# Integrally closed domain

Algebraic structures |
---|

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: fields, the ring of integers **Z**, unique factorization domains and regular local rings are all integrally closed.

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**

## Basic properties

Let *A* be an integrally closed domain with field of fractions *K* and let *L* be a field extension of *K*. Then *x*∈*L* is integral over *A* if and only if it is algebraic over *K* and its minimal polynomial over *K* has coefficients in *A*.^{[1]} 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 *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.

- 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 [math]\displaystyle{ k }[/math] be a field of characteristic not 2 and [math]\displaystyle{ S = k[x_1, \dots, x_n] }[/math] a polynomial ring over it. If [math]\displaystyle{ f }[/math] is a square-free nonconstant polynomial in [math]\displaystyle{ S }[/math], then [math]\displaystyle{ S[y]/(y^2 - f) }[/math] is an integrally closed domain.
^{[2]}In particular, [math]\displaystyle{ k[x_0, \dots, x_r]/(x_0^2 + \dots + x_r^2) }[/math] is an integrally closed domain if [math]\displaystyle{ r \ge 2 }[/math].^{[3]}

To give a non-example,^{[4]} let *k* be a field and [math]\displaystyle{ A = k[t^2, t^3] \subset k[t] }[/math] (*A* is the subalgebra generated by *t*^{2} and *t*^{3}.) *A* is not integrally closed: it has the field of fractions [math]\displaystyle{ k(t) }[/math], and the monic polynomial [math]\displaystyle{ X^2 - t^2 }[/math] 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 [math]\displaystyle{ Y^2 = X^3 }[/math] has a singularity at the origin.

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

## 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 [math]\displaystyle{ A_\mathfrak{p} }[/math] over prime ideals [math]\displaystyle{ \mathfrak{p} }[/math] of height 1 and (ii) the localization [math]\displaystyle{ A_\mathfrak{p} }[/math] at a prime ideal [math]\displaystyle{ \mathfrak{p} }[/math] 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,^{[5]} 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.^{[6]} In particular if *A* is a Noetherian, normal ring, then the domains in the product are integrally closed domains.^{[7]} Conversely, any finite product of integrally closed domains is normal. In particular, if [math]\displaystyle{ \operatorname{Spec}(A) }[/math] 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 [math]\displaystyle{ \mathfrak{p} }[/math],

- If [math]\displaystyle{ \mathfrak{p} }[/math] has height [math]\displaystyle{ \le 1 }[/math], then [math]\displaystyle{ A_\mathfrak{p} }[/math] is regular (i.e., [math]\displaystyle{ A_\mathfrak{p} }[/math] is a discrete valuation ring.)
- If [math]\displaystyle{ \mathfrak{p} }[/math] has height [math]\displaystyle{ \ge 2 }[/math], then [math]\displaystyle{ A_\mathfrak{p} }[/math] has depth [math]\displaystyle{ \ge 2 }[/math].
^{[8]}

Item (i) is often phrased as "regular in codimension 1". Note (i) implies that the set of associated primes [math]\displaystyle{ Ass(A) }[/math] has no embedded primes, and, when (i) is the case, (ii) means that [math]\displaystyle{ Ass(A/fA) }[/math] 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;^{[9]} e.g., *X* itself is nonsingular, then *X* is Cohen-Macaulay; i.e., the stalks [math]\displaystyle{ \mathcal{O}_p }[/math] 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 [math]\displaystyle{ d \ne 0 }[/math] such that [math]\displaystyle{ d x^n \in A }[/math] for all [math]\displaystyle{ n \ge 0 }[/math]. 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 [math]\displaystyle{ AX }[/math] is completely integrally closed.^{[10]} 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 [math]\displaystyle{ RX }[/math] is not integrally closed.^{[11]} Let *L* be a field extension of *K*. Then the integral closure of *A* in *L* is completely integrally closed.^{[12]}

An integral domain is completely integrally closed if and only if the monoid of divisors of *A* is a group.^{[13]}

See also: Krull domain.

## "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]/(t^{2}+4) is not integrally closed.

The localization of a completely integrally closed domain need not be completely integrally closed.^{[14]}

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.^{[15]}

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

- [math]\displaystyle{ \chi(T) = \sum_{p \in P} \operatorname{length}_p(T) p }[/math],

which makes sense as a formal sum; i.e., a divisor. We write [math]\displaystyle{ c(d) }[/math] for the divisor class of *d*. If [math]\displaystyle{ F, F' }[/math] are maximal submodules of *M*, then [math]\displaystyle{ c(\chi(M/F)) = c(\chi(M/F')) }[/math]^{[16]} and [math]\displaystyle{ c(\chi(M/F)) }[/math] is denoted (in Bourbaki) by [math]\displaystyle{ c(M) }[/math].

## See also

## Citations

- ↑ Matsumura, Theorem 9.2
- ↑ Hartshorne 1977, Ch. II, Exercise 6.4.
- ↑ Hartshorne 1977, Ch. II, Exercise 6.5. (a)
- ↑ Taken from Matsumura
- ↑ 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*x*^{2}=0. The annihilator ann(*x*) is contained in some maximal ideal [math]\displaystyle{ \mathfrak{m} }[/math]. Now, the image of*x*is nonzero in the localization of*R*at [math]\displaystyle{ \mathfrak{m} }[/math] since [math]\displaystyle{ x = 0 }[/math] at [math]\displaystyle{ \mathfrak{m} }[/math] means [math]\displaystyle{ xs = 0 }[/math] for some [math]\displaystyle{ s \not\in \mathfrak{m} }[/math] but then [math]\displaystyle{ s }[/math] is in the annihilator of*x*, contradiction. This shows that*R*localized at [math]\displaystyle{ \mathfrak{m} }[/math] 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 1972, Ch. VII, § 1, n. 2, Theorem 1
- ↑ An exercise in Bourbaki.
- ↑ Bourbaki 1972, Ch. VII, § 1, n. 6. Proposition 10.
- ↑ Bourbaki 1972, Ch. VII, § 4, n. 7

## References

- Bourbaki, Nicolas (1972).
*Commutative Algebra*. Paris: Hermann. https://archive.org/details/commutativealgeb0000bour. - Hartshorne, Robin (1977),
*Algebraic Geometry*, Graduate Texts in Mathematics,**52**, New York: Springer-Verlag, ISBN 978-0-387-90244-9 - Kaplansky, Irving (September 1974).
*Commutative Rings*. Lectures in Mathematics. University of Chicago Press. ISBN 0-226-42454-5. https://archive.org/details/commutativerings00irvi. - 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.

Original source: https://en.wikipedia.org/wiki/Integrally closed domain.
Read more |