Algebraically closed field - Revision history

en>K9re11: removed Category:Abstract algebra using HotCat as there is the more specific category ''Field theory''

2014-11-08T08:56:42Z

removed Category:Abstract algebra using HotCat as there is the more specific category ''Field theory''

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	~~{{Use dmy dates\|date=June 2013}}~~		Nothing to tell about me really.<br>Enjoying to be a part of this site.<br>I just hope I am useful in one way here.<br><br>Also visit my web page: [http://hemorrhoidtreatmentfix.com/hemorrhoid-relief hemorrhoid relief]
	In [[abstract algebra]], an '''algebraically closed field''' ''F'' contains a [[Zero of a function\|root]] for every [[Degree of a polynomial\|non-constant polynomial]] in ''F''[''x''], the [[ring of polynomials]] in the variable ''x'' with coefficients in ''F''.

	~~==Examples==~~
	As an example, the [[field (mathematics)\|field]] of [[real number]]s is not algebraically closed, because the polynomial equation ''x''<sup>2</sup> + 1 = 0  has no solution in real numbers, even though all its coefficients (1 and 0) are real. The same argument proves that no subfield of the real field is algebraically closed; in particular, the field of [[rational number]]s is not algebraically closed. Also, no [[finite field]] ''F'' is algebraically closed, because if ''a''<sub>1</sub>, ''a''<sub>2</sub>, …, ''a<sub>n</sub>'' are the elements of ''F'', then the polynomial (''x'' − ''a''<sub>1</sub>)(''x'' − ''a''<sub>2</sub>) ··· (''x'' − ''a''<sub>''n''<~~/sub~~>~~) + 1~~
	has no zero in ''F''. By contrast, the [[fundamental theorem of algebra]] states that the field of [[complex number]]s is algebraically closed. Another example of an algebraically closed field is the field of (complex) [[algebraic number]]s.

	~~==Equivalent properties==~~
	~~Given a field ''F'', the assertion “''F'' is algebraically closed” is equivalent~~ to ~~other assertions:~~

	~~===The only irreducible polynomials are those of degree one===~~
	~~The field ''F'' is algebraically closed if and only if the only [[irreducible polynomial]]s in the [[polynomial ring]] ''F''[''x''] are those of degree one.~~

	The assertion “the polynomials of degree one are irreducible” is trivially true for any field. If ''F'' is algebraically closed and ''p''(''x'') is an irreducible polynomial of ''F''[''x''], then it has some root ''a~~'' and therefore ''p''(''x'') is a multiple~~ of ~~''x'' − ''a''. Since ''p''(''x'') is irreducible,~~ this ~~means that ''p''(''x'') = ''k''(''x'' − ''a''), for some ''k'' ∈ ''F'' \ {0}~~. On the other hand, if ''F'' is not algebraically closed, then there is some non-constant polynomial ''p''(''x'') in ''F''[''x''] without roots in ''F''. Let ''q''(''x'') be some irreducible factor of ''p''(''x''). Since ''p''(''x'') has no roots in ''F'', ''q''(''x'') also has no roots in ''F''. Therefore, ''q''(''x'') has degree greater than one, since every first degree polynomial has one root in ''F''.

	~~===Every polynomial is a product of first degree polynomials===~~
	The field ''F'' is algebraically closed if and only if every polynomial ''p''(''x'') of degree ''n'' ≥ 1, with [[coefficient]]s in ''F'', [[factorization\|splits into linear factors]]. In other words, there are elements ''k'', ''x''<~~sub~~>1</sub>, ''x''<sub>2</sub>, …, ''x<sub>n</sub>'' of the field ''F'' such that ''p''(''x'') = ''k''(''x'' − ''x''<sub>1</sub>)(''x'' − ''x''<sub>2</sub>) ··· (''x'' − ''x<sub>n</sub>'').

	~~If ''F'' has this property, then clearly every non-constant polynomial~~ in ~~''F''[''x''] has some root in ''F''; in other words, ''F'' is algebraically closed. On the other hand, that the property stated~~ here holds for ''F'' if ''F'' is algebraically closed follows from the previous property together with the fact that, for any field ''K'', any polynomial in ''K''[''x''] can be written as a product of irreducible polynomials.

	~~===Polynomials of prime degree have roots===~~
	J. ~~Shipman showed in 2007 that if every polynomial over ''F'' of prime degree has a root in ''F'', then every non-constant polynomial has a root in ''F'', thus ''F'' is algebraically closed.~~

	~~===The field has no proper algebraic extension===~~
	~~The field ''F'' is algebraically closed if and only if it has no proper [[algebraic extension]].~~

	If ''F'' has no proper algebraic extension, let ''p''(''x'') be some irreducible polynomial in ''F''[''x'']. Then the [[quotient ring\|quotient]] of ''F''[''x''] modulo the [[ideal (ring theory)\|ideal]] generated by ''p''(''x'') is an algebraic extension of ''F'' whose [[degree of a field extension\|degree]] is equal to the degree of ''p''(''x''). Since it is not a proper extension, its degree is 1 and therefore the degree of ''p''(''x'') is 1.

	On the other hand, if ''F'' has some proper algebraic extension ''K'', then the [[Minimal polynomial (field theory)\|minimal polynomial]] of an element in ''K'' \ ''F'' is irreducible and its degree is greater than 1.

	~~===The field has no proper finite extension===~~
	The field ''F'' is algebraically closed if and only if it has no finite [[algebraic extension]] because if, within the [[Algebraically closed field#The field has no proper algebraic extension\|previous proof]], the word “algebraic” is replaced by the word “finite”, then the proof is still valid.

	~~===Every endomorphism of ''F~~<~~sup~~>n<~~/sup~~>~~'' has some eigenvector===~~
	~~The field ''F'' is algebraically closed if and only if, for each natural number ''n'', every [[linear map]] from ''F<sup>n</sup>'' into itself has some [[eigenvector]].~~

	An endomorphism of ''F<sup>n</sup>'' has an eigenvector if and only if its [[characteristic polynomial]] has some root. Therefore, when ''F'' is algebraically closed, every endomorphism of ''F<sup>n</sup>'' has some eigenvector. On the other hand, if every endomorphism of ''F<sup>n</sup>'' has an eigenvector, let ''p''(''x'') be an element of ''F''[''x'']. Dividing by its leading coefficient, we get another polynomial ''q''(''x'') which has roots if and only if ''p''(''x'') has roots. But if ''q''(''x'') = ''x<sup>n</sup>'' + ''a''<sub>''n'' − 1</sub>''x''<sup>''n'' − 1</sup>+ ··· + ''a''<sub>0</sub>, then ''q''(''x'') is the characteristic polynomial of the [[companion matrix]]
	:~~<math>\begin{pmatrix}0&0&\cdots&0&-a_0\\1&0&\cdots&0&-a_1\\0&1&\cdots&0&-a_2\\\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\cdots&1&-a_{n-1}\end{pmatrix}.</math>~~

	~~===Decomposition of rational expressions===~~
	The field ''F'' is algebraically closed if and only if every [[rational function]] in one variable ''x'', with coefficients in ''F'', can be written as the sum of a polynomial function with rational functions of the form ''a''/(''x'' − ''b'')<sup>n</sup>, where ''n'' is a natural number, and ''a'' and ''b'' are elements of ''F''.

	~~If ''F'' is algebraically closed then, since the irreducible polynomials in ''F''[''x''] are all of degree 1, the property stated above holds by the~~ [~~[Partial fraction decomposition#Statement of theorem\|theorem on partial fraction decomposition]].~~

	On the other hand, suppose that the property stated above holds for the field ''F''. Let ''p''(''x'') be an irreducible element in ''F''[''x'']. Then the rational function 1/''p'' can be written as the sum of a polynomial function ''q'' with rational functions of the form ''a''/(''x'' − ''b'')<sup>n</sup>. Therefore, the rational expression
	:~~<math>\frac1{p(x)}-q(x)=\frac{1-p(x)q(x)}{p(x)}<~~/~~math>~~
	can be written as a quotient of two polynomials in which the denominator is a product of first degree polynomials. Since ''p''(''x'') is irreducible, it must divide this product and, therefore, it must also be a first degree polynomial.

	~~===Relatively prime polynomials and roots===~~
	For any field ''F'', if two polynomials ''p''(''x''),''q''(''x'') ∈ ''F''[''x''] are [[coprime\|relatively prime]] then they do not have a common root, for if ''a'' ∈ ''F'' was a common root, then ''p''(''x'') and  ''q''(''x'') would both be multiples of ''x'' − ''a'' and therefore they would not be relatively prime. The fields for which the reverse implication holds (that is, the fields such that whenever two polynomials have no common root then they are relatively prime) are precisely the algebraically closed fields.

	If the field ''F'' is algebraically closed, let ''p''(''x'') and ''q''(''x'') be two polynomials which are not relatively prime and let ''r''(''x'') be their [[greatest common divisor]]. Then, since ''r''(''x'') is not constant, it will have some root ''a'', which will be then a common root of ''p''(''x'') and ''q''(''x'').

	If ''F'' is not algebraically closed, let ''p''(''x'') be a polynomial whose degree is at least 1 without roots. Then ''p''(''x'') and ''p''(''x'') are not relatively prime, but they have no common roots (since none of them has roots).

	~~==Other properties==~~
	If ''F'' is an algebraically closed field and ''n'' is a natural number, then ''F'' contains all ''n''th roots of unity, because these are (by definition) the ''n'' (not necessarily distinct) zeroes of the polynomial ''x<sup>n</sup>'' − 1. A field extension that is contained in an extension generated by the roots of unity is a ''cyclotomic extension'', and the extension of a field generated by all roots of unity is sometimes called its ''cyclotomic closure''. Thus algebraically closed fields are cyclotomically closed. The converse is not true. ~~Even assuming that every polynomial of the form ''x<sup>n<~~/~~sup>'' − ''a'' splits into linear factors is not enough to assure that the field is algebraically closed.~~

	~~If a proposition which can be expressed in the language of [[first~~-order logic]] is true for an algebraically closed field, then it is true for every algebraically closed field with the same [[Characteristic (algebra)\|characteristic]]. Furthermore, if such a proposition is valid for an algebraically closed field with characteristic 0, then not only is it valid for all other algebraically closed fields with characteristic 0, but there is some natural number ''N'' such that the proposition is valid for every algebraically closed field with characteristic ''p'' when ''p'' > ''N''.<ref>See subsections ''Rings and fields'' and ''Properties of mathematical theories'' in §2 of J. Barwise's "An introduction to first-order logic".</ref>

	Every field ''F'' has some extension which is algebraically closed. Among all such extensions there is one and ([[Up to\|up to isomorphism]], but not [[essentially unique\|unique isomorphism]]) only one which is an [[algebraic extension]] of ''F'';<ref>See Lang's ''Algebra'', §VII.2 or van der Waerden's ''Algebra I'', §10.1.</ref> it is called the [[algebraic closure]] of ''F''.

	~~The theory of algebraically closed fields has [[quantifier elimination]].~~

	~~==Notes==~~
	~~{{Reflist}}~~

	~~==References==~~
	* {{Citation \| last = Barwise \| first = Jon \| author-link = Jon Barwise \| year = 1978 \| contribution = An introduction to first-order logic \| editor-last = Barwise \| editor-first = Jon \| title = Handbook of Mathematical Logic \| series = Studies in Logic and the Foundations of Mathematics \| publisher = North Holland \| isbn = 0-7204-2285-X}}
	* {{Lang Algebra}}
	* {{citation\|last = Shipman\|first = Joseph\|year = 2007\|title = Improving the Fundamental Theorem of Algebra\|periodical = Mathematical Intelligencer\|volume = 29\|issue = 4\|pages = 9–14\|doi=10.1007/BF02986170\|issn = 0343-6993}}
	* {{Citation \| last = van der Waerden \| first = Bartel Leendert \| author-link = Bartel Leendert van der Waerden \| title = Algebra \| volume = I \| edition = 7th \| year = 2003 \| publisher = Springer-Verlag \| isbn = 0-387-40624-7}}

	~~{{DEFAULTSORT:Algebraically Closed Field}}~~
	~~[[Category:Abstract algebra]]~~
	~~[[Category:Field theory]~~]

en>ChrisGualtieri: DMY/MDY Tagging on Date O/I/A using AWB

2013-06-28T01:09:28Z

DMY/MDY Tagging on Date O/I/A using AWB

← Older revision		Revision as of 03:09, 28 June 2013
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	~~Hi there!~~ :) ~~My name~~ is ~~Freda~~, I'm a ~~student studying Industrial~~ and ~~Labor Relations~~ from ~~Bartelshagen Ii~~, ~~Germany~~.<br><br>~~my webpage~~ [~~http:~~//~~Www~~.~~Chrisp-Voc~~.de/~~gaestebuch~~.~~html?180%FFm88%FFnofollow%FFSims~~+4+~~Crack%FFnofollow%FFsuperfit+cla+fat+burner+review%FF%FFThe+Sims~~+4~~+Crack%FF FIFA coin generator~~]		{{Use dmy dates\|date=June 2013}}
			In [[abstract algebra]], an '''algebraically closed field''' ''F'' contains a [[Zero of a function\|root]] for every [[Degree of a polynomial\|non-constant polynomial]] in ''F''[''x''], the [[ring of polynomials]] in the variable ''x'' with coefficients in ''F''.

			==Examples==
			As an example, the [[field (mathematics)\|field]] of [[real number]]s is not algebraically closed, because the polynomial equation ''x''<sup>2</sup> + 1 = 0  has no solution in real numbers, even though all its coefficients (1 and 0) are real. The same argument proves that no subfield of the real field is algebraically closed; in particular, the field of [[rational number]]s is not algebraically closed. Also, no [[finite field]] ''F'' is algebraically closed, because if ''a''<sub>1</sub>, ''a''<sub>2</sub>, …, ''a<sub>n</sub>'' are the elements of ''F'', then the polynomial (''x'' − ''a''<sub>1</sub>)(''x'' − ''a''<sub>2</sub>) ··· (''x'' − ''a''<sub>''n''</sub>) + 1
			has no zero in ''F''. By contrast, the [[fundamental theorem of algebra]] states that the field of [[complex number]]s is algebraically closed. Another example of an algebraically closed field is the field of (complex) [[algebraic number]]s.

			==Equivalent properties==
			Given a field ''F'', the assertion “''F'' is algebraically closed” is equivalent to other assertions:

			===The only irreducible polynomials are those of degree one===
			The field ''F'' is algebraically closed if and only if the only [[irreducible polynomial]]s in the [[polynomial ring]] ''F''[''x''] are those of degree one.

			The assertion “the polynomials of degree one are irreducible” is trivially true for any field. If ''F'' is algebraically closed and ''p''(''x'') is an irreducible polynomial of ''F''[''x''], then it has some root ''a'' and therefore ''p''(''x'') is a multiple of ''x'' − ''a''. Since ''p''(''x'') is irreducible, this means that ''p''(''x'') = ''k''(''x'' − ''a''), for some ''k'' ∈ ''F'' \ {0}. On the other hand, if ''F'' is not algebraically closed, then there is some non-constant polynomial ''p''(''x'') in ''F''[''x''] without roots in ''F''. Let ''q''(''x'') be some irreducible factor of ''p''(''x''). Since ''p''(''x'') has no roots in ''F'', ''q''(''x'') also has no roots in ''F''. Therefore, ''q''(''x'') has degree greater than one, since every first degree polynomial has one root in ''F''.

			===Every polynomial is a product of first degree polynomials===
			The field ''F'' is algebraically closed if and only if every polynomial ''p''(''x'') of degree ''n'' ≥ 1, with [[coefficient]]s in ''F'', [[factorization\|splits into linear factors]]. In other words, there are elements ''k'', ''x''<sub>1</sub>, ''x''<sub>2</sub>, …, ''x<sub>n</sub>'' of the field ''F'' such that ''p''(''x'') = ''k''(''x'' − ''x''<sub>1</sub>)(''x'' − ''x''<sub>2</sub>) ··· (''x'' − ''x<sub>n</sub>'').

			If ''F'' has this property, then clearly every non-constant polynomial in ''F''[''x''] has some root in ''F''; in other words, ''F'' is algebraically closed. On the other hand, that the property stated here holds for ''F'' if ''F'' is algebraically closed follows from the previous property together with the fact that, for any field ''K'', any polynomial in ''K''[''x''] can be written as a product of irreducible polynomials.

			===Polynomials of prime degree have roots===
			J. Shipman showed in 2007 that if every polynomial over ''F'' of prime degree has a root in ''F'', then every non-constant polynomial has a root in ''F'', thus ''F'' is algebraically closed.

			===The field has no proper algebraic extension===
			The field ''F'' is algebraically closed if and only if it has no proper [[algebraic extension]].

			If ''F'' has no proper algebraic extension, let ''p''(''x'') be some irreducible polynomial in ''F''[''x'']. Then the [[quotient ring\|quotient]] of ''F''[''x''] modulo the [[ideal (ring theory)\|ideal]] generated by ''p''(''x'') is an algebraic extension of ''F'' whose [[degree of a field extension\|degree]] is equal to the degree of ''p''(''x''). Since it is not a proper extension, its degree is 1 and therefore the degree of ''p''(''x'') is 1.

			On the other hand, if ''F'' has some proper algebraic extension ''K'', then the [[Minimal polynomial (field theory)\|minimal polynomial]] of an element in ''K'' \ ''F'' is irreducible and its degree is greater than 1.

			===The field has no proper finite extension===
			The field ''F'' is algebraically closed if and only if it has no finite [[algebraic extension]] because if, within the [[Algebraically closed field#The field has no proper algebraic extension\|previous proof]], the word “algebraic” is replaced by the word “finite”, then the proof is still valid.

			===Every endomorphism of ''F<sup>n</sup>'' has some eigenvector===
			The field ''F'' is algebraically closed if and only if, for each natural number ''n'', every [[linear map]] from ''F<sup>n</sup>'' into itself has some [[eigenvector]].

			An endomorphism of ''F<sup>n</sup>'' has an eigenvector if and only if its [[characteristic polynomial]] has some root. Therefore, when ''F'' is algebraically closed, every endomorphism of ''F<sup>n</sup>'' has some eigenvector. On the other hand, if every endomorphism of ''F<sup>n</sup>'' has an eigenvector, let ''p''(''x'') be an element of ''F''[''x'']. Dividing by its leading coefficient, we get another polynomial ''q''(''x'') which has roots if and only if ''p''(''x'') has roots. But if ''q''(''x'') = ''x<sup>n</sup>'' + ''a''<sub>''n'' − 1</sub>''x''<sup>''n'' − 1</sup>+ ··· + ''a''<sub>0</sub>, then ''q''(''x'') is the characteristic polynomial of the [[companion matrix]]
			:<math>\begin{pmatrix}0&0&\cdots&0&-a_0\\1&0&\cdots&0&-a_1\\0&1&\cdots&0&-a_2\\\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\cdots&1&-a_{n-1}\end{pmatrix}.</math>

			===Decomposition of rational expressions===
			The field ''F'' is algebraically closed if and only if every [[rational function]] in one variable ''x'', with coefficients in ''F'', can be written as the sum of a polynomial function with rational functions of the form ''a''/(''x'' − ''b'')<sup>n</sup>, where ''n'' is a natural number, and ''a'' and ''b'' are elements of ''F''.

			If ''F'' is algebraically closed then, since the irreducible polynomials in ''F''[''x''] are all of degree 1, the property stated above holds by the [[Partial fraction decomposition#Statement of theorem\|theorem on partial fraction decomposition]].

			On the other hand, suppose that the property stated above holds for the field ''F''. Let ''p''(''x'') be an irreducible element in ''F''[''x'']. Then the rational function 1/''p'' can be written as the sum of a polynomial function ''q'' with rational functions of the form ''a''/(''x'' − ''b'')<sup>n</sup>. Therefore, the rational expression
			:<math>\frac1{p(x)}-q(x)=\frac{1-p(x)q(x)}{p(x)}</math>
			can be written as a quotient of two polynomials in which the denominator is a product of first degree polynomials. Since ''p''(''x'') is irreducible, it must divide this product and, therefore, it must also be a first degree polynomial.

			===Relatively prime polynomials and roots===
			For any field ''F'', if two polynomials ''p''(''x''),''q''(''x'') ∈ ''F''[''x''] are [[coprime\|relatively prime]] then they do not have a common root, for if ''a'' ∈ ''F'' was a common root, then ''p''(''x'') and  ''q''(''x'') would both be multiples of ''x'' − ''a'' and therefore they would not be relatively prime. The fields for which the reverse implication holds (that is, the fields such that whenever two polynomials have no common root then they are relatively prime) are precisely the algebraically closed fields.

			If the field ''F'' is algebraically closed, let ''p''(''x'') and ''q''(''x'') be two polynomials which are not relatively prime and let ''r''(''x'') be their [[greatest common divisor]]. Then, since ''r''(''x'') is not constant, it will have some root ''a'', which will be then a common root of ''p''(''x'') and ''q''(''x'').

			If ''F'' is not algebraically closed, let ''p''(''x'') be a polynomial whose degree is at least 1 without roots. Then ''p''(''x'') and ''p''(''x'') are not relatively prime, but they have no common roots (since none of them has roots).

			==Other properties==
			If ''F'' is an algebraically closed field and ''n'' is a natural number, then ''F'' contains all ''n''th roots of unity, because these are (by definition) the ''n'' (not necessarily distinct) zeroes of the polynomial ''x<sup>n</sup>'' − 1. A field extension that is contained in an extension generated by the roots of unity is a ''cyclotomic extension'', and the extension of a field generated by all roots of unity is sometimes called its ''cyclotomic closure''. Thus algebraically closed fields are cyclotomically closed. The converse is not true. Even assuming that every polynomial of the form ''x<sup>n</sup>'' − ''a'' splits into linear factors is not enough to assure that the field is algebraically closed.

			If a proposition which can be expressed in the language of [[first-order logic]] is true for an algebraically closed field, then it is true for every algebraically closed field with the same [[Characteristic (algebra)\|characteristic]]. Furthermore, if such a proposition is valid for an algebraically closed field with characteristic 0, then not only is it valid for all other algebraically closed fields with characteristic 0, but there is some natural number ''N'' such that the proposition is valid for every algebraically closed field with characteristic ''p'' when ''p'' > ''N''.<ref>See subsections ''Rings and fields'' and ''Properties of mathematical theories'' in §2 of J. Barwise's "An introduction to first-order logic".</ref>

			Every field ''F'' has some extension which is algebraically closed. Among all such extensions there is one and ([[Up to\|up to isomorphism]], but not [[essentially unique\|unique isomorphism]]) only one which is an [[algebraic extension]] of ''F'';<ref>See Lang's ''Algebra'', §VII.2 or van der Waerden's ''Algebra I'', §10.1.</ref> it is called the [[algebraic closure]] of ''F''.

			The theory of algebraically closed fields has [[quantifier elimination]].

			==Notes==
			{{Reflist}}

			==References==
			* {{Citation \| last = Barwise \| first = Jon \| author-link = Jon Barwise \| year = 1978 \| contribution = An introduction to first-order logic \| editor-last = Barwise \| editor-first = Jon \| title = Handbook of Mathematical Logic \| series = Studies in Logic and the Foundations of Mathematics \| publisher = North Holland \| isbn = 0-7204-2285-X}}
			* {{Lang Algebra}}
			* {{citation\|last = Shipman\|first = Joseph\|year = 2007\|title = Improving the Fundamental Theorem of Algebra\|periodical = Mathematical Intelligencer\|volume = 29\|issue = 4\|pages = 9–14\|doi=10.1007/BF02986170\|issn = 0343-6993}}
			* {{Citation \| last = van der Waerden \| first = Bartel Leendert \| author-link = Bartel Leendert van der Waerden \| title = Algebra \| volume = I \| edition = 7th \| year = 2003 \| publisher = Springer-Verlag \| isbn = 0-387-40624-7}}

			{{DEFAULTSORT:Algebraically Closed Field}}
			[[Category:Abstract algebra]]
			[[Category:Field theory]]

96.238.44.157: /* Other properties */

2012-07-12T23:02:03Z

Other properties

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