Initial value theorem: Difference between revisions

Revision as of 21:43, 9 April 2013

In commutative algebra, an integrally closed domain A is an integral domain whose integral closure in its field of fractions is A itself. Many well-studied domains are integrally closed: Fields, the ring of integers Z, unique factorization domains and regular local rings are all integrally closed.

To give a non-example,^[1] let $A=k[t^{2},t^{3}]\subset B=k[t]$ (k a field). A and B have the same field of fractions, and B is the integral closure of A (since B is a UFD.) In other words, A is not integrally closed. This is related to the fact that the plane curve $Y^{2}=X^{3}$ has a singularity at the origin.

Let A be an integrally closed domain with field of fractions K and let L be a finite extension of K. Then x in L is integral over A if and only if its minimal polynomial over K has coefficients in A.^[2] This implies in particular that an integral element over an integrally closed domain A has a minimal polynomial over A. This is stronger than the statement that any integral element satisfies some monic polynomial. In fact, the statement is false without "integrally closed" (consider $A=\mathbb {Z} [{\sqrt {5}}].$ )

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.

Note that integrally closed domain appear in the following chain of class inclusions:

Commutative rings ⊃ integral domains ⊃ integrally closed domains ⊃ unique factorization domains ⊃ principal ideal domains ⊃ Euclidean domains ⊃ fields

Examples

The following are integrally closed domains.

Any principal ideal domain (in particular, any field).
Any unique factorization domain (in particular, any polynomial ring over a unique factorization domain.)
Any GCD domain (in particular, any Bézout domain or valuation domain).
Any Dedekind domain.
Any symmetric algebra over a field (since every symmetric algebra is isomorphic to a polynomial ring in several variables over a field).

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

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In 12 months 2013, c ommercial retails, shoebox residences and mass market properties continued to be the celebrities of the property market. Models are snapped up in report time and at document breaking prices. Builders are having fun with overwhelming demand and patrons need more. We feel that these segments of the property market are booming is a repercussion of the property cooling measures no.6 and no. 7. With additional buyer's stamp responsibility imposed on residential properties, buyers change their focus to commercial and industrial properties. I imagine every property purchasers need their property funding to understand in value. 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,^[3] 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.^[4] In particular if A is a Noetherian, normal ring, then the domains in the product are integrally closed domains.^[5] 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 A is normal if and only if it satisfies the following: for any prime ideal ${\mathfrak {p}}$ ,

(i) If ${\mathfrak {p}}$ has height $\leq 1$ , then $A_{\mathfrak {p}}$ is regular (i.e., $A_{\mathfrak {p}}$ is a discrete valuation ring.)
(ii) If ${\mathfrak {p}}$ has height $\geq 2$ , then $A_{\mathfrak {p}}$ has depth $\geq 2$ .^[6]

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 nonzero zero-divisor 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;^[7] 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. x in K is said to be almost integral over A if there is a $d\neq 0$ 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.^[8] 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.^[9] Let L be a field extension of K. Then the integral closure of A in L is completely integrally closed.^[10]

"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 need not be completely integrally closed.^[11]

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

Modules over an integrally closed domain

Template:Expand section

References

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Bourbaki, Commutative algebra.
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Matsumura, Hideyuki (1989), Commutative Ring Theory, Cambridge Studies in Advanced Mathematics (2nd ed.), Cambridge University Press, ISBN 978-0-521-36764-6.
Matsumura, Hideyuki (1970) Commutative algebra ISBN 0-8053-7026-9.

↑ Taken from Matsumura
↑ Matsumura, Theorem 9.2
↑ 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.
↑ An exercise in Bourbaki.

[1] Taken from Matsumura

[2] Matsumura, Theorem 9.2

[3] 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.

[4] Kaplansky, Theorem 168, pg 119.

[5] Matsumura 1989, p. 64

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

[7] ver an algebraically closed field

[8] An exercise in Matsumura.

[9] Matsumura, Exercise 10.4

[10] An exercise in Bourbaki.

[11] An exercise in Bourbaki.

[1]

[2]

[3]

[4]

[5]

[6]

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[11]

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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. Many well-studied domains are integrally closed: [[Field (mathematics)|Field]]s, the ring of integers '''Z''', [[unique factorization domain]]s and regular local rings are all integrally closed.
+To give a non-example,<ref>Taken from Matsumura</ref> let <math>A = k[t^2, t^3] \subset B = k[t]</math> (''k'' a field). ''A'' and ''B'' have the same field of fractions, and ''B'' is the integral closure of ''A'' (since ''B'' is a UFD.) In other words, ''A'' is not integrally closed. This is related to the fact that the plane curve <math>Y^2 = X^3</math> has a singularity at the origin.
+Let ''A'' be an integrally closed domain with field of fractions ''K'' and let ''L'' be a finite extension of ''K''. Then ''x'' in ''L'' is integral over ''A'' if and only if its minimal polynomial over ''K'' has coefficients in ''A''.<ref>Matsumura, Theorem 9.2</ref> This implies in particular that an integral element over an integrally closed domain ''A'' has a minimal polynomial over ''A''. This is stronger than the statement that any integral element satisfies some monic polynomial. In fact, the statement is false without "integrally closed" (consider <math>A = \mathbb{Z}[\sqrt{5}].</math>)
+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 up and going down|going-down property]] holds for the extension ''A''⊆''B''.
+Note that integrally closed domain appear in the following chain of [[subclass (set theory)|class inclusions]]:
+: '''[[Commutative ring]]s''' ⊃  '''[[integral domain]]s''' ⊃ '''integrally closed domains''' ⊃ '''[[unique factorization domain]]s''' ⊃ '''[[principal ideal domain]]s''' ⊃ '''[[Euclidean domain]]s''' ⊃ '''[[field (mathematics)|field]]s'''
+== Examples ==
+The following are integrally closed domains.
+*Any principal ideal domain (in particular, any field).
+*Any [[unique factorization domain]] (in particular, any polynomial ring over a unique factorization domain.)
+*Any [[GCD domain]] (in particular, any [[Bézout domain]] or [[valuation domain]]).
+*Any [[Dedekind domain]].
+*Any [[symmetric algebra]] over a field (since every symmetric algebra is isomorphic to a polynomial ring in several variables over a field).
+== 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 <math>A_\mathfrak{p}</math> over prime ideals <math>\mathfrak{p}</math> of height 1 and (ii) the localization <math>A_\mathfrak{p}</math> at a prime ideal <math>\mathfrak{p}</math> 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 ring]]s containing it.
+== Normal rings ==
+{{See also|normal variety}}
+Authors including [[Jean-Pierre Serre|Serre]], [[Alexander Grothendieck|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]],<ref>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 ''x''<sup>2</sup>=0. The [[annihilator]] ann(''x'') is contained in some maximal ideal <math>\mathfrak{m}</math>. Now, the image of ''x'' is nonzero in the localization of ''R'' at <math>\mathfrak{m}</math> since <math>x = 0</math> at <math>\mathfrak{m}</math> means <math>xs = 0</math> for some <math>s \not\in \mathfrak{m}</math> but then <math>s</math> is in the annihilator of ''x'', contradiction. This shows that ''R'' localized at <math>\mathfrak{m}</math> is not reduced.</ref> and this is sometimes included in the definition. In general, if ''A'' is a [[Noetherian ring|Noetherian]] ring whose localizations at maximal ideals are all domains, then ''A'' is a finite product of domains.<ref>Kaplansky, Theorem 168, pg 119.</ref> In particular if ''A'' is a Noetherian, normal ring, then the domains in the product are integrally closed domains.<ref>Matsumura 1989, p. 64</ref> Conversely, any finite product of integrally closed domains is normal. In particular, if <math>\operatorname{Spec}(A)</math> is noetherian, normal and connected, then ''A'' is an integrally closed domain. (cf. [[smooth variety]])
+Let ''A'' be a noetherian ring. Then ''A'' is normal if and only if it satisfies the following: for any prime ideal <math>\mathfrak{p}</math>,
+*(i) If <math>\mathfrak{p}</math> has height <math>\le 1</math>, then <math>A_\mathfrak{p}</math> is [[regular local ring|regular]] (i.e., <math>A_\mathfrak{p}</math> is a [[discrete valuation ring]].)
+*(ii) If <math>\mathfrak{p}</math> has height <math>\ge 2</math>, then <math>A_\mathfrak{p}</math> has depth <math>\ge 2</math>.<ref>Matsumura, Commutative algebra, pg. 125. For a domain, the theorem is due to Krull (1931). The general case is due to Serre.</ref>
+Item (i) is often phrased as "regular in codimension 1". Note (i) implies that the set of [[associated prime]]s <math>Ass(A)</math> has no [[embedded prime]]s, and, when (i) is the case, (ii) means that <math>Ass(A/fA)</math> has no embedded prime for any nonzero zero-divisor ''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;<ref>over an algebraically closed field</ref> e.g., ''X'' itself is nonsingular, then ''X'' is Cohen-Macaulay; i.e., the stalks <math>\mathcal{O}_p</math> of the structure sheaf are Cohen-Macaulay for all prime ideals p. Then we can say: ''X'' is [[normal scheme|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. ''x'' in ''K'' is said to be '''almost integral over ''A'' ''' if there is a <math>d \ne 0</math> such that <math>d x^n \in A</math> for all <math>n \ge 0</math>. 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 <math>A[[X]]</math> is completely integrally closed.<ref>An exercise in Matsumura.</ref> 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 <math>R[[X]]</math> is not integrally closed.<ref>Matsumura, Exercise 10.4</ref> Let ''L'' be a field extension of ''K''. Then the integral closure of ''A'' in ''L'' is completely integrally closed.<ref>An exercise in Bourbaki.</ref>
+== "Integrally closed" under constructions ==
+The following conditions are equivalent for an integral domain ''A'':
+# ''A'' is integrally closed;
+# ''A''<sub>''p''</sub> (the localization of ''A'' with respect to ''p'') is integrally closed for every [[prime ideal]] ''p'';
+# ''A''<sub>''m''</sub> is integrally closed for every [[maximal ideal]] ''m''.
+→ 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 [[Localization of a module#Flatness|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]/(t<sup>2</sup>+4) is not integrally closed.
+The localization of a completely integrally closed need not be completely integrally closed.<ref>An exercise in Bourbaki.</ref>
+A direct limit of integrally closed domains is an integrally closed domain.
+== Modules over an integrally closed domain ==
+{{expand section|date=February 2013}}
+== See also ==
+*[[Unibranch local ring]]
+== References ==
+{{reflist}}
+*Bourbaki, Commutative algebra.
+* {{cite book | last = Kaplansky | first = Irving | title = Commutative Rings
+  | series = Lectures in Mathematics |date=September 1974
+  | publisher = [[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 978-0-521-36764-6.
+*Matsumura, Hideyuki (1970) ''Commutative algebra'' ISBN 0-8053-7026-9.
+[[Category:Commutative algebra]]

Initial value theorem: Difference between revisions

Revision as of 21:43, 9 April 2013

Contents

Examples

Noetherian integrally closed domain

Normal rings

Completely integrally closed domains

"Integrally closed" under constructions

Modules over an integrally closed domain

See also

References

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Initial value theorem: Difference between revisions

Revision as of 21:43, 9 April 2013

Examples

Noetherian integrally closed domain

Normal rings

Completely integrally closed domains

"Integrally closed" under constructions

Modules over an integrally closed domain

See also

References

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