Integrally closed domain

In commutative algebra, an integrally closed domain A is an integral domain whose integral closure in its field of fractions is A itself. Many well-studied domains are integrally closed: Fields, the ring of integers Z, unique factorization domains and regular local rings are all integrally closed.

To give a non-example,[1] let k be a field and A = k[t^2, t^3] \subset B = k[t] (A is the subalgebra generated by t2 and t3.) A and B have the same field of fractions, and B is the integral closure of A (since B is a UFD.) In other words, A is not integrally closed. This is related to the fact that the plane curve Y^2 = X^3 has a singularity at the origin.

Let A be an integrally closed domain with field of fractions K and let L be a finite extension of K. Then x in L is integral over A if and only if its minimal polynomial over K has coefficients in A.[2] This implies in particular that an integral element over an integrally closed domain A has a minimal polynomial over A. This is stronger than the statement that any integral element satisfies some monic polynomial. In fact, the statement is false without "integrally closed" (consider A = \mathbb{Z}[\sqrt{5}].)

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.

Note that integrally closed domains appear in 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

Examples

The following are integrally closed domains.

Noetherian integrally closed domain

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

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

See also: normal variety

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,[3] 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.[4] In particular if A is a Noetherian, normal ring, then the domains in the product are integrally closed domains.[5] Conversely, any finite product of integrally closed domains is normal. In particular, if \operatorname{Spec}(A) 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 \mathfrak{p},

Item (i) is often phrased as "regular in codimension 1". Note (i) implies that the set of associated primes Ass(A) has no embedded primes, and, when (i) is the case, (ii) means that Ass(A/fA) 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;[7] e.g., X itself is nonsingular, then X is Cohen-Macaulay; i.e., the stalks \mathcal{O}_p 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 d \ne 0 such that d x^n \in A for all n \ge 0. 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 A[[X]] is completely integrally closed.[8] 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 R[[X]] is not integrally closed.[9] Let L be a field extension of K. Then the integral closure of A in L is completely integrally closed.[10]

An integral domain is completely integrally closed if and only if the monoid of divisors of A is a group.[11]

See also: Krull domain.

"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 need not be completely integrally closed.[12]

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.[13]

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

\chi(T) = \sum_{p \in P} \operatorname{length}_p(T) p,

which makes sense as a formal sum; i.e., a divisor. We write c(d) for the divisor class of d. If F, F' are maximal submodules of M, then c(\chi(M/F)) = c(\chi(M/F'))[14] and c(\chi(M/F)) is denoted (in Bourbaki) by c(M).

See also

References

  1. ↑ Taken from Matsumura
  2. ↑ Matsumura, Theorem 9.2
  3. ↑ 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 \mathfrak{m}. Now, the image of x is nonzero in the localization of R at \mathfrak{m} since x = 0 at \mathfrak{m} means xs = 0 for some s \not\in \mathfrak{m} but then s is in the annihilator of x, contradiction. This shows that R localized at \mathfrak{m} is not reduced.
  4. ↑ Kaplansky, Theorem 168, pg 119.
  5. ↑ Matsumura 1989, p. 64
  6. ↑ Matsumura, Commutative algebra, pg. 125. For a domain, the theorem is due to Krull (1931). The general case is due to Serre.
  7. ↑ over an algebraically closed field
  8. ↑ An exercise in Matsumura.
  9. ↑ Matsumura, Exercise 10.4
  10. ↑ An exercise in Bourbaki.
  11. ↑ Bourbaki, Ch. VII, § 1, n. 2, Theorem 1
  12. ↑ An exercise in Bourbaki.
  13. ↑ Bourbaki Ch. VII, § 1, n. 6. Proposition 10.
  14. ↑ Bourbaki Ch. VII, § 4, n. 7
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