Fact-checked by Grok 2 weeks ago

Integral element

In commutative algebra, an integral element of an extension ring S over a commutative ring with unity R is an element s \in S that satisfies a monic polynomial equation s^n + r_{n-1} s^{n-1} + \dots + r_1 s + r_0 = 0 with coefficients r_i \in R.^[1] This concept generalizes the notion of algebraic integers, where elements of algebraic number fields that are roots of monic polynomials over \mathbb{Z} are precisely the integers of those fields.^[2] The collection of all elements in S that are integral over R forms a subring of S containing R, known as the integral closure of R in S.^[2] A ring extension S/R is called integral if every element of S is integral over R; in such cases, S is finitely generated as an R-module whenever it is finitely generated as an R-algebra.^[3] Integral elements satisfy key closure properties: if \alpha and \beta are integral over R, then so are their sum and product.^[2] Integral extensions preserve significant structural features of rings, such as the lying-over theorem, which ensures that prime ideals in R extend to prime ideals in S in a surjective manner on the spectra.^[3] They are foundational in algebraic geometry for studying morphisms of schemes and in number theory for analyzing Dedekind domains and unique factorization.^[2] For instance, the ring of integers in a number field is the integral closure of \mathbb{Z} in that field, highlighting the role of integrality in arithmetic.^[2]

Definitions and Equivalents

Definition

In commutative algebra, let R be a commutative ring with identity and A an R-algebra; although A need not be commutative in general, the notion of an integral element is typically studied when A is commutative. An element \alpha \in A is integral over R if there exists a positive integer n and elements r_0, r_1, \dots, r_{n-1} \in R such that

\alpha^n + r_{n-1} \alpha^{n-1} + \dots + r_1 \alpha + r_0 = 0.

^[3] This equation means that \alpha satisfies a monic polynomial of degree n with coefficients in R, i.e., f(x) = x^n + r_{n-1} x^{n-1} + \dots + r_0 \in R such that f(\alpha) = 0.^[4] The requirement that the polynomial be monic, with leading coefficient $1, ensures the definition captures elements that generate R-submodules of finite type in a normalized way, independent of scaling by units in R; a non-monic [polynomial](/page/Polynomial) with leading coefficient in Rwould not suffice if that coefficient is not a [unit](/page/Unit), potentially failing to preserve ring-like properties overR.[3] This condition generalizes the classical notion of algebraic integers, where roots of monic polynomials over \mathbb{Z}$ form the integers of number fields. The concept of an integral element was introduced by Richard Dedekind in the context of algebraic integers during the 1870s, particularly in his supplements to Dirichlet's Vorlesungen über Zahlentheorie.^[5]

Equivalent Definitions

An element \alpha \in A, where R is a subring of the commutative ring A, is integral over R if and only if the subring R[\alpha] is finitely generated as an R-module.^[6] This equivalence holds because the monic polynomial condition implies that higher powers of \alpha can be reduced, making \{1, \alpha, \dots, \alpha^{n-1}\} an R-module basis for R[\alpha] where n is the degree of the polynomial, and conversely, module-finiteness allows construction of a monic relation via linear algebra.^[6] More precisely, R[\alpha] being finitely generated as an R-module is equivalent to the existence of elements \beta_1, \dots, \beta_m \in A such that

\{1, \alpha, \alpha^2, \dots, \alpha^{m-1}\} \subseteq \sum_{j=1}^m R \beta_j,

meaning there exist r_{ij} \in R satisfying \alpha^j = \sum_{i=1}^m r_{ij} \beta_i for $0 \leq j < m.^[6] This spanning condition captures the finite dependence of powers of \alpha over R. Since R[\alpha] is generated as a ring by adjoining \alpha to R, \alpha is integral over R if and only if the subring R[\alpha] is an integral extension of R.^[6] To see the equivalence between the monic polynomial condition and module-finiteness, suppose R[\alpha] is generated as an R-module by \beta_1, \dots, \beta_m. Multiplication by \alpha defines an R-linear endomorphism of the R-module spanned by the \beta_j, represented by a matrix T = (a_{ij}) with entries in R such that \alpha \beta_j = \sum_i a_{ij} \beta_i. The characteristic polynomial \chi(T; X) = \det(XI - T) is monic of degree m with coefficients in R. By the Cayley-Hamilton theorem, \chi(T; \alpha) = 0, so \alpha satisfies the monic polynomial \chi(T; X), proving integrality. The converse direction follows directly from the polynomial reducing higher powers.^[6] Another characterization, particularly useful in number-theoretic settings, involves traces: in a Dedekind domain R with quotient field K and finite-dimensional K-algebra L containing \alpha \in L, if the traces \operatorname{Tr}_{L/K}(\alpha^i) lie in R for n consecutive powers i starting from a sufficiently large exponent a (with a bounded above by O(n \log n), where n = [L:K]), then \alpha is integral over R. This condition ensures the ideal generated by such traces contains the unit ideal, aligning with integrality in these contexts.^[7]

Fundamental Properties

Integral Closure as a Ring

The integral closure \overline{R} of a subring R in an R-algebra A is the set of all elements in A that are integral over R, where an element \alpha \in A is integral over R if it satisfies a monic polynomial equation with coefficients in R: \alpha^n + a_{n-1} \alpha^{n-1} + \cdots + a_0 = 0 for some n \geq 1 and a_i \in R.^[8]^[9] This set \overline{R} forms a subring of A containing R, and moreover, \overline{R} is an \overline{R}-algebra under the natural structure inherited from A.^[10]^[8] To establish that \overline{R} is a ring, it suffices to verify closure under addition and multiplication, along with the presence of additive inverses and the multiplicative identity. First, R \subseteq \overline{R}, since every element of R satisfies the monic polynomial x - r = 0 for r \in R. The multiplicative identity $1_A \in A is integral over R via the polynomial x - 1 = 0, so $1_A \in \overline{R}. For additive inverses, if \alpha \in \overline{R} satisfies \alpha^n + a_{n-1} \alpha^{n-1} + \cdots + a_0 = 0, then -\alpha satisfies the same equation after multiplying by (-1)^n, confirming -\alpha \in \overline{R}.^[9]^[10] The key properties are closure under sums and products. Suppose \alpha, \beta \in \overline{R}. Then R[\alpha] and R[\beta] are finitely generated as R-modules, since integrality is equivalent to R[\alpha] being a finite R-module. Thus, R[\alpha, \beta] = R[\alpha][\beta] is also a finitely generated R-module. Now consider \gamma = \alpha + \beta. The ring R[\gamma] embeds into R[\alpha, \beta], and R[\alpha, \beta] is a faithful R[\gamma]-module (as it contains $1). By the Cayley-Hamilton theorem applied to the regular representation of \gamma over this module, \gamma satisfies a monic polynomial over R, so \gamma \in \overline{R}. Similarly, for \delta = \alpha \beta, R[\delta] embeds into R[\alpha, \beta], which is faithful as an R[\delta]-module, yielding the same conclusion via Cayley-Hamilton. This module-finiteness approach shows both the sum and product are integral over R.^[9]^[10]^[8] Furthermore, if \alpha, \beta \in \overline{R} and $1 + \beta is a unit in A, then \alpha / (1 + \beta) is integral over R. This follows because $1 + \beta is integral over R, its inverse is also integral over R (as the inverse of an integral unit is integral), and the product of integral elements is integral.^[2] In general, \overline{R} properly contains R unless A is already integral over R, in which case \overline{R} = A. For instance, if A is not integral over R, elements outside R but integral over it populate \overline{R}, making it a strict extension. As an \overline{R}-algebra, the multiplication in A restricts to make \overline{R} closed under the operations, preserving the ring structure.^[9]^[8]

Transitivity of Integrality

One key property of integral elements is their transitivity across ring extensions. Specifically, if R \subseteq S \subseteq A are commutative rings and an element \alpha \in A is integral over S, while S is an integral extension of R (meaning every element of S is integral over R), then \alpha is integral over R. This theorem ensures that integrality composes, allowing properties to propagate through chains of extensions.^[11]^[12] The proof relies on the equivalent characterization of integrality in terms of module finiteness: \alpha is integral over S if and only if S[\alpha] is finitely generated as an S-module. Since S is integral over R, it is finitely generated as an R-module, say by elements s_1, \dots, s_m. Similarly, S[\alpha] is finitely generated as an S-module by elements t_1, \dots, t_n. To show S[\alpha] is finitely generated as an R-module, consider the set of products \{ s_i t_j \mid 1 \leq i \leq m, 1 \leq j \leq n \}; any element of S[\alpha] can be expressed as an R-linear combination of these products, establishing finite generation over R. Thus, \alpha is integral over R.^[11] This transitivity has important implications for towers of ring extensions. In a tower R = R_0 \subseteq R_1 \subseteq \cdots \subseteq R_k = A, if each consecutive pair R_i \subseteq R_{i+1} is integral, then the entire tower is an integral extension of R over A, preserving integrality throughout the chain. Such towers arise frequently in algebraic number theory and geometry, facilitating the study of global properties from local ones.^[11]^[12]

Integral Closedness in Fraction Fields

An integral domain R with fraction field K is integrally closed (or normal) if it equals its integral closure in K, meaning every element of K that satisfies a monic polynomial equation with coefficients in R actually lies in R.^[13] This property holds for principal ideal domains, as any fraction a/b in reduced form that is integral over the domain must have b a unit, placing it in the domain itself.^[13] More broadly, unique factorization domains are integrally closed, since factorization properties ensure that integral elements over the domain remain within it.^[6] Not all domains exhibit this closedness; for example, the ring \mathbb{Z}[\sqrt{5}] is not integrally closed in its fraction field \mathbb{Q}(\sqrt{5}), as \frac{1 + \sqrt{5}}{2} satisfies the monic equation X^2 - X - 1 = 0 but does not belong to \mathbb{Z}[\sqrt{5}].^[6] A significant transitivity result characterizes integral closedness in this setting: if R is an integrally closed domain with fraction field K, and S is a subring of K containing R that is integral over R, then the integral closure T of R in S is integrally closed in K (hence also in \mathrm{Frac}(S)). To see this, suppose y \in K is integral over T. By the transitivity of integrality, y is integral over R. Since R is integrally closed in K, it follows that y \in R \subseteq T. Thus, T contains all elements of K integral over it.^[14] The integral closure of a domain in its fraction field, often called its normalization, is always an integrally closed domain by the idempotence of integral closure. This normalization provides the "normal model" of the domain, resolving singularities in algebraic geometry contexts where applicable.^[13]

Relation to Finiteness Conditions

In commutative algebra, integrality is closely linked to finiteness properties of ring extensions and modules. A key result is that if S is an integral extension of a ring R and S is finitely generated as an R-algebra, then S is finitely generated as an R-module; moreover, if R is Noetherian, then S is Noetherian.^[6] This finiteness as a module follows from the fact that each generator of the algebra satisfies a monic polynomial over R, allowing the powers to be expressed linearly in terms of a finite basis.^[15] For the integral closure \bar{R} of a Noetherian ring R in a finite separable extension of its total ring of fractions, \bar{R} is finitely generated as an R-module.^[16] In particular, when R is a Noetherian normal domain with fraction field K and L/K is a finite separable extension, the integral closure of R in L is a finite R-module.^[16] Integral extensions also relate to other finiteness conditions, such as torsion-freeness and projectivity of modules. If R is an integral domain, then any integral extension S of R is torsion-free as an R-module, meaning no nonzero element of S is annihilated by a nonzero element of R.^[6] In special cases, such as when R is a principal ideal domain (e.g., \mathbb{Z} or k for a field k) and S is the integral closure in a finite separable extension of the fraction field, S is a free (hence projective) R-module of rank equal to the degree of the field extension.^[6] However, integrality does not imply flatness in general. While torsion-freeness holds over domains, flatness requires the extension to preserve exact sequences, which fails in many cases where the module is finite but not projective over R.^[15] For instance, finite integral extensions that are not locally free of constant rank provide counterexamples to flatness.^[15] In non-Noetherian settings, these finiteness properties can fail dramatically. For example, consider the ring R = k + x k[x^q \mid q \in \mathbb{Q}^+] over a field k; this ring is not Noetherian, and for $0 < \alpha < 1, the element x^\alpha is integral over R, but the ideal I_\alpha = x k[x^q \mid q \in \mathbb{Q}^+] annihilating the relation is not finitely generated as an R-module.^[17] Thus, the integral closure of R is not finitely generated as an R-module.^[17]

Examples

In Algebraic Number Theory

In algebraic number theory, the integral closure of the rational integers \mathbb{Z} in the field of rational numbers \mathbb{Q} is precisely \mathbb{Z} itself, as every element of \mathbb{Q} that satisfies a monic polynomial with coefficients in \mathbb{Z} must already lie in \mathbb{Z}.^[18] A fundamental example arises in quadratic number fields. For a quadratic extension K = \mathbb{Q}(\sqrt{d}) where d is a square-free integer not equal to 0 or 1, the ring of integers \mathcal{O}_K, which is the integral closure of \mathbb{Z} in K, takes the form \mathbb{Z}[\sqrt{d}] if d \equiv 2, 3 \pmod{4}, and \mathbb{Z}\left[\frac{1 + \sqrt{d}}{2}\right] if d \equiv 1 \pmod{4}.^[19] For instance, consider \sqrt{2} \in \mathbb{Q}(\sqrt{2}); it satisfies the monic polynomial equation x^2 - 2 = 0 with integer coefficients, confirming its integrality over \mathbb{Z}, and \mathbb{Z}[\sqrt{2}] forms the full ring of integers in this field.^[19] In cyclotomic fields, the integral closure of \mathbb{Z} in K = \mathbb{Q}(\zeta_n), where \zeta_n is a primitive nth root of unity, is the cyclotomic ring \mathbb{Z}[\zeta_n].^[20] This ring consists of all algebraic integers within the field and plays a central role in the study of units and ideal class groups for these extensions. The ring of all algebraic integers, denoted \overline{\mathbb{Z}}, is the maximal integral closure of \mathbb{Z} in the complex numbers \mathbb{C}, comprising every complex number that satisfies a monic polynomial over \mathbb{Z}.^[21] It serves as the universal domain for algebraic integers across all number fields. In the local setting, the p-adic integers \mathbb{Z}_p form the integral closure of \mathbb{Z} in the field of p-adic numbers \mathbb{Q}_p, for a prime p, consisting of those elements with p-adic valuation at least 0 that are integral over \mathbb{Z}.^[22]

In Algebraic Geometry

In algebraic geometry, integral elements play a central role in the study of affine varieties through the integral closure of their coordinate rings. For an affine variety X = \mathrm{Spec}(A) over an algebraically closed field k, where A is the coordinate ring, an element in the function field k(X) is integral over A if it satisfies a monic polynomial with coefficients in A. The integral closure A^\nu of A in k(X) is then the ring consisting of all such integral elements, and \mathrm{Spec}(A^\nu) provides the normalization \tilde{X} \to X, a finite birational morphism that resolves singularities in a specific way. This process is particularly significant for curves, where the normalization yields a smooth model of the variety.^[23]^[24] A classic example is the cuspidal curve defined by y^2 = x^3 in \mathbb{A}^2_k, with coordinate ring A = k[x, y]/(y^2 - x^3). This ring is not integrally closed, as the element t = y/x in the function field satisfies the monic equation t^2 - x = 0 over A, making t integral over A. The integral closure is A^\nu = k, with the parametrization x = t^2, y = t^3, realizing the normalization as the affine line \mathbb{A}^1_k. Geometrically, this map \tilde{C} \to C is an isomorphism away from the cusp at the origin, where the singularity is resolved by "unfolding" the curve into a smooth line. In contrast, the nodal curve y^2 = x^3 + x^2 = x^2(x + 1) has coordinate ring B = k[x, y]/(y^2 - x^3 - x^2), featuring a node at the origin with two transverse branches. Adjoining t = y/x yields t^2 = x + 1, so x = t^2 - 1, y = t(t^2 - 1), and the integral closure is B^\nu = k, again normalizing to \mathbb{A}^1_k. Here, the normalization separates the branches, mapping two points on the line to the node.^[25]^[25] The relation to resolution of singularities is that integrally closed domains correspond to normal varieties, which for curves (dimension one) are precisely the smooth ones, as their local rings are discrete valuation rings. Thus, normalization provides a minimal desingularization for plane curves, finite and birational, transforming singular affine varieties into normal ones via the integral closure. Geometrically, integral elements over the coordinate ring parametrize morphisms from normal varieties to the original X; specifically, the normalization \tilde{X} \to X is universal among finite birational morphisms from normal schemes, meaning any such morphism from another normal variety factors uniquely through it. This property underscores the role of integrality in capturing the "integral" or "non-fractional" parametrizations of maps between varieties.^[24]^[23]

Other Integrality Examples

In the context of function fields, consider a field k and the rational function field k(x). An element y algebraic over k(x) satisfies a monic irreducible polynomial f(y) = 0 with coefficients in k, making y integral over k. For instance, if y^2 = x^3 + 1, then y is integral over k because it roots the monic polynomial t^2 - (x^3 + 1), and the ring k embeds into the algebraic closure of k(x).^[10] Such algebraic functions highlight integrality bridging polynomial and rational structures without invoking geometric interpretations. Discrete valuation rings provide a key example of integrally closed domains. A discrete valuation ring (DVR) (R, \mathfrak{m}) of Krull dimension 1, where \mathfrak{m} is principal and generated by a uniformizer \pi, is integrally closed in its fraction field K = \operatorname{Frac}(R).^[8] To see this, suppose \alpha \in K is integral over R, so \alpha satisfies a monic polynomial t^n + a_{n-1} t^{n-1} + \cdots + a_0 = 0 with a_i \in R. Writing \alpha = u \pi^k with u \in R^\times and k \in \mathbb{Z}, the valuation v(\alpha) = k must be non-negative, as otherwise the constant term would force a contradiction in valuations, placing \alpha \in R.^[26] Thus, every DVR is its own integral closure, exemplifying perfect integrality in valuation-theoretic settings. A basic polynomial example occurs in \mathbb{Z} over \mathbb{Z}. The indeterminate x is not integral over \mathbb{Z}, as any monic polynomial t^n + a_{n-1} t^{n-1} + \cdots + a_0 = 0 with a_i \in \mathbb{Z} satisfied by x would imply x^n = -a_{n-1} x^{n-1} - \cdots - a_0, but the left side has no constant term while the right does unless all a_i = 0, contradicting the monic assumption for n \geq 1.^[12] However, roots of monic polynomials over \mathbb{Z} within \mathbb{Z}, such as \sqrt{2} satisfying t^2 - 2 = 0, are integral over \mathbb{Z}, adjoining them via module-finiteness. In power series rings, integrality imposes relations on coefficients. For a power series ring R[] over a domain R, an element f(t) = \sum b_i t^i \in R[] integral over R requires its coefficients b_i to satisfy monic polynomial equations with coefficients in R, often leading to recursive integral dependencies among the b_i.^[27] For example, if f(t) is algebraic over R(t), its minimal monic polynomial over R ensures that the coefficients of f(t) generate a finitely generated module over R, constraining lower-order terms integrally from higher ones. This contrasts with free power series, where coefficients lack such relations.

Integral Extensions

Definition of Integral Extensions

In commutative algebra, a ring homomorphism \phi: R \to S is called an integral extension (or integral ring map) if every element of S is integral over the image \phi(R), meaning that for each s \in S, there exists a monic polynomial P(x) = x^n + a_{n-1} x^{n-1} + \cdots + a_0 \in \phi(R) such that P(s) = 0.^[15] Equivalently, when viewing R as a subring of S via \phi, the extension R \subseteq S is integral if every element of S satisfies such a monic equation with coefficients in R.^[6] Integral extensions are necessarily ring homomorphisms, and they satisfy the property that the composition of integral extensions is again integral; specifically, if R \to S and S \to T are integral, then R \to T is integral, which follows from the transitivity of integrality for elements.^[28] Moreover, if S is integral over R, then S can be expressed as the union of all R-subalgebras R[\alpha_1, \dots, \alpha_k] generated by finite sets of elements \alpha_i \in S, each of which is integral over R.^[6] While every integral ring extension induces an algebraic extension of fraction fields (assuming integral domains), the converse does not hold: there exist ring extensions where elements are algebraic over the fraction field of the base ring but not integral over the base ring itself. For instance, in the extension \mathbb{Z} \subseteq \mathbb{Z}[1/2], the element $1/2 satisfies the equation $2x - 1 = 0 (hence algebraic over \mathbb{Q} = \operatorname{Frac}(\mathbb{Z})) but no monic polynomial over \mathbb{Z}.

Cohen-Seidenberg Theorems

The Cohen-Seidenberg theorems provide fundamental results on the behavior of prime ideals under integral ring extensions. These theorems, established in the context of commutative rings with identity, describe how prime ideals in the base ring R "lie over" to the extension ring S, where S is integral over R via a ring homomorphism \phi: R \to S. Specifically, they ensure surjectivity and controlled chaining of the contraction map on spectra, \operatorname{Spec}(S) \to \operatorname{Spec}(R) given by Q \mapsto \phi^{-1}(Q).^[29] The lying-over theorem states that for an integral extension R \to S, every prime ideal P of R admits at least one prime ideal Q of S such that Q \cap R = P. Equivalently, the map \operatorname{Spec}(S) \to \operatorname{Spec}(R) is surjective. To sketch the proof, localize at the multiplicative set S = R \setminus P, yielding the local ring S_P = S^{-1}S. Since S is integral over R, S_P is integral over R_P, and P S_P is a proper ideal (as ring homomorphisms preserve the unit, preventing the image of R_P / P R_P from being zero). This proper ideal is contained in some maximal ideal M of S_P (by Zorn's lemma). The contraction Q = M \cap S is then a prime ideal of S lying over P.^[29]^[8] The going-up theorem asserts that if P \subseteq P' are prime ideals in R and Q is a prime in S with Q \cap R = P, then there exists a prime Q' in S such that Q \subseteq Q' and Q' \cap R = P'. This allows chains of prime ideals in R to lift to chains in S of the same length. The proof proceeds by applying the lying-over theorem in the quotient setting: consider the integral extension R/P \to S/Q, where the image of P' lies over the zero ideal in R/P (noting that integrality passes to such quotients), yielding a prime in S/Q that pulls back to the desired Q'.^[29]^[30] Complementing these, the incomparability theorem guarantees that if Q_1 and Q_2 are distinct primes in S both lying over the same prime P in R, then neither Q_1 \subseteq Q_2 nor Q_2 \subseteq Q_1. Thus, there are no primes strictly between a lying-over pair. The proof uses the fiber ring S \otimes_R k(P) (equivalently, S_P / P S_P), which is integral over the field k(P); hence, it is 0-dimensional, with all its prime ideals maximal. This implies that the primes over P in \operatorname{Spec}(S) cannot be comparable, as their images in the fiber would form a strict chain of primes in a 0-dimensional ring.^[29]^[31] These theorems imply that integral extensions preserve the Krull dimension: \dim S = \dim R. Chains of primes in R extend equivalently to S via going-up, while lying-over and incomparability ensure no lengthening or shortening occurs, maintaining the supremum length of such chains. This dimension equality holds without further assumptions on normality or domains, unlike the going-down theorem, which requires additional conditions for descent.^[29]^[8]

Geometric Interpretations

In scheme theory, an integral ring extension R \to S corresponds geometrically to a surjective morphism of affine schemes \operatorname{Spec}(S) \to \operatorname{Spec}(R), ensuring that every prime ideal in R lies under at least one prime ideal in S, which manifests as non-empty fibers over every point in \operatorname{Spec}(R).^[32] This surjectivity arises from the lying-over theorem in commutative algebra, translated to the geometric setting where the map covers the base scheme completely. The going-up theorem further interprets integral extensions geometrically by preserving dimensions in the fibers: chains of prime ideals in R of a given length map to chains in S of the same length, implying that the fibers of the morphism \operatorname{Spec}(S) \to \operatorname{Spec}(R) have dimension zero at generic points while maintaining overall dimension equality between source and target for dominant maps.^[32] Integral morphisms are thus universally closed, meaning they are closed in the Zariski topology and remain so under arbitrary base changes, which reflects the stability of integral dependence under localization and completion. This closedness ensures that images of closed subsets remain closed, providing a robust framework for studying geometric closures and resolutions. Integral extensions are stable under base change, so if R \to S is integral, then for any ring map R \to R', the induced map R' \to S \otimes_R R' is also integral, preserving the geometric properties of the morphism in fiber products.^[32] A prominent example is the normalization map for an integral scheme X: the normalization \tilde{X} \to X is an integral morphism that is birational, meaning it induces an isomorphism on dense open subsets, and becomes an isomorphism if X is already normal.^[23] This map resolves singularities while maintaining the birational equivalence essential for studying varieties up to rational maps.

Galois Actions on Integral Extensions

In a Galois extension L/K of number fields, with rings of integers \mathcal{O}_K and \mathcal{O}_L, the Galois group \mathrm{Gal}(L/K) acts on \mathcal{O}_L by field automorphisms that fix K pointwise, thereby mapping integral elements over \mathcal{O}_K to other integral elements, preserving the ring structure.^[33] This action extends naturally to the integral closure of \mathcal{O}_K in L, which coincides with \mathcal{O}_L when \mathcal{O}_K is integrally closed.^[34] A key result states that if \mathcal{O}_K is integrally closed in its fraction field K, then for a finite Galois extension [L/K](/page/Galois_extension), the ring \mathcal{O}_L is precisely the integral closure of \mathcal{O}_K in L, and the action of \mathrm{Gal}(L/K) on this closure is well-defined and faithful in the sense that it respects the integrality condition.^[33] This theorem ensures that the arithmetic structure of the extension is preserved under the Galois symmetries, facilitating the study of ideals and units in \mathcal{O}_L. An important application arises in the analysis of ramification via Dedekind's discriminant theorem: in such an extension, a prime ideal \mathfrak{p} of \mathcal{O}_K ramifies in \mathcal{O}_L if and only if \mathfrak{p} divides the discriminant ideal \mathfrak{D}_{\mathcal{O}_L / \mathcal{O}_K}, where the discriminant is computed using the Galois action on an integral basis of \mathcal{O}_L over \mathcal{O}_K.^[35] This criterion leverages the transitive action of \mathrm{Gal}(L/K) on the prime ideals of \mathcal{O}_L lying over \mathfrak{p} to characterize the ramification behavior.^[34] The subring fixed by the full Galois group action, \mathcal{O}_L^{\mathrm{Gal}(L/K)}, recovers exactly \mathcal{O}_K, reflecting the invariance of the base ring under the symmetries of the extension.^[33] Furthermore, the trace and norm maps induced by the Galois action are compatible with integrality: for \alpha \in \mathcal{O}_L, the trace \mathrm{Tr}_{L/K}(\alpha) \in \mathcal{O}_K and the norm N_{L/K}(\alpha) \in \mathcal{O}_K, as these are sums and products over the Galois conjugates, each of which remains integral.^[34] In equation form, if \{\sigma_i\} enumerates \mathrm{Gal}(L/K),

\mathrm{Tr}_{L/K}(\alpha) = \sum_i \sigma_i(\alpha), \quad N_{L/K}(\alpha) = \prod_i \sigma_i(\alpha),

both landing in \mathcal{O}_K when \alpha is integral over \mathcal{O}_K.^[33]

Integral Closure and Finiteness

Integral Closure

In commutative algebra, given a commutative ring R with unity and an R-algebra A, the integral closure of R in A, denoted \overline{R} or A^i, is defined as the subring consisting of all elements \alpha \in A that are integral over R, meaning each such \alpha satisfies a monic polynomial with coefficients in R.^[36] This construction forms a ring containing R, and integrality is preserved under localization: if S is a multiplicative subset of R, then the integral closure of S^{-1}R in S^{-1}A coincides with S^{-1}\overline{R}.^[37] For integral domains, the integral closure takes on a particularly significant role as the normalization of the domain. Specifically, if R is an integral domain with fraction field K, then the integral closure of R in K is the largest subring of K consisting of elements integral over R; a domain R is called normal if it equals this integral closure.^[37] Every unique factorization domain is normal, as its elements satisfy monic polynomials derived from their factorizations.^[37] A notable property in the context of Dedekind domains is that the integral closure of a Dedekind domain in a finite extension of its fraction field is again a Dedekind domain, preserving the structure of being Noetherian, integrally closed, and one-dimensional.^[38] Computing the integral closure, especially for affine domains or reduced Noetherian rings, relies on normalization algorithms that identify integral elements by solving for roots of monic polynomials or leveraging the Rees algebra to detect integrality.^[39] These methods, often implemented in systems like Singular, proceed by iteratively adjoining integral elements until the ring stabilizes, with conductors providing bounds on the process by measuring the "distance" to normality without fully resolving the extension.^[40] For polynomial rings over fields, such algorithms efficiently handle the finite generation under Noetherian hypotheses.

Finiteness of Integral Closure

A fundamental result in commutative algebra states that if R is a Noetherian ring and A is an integral extension of R, then the integral closure \overline{R} of R in A is finitely generated as an R-module.^[41] This theorem ensures that the process of taking the integral closure preserves the Noetherian property in a controlled manner, allowing \overline{R} to be viewed as a finite R-module extension. The proof relies on the Artin-Rees lemma to manage filtrations associated with ideals in the Rees algebra R[It], where I is an ideal of R. For the complete local Noetherian case, the Cohen structure theorem embeds R into a power series ring, and separability arguments (using traces in separable extensions) show that \overline{R} is contained in a finitely generated submodule. In the general Noetherian setting, localization at maximal ideals reduces the problem to the local case, with Artin-Rees ensuring that the powers of ideals stabilize sufficiently to yield finite generation globally.^[41] An element \alpha \in A contributes to this generation via its integral dependence relation, satisfying a monic polynomial equation

\alpha^n + a_{n-1} \alpha^{n-1} + \cdots + a_1 \alpha + a_0 = 0,

where each a_i \in R; the powers $1, \alpha, \dots, \alpha^{n-1} span a finite R-submodule, and the Noetherian condition limits the number of such relations needed to cover \overline{R}.^[42] In non-Noetherian rings, this finiteness fails. For instance, consider a valuation domain with value group \mathbb{Q}, which is not Noetherian; its integral closure in an extension may require infinitely many generators due to the dense ordering of valuations. Another counterexample is the ring R = k[X_1, X_2, \dots] / (X_1 - X_n^n \mid n \geq 2) over a field k, where ascending chains of ideals do not stabilize, and \overline{R} is not finitely generated over R.^[41] In algebraic geometry, this finiteness theorem implies that the normalization morphism \operatorname{Spec}(\overline{R}) \to \operatorname{Spec}(R) is finite, meaning the normalization of an affine scheme is a finite cover. This property is crucial for resolving singularities, as it allows normalization to be computed effectively and preserves properties like dimension and irreducibility in birational geometry.^[16]

Conductor

In commutative algebra, for an integral domain R with field of fractions K and integral closure \bar{R} in K, the conductor ideal C of R is defined as the annihilator ideal \operatorname{Ann}_R(\bar{R}/R), consisting of all elements r \in R such that r \bar{R} \subseteq R.^[41] This ideal captures the extent to which R fails to be integrally closed, serving as a measure of how "integral" R is relative to its closure.^[41] Equivalently, C is the largest ideal of R that is also an ideal in \bar{R}.^[2] The conductor C is an ideal in both R and \bar{R}, and if \bar{R} is finitely generated as an R-module, then C contains a non-zerodivisor of R.^[41] In the case of one-dimensional Noetherian analytically unramified local rings, C is m-primary, where m is the maximal ideal.^[41] This structure highlights C's role in assessing the deviation from normality, with R = \bar{R} implying C = R.^[41] The annihilator characterization implies that r \in C if and only if r(\alpha - \beta) = 0 for all \alpha \in \bar{R} and \beta \in R, since elements of \bar{R}/R are cosets \alpha + R.^[41] This condition underscores C's connection to the module structure of the extension. In algebraic number theory, for an order O in the ring of integers \mathcal{O}_K of a number field K, the conductor c relates to the discriminant ideal D_{O/\mathbb{Z}} and the different ideal D_{\mathcal{O}_K/\mathbb{Z}} via the formula D_{O/\mathbb{Z}} = N_{\mathcal{O}_K/\mathbb{Z}}(c) \cdot D_{\mathcal{O}_K/\mathbb{Z}}, where N denotes the norm.^[35] This linkage shows how the conductor influences ramification and the scaling of discriminants in subrings. For quadratic orders, such as those in imaginary quadratic fields \mathbb{Q}(\sqrt{d}) with d < 0 square-free, the conductor ideal takes the explicit form C = (f), where f \in \mathbb{Z}_{\geq 0} is the conductor of the order, and the discriminant of the order is f^2 d_K with d_K the field discriminant.^[43] This computation facilitates explicit analysis of ideal class groups and prime splitting in such extensions.^[43]

Advanced Topics

Noether's Normalization Lemma

Noether's normalization lemma asserts that if k is a field and A is a finitely generated k-algebra, then there exist algebraically independent elements z_1, \dots, z_d \in A (where d is the Krull dimension of A) such that the subring B = k[z_1, \dots, z_d] is a polynomial ring and A is integral over B, meaning A is a finitely generated module over B.^[44] This result, originally proved by Emmy Noether in 1926 under the assumption that k is infinite, was later extended to finite fields by Akizuki and Nagata.^[45] The proof proceeds by induction on the number of generators of A. Suppose A = k[x_1, \dots, x_n]/I for some ideal I. If the images of the x_i are algebraically independent, then d = n and the lemma holds trivially. Otherwise, there exists a nonzero polynomial f \in I of minimal degree e \geq 1. Choose exponents a_i = e^{n-i} for i = 1, \dots, n-1 and define new elements y_i = x_i - \alpha x_n^{a_i} for a generic \alpha \in k (ensuring the leading term of f involves a power of x_n that makes the relation monic in x_n). This substitution yields a monic polynomial equation in x_n over k[y_1, \dots, y_{n-1}], showing that A is integral over the subalgebra generated by the y_i. By induction, this subalgebra contains a polynomial subring over which A is integral, and transitivity of integral extensions completes the argument.^[44]^[45] Geometrically, the lemma implies that the affine variety corresponding to A admits a finite morphism to affine d-space \mathbb{A}^d_k, where the map is given by the inclusion k[z_1, \dots, z_d] \hookrightarrow A. This homomorphism \phi: k[z_1, \dots, z_d] \to A satisfies the property that A is module-finite over its image, capturing the finite-type nature of the extension.^[44] Applications of the lemma abound in dimension theory: since integral extensions preserve Krull dimension, \dim A = d = \dim B, providing a concrete realization of the dimension as the transcendence degree of the fraction field of A over k.^[44] It also plays a key role in proving Hilbert's Nullstellensatz; for example, if A is a field (so \dim A = 0), then A is a finite algebraic extension of k, implying that maximal ideals in finitely generated k-algebras correspond to points in affine space over algebraic closures of k.^[45]

Integral Morphisms

In algebraic geometry, an integral morphism of schemes is defined as follows: given a morphism f: X \to S, it is integral if f is affine and, for every affine open subscheme \operatorname{Spec}(R) \subset S, the preimage f^{-1}(\operatorname{Spec}(R)) = \operatorname{Spec}(A) where the ring homomorphism R \to A makes A an integral extension of R, meaning every element of A satisfies a monic polynomial equation with coefficients in R.^[46] This condition ensures that the fibers of f behave like integral ring extensions locally on the base.^[46] Integral morphisms possess several key categorical properties. By definition, they are affine morphisms.^[47] They are stable under base change: if f: X \to S is integral and S' \to S is any morphism, then the base-changed morphism X \times_S S' \to S' is also integral. Composition preserves integrality: the composite of two integral morphisms is integral. Moreover, integral morphisms are universally closed, meaning that for any base change, the resulting morphism is closed (maps closed sets to closed sets).^[48] An integral morphism that is locally of finite type is finite. In the context of scheme theory, finite morphisms—those where the structure sheaf pushforward f_* \mathcal{O}_X is locally finitely generated as an \mathcal{O}_S-module—are a special case of integral morphisms.^[47] Conversely, an integral morphism that is locally of finite type is finite. Finite morphisms are proper under suitable conditions, such as when the target scheme is locally Noetherian.^[49] A notable cohomological property is that for a finite morphism f: X \to S, the pushforward sheaf f_* \mathcal{O}_X is coherent whenever f is of finite presentation. These properties make integral morphisms fundamental in studying families of schemes and their geometric invariants.

Absolute Integral Closure

The absolute integral closure of an integral domain R, denoted R^+, is defined as the integral closure of R in an algebraic closure of its fraction field \mathrm{Frac}(R). This construction provides a universal integral extension that incorporates all elements algebraic over \mathrm{Frac}(R) while remaining integral over R. For domains, R^+ is unique up to non-canonical isomorphism.^[50] In positive characteristic p, R^+ exhibits special structure related to the Frobenius endomorphism. Specifically, for a reduced Noetherian domain R, R^+ coincides with its perfect closure, which is the union \bigcup_{n \geq 0} R^{1/p^n}, where R^{1/p^n} consists of the p^n-th roots of elements in \mathrm{Frac}(R) that are integral over R. Moreover, if R is a local Noetherian domain that is an image of a Cohen-Macaulay local ring, then R^+ is a big Cohen-Macaulay algebra.^[51]^[52] In mixed characteristic, the absolute integral closure of a Henselian local domain inherits desirable homological properties from its base ring. For instance, if R is an analytically irreducible Henselian local ring, the completion of R^+ remains an integral domain and satisfies Cohen-Macaulayness. This contrasts with pure characteristic 0, where R^+ may require careful construction to ensure it forms a ring without additional assumptions, as the lack of Frobenius action complicates the closure process. For example, in characteristic 0, the absolute integral closure of k[] (with k a field) is \bigcup_{n \geq 1} k[[t^{1/n}]].^[53]^[54] A notable example occurs when R = \mathbb{Z}, where \mathbb{Z}^+ is the ring of all algebraic integers, and every finitely generated ideal in \mathbb{Z}^+ is principal, making it a Bézout domain.^[54] The absolute integral closure finds applications in anabelian geometry and étale cohomology, particularly in demonstrating that cohomology classes with coefficients in finite flat group schemes over a base scheme can be annihilated by finite covers of the base.^[54]

References

[1]
Integral Element -- from Wolfram MathWorld
Given a commutative unit ring R and an extension ring S, an element s of S is called integral over R if it is one of the roots of a monic polynomial with ...
[2]
[PDF] A Primer of Commutative Algebra - James Milne
The integral closure of A in B is the subring of B consisting of the elements integral over A. When A is an integral domain, the integral closure of A in ...
[3]
[PDF] Integrality and valuation rings - UChicago Math
As stated in the introduction to the chapter, integrality is a condition on rings parallel to that of algebraicity for field extensions. Definition 1.1 Let R be ...
[4]
[PDF] math 131b: algebra ii part b: commutative algebra - Brandeis
(Equivalence proved later.) (1) α is the root of a monic polynomial with coefficients in R, i.e., ... Any element of R[α, β] is integral over R[β] by Lemma 1.4.
[5]
Richard Dedekind (1831 - 1916) - Biography - MacTutor
Dedekind formulated his theory in the ring of integers of an algebraic number field. The general term 'ring' does not appear, it was introduced later by ...
[6]
[PDF] Integral Ring Extensions
Proof (i) ⇔ (ii): If b is a root of a monic, degree n polynomial over A, then A[b] is spanned as an. A-module by {1, b, b2,...,bn−1}. Conversely, if A[b] is ...Missing: finiteness | Show results with:finiteness
[7]
[PDF] Hideyuki Matsumura - Commutative Algebra
Part I is a self-contained exposition of basis concepts such as flatness, dimen- sion, depth, normal rings, and regular local rings. Part II deals with the ...
[8]
[PDF] Analysis of the Noether Normalization Lemma in Atiyah and ...
Lemma 2.3. (Integral Closure is a ring): The set C of elements of B which are integral over A (the integral closure of A in B) is a subring of B containing A. ...
[9]
[PDF] A Primer of Commutative Algebra - James Milne
A unit of a ring is an element admitting an inverse. The units of a ring. A form a group, which we denote by2 A . Throughout “ring” means “commutative ring”.Missing: textbook | Show results with:textbook
[10]
[PDF] INTEGRAL EXTENSIONS 1. Integral Dependence Let A and B be ...
If A ⊆ B ⊆ C are commutative rings, and B is integral over A and C is integral over B, then C is integral over A. Thus integrality is transitive. Proof. Let c ∈ ...
[11]
[PDF] Math101b_notesB1: Integrality - Brandeis
MATH 101B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA. Proof. Suppose that x = a/b ∈ Q is integral over Z where a, b ∈ Z are relatively prime. Then there are ...
[12]
Section 10.37 (037B): Normal rings—The Stacks project
A domain R is called normal if it is integrally closed in its field of fractions. Lemma 10.37.2. Let R \to S be a ring map. If ...
[13]
[PDF] Faye Jackson MATH 327 - TOC The Poetry of Number Theory
May 6, 2024 · A domain is called integrally closed if it is own integral closure in its fraction field. It follows from. “transitivity of integrality” that ...
[14]
10.36 Finite and integral ring extensions - Stacks Project
An element s in S is integral over R if P^\varphi (s) = 0 for a monic polynomial P(x) in R[x]. A ring map is integral if every s in S is integral over R.
[15]
Lemma 10.161.8 (032L)—The Stacks project
If is not noetherian, the integral closure is still contained in a finite -submodule of . This can be useful, and moreover it is the key point in the proof ...
[16]
[PDF] On Complete Integral Closure of Integral Domains - Clemson OPEN
The set of all elements in the quotient field of D which are integral over D forms a ring ¯D and is called the integral closure of D. If ¯D = D, we say that D ...
[17]
[PDF] lecture 2 - UCSB Math
Sep 28, 2021 · K is the integral closure of Z in K, denoted OK. 3. Page 4. Examples: The ring of algebraic integers in Q is Z since Z is a PID. Z[i] is the ...
[18]
[PDF] Math 210B. Quadratic integer rings 1. Computing the integral ...
Computing the integral closure of Z. Let d ∈ Z − {0,1} be squarefree, and K = Q(. √ d). In this handout, we aim to compute the integral closure OK of Z in ...
[19]
[PDF] Algebraic Number Theory Notes - Math (Princeton)
Sep 28, 2025 · The ring of integers of K is Z[ζ]; i.e. OK = Z[ζ]. Proof. We will use the fact from Theorem 3.3 that the ideal (1−ζ) is prime. Let us take a ...
[20]
[PDF] integral dependence - UiO
The ring Z is integrally closed in Q. Example. The integral closure of Z in C is called the ring of algebraic integers,. i.e. a complex number z is an ...<|control11|><|separator|>
[21]
[PDF] Chapter 3: p-adic integration - Harvard Mathematics Department
Definition 1.11. A p-adic field K is a finite extension of Qp. The ring of integers OK ⊂ K is the integral closure of Zp in K. Lemma 1.12.
[22]
Section 29.54 (035E): Normalization—The Stacks project
29.54 Normalization. Next, we come to the normalization of a scheme X. We only define/construct it when X has locally finitely many irreducible components.Missing: cusp flat
[23]
[PDF] Algebraic Geometry I (Math 6130) Utah/Fall 2020 7. Local Properties ...
Thus the integral closure yields a finite and surjective regular map f : Y → X of affine varieties. Moreover, Y is a normal variety.
[24]
[PDF] introduction to algebraic geometry, class 20
Normalization, and desingularization of curves. Note to myself: perhaps normalization should be done earlier, as soon as integral closure is introduced. That ...
[25]
[PDF] 9. Discrete valuation rings - Brandeis
(5) A is a discrete valuation ring. (6) A is integrally closed. Proof. It is easy to see that the first 4 conditions are equivalent. Given.Missing: source | Show results with:source
[26]
[PDF] Algebraic and Integral Closure of a Polynomial Ring in its Power ...
Feb 1, 2024 · Let R be a domain. We look at the algebraic and integral closure of a polynomial ring, R[x], in its power series ring, R[[x]]. A power series α ...
[27]
[PDF] 9. Integral Ring Extensions
In this case the ring R is not normal: the rational function ϕ = y x. ∈ Quot(R)\R satisfies the monic equation ϕ2 −x−1 = y2 x2 −x−1 = x3+x2 x2. −x−1 ...
[28]
PRIME IDEALS AND INTEGRAL DEPENDENCE - Project Euclid
PRIME IDEALS AND INTEGRAL DEPENDENCE. I. S. COHEN AND A. SEIDENBERG. Let 9t and © be commutative rings such that © contains, and has the same identity element ...
[29]
[PDF] The Going Up and Going Down Theorems
They were proved by Cohen and Seidenberg in 1946. They have geometric meaning in terms of dimension. All rings are commutative with 1. The Going Up Theorem.
[30]
[PDF] Cohen-Seidenberg Theorem Lemma 1. Let B/A be an extension of ...
Theorem 5. (Cohen-Seidenberg Theorem) Let B/A be an integral ex- tension of commutative rings. Then the following are true: 1. Page 2. 2. (1) Incomparability ...
[31]
29.44 Integral and finite morphisms - Stacks Project
It is clear that integral/finite morphisms are separated and quasi-compact. It is also clear that a finite morphism is a morphism of finite type.
[32]
[PDF] Algebraic Number Theory - UCSB Math
It is a very sad moment for me to write this "Geleitwort" to the English translation of Jurgen Neukirch's book on Algebraic Number Theory. ... ring of integers of ...
[33]
[PDF] Algebraic Number Theory - James Milne
An abelian extension of a field is a Galois extension of the field with abelian Galois group. Class field theory describes the abelian extensions of a number ...
[34]
[PDF] 12 The different and the discriminant
Oct 20, 2016 · There is only one prime q above p, so we also have q|DB/A if and only if q is ramified. We now note an important corollary of Theorem 12.19.
[35]
Integral Closure -- from Wolfram MathWorld
The integral closure of a commutative unit ring R in an extension ring S is the set of all elements of S which are integral over R.
[36]
integral closure in nLab
### Summary of Integral Closure (ncatlab.org/nlab/show/integral+closure)
[37]
[PDF] 5 Dedekind extensions - MIT Mathematics
Sep 22, 2016 · In this lecture we prove that the integral closure of a Dedekind domain in a finite extension of its fraction field is also a Dedekind domain; ...
[38]
[0904.3561] Normalization of Rings - arXiv
Apr 22, 2009 · We present a new algorithm to compute the integral closure of a reduced Noetherian ring in its total ring of fractions.
[39]
Integral closure of rings and ideals - Singular
Integral closure of rings and ideals. In this section we present an algorithm to compute, for a reduced affine ring $ A$ , the normalization, that is, the ...
[40]
[PDF] Integral Closure of Ideals, Rings, and Modules - Purdue Math
2.) Page 11. Notation and basic definitions. Except where otherwise noted, all rings in this book are commutative with identity and most are Noetherian. An ...<|control11|><|separator|>
[41]
[PDF] Math 145. Integral closure This handout aims to show the following ...
Let L be a finite extension of the fraction field K of A. Then the integral closure of A in L is a finite A-module. In particular, the integral closure of A in ...
[42]
[PDF] An Introduction to Orders of Number Fields - Kiran S. Kedlaya
May 3, 2002 · ring of integers, so is not integrally closed. Further, by ring theory we know that any U.F.D. is integrally closed (see Lang[2], Prop.7, p ...
[43]
[PDF] 8.4. Noether Normalisation. - DPMMS
If A is an integral domain then its integral closure in Q(A) is the intersection of all valuation rings in Q(A) containing A. Proof. Since valuation rings are ...<|control11|><|separator|>
[44]
[PDF] Commutative Algebra `a la A. J. de Jong - Columbia Math Department
We now have enough machinery to prove Noether normalization. Proof. Let A be as in the theorem. We write A = k[x1,...,xn]/I. We proceed.
[45]
https://www.math.columbia.edu/~dejong/courses/deJongNotes.pdf
[46]
https://stacks.math.columbia.edu/tag/01WH
[47]
Lemma 29.44.7 (01WM)—The Stacks project
An integral morphism is affine by definition. A base change of an integral morphism is integral so in order to prove (2) it suffices to show that an integral ...Missing: geometric | Show results with:geometric
[48]
https://stacks.math.columbia.edu/tag/01WM
[49]
[PDF] Absolute integral closures of commutative rings - arXiv
Jan 17, 2023 · 2 First properties. First, we establish that the new definition of absolute integral closure agrees with the existing one for domains ...
[50]
Homological properties of the perfect and absolute integral closures ...
Jan 14, 2010 · For a Noetherian local domain R let R + be the absolute integral closure of R and let R ∞ be the perfect closure of R, when R has prime
[51]
On the vanishing of local cohomology of the absolute integral ...
Jun 15, 2016 · The absolute integral closure of R, denoted , is the integral closure of R in a fixed algebraic closure of K. A famous result of M. Hochster ...
[52]
R+ˆ is surprisingly an integral domain - ScienceDirect
More generally, this theorem remains true if the completeness assumption is relaxed to allow R to be an analytically irreducible Henselian local ring. It is ...
[53]
(PDF) Absolute Integral Closure - ResearchGate
The main object we will study is given in the following deﬁnition. Definition 2.2.Let Rbe an integral domain with fraction ﬁeld K. Let Kbe a ﬁxed algebraic.