Integrally closed domain

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In commutative algebra, an integrally closed domain A is an integral domain whose integral closure in its field of fractions is A itself. Spelled out, this means that if x is an element of the field of fractions of A that is a root of a monic polynomial with coefficients in A, then x is itself an element of A. Many well-studied domains are integrally closed, as shown by the following chain of class inclusions:

commutative rings ⊃ integral domains ⊃ integrally closed domains ⊃ GCD domains ⊃ unique factorization domains ⊃ principal ideal domains ⊃ Euclidean domains ⊃ fields ⊃ finite fields

An explicit example is the ring of integers Z, a Euclidean domain. All regular local rings are integrally closed as well.

Basic properties

Let A be an integrally closed domain with field of fractions K and let L be a field extension of K. Then x∈L is integral over A if and only if it is algebraic over K and its minimal polynomial over K has coefficients in A.[1] In particular, this means that any element of L integral over A is root of a monic polynomial in A[X] that is irreducible in K[X].

If A is a domain contained in a field K, we can consider the integral closure of A in K (i.e. the set of all elements of K that are integral over A). This integral closure is an integrally closed domain.

Integrally closed domains also play a role in the hypothesis of the Going-down theorem. The theorem states that if A⊆B is an integral extension of domains and A is an integrally closed domain, then the going-down property holds for the extension A⊆B.

Examples

The following are integrally closed domains.

  • A principal ideal domain (in particular: the integers and any field).
  • A unique factorization domain (in particular, any polynomial ring over a field, over the integers, or over any unique factorization domain).
  • A GCD domain (in particular, any Bézout domain or valuation domain).
  • A Dedekind domain.
  • A symmetric algebra over a field (since every symmetric algebra is isomorphic to a polynomial ring in several variables over a field).
  • Let [math]\displaystyle{ k }[/math] be a field of characteristic not 2 and [math]\displaystyle{ S = k[x_1, \dots, x_n] }[/math] a polynomial ring over it. If [math]\displaystyle{ f }[/math] is a square-free nonconstant polynomial in [math]\displaystyle{ S }[/math], then [math]\displaystyle{ S[y]/(y^2 - f) }[/math] is an integrally closed domain.[2] In particular, [math]\displaystyle{ k[x_0, \dots, x_r]/(x_0^2 + \dots + x_r^2) }[/math] is an integrally closed domain if [math]\displaystyle{ r \ge 2 }[/math].[3]

To give a non-example,[4] let k be a field and [math]\displaystyle{ A = k[t^2, t^3] \subset k[t] }[/math] (A is the subalgebra generated by t2 and t3.) A is not integrally closed: it has the field of fractions [math]\displaystyle{ k(t) }[/math], and the monic polynomial [math]\displaystyle{ X^2 - t^2 }[/math] in the variable X has root t which is in the field of fractions but not in A. This is related to the fact that the plane curve [math]\displaystyle{ Y^2 = X^3 }[/math] has a singularity at the origin.

Another domain that is not integrally closed is [math]\displaystyle{ A = \mathbb{Z}[\sqrt{5}] }[/math]; it does not contain the element [math]\displaystyle{ \frac{\sqrt{5}+1}{2} }[/math] of its field of fractions, which satisfies the monic polynomial [math]\displaystyle{ X^2-X-1 = 0 }[/math].

Noetherian integrally closed domain

For a noetherian local domain A of dimension one, the following are equivalent.

  • A is integrally closed.
  • The maximal ideal of A is principal.
  • A is a discrete valuation ring (equivalently A is Dedekind.)
  • A is a regular local ring.

Let A be a noetherian integral domain. Then A is integrally closed if and only if (i) A is the intersection of all localizations [math]\displaystyle{ A_\mathfrak{p} }[/math] over prime ideals [math]\displaystyle{ \mathfrak{p} }[/math] of height 1 and (ii) the localization [math]\displaystyle{ A_\mathfrak{p} }[/math] at a prime ideal [math]\displaystyle{ \mathfrak{p} }[/math] of height 1 is a discrete valuation ring.

A noetherian ring is a Krull domain if and only if it is an integrally closed domain.

In the non-noetherian setting, one has the following: an integral domain is integrally closed if and only if it is the intersection of all valuation rings containing it.

Normal rings

Authors including Serre, Grothendieck, and Matsumura define a normal ring to be a ring whose localizations at prime ideals are integrally closed domains. Such a ring is necessarily a reduced ring,[5] and this is sometimes included in the definition. In general, if A is a Noetherian ring whose localizations at maximal ideals are all domains, then A is a finite product of domains.[6] In particular if A is a Noetherian, normal ring, then the domains in the product are integrally closed domains.[7] Conversely, any finite product of integrally closed domains is normal. In particular, if [math]\displaystyle{ \operatorname{Spec}(A) }[/math] is noetherian, normal and connected, then A is an integrally closed domain. (cf. smooth variety)

Let A be a noetherian ring. Then (Serre's criterion) A is normal if and only if it satisfies the following: for any prime ideal [math]\displaystyle{ \mathfrak{p} }[/math],

  1. If [math]\displaystyle{ \mathfrak{p} }[/math] has height [math]\displaystyle{ \le 1 }[/math], then [math]\displaystyle{ A_\mathfrak{p} }[/math] is regular (i.e., [math]\displaystyle{ A_\mathfrak{p} }[/math] is a discrete valuation ring.)
  2. If [math]\displaystyle{ \mathfrak{p} }[/math] has height [math]\displaystyle{ \ge 2 }[/math], then [math]\displaystyle{ A_\mathfrak{p} }[/math] has depth [math]\displaystyle{ \ge 2 }[/math].[8]

Item (i) is often phrased as "regular in codimension 1". Note (i) implies that the set of associated primes [math]\displaystyle{ Ass(A) }[/math] has no embedded primes, and, when (i) is the case, (ii) means that [math]\displaystyle{ Ass(A/fA) }[/math] has no embedded prime for any non-zerodivisor f. In particular, a Cohen-Macaulay ring satisfies (ii). Geometrically, we have the following: if X is a local complete intersection in a nonsingular variety;[9] e.g., X itself is nonsingular, then X is Cohen-Macaulay; i.e., the stalks [math]\displaystyle{ \mathcal{O}_p }[/math] of the structure sheaf are Cohen-Macaulay for all prime ideals p. Then we can say: X is normal (i.e., the stalks of its structure sheaf are all normal) if and only if it is regular in codimension 1.

Completely integrally closed domains

Let A be a domain and K its field of fractions. An element x in K is said to be almost integral over A if the subring A[x] of K generated by A and x is a fractional ideal of A; that is, if there is a nonzero [math]\displaystyle{ d \in A }[/math] such that [math]\displaystyle{ d x^n \in A }[/math] for all [math]\displaystyle{ n \ge 0 }[/math]. Then A is said to be completely integrally closed if every almost integral element of K is contained in A. A completely integrally closed domain is integrally closed. Conversely, a noetherian integrally closed domain is completely integrally closed.

Assume A is completely integrally closed. Then the formal power series ring [math]\displaystyle{ AX }[/math] is completely integrally closed.[10] This is significant since the analog is false for an integrally closed domain: let R be a valuation domain of height at least 2 (which is integrally closed.) Then [math]\displaystyle{ RX }[/math] is not integrally closed.[11] Let L be a field extension of K. Then the integral closure of A in L is completely integrally closed.[12]

An integral domain is completely integrally closed if and only if the monoid of divisors of A is a group.[13]


"Integrally closed" under constructions

The following conditions are equivalent for an integral domain A:

  1. A is integrally closed;
  2. Ap (the localization of A with respect to p) is integrally closed for every prime ideal p;
  3. Am is integrally closed for every maximal ideal m.

1 → 2 results immediately from the preservation of integral closure under localization; 2 → 3 is trivial; 3 → 1 results from the preservation of integral closure under localization, the exactness of localization, and the property that an A-module M is zero if and only if its localization with respect to every maximal ideal is zero.

In contrast, the "integrally closed" does not pass over quotient, for Z[t]/(t2+4) is not integrally closed.

The localization of a completely integrally closed domain need not be completely integrally closed.[14]

A direct limit of integrally closed domains is an integrally closed domain.

Modules over an integrally closed domain

Let A be a Noetherian integrally closed domain.

An ideal I of A is divisorial if and only if every associated prime of A/I has height one.[15]

Let P denote the set of all prime ideals in A of height one. If T is a finitely generated torsion module, one puts:

[math]\displaystyle{ \chi(T) = \sum_{p \in P} \operatorname{length}_p(T) p }[/math],

which makes sense as a formal sum; i.e., a divisor. We write [math]\displaystyle{ c(d) }[/math] for the divisor class of d. If [math]\displaystyle{ F, F' }[/math] are maximal submodules of M, then [math]\displaystyle{ c(\chi(M/F)) = c(\chi(M/F')) }[/math][16] and [math]\displaystyle{ c(\chi(M/F)) }[/math] is denoted (in Bourbaki) by [math]\displaystyle{ c(M) }[/math].

See also

Citations

  1. ↑ Matsumura, Theorem 9.2
  2. ↑ Hartshorne 1977, Ch. II, Exercise 6.4.
  3. ↑ Hartshorne 1977, Ch. II, Exercise 6.5. (a)
  4. ↑ Taken from Matsumura
  5. ↑ If all localizations at maximal ideals of a commutative ring R are reduced rings (e.g. domains), then R is reduced. Proof: Suppose x is nonzero in R and x2=0. The annihilator ann(x) is contained in some maximal ideal [math]\displaystyle{ \mathfrak{m} }[/math]. Now, the image of x is nonzero in the localization of R at [math]\displaystyle{ \mathfrak{m} }[/math] since [math]\displaystyle{ x = 0 }[/math] at [math]\displaystyle{ \mathfrak{m} }[/math] means [math]\displaystyle{ xs = 0 }[/math] for some [math]\displaystyle{ s \not\in \mathfrak{m} }[/math] but then [math]\displaystyle{ s }[/math] is in the annihilator of x, contradiction. This shows that R localized at [math]\displaystyle{ \mathfrak{m} }[/math] is not reduced.
  6. ↑ Kaplansky, Theorem 168, pg 119.
  7. ↑ Matsumura 1989, p. 64
  8. ↑ Matsumura, Commutative algebra, pg. 125. For a domain, the theorem is due to Krull (1931). The general case is due to Serre.
  9. ↑ over an algebraically closed field
  10. ↑ An exercise in Matsumura.
  11. ↑ Matsumura, Exercise 10.4
  12. ↑ An exercise in Bourbaki.
  13. ↑ Bourbaki 1972, Ch. VII, § 1, n. 2, Theorem 1
  14. ↑ An exercise in Bourbaki.
  15. ↑ Bourbaki 1972, Ch. VII, § 1, n. 6. Proposition 10.
  16. ↑ Bourbaki 1972, Ch. VII, § 4, n. 7

References



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