Integrally closed domain

In commutative algebra, an integrally closed domain A is an integral domain whose integral closure in the field of fractions of A 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 $A = k[t^2, t^3] \subset B = k[t]$ (k a field). 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 Y² = X³ has a singularity at the origin.

Let A be an integrally closed domain with field of fractions K and a finite extension L 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 has a minimal polynomial over A: this is stronger than that an integral element satisfying 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.

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 Bezout 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

If A is a noetherian integrally closed domain, then A is the intersection of all localizations $A_\mathfrak{p}$ over prime ideals $\mathfrak{p}$ of height 1.

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.

Related notions

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.

Let A be a noetherian ring. Then 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 nonzero zero-divisor 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.

Let A be a domain and K its field of fractions. x in K is said to be almost integral over A 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]

"Integrally closed" under constructions

The following conditions are equivalent for an integral domain A:

1. A is integrally closed;
2. A_p (the localization of A with respect to p) is integrally closed for every prime ideal p;
3. 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.^[11]

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. ^ An exercise in Bourbaki.

• Bourbaki, Commutative algebra.
• Kaplansky, Irving (September 1974). Commutative Rings. Lectures in Mathematics. University of Chicago Press. ISBN 0226424545. 
• 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.