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.[1] 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 ${\displaystyle k}$ be a field of characteristic not 2 and ${\displaystyle S=k[x_{1},\dots ,x_{n}]}$ a polynomial ring over it. If ${\displaystyle f}$ is a square-free nonconstant polynomial in ${\displaystyle S}$, then ${\displaystyle S[y]/(y^{2}-f)}$ is an integrally closed domain.[2] In particular, ${\displaystyle k[x_{0},\dots ,x_{r}]/(x_{0}^{2}+\dots +x_{r}^{2})}$ is an integrally closed domain if ${\displaystyle r\geq 2}$.[3]

To give a non-example,[4] let k be a field and ${\displaystyle A=k[t^{2},t^{3}]\subset k[t]}$ (A is the subalgebra generated by t² and t³.) A is not integrally closed: it has the field of fractions ${\displaystyle k(t)}$, and the monic polynomial ${\displaystyle X^{2}-t^{2}}$ 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 ${\displaystyle Y^{2}=X^{3}}$ has a singularity at the origin.

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

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 ${\displaystyle A_{\mathfrak {p}}}$ 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.

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,[5] 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.[6] In particular if A 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 ${\displaystyle \operatorname {Spec} (A)}$ 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 ${\displaystyle {\mathfrak {p}}}$,

1. 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.)
2. If ${\displaystyle {\mathfrak {p}}}$ has height ${\displaystyle \geq 2}$, then ${\displaystyle A_{\mathfrak {p}}}$ has depth ${\displaystyle \geq 2}$.[8]

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;[9] 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.

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.[10] 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.[11] Let L be a field extension of K. Then the integral closure of A in L is completely integrally closed.[12]

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

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.[14]

A direct limit of integrally closed domains is 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.[15]

Let P denote 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 ${\displaystyle c(\chi (M/F))=c(\chi (M/F'))}$[16] and ${\displaystyle c(\chi (M/F))}$ is denoted (in Bourbaki) by ${\displaystyle c(M)}$.

• Unibranch local ring