# 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,[1] let k be a field and ${\displaystyle A=k[t^{2},t^{3}]\subset B=k[t]}$ (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 ${\displaystyle Y^{2}=X^{3}}$ 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 ${\displaystyle A=\mathbb {Z} [{\sqrt {5}}].}$)

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 domains appear in the following chain of class inclusions:

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

## 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 ${\displaystyle A_{\mathfrak {p}}}$ over prime ideals ${\displaystyle {\mathfrak {p}}}$ of height 1 and (ii) the localization ${\displaystyle A_{\mathfrak {p}}}$ at a prime ideal ${\displaystyle {\mathfrak {p}}}$ 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 ${\displaystyle \operatorname {Spec} (A)}$ 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 ${\displaystyle {\mathfrak {p}}}$,

• (i) If ${\displaystyle {\mathfrak {p}}}$ has height ${\displaystyle \leq 1}$, then ${\displaystyle A_{\mathfrak {p}}}$ is regular (i.e., ${\displaystyle A_{\mathfrak {p}}}$ is a discrete valuation ring.)
• (ii) If ${\displaystyle {\mathfrak {p}}}$ has height ${\displaystyle \geq 2}$, then ${\displaystyle A_{\mathfrak {p}}}$ has depth ${\displaystyle \geq 2}$.[6]

Item (i) is often phrased as "regular in codimension 1". Note (i) implies that the set of associated primes ${\displaystyle Ass(A)}$ has no embedded primes, and, when (i) is the case, (ii) means that ${\displaystyle Ass(A/fA)}$ 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;[7] e.g., X itself is nonsingular, then X is Cohen-Macaulay; i.e., the stalks ${\displaystyle {\mathcal {O}}_{p}}$ 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 ${\displaystyle d\neq 0}$ such that ${\displaystyle dx^{n}\in A}$ for all ${\displaystyle n\geq 0}$. 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 ${\displaystyle A[[X]]}$ 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 ${\displaystyle R[[X]]}$ 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:

${\displaystyle \chi (T)=\sum _{p\in P}\operatorname {length} _{p}(T)p}$,

which makes sense as a formal sum; i.e., a divisor. We write ${\displaystyle c(d)}$ for the divisor class of d. If ${\displaystyle F,F'}$ are maximal submodules of M, then ${\displaystyle c(\chi (M/F))=c(\chi (M/F'))}$[14] and ${\displaystyle c(\chi (M/F))}$ is denoted (in Bourbaki) by ${\displaystyle c(M)}$.

## 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 ${\displaystyle {\mathfrak {m}}}$. Now, the image of x is nonzero in the localization of R at ${\displaystyle {\mathfrak {m}}}$ since ${\displaystyle x=0}$ at ${\displaystyle {\mathfrak {m}}}$ means ${\displaystyle xs=0}$ for some ${\displaystyle s\not \in {\mathfrak {m}}}$ but then ${\displaystyle s}$ is in the annihilator of x, contradiction. This shows that R localized at ${\displaystyle {\mathfrak {m}}}$ 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