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. Spelled out, this means that if x is an element of the field of fractions of A that 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:

An explicit example is the ring of integers Z, a Euclidean domain. All regular local rings are integrally closed as well.

A ring whose localizations at all prime ideals are integrally closed domains is a normal ring.

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&isin;L 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 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 $$k$$ 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 $$r \ge 2$$.

To give a non-example, let k be a field and $$A = k[t^2, t^3] \subset k[t]$$, the subalgebra generated by t2 and t3. Then 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 that is not integrally closed is $$A = \mathbb{Z}[\sqrt{5}]$$; its field of fractions contains the element $$\frac{\sqrt{5}+1}{2}$$, which is not in A but 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 $$\mathfrak{p}$$, If $$\mathfrak{p}$$ has height $$\le 1$$, then $$A_\mathfrak{p}$$ is regular (i.e., $$A_\mathfrak{p}$$ is a discrete valuation ring.) If $$\mathfrak{p}$$ has height $$\ge 2$$, then $$A_\mathfrak{p}$$ has depth $$\ge 2$$.  

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 nonzero $$d \in A$$ 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 $$AX$$ 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 $$RX$$ 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
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 $$c(M)$$.