Incommutative algebra, anintegrally closed domainA is anintegral domain whoseintegral closure in itsfield of fractions isA itself. Spelled out, this means that ifx is an element of the field of fractions ofA that is a root of amonic polynomial withcoefficients inA, thenx is itself an element ofA. Many well-studied domains are integrally closed, as shown by the following chain ofclass inclusions:

rngs ⊃rings ⊃commutative rings ⊃integral domains ⊃integrally closed domains ⊃GCD domains ⊃unique factorization domains ⊃principal ideal domains ⊃Euclidean domains ⊃fields ⊃algebraically closed fields

An explicit example is thering of integersZ, aEuclidean domain. Allregular local rings are integrally closed as well.

A ring whoselocalizations at allprime ideals are integrally closed domains is anormal ring.

LetA be an integrally closed domain with field of fractionsK and letL be afield extension ofK. Thenx∈L isintegral overA if and only if it isalgebraic overK and itsminimal polynomial overK has coefficients inA.^[1] In particular, this means that any element ofL integral overA is root of a monic polynomial inA[X] that isirreducible inK[X].

IfA is a domain contained in afieldK, we can consider theintegral closure ofA inK (i.e. the set of all elements ofK that are integral overA). This integral closure is an integrally closed domain.

Integrally closed domains also play a role in the hypothesis of theGoing-down theorem. The theorem states that ifA⊆B is anintegral extension of domains andA is an integrally closed domain, then thegoing-down property holds for the extensionA⊆B.

The following are integrally closed domains.

• Aprincipal ideal domain (in particular: the integers and any field).
• Aunique factorization domain (in particular, any polynomial ring over a field, over the integers, or over any unique factorization domain).
• AGCD domain (in particular, anyBézout domain orvaluation domain).
• ADedekind domain.
• Asymmetric algebra over a field (since every symmetric algebra is isomorphic to a polynomial ring in several variables over a field).
• Let${\displaystyle k}$ be a field ofcharacteristic not 2 and${\displaystyle S=k[x_1,\dots ,x_n]}$ a polynomial ring over it. If${\displaystyle f}$ is asquare-free nonconstant polynomial in${\displaystyle S}$, then${\displaystyle S[y]/(y^2-f)}$ is an integrally closed domain.^[2] In particular,${\displaystyle k[x_0,\dots ,x_r]/(x_0^2+\cdots +x_r^2)}$ is an integrally closed domain if${\displaystyle r\ge 2}$.^[3]

To give a non-example,^[4] letk be a field and${\displaystyle A=k[t^2,t^3]\subset k[t]}$, the subalgebra generated byt² andt³. ThenA is not integrally closed: it has the field of fractions${\displaystyle k(t)}$, and the monic polynomial${\displaystyle X^2-t^2}$ in the variableX has roott which is in the field of fractions but not inA. This is related to the fact that theplane curve${\displaystyle Y^2=X^3}$ has asingularity at the origin.

Another domain that is not integrally closed is${\displaystyle A=\mathbb{Z}[\sqrt{5}]}$; its field of fractions contains the element${\displaystyle \frac{\sqrt{5}+1}{2}}$, which is not inA but satisfies the monic polynomial${\displaystyle X^2-X-1=0}$.

For anoetherianlocal domainA of dimension one, the following are equivalent.

• A is integrally closed.
• Themaximal ideal ofA is principal.
• A is adiscrete valuation ring (equivalentlyA is Dedekind.)
• A is a regular local ring.

LetA be a noetherian integral domain. ThenA 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 aKrull 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 allvaluation rings containing it.

Authors includingSerre,Grothendieck, andMatsumura define anormal ring to be a ring whoselocalizations at prime ideals are integrally closed domains. Such a ring is necessarily areduced ring,^[5] and this is sometimes included in the definition. In general, ifA is aNoetherian ring whose localizations at maximal ideals are all domains, thenA is a finite product of domains.^[6] In particular ifA 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${\displaystyle \mathrm{Spec}(A)}$ is noetherian, normal and connected, thenA is an integrally closed domain. (cf.smooth variety)

LetA be a noetherian ring. Then (Serre's criterion)A is normal if and only if it satisfies the following: for any prime ideal${\displaystyle \mathfrak{p}}$,

1. If${\displaystyle \mathfrak{p}}$ has height${\displaystyle \le 1}$, then${\displaystyle A_{\mathfrak{p}}}$ isregular (i.e.,${\displaystyle A_{\mathfrak{p}}}$ is adiscrete valuation ring.)
2. If${\displaystyle \mathfrak{p}}$ has height${\displaystyle \ge 2}$, then${\displaystyle A_{\mathfrak{p}}}$ has depth${\displaystyle \ge 2}$.^[8]

Item (i) is often phrased as "regular in codimension 1". Note (i) implies that the set ofassociated primes${\displaystyle Ass(A)}$ has noembedded primes, and, when (i) is the case, (ii) means that${\displaystyle Ass(A/fA)}$ has no embedded prime for any non-zerodivisorf. In particular, aCohen-Macaulay ring satisfies (ii). Geometrically, we have the following: ifX is alocal complete intersection in a nonsingular variety;^[9] e.g.,X itself is nonsingular, thenX 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 isnormal (i.e., the stalks of its structure sheaf are all normal) if and only if it is regular in codimension 1.

LetA be a domain andK its field of fractions. An elementx inK is said to bealmost integral overA if the subringA[x] ofK generated byA andx is afractional ideal ofA; that is, if there is a nonzero${\displaystyle d\in A}$ such that${\displaystyle d{x}^n\in A}$ for all${\displaystyle n\ge 0}$. ThenA is said to becompletely integrally closed if every almost integral element ofK is contained inA. A completely integrally closed domain is integrally closed. Conversely, a noetherian integrally closed domain is completely integrally closed.

AssumeA is completely integrally closed. Then the formalpower series ring${\displaystyle A[[X]]}$ is completely integrally closed.^[10] This is significant since the analog is false for an integrally closed domain: letR be a valuation domain of height at least 2 (which is integrally closed). Then${\displaystyle R[[X]]}$ is not integrally closed.^[11] LetL be a field extension ofK. Then the integral closure ofA inL is completely integrally closed.^[12]

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

The following conditions are equivalent for an integral domainA:

1. A is integrally closed;
2. A_p (the localization ofA with respect top) is integrally closed for everyprime idealp;
3. A_m is integrally closed for everymaximal idealm.

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, theexactness of localization, and the property that anA-moduleM 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, forZ[t]/(t²+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.

This sectionneeds expansion. You can help byadding to it.(February 2013)

LetA be a Noetherian integrally closed domain.

An idealI ofA isdivisorial if and only if everyassociated prime ofA/I has height one.^[15]

LetP denote the set of all prime ideals inA of height one. IfT is a finitely generated torsion module, one puts:

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

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

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