# Integrally closed domain

(Redirected from Normal ring)

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,[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 $Y^2 = X^3$ has a singularity at the origin.

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.[2] 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" (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 AB is an integral extension of domains and A is an integrally closed domain, then the going-down property holds for the extension AB.

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

Commutative ringsintegral domainsintegrally closed domainsunique factorization domainsprincipal ideal domainsEuclidean domainsfields

## Examples

The following are integrally closed domains.

## 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,[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. 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 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.

## Completely integrally closed domains

Let A be a domain and K its field of fractions. 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; that is, 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]

An integral domain is completely integrally closed if and only if the monoid of divisors of A is a group.[11]

## "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.[12]

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.[13]

Let P denotes 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 $c(\chi(M/F)) = c(\chi(M/F'))$[14] and $c(\chi(M/F))$ is denoted (in Bourbaki) by $c(M)$.

## 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. ^ Bourbaki, Ch. VII, § 1, n. 2, Theorem 1
12. ^ An exercise in Bourbaki.
13. ^ Bourbaki Ch. VII, § 1, n. 6. Proposition 10.
14. ^ Bourbaki Ch. VII, § 4, n. 7