integrally closed domain

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.Matsumura, Theorem 9.2 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.

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 $S\; =\; k[x\_1,\; \backslash dots,\; x\_n]$ 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.{{harvnb|Hartshorne|1977|loc=Ch. II, Exercise 6.4.}} In particular, $k[x\_0,\; \backslash dots,\; x\_r]/(x\_0^2\; +\; \backslash dots\; +\; x\_r^2)$ is an integrally closed domain if $r\; \backslash ge\; 2$.{{harvnb|Hartshorne|1977|loc=Ch. II, Exercise 6.5. (a)}}

To give a non-example,Taken from Matsumura let k be a field and $A\; =\; k[t^2,\; t^3]\; \backslash subset\; k[t]$, 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 $A\; =\; \backslash mathbb\{Z\}[\backslash sqrt\{5\}\backslash ,]$; its field of fractions contains the element $\backslash frac\{\backslash sqrt\{5\}+1\}\{2\}$, which is not in A but satisfies the monic polynomial $X^2-X-1\; =\; 0$.

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 $A\_\backslash mathfrak\{p\}$ over prime ideals $\backslash mathfrak\{p\}$ of height 1 and (ii) the localization $A\_\backslash mathfrak\{p\}$ at a prime ideal $\backslash mathfrak\{p\}$ 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.

{{redirect|normal domain|the concept in multiple integration|Multiple integral#normal domains}}

{{See also|normal variety}}

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,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 $\backslash mathfrak\{m\}$. Now, the image of x is nonzero in the localization of R at $\backslash mathfrak\{m\}$ since $x\; =\; 0$ at $\backslash mathfrak\{m\}$ means $xs\; =\; 0$ for some $s\; \backslash not\backslash in\; \backslash mathfrak\{m\}$ but then $s$ is in the annihilator of x, contradiction. This shows that R localized at $\backslash mathfrak\{m\}$ is not reduced. 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.Kaplansky, Theorem 168, pg 119. In particular if A is a Noetherian, normal ring, then the domains in the product are integrally closed domains.Matsumura 1989, p. 64 Conversely, any finite product of integrally closed domains is normal. In particular, if $\backslash operatorname\{Spec\}(A)$ 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 $\backslash mathfrak\{p\}$,

1. If $\backslash mathfrak\{p\}$ has height $\backslash le\; 1$, then $A\_\backslash mathfrak\{p\}$ is regular (i.e., $A\_\backslash mathfrak\{p\}$ is a discrete valuation ring.)
2. If $\backslash mathfrak\{p\}$ has height $\backslash ge\; 2$, then $A\_\backslash mathfrak\{p\}$ has depth $\backslash ge\; 2$.Matsumura, Commutative algebra, pg. 125. For a domain, the theorem is due to Krull (1931). The general case is due to Serre.

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;over an algebraically closed field e.g., X itself is nonsingular, then X is Cohen-Macaulay; i.e., the stalks $\backslash mathcal\{O\}\_p$ 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.

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 $d\; \backslash in\; A$ such that $d\; x^n\; \backslash in\; A$ for all $n\; \backslash ge\; 0$. 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.An exercise in Matsumura. 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.Matsumura, Exercise 10.4 Let L be a field extension of K. Then the integral closure of A in L is completely integrally closed.An exercise in Bourbaki.

An integral domain is completely integrally closed if and only if the monoid of divisors of A is a group.{{harvnb|Bourbaki|1972|loc=Ch. VII, § 1, n. 2, Theorem 1}}

{{See also|Krull domain}}

The following conditions are equivalent for an integral domain A:

1. A is integrally closed;
2. A_p (the localization of A with respect to p) is integrally closed for every prime ideal p;
3. 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]/(t²+4) is not integrally closed.

The localization of a completely integrally closed domain need not be completely integrally closed.An exercise in Bourbaki.

A direct limit of integrally closed domains is an integrally closed domain.

{{expand section|date=February 2013}}

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.{{harvnb|Bourbaki|1972|loc=Ch. VII, § 1, n. 6. Proposition 10.}}

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

:$\backslash chi(T)\; =\; \backslash sum\_\{p\; \backslash in\; P\}\; \backslash operatorname\{length\}\_p(T)\; p$,

which makes sense as a formal sum; i.e., a divisor. We write $c(d)$ for the divisor class of d. If $F,\; F\text{'}$ are maximal submodules of M, then $c(\backslash chi(M/F))\; =\; c(\backslash chi(M/F\text{'}))${{harvnb|Bourbaki|1972|loc=Ch. VII, § 4, n. 7}} and $c(\backslash chi(M/F))$ is denoted (in Bourbaki) by $c(M)$.

• Unibranch local ring

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• {{cite book |first=Nicolas |last=Bourbaki |authorlink = Nicolas Bourbaki|title=Commutative Algebra |date=1972 |url=https://archive.org/details/commutativealgeb0000bour |url-access=registration |location=Paris |publisher=Hermann }}
• {{Hartshorne AG}}
• {{cite book | last = Kaplansky | first = Irving |authorlink = Irving Kaplansky| title = Commutative Rings | series = Lectures in Mathematics | date = September 1974 | publisher = University of Chicago Press | isbn = 0-226-42454-5 | url-access = registration | url = https://archive.org/details/commutativerings00irvi }}
• {{cite book |last=Matsumura |first=Hideyuki |authorlink = Hideyuki Matsumura|year=1989 |title=Commutative Ring Theory |series=Cambridge Studies in Advanced Mathematics |edition=2nd |publisher=Cambridge University Press |isbn=0-521-36764-6 }}
• {{cite book |last=Matsumura |first=Hideyuki |year=1970 |title=Commutative Algebra |isbn=0-8053-7026-9 }}

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Category:Commutative algebra