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 domain appear in the following chain of class inclusions:

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

Examples

The following are integrally closed domains.

Any principal ideal domain (in particular, any field).
Any unique factorization domain (in particular, any polynomial ring over a unique factorization domain.)
Any GCD domain (in particular, any Bézout domain or valuation domain).
Any Dedekind domain.
Any symmetric algebra over a field (since every symmetric algebra is isomorphic to a polynomial ring in several variables over a field).

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 $A_\mathfrak{p}$ 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

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 $\mathfrak{p}$ ,

(i) If $\mathfrak{p}$ has height $\le 1$ , then $A_\mathfrak{p}$ is regular (i.e., $A_\mathfrak{p}$ is a discrete valuation ring.)
(ii) If $\mathfrak{p}$ has height $\ge 2$ , then $A_\mathfrak{p}$ has depth $\ge 2$ .^[6]

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. 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; 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]

"Integrally closed" under constructions

The following conditions are equivalent for an integral domain A:

A is integrally closed;
A_p (the localization of A with respect to p) is integrally closed for every prime ideal p;
A_m 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 $c(M)$ .

References

↑ Taken from Matsumura
↑ Matsumura, Theorem 9.2
↑ 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.
↑ Kaplansky, Theorem 168, pg 119.
↑ Matsumura 1989, p. 64
↑ Matsumura, Commutative algebra, pg. 125. For a domain, the theorem is due to Krull (1931). The general case is due to Serre.
↑ over an algebraically closed field
↑ An exercise in Matsumura.
↑ Matsumura, Exercise 10.4
↑ An exercise in Bourbaki.
↑ Bourbaki, Ch. VII, § 1, n. 2, Theorem 1
↑ An exercise in Bourbaki.
↑ Bourbaki Ch. VII, § 1, n. 6. Proposition 10.
↑ Bourbaki Ch. VII, § 4, n. 7

Bourbaki, Commutative algebra.
Kaplansky, Irving (September 1974). Commutative Rings. Lectures in Mathematics. University of Chicago Press. ISBN 0-226-42454-5.
Matsumura, Hideyuki (1989), Commutative Ring Theory, Cambridge Studies in Advanced Mathematics (2nd ed.), Cambridge University Press, ISBN 978-0-521-36764-6.
Matsumura, Hideyuki (1970) Commutative algebra ISBN 0-8053-7026-9.