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. Spelled out, this means that if x is an element of the field of fractions of A that is a root of a monic polynomial with coefficients in A, then x is itself an element of A. Many well-studied domains are integrally closed, as shown by the following chain of class 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 the ring of integers Z, a Euclidean domain. All regular local rings are integrally closed as well.

A ring whose localizations at all prime ideals are integrally closed domains is a normal ring.

Basic properties

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.

Examples

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 $S=k[x_{1},\dots ,x_{n}]$ 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.^[2] In particular, $k[x_{0},\dots ,x_{r}]/(x_{0}^{2}+\dots +x_{r}^{2})$ is an integrally closed domain if $r\geq 2$ .^[3]

To give a non-example,^[4] let k be a field and $A=k[t^{2},t^{3}]\subset k[t]$ , 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 $A=\mathbb {Z} [{\sqrt {5}}\,]$ ; its field of fractions contains the element ${\frac {{\sqrt {5}}+1}{2}}$ , which is not in A but satisfies the monic polynomial $X^{2}-X-1=0$ .

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,^[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 $\operatorname {Spec} (A)$ 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 ${\mathfrak {p}}$ ,

If ${\mathfrak {p}}$ has height $\leq 1$ , then $A_{\mathfrak {p}}$ is regular (i.e., $A_{\mathfrak {p}}$ is a discrete valuation ring.)
If ${\mathfrak {p}}$ has height $\geq 2$ , then $A_{\mathfrak {p}}$ has depth $\geq 2$ .^[8]

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;^[9] 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. 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 $d\in A$ such that $dx^{n}\in A$ for all $n\geq 0$ . 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 $A[[X]]$ 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 $R[[X]]$ 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]

"Integrally closed" under constructions

The following conditions are equivalent for an integral domain A:

A is integrally closed;
A_p (the localization of A with respect to p) is integrally closed for every prime ideal p;
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.

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.^[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:

\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 $c(\chi (M/F))=c(\chi (M/F'))$ ^[16] and $c(\chi (M/F))$ is denoted (in Bourbaki) by $c(M)$ .

Citations

^ Matsumura, Theorem 9.2
^ Hartshorne 1977, Ch. II, Exercise 6.4.
^ Hartshorne 1977, Ch. II, Exercise 6.5. (a)
^ Taken from Matsumura
^ 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.
^ Kaplansky, Theorem 168, pg 119.
^ Matsumura 1989, p. 64
^ Matsumura, Commutative algebra, pg. 125. For a domain, the theorem is due to Krull (1931). The general case is due to Serre.
^ over an algebraically closed field
^ An exercise in Matsumura.
^ Matsumura, Exercise 10.4
^ An exercise in Bourbaki.
^ Bourbaki 1972, Ch. VII, § 1, n. 2, Theorem 1
^ An exercise in Bourbaki.
^ Bourbaki 1972, Ch. VII, § 1, n. 6. Proposition 10.
^ Bourbaki 1972, Ch. VII, § 4, n. 7

References

Bourbaki, Nicolas (1972). Commutative Algebra. Paris: Hermann.
Hartshorne, Robin (1977), Algebraic Geometry, Graduate Texts in Mathematics, vol. 52, New York: Springer-Verlag, ISBN 978-0-387-90244-9, MR 0463157
Kaplansky, Irving (September 1974). Commutative Rings. Lectures in Mathematics. University of Chicago Press. ISBN 0-226-42454-5.
Matsumura, Hideyuki (1989). Commutative Ring Theory. Cambridge Studies in Advanced Mathematics (2nd ed.). Cambridge University Press. ISBN 0-521-36764-6.
Matsumura, Hideyuki (1970). Commutative Algebra. ISBN 0-8053-7026-9.

[1] Matsumura, Theorem 9.2

[2] Hartshorne 1977, Ch. II, Exercise 6.4.

[3] Hartshorne 1977, Ch. II, Exercise 6.5. (a)

[4] Taken from Matsumura

[5] 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.

[6] Kaplansky, Theorem 168, pg 119.

[7] Matsumura 1989, p. 64

[8] Matsumura, Commutative algebra, pg. 125. For a domain, the theorem is due to Krull (1931). The general case is due to Serre.

[9] ver an algebraically closed field

[10] An exercise in Matsumura.

[11] Matsumura, Exercise 10.4

[12] An exercise in Bourbaki.

[13] Bourbaki 1972, Ch. VII, § 1, n. 2, Theorem 1

[14] An exercise in Bourbaki.

[15] Bourbaki 1972, Ch. VII, § 1, n. 6. Proposition 10.

[16] Bourbaki 1972, Ch. VII, § 4, n. 7

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