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 which 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:

rngsringscommutative ringsintegral domainsintegrally closed domainsGCD domainsunique factorization domainsprincipal ideal domainsEuclidean domainsfieldsalgebraically closed fields

An explicit example is the ring of integers Z, an 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 xL is integral over A if and only if it is algebraic over K and its minimal polynomial over K has coefficients in A. 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 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: 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 $k$ 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 in $S$ , 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 $r\geq 2$ .

To give a non-example, let k be a field and $A=k[t^{2},t^{3}]\subset k[t]$ (A is the subalgebra generated by t2 and t3.) A is not integrally closed: it has the field of fractions $k(t)$ , and the monic polynomial $X^{2}-t^{2)$ 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 $Y^{2}=X^{3)$ has a singularity at the origin.

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

## 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 ${\mathfrak {p))$ ,

1. If ${\mathfrak {p))$ has height $\leq 1$ , then $A_{\mathfrak {p))$ is regular (i.e., $A_{\mathfrak {p))$ is a discrete valuation ring.)
2. 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 domain need not be completely integrally closed.

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

## Modules over an integrally closed domain

This section needs expansion. You can help by adding to it. (February 2013)

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 $c(M)$ .

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 ${\mathfrak {m))$ . 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.
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
• Bourbaki, Nicolas (1972). Commutative Algebra. Paris: Hermann.
• Hartshorne, Robin (1977), Algebraic Geometry, Graduate Texts in Mathematics, vol. 52, New York: Springer-Verlag, ISBN 978-0-387-90244-9, MR 0463157
• 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 0-521-36764-6.
• Matsumura, Hideyuki (1970). Commutative Algebra. ISBN 0-8053-7026-9.