# 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, 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.

Note that integrally closed domains appear in the following chain of class inclusions:

commutative ringsintegral domainsintegrally closed domainsGCD domainsunique factorization domainsprincipal ideal domainsEuclidean domainsfieldsfinite fields

## Basic properties

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. 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". (For example, consider $A=\mathbb {Z} [{\sqrt {5}}]$ , which is not integrally closed over $\mathbb {Z}$  because it does not for example contain the element ${\frac {{\sqrt {5}}+1}{2}}$  of its field of fractions, which satisfies the monic integral polynomial $X^{2}-X-1=0$ ).

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

## Examples

The following are integrally closed domains.

• A principal ideal domain (in particular, any field).
• A unique factorization domain (in particular, any polynomial ring over a 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 $k$  be a field of characteristic not 2 and $S=k[x_{1},\dots ,x_{n}]$  a polynomial ring over it. If $f$  is a square-free nonconstant polynomial, then $S[y]/(y^{2}-f)$  is an integrally closed domain. In particular, $k[x_{0},\dots ,x_{r}]/(x_{0}^{2}+\dots +x_{r}^{2})$  is an integrally closed domain if $r\geq 2$ .

## 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, 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. In particular if A is a Noetherian, normal ring, then the domains in the product are integrally closed domains. 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 $\leq 1$ , then $A_{\mathfrak {p}}$  is regular (i.e., $A_{\mathfrak {p}}$  is a discrete valuation ring.)
• (ii) If ${\mathfrak {p}}$  has height $\geq 2$ , then $A_{\mathfrak {p}}$  has depth $\geq 2$ .

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; 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 $d\neq 0$  such that $dx^{n}\in A$  for all $n\geq 0$ . 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. 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. Let L be a field extension of K. Then the integral closure of A in L is completely integrally closed.

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

## "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.

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.

Let P denote 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 $c(\chi (M/F))=c(\chi (M/F'))$  and $c(\chi (M/F))$  is denoted (in Bourbaki) by $c(M)$ .