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130.159.122.146 at 17:28, 6 January 2015

2015-01-06T17:28:56Z

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	In [[mathematics]], more specifically [[Field theory (mathematics)\|field theory]], the '''degree of a field extension''' is a rough measure of the "size" of the extension. The concept plays an important role in many parts of [[mathematics]], including [[abstract algebra\|algebra]] and [[number theory]] — indeed in any area where [[field (mathematics)\|fields]] appear prominently.		40 yrs old Fleet Manager Arron from Terrebonne, usually spends time with pastimes which includes ice skating, ganhando dinheiro na internet and sleeping. Discovers the beauty in going to places around the world, of late just coming back from Serra da Capivara National Park.<br><br>my web blog; [http://ganhandodinheironainternet.comoganhardinheiro101.com/ ganhar dinheiro]

	~~== Definition and notation ==~~

	Suppose that ''E''/''F'' is a [[field extension]]. Then ''E'' may be considered as a [[vector space]] over ''F'' (the field of scalars). The [[dimension (vector space)\|dimension]] of this vector space is called the '''degree of the field extension''', ~~and it is denoted by [E:F].~~

	The degree may be finite or infinite, the field being called a '''finite extension''' or '''infinite extension''' accordingly. An extension ''E''/''F'' is also sometimes said to be simply '''finite''' if it is a finite extension; this should not be confused with ~~the fields themselves being [[finite field]]s (fields with finitely many elements).~~

	The degree should not be confused with the [[transcendence degree]] of a field; for example, the field '''Q'''(''X'') of [[rational function]]s has infinite degree over '''Q''', but transcendence degree only equal to 1.

	~~== The multiplicativity formula for degrees ==~~

	~~Given three fields arranged in a [[tower of fields\|tower]], say ''K'' a subfield of ''L''~~ which ~~is in turn a subfield of ''M''~~, ~~there is a simple relation between the degrees of the three extensions ''L''/''K'', ''M''/''L''~~ and ~~''M''/''K'':~~
	~~: <math>[M:K] = [M:L] \cdot [L:K]~~.~~</math>~~
	~~In other words,~~ the ~~degree~~ going ~~from the "bottom"~~ to the "top" field is just the product of the degrees going from the "bottom" to the "middle" and then from the "middle" to the "top". It is quite analogous to [[Lagrange's theorem (group theory)\|Lagrange's theorem]] in [[group theory]], ~~which relates the order of a group to the order and [[Index of a subgroup\|index]]~~ of ~~a subgroup — indeed [[Galois theory]] shows that this analogy is more than~~ just ~~a coincidence.~~

	The formula holds for both finite and infinite degree extensions. In the infinite case, the product is interpreted in the sense of products of [[cardinal number]]s. In particular, this means that if ''M''/''K'' is finite, then both ''M''/''L'' and ''L''/''K'' are finite.

	If ''M''/''K'' is finite, then the formula imposes strong restrictions on the kinds of fields that can occur between ''M'' and ''K'', via simple arithmetical considerations. For example, if the degree [''M'':''K''] is a [[prime number]] ''p'', then for any intermediate field ''L'', one of two things can happen: either [''M'':''L''] = ''p'' and [''L'':''K''] = 1, in which case ''L'' is equal to ''K'', or [''M'':''L''] = 1 and [''L'':''K''] = ''p'', in which case ''L'' is equal to ''M''. Therefore there are no intermediate fields (apart from ''M'' and ''K'' themselves).

	~~=== Proof of the multiplicativity formula in the finite case ===~~

	Suppose that ''K'', ''L'' and ''M'' form a tower of fields as in the degree formula above, and that both ''d'' = [''L'':''K''] and ''e'' = [''M'':''L''] are finite. This means that we may select a [[basis (linear algebra)\|basis]] {''u''<sub>1</sub>, ..., ''u''<sub>''d''</sub>} for ''L'' over ''K'', and a basis {''w''<sub>1</sub>, ..., ''w''<sub>''e''</sub>} for ''M'' over ''L''. We will show that the elements ''u''<sub>''m''</sub>''w''<sub>''n''</sub>, for ''m'' ranging through 1, 2, ..., ''d'' and ''n'' ranging through 1, 2, ..., ''e'', form a basis for ''M''/''K''; since there are precisely ''de'' of them, this proves that the dimension of ''M''/''K'' is ''de'', which is the desired result.

	First we check that they [[linear span\|span]] ''M''/''K''. If ''x'' is any element of ''M'', then since the ''w''<sub>''n''</sub> form a basis for ''M'' over ''L'', we can find elements ''a''<sub>''n''</sub> in ''L'' such that
	~~: <math> x = \sum_{n=1}^e a_n w_n = a_1 w_1 + \cdots + a_e w_e.</math>~~
	~~Then, since the ''u''<sub>''m''</sub> form a basis for ''L'' over ''K'', we can find elements ''b''<sub>''m'',''n''</sub> in ''K'' such that for each ''n'',~~
	~~: <math> a_n = \sum_{m=1}^d b_{m,n} u_m = b_{1,n} u_1 + \cdots + b_{d,n} u_d.</math>~~
	~~Then using the [[distributive law]] and [[associativity]] of multiplication in ''M'' we have~~
	~~: <math> x = \sum_{n=1}^e \left(\sum_{m=1}^d b_{m,n} u_m\right) w_n = \sum_{n=1}^e \sum_{m=1}^d b_{m,n} (u_m w_n),</math>~~
	~~which shows that ''x'' is a linear combination of the ''u''<sub>''m''</sub>''w''<sub>''n''</sub> with coefficients~~ from ~~''K''; in other words they span ''M'' over ''K''.~~

	~~Secondly we must check that they are [[linear independence\|linearly independent]] over ''K''. So assume that~~
	~~: <math> 0 = \sum_{n=1}^e \sum_{m=1}^d b_{m,n} (u_m w_n)</math>~~
	~~for some coefficients ''b''<sub>''m'',''n''</sub> in ''K''~~. ~~Using distributivity and associativity again, we can group the terms as~~
	: <~~math~~> ~~0 = \sum_{n=1}^e \left(\sum_{m=1}^d b_{m,n} u_m\right) w_n,~~<~~/math~~>
	~~and we see that the terms in parentheses must be zero, because they are elements of ''L'', and the ''w''<sub>''n''</sub> are linearly independent over ''L''. That is,~~
	~~: <math> 0 = \sum_{m=1}^d b_{m,n} u_m </math>~~
	for each ''n''. Then, since the ''b''<sub>''m'',''n''</sub> coefficients are in ''K'', and the ''u''<sub>''m''</sub> are linearly independent over ''K'', we must have that ''b''<sub>''m'',''n''</sub> = 0 for all ''m'' and all ''n''. This shows that the elements ''u''<sub>''m''</sub>''w''<sub>''n''</sub> are linearly independent over ''K''. This concludes the proof.

	~~=== Proof of the formula in the infinite case ===~~

	In this case, we start with bases ''u''<sub>α</sub> and ''w''<sub>β</sub> of ''L''/''K'' and ''M''/''L'' respectively, where α is taken from an indexing set ''A'', and β from an indexing set ''B''. Using an entirely similar argument as the one above, we find that the products ''u''<sub>α</sub>''w''<sub>β~~</sub> form a basis for ''M''/''K''. These are indexed by the [[cartesian product]] ''A'' × ''B'', which by definition has~~ [~~[cardinality]] equal to the product of the cardinalities of ''A'' and ''B''.~~

	~~== Examples ==~~

	* The [[complex number]]s are a field extension over the [[real number]]s with degree ['''C''':~~'''R'''] = 2, and thus there are no non-trivial [[field (mathematics)\|field]]s between them.~~
	* The field extension '''Q'''(√2, √3), obtained by adjoining √2 and √3 to the field '''Q''' of [[rational number]]s, has degree 4, that is, ['''Q'''(√2, √3):'''Q'''] = 4. The intermediate field '''Q'''(√2) has degree 2 over '''Q'''; we conclude from the multiplicativity formula that ['''Q'''(√2, √3):'''Q'''(√2)] = 4/2 = 2.
	* The [[finite field]] '''GF'''(125) = '''GF'''(5<sup>3</sup>) has degree 3 over its subfield '''GF'''(5). More generally, if ''p'' is a prime and ''n'', ''m'' are positive integers with ''n'' dividing ''m'', then ['''GF'''(''p''<sup>''m''</sup>):'''GF'''(''p''<sup>''n''</sup>)] = ''m''/''n''.
	* The field extension '''C'''(''T'')/'''C''', where '''C'''(''T'') is the field of [[rational function]]s over '''C''', has infinite degree (indeed it is a [[purely transcendental]] extension). This can be seen by observing that the elements 1, ''T'', ''T''<sup>2</sup>, etc., are linearly independent over '''C'''.
	* The field extension '''C'''(''T''<sup>2</sup>) also has infinite degree over '''C'''. However, if we view '''C'''(''T''<sup>2</sup>) as a subfield of '''C'''(''T''), then in fact ['''C'''(''T''):'''C'''(''T''<sup>2</~~sup>)] = 2~~. More generally, if ''X'' and ''Y'' are [[algebraic curve]]s over a field ''K'', and ''F'' : ''X'' → ''Y'' is a surjective morphism between them of degree ''d'', then the [[function field of an algebraic variety\|function field]]s ''K''(''X'') and ''K''(''Y'') are both of infinite degree over ''K'', but the degree [''K''(''X''):''K''(''Y'')] turns out to be equal to ''d''.

	~~== Generalization ==~~

	Given two [[division ring]]s ''E'' and ''F'' with ''F'' contained in ''E'' and the multiplication and addition of ''F'' being the restriction of the operations in ''E'', we can consider ''E'' as a vector space over ''F'' in two ways: having the scalars act on the left, giving a dimension [''E'':''F'']<sub>l</sub>, and having them act on the right, giving a dimension [''E'':''F'']<sub>r</sub>. The two dimensions need not agree. Both dimensions however satisfy a multiplication formula for towers of division rings; the proof above applies to left-acting scalars without change.

	~~== References ==~~

	* page 215, {{cite book \| author=Jacobson, N. \| authorlink=Nathan Jacobson\| title=Basic Algebra I \| publisher=W. H. Freeman and Company \| year=1985 \| isbn=0-7167-1480-9 }} Proof of the multiplicativity formula.
	* page 465, {{cite book \| author=Jacobson, N. \| authorlink=Nathan Jacobson\| title=Basic Algebra II \| publisher=W. H. Freeman and Company \| year=1989 \| isbn=0-7167-1933-9 }} Briefly discusses the infinite dimensional case.

	~~[[Category:Field extensions]~~]

en>David Eppstein: /* References */ link Liao Shijun

2013-12-03T04:21:59Z

References: link Liao Shijun

← Older revision		Revision as of 06:21, 3 December 2013
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	~~Jayson Berryhill~~ is ~~how~~ [~~http~~://~~findyourflirt~~.~~net~~/index.~~php?m~~=~~member_profile&~~p=~~profile&id~~=~~117823 psychic chat online~~] I'm ~~called~~ and ~~my spouse doesn~~'~~t like it at all~~. ~~The preferred pastime~~ for ~~him and his kids~~ is ~~fashion and he~~'~~ll be starting something else along with it~~. ~~My spouse and I reside~~ in ~~Mississippi~~ and ~~I love each~~ [~~http~~://~~www~~.~~sirudang~~.~~com~~/~~siroo_Notice~~/~~2110 live psychic reading] day residing~~ right ~~here~~. ~~Office supervising~~ is ~~exactly~~ where ~~her primary earnings arrives~~ from ~~but she~~'~~s currently applied for~~ an ~~additional 1~~.<br><br>~~my webpage~~; [~~http~~://~~cartoonkorea~~.~~com~~/~~ce002~~/~~1093612~~ are ~~psychics real~~]		In [[mathematics]], more specifically [[Field theory (mathematics)\|field theory]], the '''degree of a field extension''' is a rough measure of the "size" of the extension. The concept plays an important role in many parts of [[mathematics]], including [[abstract algebra\|algebra]] and [[number theory]] — indeed in any area where [[field (mathematics)\|fields]] appear prominently.

			== Definition and notation ==

			Suppose that ''E''/''F'' is a [[field extension]]. Then ''E'' may be considered as a [[vector space]] over ''F'' (the field of scalars). The [[dimension (vector space)\|dimension]] of this vector space is called the '''degree of the field extension''', and it is denoted by [E:F].

			The degree may be finite or infinite, the field being called a '''finite extension''' or '''infinite extension''' accordingly. An extension ''E''/''F'' is also sometimes said to be simply '''finite''' if it is a finite extension; this should not be confused with the fields themselves being [[finite field]]s (fields with finitely many elements).

			The degree should not be confused with the [[transcendence degree]] of a field; for example, the field '''Q'''(''X'') of [[rational function]]s has infinite degree over '''Q''', but transcendence degree only equal to 1.

			== The multiplicativity formula for degrees ==

			Given three fields arranged in a [[tower of fields\|tower]], say ''K'' a subfield of ''L'' which is in turn a subfield of ''M'', there is a simple relation between the degrees of the three extensions ''L''/''K'', ''M''/''L'' and ''M''/''K'':
			: <math>[M:K] = [M:L] \cdot [L:K].</math>
			In other words, the degree going from the "bottom" to the "top" field is just the product of the degrees going from the "bottom" to the "middle" and then from the "middle" to the "top". It is quite analogous to [[Lagrange's theorem (group theory)\|Lagrange's theorem]] in [[group theory]], which relates the order of a group to the order and [[Index of a subgroup\|index]] of a subgroup — indeed [[Galois theory]] shows that this analogy is more than just a coincidence.

			The formula holds for both finite and infinite degree extensions. In the infinite case, the product is interpreted in the sense of products of [[cardinal number]]s. In particular, this means that if ''M''/''K'' is finite, then both ''M''/''L'' and ''L''/''K'' are finite.

			If ''M''/''K'' is finite, then the formula imposes strong restrictions on the kinds of fields that can occur between ''M'' and ''K'', via simple arithmetical considerations. For example, if the degree [''M'':''K''] is a [[prime number]] ''p'', then for any intermediate field ''L'', one of two things can happen: either [''M'':''L''] = ''p'' and [''L'':''K''] = 1, in which case ''L'' is equal to ''K'', or [''M'':''L''] = 1 and [''L'':''K''] = ''p'', in which case ''L'' is equal to ''M''. Therefore there are no intermediate fields (apart from ''M'' and ''K'' themselves).

			=== Proof of the multiplicativity formula in the finite case ===

			Suppose that ''K'', ''L'' and ''M'' form a tower of fields as in the degree formula above, and that both ''d'' = [''L'':''K''] and ''e'' = [''M'':''L''] are finite. This means that we may select a [[basis (linear algebra)\|basis]] {''u''<sub>1</sub>, ..., ''u''<sub>''d''</sub>} for ''L'' over ''K'', and a basis {''w''<sub>1</sub>, ..., ''w''<sub>''e''</sub>} for ''M'' over ''L''. We will show that the elements ''u''<sub>''m''</sub>''w''<sub>''n''</sub>, for ''m'' ranging through 1, 2, ..., ''d'' and ''n'' ranging through 1, 2, ..., ''e'', form a basis for ''M''/''K''; since there are precisely ''de'' of them, this proves that the dimension of ''M''/''K'' is ''de'', which is the desired result.

			First we check that they [[linear span\|span]] ''M''/''K''. If ''x'' is any element of ''M'', then since the ''w''<sub>''n''</sub> form a basis for ''M'' over ''L'', we can find elements ''a''<sub>''n''</sub> in ''L'' such that
			: <math> x = \sum_{n=1}^e a_n w_n = a_1 w_1 + \cdots + a_e w_e.</math>
			Then, since the ''u''<sub>''m''</sub> form a basis for ''L'' over ''K'', we can find elements ''b''<sub>''m'',''n''</sub> in ''K'' such that for each ''n'',
			: <math> a_n = \sum_{m=1}^d b_{m,n} u_m = b_{1,n} u_1 + \cdots + b_{d,n} u_d.</math>
			Then using the [[distributive law]] and [[associativity]] of multiplication in ''M'' we have
			: <math> x = \sum_{n=1}^e \left(\sum_{m=1}^d b_{m,n} u_m\right) w_n = \sum_{n=1}^e \sum_{m=1}^d b_{m,n} (u_m w_n),</math>
			which shows that ''x'' is a linear combination of the ''u''<sub>''m''</sub>''w''<sub>''n''</sub> with coefficients from ''K''; in other words they span ''M'' over ''K''.

			Secondly we must check that they are [[linear independence\|linearly independent]] over ''K''. So assume that
			: <math> 0 = \sum_{n=1}^e \sum_{m=1}^d b_{m,n} (u_m w_n)</math>
			for some coefficients ''b''<sub>''m'',''n''</sub> in ''K''. Using distributivity and associativity again, we can group the terms as
			: <math> 0 = \sum_{n=1}^e \left(\sum_{m=1}^d b_{m,n} u_m\right) w_n,</math>
			and we see that the terms in parentheses must be zero, because they are elements of ''L'', and the ''w''<sub>''n''</sub> are linearly independent over ''L''. That is,
			: <math> 0 = \sum_{m=1}^d b_{m,n} u_m </math>
			for each ''n''. Then, since the ''b''<sub>''m'',''n''</sub> coefficients are in ''K'', and the ''u''<sub>''m''</sub> are linearly independent over ''K'', we must have that ''b''<sub>''m'',''n''</sub> = 0 for all ''m'' and all ''n''. This shows that the elements ''u''<sub>''m''</sub>''w''<sub>''n''</sub> are linearly independent over ''K''. This concludes the proof.

			=== Proof of the formula in the infinite case ===

			In this case, we start with bases ''u''<sub>α</sub> and ''w''<sub>β</sub> of ''L''/''K'' and ''M''/''L'' respectively, where α is taken from an indexing set ''A'', and β from an indexing set ''B''. Using an entirely similar argument as the one above, we find that the products ''u''<sub>α</sub>''w''<sub>β</sub> form a basis for ''M''/''K''. These are indexed by the [[cartesian product]] ''A'' × ''B'', which by definition has [[cardinality]] equal to the product of the cardinalities of ''A'' and ''B''.

			== Examples ==

			* The [[complex number]]s are a field extension over the [[real number]]s with degree ['''C''':'''R'''] = 2, and thus there are no non-trivial [[field (mathematics)\|field]]s between them.
			* The field extension '''Q'''(√2, √3), obtained by adjoining √2 and √3 to the field '''Q''' of [[rational number]]s, has degree 4, that is, ['''Q'''(√2, √3):'''Q'''] = 4. The intermediate field '''Q'''(√2) has degree 2 over '''Q'''; we conclude from the multiplicativity formula that ['''Q'''(√2, √3):'''Q'''(√2)] = 4/2 = 2.
			* The [[finite field]] '''GF'''(125) = '''GF'''(5<sup>3</sup>) has degree 3 over its subfield '''GF'''(5). More generally, if ''p'' is a prime and ''n'', ''m'' are positive integers with ''n'' dividing ''m'', then ['''GF'''(''p''<sup>''m''</sup>):'''GF'''(''p''<sup>''n''</sup>)] = ''m''/''n''.
			* The field extension '''C'''(''T'')/'''C''', where '''C'''(''T'') is the field of [[rational function]]s over '''C''', has infinite degree (indeed it is a [[purely transcendental]] extension). This can be seen by observing that the elements 1, ''T'', ''T''<sup>2</sup>, etc., are linearly independent over '''C'''.
			* The field extension '''C'''(''T''<sup>2</sup>) also has infinite degree over '''C'''. However, if we view '''C'''(''T''<sup>2</sup>) as a subfield of '''C'''(''T''), then in fact ['''C'''(''T''):'''C'''(''T''<sup>2</sup>)] = 2. More generally, if ''X'' and ''Y'' are [[algebraic curve]]s over a field ''K'', and ''F'' : ''X'' → ''Y'' is a surjective morphism between them of degree ''d'', then the [[function field of an algebraic variety\|function field]]s ''K''(''X'') and ''K''(''Y'') are both of infinite degree over ''K'', but the degree [''K''(''X''):''K''(''Y'')] turns out to be equal to ''d''.

			== Generalization ==

			Given two [[division ring]]s ''E'' and ''F'' with ''F'' contained in ''E'' and the multiplication and addition of ''F'' being the restriction of the operations in ''E'', we can consider ''E'' as a vector space over ''F'' in two ways: having the scalars act on the left, giving a dimension [''E'':''F'']<sub>l</sub>, and having them act on the right, giving a dimension [''E'':''F'']<sub>r</sub>. The two dimensions need not agree. Both dimensions however satisfy a multiplication formula for towers of division rings; the proof above applies to left-acting scalars without change.

			== References ==

			* page 215, {{cite book \| author=Jacobson, N. \| authorlink=Nathan Jacobson\| title=Basic Algebra I \| publisher=W. H. Freeman and Company \| year=1985 \| isbn=0-7167-1480-9 }} Proof of the multiplicativity formula.
			* page 465, {{cite book \| author=Jacobson, N. \| authorlink=Nathan Jacobson\| title=Basic Algebra II \| publisher=W. H. Freeman and Company \| year=1989 \| isbn=0-7167-1933-9 }} Briefly discusses the infinite dimensional case.

			[[Category:Field extensions]]

en>Chris the speller: sp

2012-08-16T14:47:58Z

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