Integrally closed domain
Algebraic structures |
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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
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
Integrally closed domains also play a role in the hypothesis of the
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 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 be a field of characteristic not 2 and a polynomial ring over it. If is a square-free nonconstant polynomial in , then is an integrally closed domain.[2] In particular, is an integrally closed domain if .[3]
To give a non-example,[4] let k be a field and , the subalgebra generated by t2 and t3. Then A is not integrally closed: it has the field of fractions , and the monic polynomial 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 has a singularity at the origin.
Another domain that is not integrally closed is ; its field of fractions contains the element , which is not in A but satisfies the monic polynomial .
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 over prime ideals of height 1 and (ii) the localization at a prime ideal of height 1 is a discrete valuation ring.
A noetherian ring is a
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 is noetherian, normal and connected, then A is an integrally closed domain. (cf.
Let A be a noetherian ring. Then (
- If has height , then is regular (i.e., is a discrete valuation ring.)
- If has height , then has depth .[8]
Item (i) is often phrased as "regular in codimension 1". Note (i) implies that the set of associated primes has no
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 such that for all . 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 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 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;
- Ap (the localization of A with respect to p) is integrally closed for every prime ideal p;
- 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
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
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Let A be a Noetherian integrally closed domain.
An ideal I of A is
Let P denote the set of all prime ideals in A of height one. If T is a finitely generated torsion module, one puts:
- ,
which makes sense as a formal sum; i.e., a divisor. We write for the divisor class of d. If are maximal submodules of M, then [16] and is denoted (in Bourbaki) by .
See also
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 . Now, the image of x is nonzero in the localization of R at since at means for some but then is in the annihilator of x, contradiction. This shows that R localized at 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.
- MR 0463157
- ISBN 0-226-42454-5.
- ISBN 0-521-36764-6.
- Matsumura, Hideyuki (1970). Commutative Algebra. ISBN 0-8053-7026-9.