one uses the following lemma of Krull: in a Noetherian local ring of dimension $d$, every ideal generated by $d$ elements is primary for the maximal ideal if its radical is the maximal ideal; applying it successively exhibits a chain of length at most the number of generators of $\mathrm{P}$, giving a contradiction when a chain of length $2$ is assumed with one generator. The full argument is in the standard references.

Theorem (generalised principal ideal theorem). Let $R$ be a Noetherian commutative ring and let $I = (a_1, \ldots, a_n)$ be generated by $n$ elements. Then every prime ideal minimal over $I$ has height at most $n$:

$$ \operatorname{ht}(\mathrm{P}) \leq n. $$

Proof sketch. Induct on $n$. The case $n = 1$ is the principal ideal theorem. For $n > 1$, let $\mathrm{P}$ be minimal over $(a_1, \ldots, a_n)$ and localise at $\mathrm{P}$; choose a prime $\mathrm{Q} \subsetneq \mathrm{P}$ maximal among the primes containing $(a_1, \ldots, a_{n-1})$. By induction $\operatorname{ht}(\mathrm{Q}) \leq n-1$. The image of $a_n$ in $R/\mathrm{Q}$ generates an ideal whose radical is $\mathrm{P}/\mathrm{Q}$, and the principal ideal theorem applied in the Noetherian ring $R/\mathrm{Q}$ gives $\operatorname{ht}(\mathrm{P}/\mathrm{Q}) \leq 1$. Adding the two heights gives $\operatorname{ht}(\mathrm{P}) \leq n$.

Consequences

Corollary (height is bounded by generators). If $\mathrm{P}$ is a prime ideal of a Noetherian ring $R$, then $\operatorname{ht}(\mathrm{P})$ is finite, bounded by the minimal number of generators of any ideal whose radical is $\mathrm{P}$. In particular a Noetherian local ring $(R, \mathrm{M})$ satisfies $\operatorname{ht}(\mathrm{M}) \leq \dim_{R/\mathrm{M}} \mathrm{M}/\mathrm{M}^2$, the embedding dimension.

Corollary (dimension and generators). A Noetherian local ring of dimension $d$ has its maximal ideal generated by at least $d$ elements, and if $\mathrm{M}$ can be generated by $d$ elements then $R$ is regular, with $\dim \mathrm{M}/\mathrm{M}^2 = d$. Regular local rings are the local models of smooth points; they are unique factorisation domains, a theorem of Auslander and Buchsbaum, and their theory is developed through the homological methods that belong to a later Part.

Corollary (dimension one). A Noetherian domain of dimension one is characterised by: every nonzero prime ideal is maximal. A Dedekind domain is a Noetherian integrally closed domain of dimension one.

Corollary. The following conditions on a Noetherian ring $R$ are equivalent.

(a) $R$ has Krull dimension $0$.

(b) Every prime ideal is maximal.

(c) $R$ is Artinian.

Proof. The equivalence of (a) and (b) is the definition, and (c) is equivalent to (a) by the structure theory of Noetherian and Artinian Rings.

Worked Computations

Example. In $k[x,y]$ the prime ideal $(x)$ has height $1$: it is minimal over the principal ideal $(x)$, so the principal ideal theorem gives $\operatorname{ht} \leq 1$, and $(0) \subsetneq (x)$ gives $\operatorname{ht} \geq 1$. The maximal ideal $(x,y)$ has height $2$, since it is minimal over the two-generator ideal $(x,y)$ so $\operatorname{ht} \leq 2$, while the chain $(0) \subsetneq (x) \subsetneq (x,y)$ has length $2$.

Example. The ideal $(x^2, xy)$ of $k[x,y]$ has minimal primes $(x)$ and $(x,y)$, of heights $1$ and $2$. It is generated by two elements, consistent with the generalised principal ideal theorem, which allows height at most $2$.

Example. In a principal ideal domain, every nonzero prime is maximal, so $\dim = 1$; the principal ideal theorem is vacuous in the height direction but correct, since a minimal prime over a principal ideal is either height $0$ or height $1$.

Summary

An element $\alpha$ of a ring extension $S/R$ is integral over $R$ if it satisfies a monic polynomial equation over $R$; this holds exactly when $R[\alpha]$ is a finitely generated $R$-submodule of $S$. The integral elements form a subring containing $R$, the integral closure, and the property is transitive. An integral domain equal to its integral closure in its fraction field is integrally closed, also called normal; every unique factorisation domain is integrally closed, and the classical example is that $\mathbb{Z}$ is integrally closed with integral closure in $\mathbb{Q}$ equal to $\mathbb{Z}$.

For an integral extension $R \subseteq S$ the four Cohen–Seidenberg theorems govern the primes: lying over (every prime of $R$ is the contraction of a prime of $S$), incomparability (distinct primes of $S$ over the same prime of $R$ are incomparable), going up (a chain of primes of $R$ lifts above a given prime of $S$), and going down (below a given prime of $S$, provided $R$ is integrally closed). Consequently an integral extension induces a surjection on primes and preserves Krull dimension.

Krull dimension is the supremum of the lengths of chains of prime ideals; height is dimension after localising at a prime. A field has dimension $0$, a principal ideal domain dimension $1$, and the polynomial ring $k[x_1, \ldots, x_n]$ dimension $n$. Noether normalisation expresses a finitely generated algebra over a field as an integral extension of a polynomial ring, which yields Zariski's lemma and the dimension formula for affine algebras. Krull's principal ideal theorem bounds the height of a minimal prime over an $n$-generator ideal by $n$, and its $n = 1$ case is the statement that a minimal prime over a principal ideal has height at most $1$.

Summary of Notation

Symbol	Meaning
$R \subseteq S$	Extension of commutative rings
$\alpha$ integral over $R$	Root of a monic polynomial in $R[x]$
$R[\alpha]$, $R[\alpha_1, \ldots, \alpha_n]$	Subring generated over $R$
$\operatorname{Frac}(R)$	Fraction field of an integral domain
integrally closed / normal	Equal to its integral closure in its fraction field
$\mathrm{P}, \mathrm{Q}$	Prime ideals of $R$ and $S$, with $\mathrm{Q} \cap R$ the contraction
$\mathrm{Q}$ lies over $\mathrm{P}$	$\mathrm{Q} \cap R = \mathrm{P}$
$\dim R$	Krull dimension: supremum of lengths of prime chains
$\operatorname{ht}(\mathrm{P})$	Height of $\mathrm{P}$, equal to $\dim R_\mathrm{P}$
$k[x_1, \ldots, x_n]$	Polynomial ring of dimension $n$
Noether normalisation	Integral extension $k[y_1, \ldots, y_d] \subseteq A$ with the $y_i$ algebraically independent
$\operatorname{tr.deg}_k$	Transcendence degree
$\mathrm{M}/\mathrm{M}^2$	Cotangent space of a local ring, of dimension the embedding dimension
algebraic integer	Element of $\overline{\mathbb{Q}}$ integral over $\mathbb{Z}$

Integral Extensions and Krull Dimension

Introduction

Integral Elements

Definition and Elementary Properties

Examples

The Cohen–Seidenberg Theorems

Lying Over and Incomparability

Going Up and Going Down

Krull Dimension

Definition and Elementary Properties

Integral Extensions Preserve Dimension

Noether Normalisation and the Dimension of a Polynomial Ring

Noether Normalisation

Dimension of a Polynomial Ring

Krull's Principal Ideal Theorem

The Theorem

Consequences

Worked Computations

Summary

Summary of Notation

Further Reading