Edge detection - Revision history

128.187.177.102 at 23:52, 8 December 2014

2014-12-08T23:52:20Z

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	~~A '''homomorphism''' between two algebras~~, '~~'A'' and ''B''~~, ~~[[Algebra over a field\|over a field]] (or [[Algebra (ring theory)\|ring]]) ''K''~~, is ~~a [[Function (mathematics)\|map]]~~ <~~math~~>F:~~A\rightarrow B<~~/~~math> such that for all ''k'' in ''K'' and ''x'',''y'' in ''A'',~~		"Era la notte prima di Natale e tutta la casa era in silenzio. Oh, I forgot, my articles are supposed to be in English. My bad! Today I wanted to tell y'all about my Christmas adventure in Italy, nevertheless I is certain to keep this short article in my native language, which naturally is English - Southern fashion.<br><br>For Men we [http://safedietplans.com/bmr-calculator bmr calculator] would add 66 + (6.3 x whatever the fat is inside pounds) + (12.9 x the height in inches) - (6.8 x any a age is inside years) following the same sequence plus formula because above however, utilizing these numbers.This particular formula will be applied to adults just, for kids it is a bit different.<br><br>If the basal metabolic rate is 2,300 calories a day, 1,200 calories of food consumption is not enough food for we to function properly. Doing this for too long can damage a metabolism! Having a healthy body at a healthy weight plus acceptable body fat percentage are the most crucial elements. You wish a healthy, lean body which may provide we the best chance to live healthy for a long time.<br><br>Horsegram is powdered to a good consistency. Heat sour buttermilk plus add 100 gm of horsegram powder for this and create a consistent paste. Apply this paste onto fat deposits found on the body and massage vigorously in upward strokes. Horsegram is acknowledged to minimize body fat fairly effectively.Take hot water bathtub following half an hr. Use 'eladhi choornam' instead of soap. Add a few drops of water to this choornam and prepare a thick paste and employ it for bathing reasons.<br><br>Same goes to our body. Do we understand why certain folks eat so little yet not seeing any encouraging fat loss results? Every 1 of you has our own battery indicator, the bmr that is the amount of calories required by our body to let you have the stamina to do daily activities like strolling, doing home chores, working, driving plus etc. BMR furthermore based on the individual person's age, muscle, physic rating, body fat %, visceral fat plus body water %.<br><br>If you cut back on calories too much, your body can go into "starvation mode' in that it begins storing everything we consume rather of burning from it for fuel. If you cut a calories too drastically you may see no weight loss in addition to be tired or sluggish. To rev up a metabolism and let your body run effectively you must eat throughout the day plus make certain you may be eating enough.<br><br>I hope we are inside the normal range, nevertheless when you are overweight we can program several weight reduction program considering the BMR and present activity level and hopefully enhance a wellness.

	* ''F''(~~''kx'') = ''kF''(''~~x'')

	* ''F''(''x'' + ~~''y'') = ''F''~~(''x'') ~~+ ''F''~~(~~''y'')~~

	* ''F''(''xy'') = ''F''(''x'')~~''F''(''y'')<ref>{{cite book \| last1=Dummit \| first1=David S~~. \| last2=Foote \| first2=Richard M. \| title=Abstract Algebra \| publisher=[[John Wiley & Sons]] \| year=2004 \| edition=3rd \| isbn=0-471-43334-9}}</ref><ref>{{cite book \| last=Lang \| first=Serge \| authorlink=Serge Lang \| title=Algebra \| publisher=[[Springer Science+Business Media\|Springer]] \| series=[[Graduate Texts in Mathematics]] \| year=2002 \| isbn=0-387-95385-X}}</ref>

	~~If ''F'' is [[bijective]] then ''F'' is said~~ to ~~be an '''isomorphism''' between ''A'' and ''B''.~~

	~~A common abbreviation~~ for ~~"homomorphism between algebras" is "algebra homomorphism" or "algebra map". Every algebra homomorphism~~ is a ~~homomorphism of ''K''-modules~~.

	~~== Unital algebra homomorphisms ==~~

	~~If ''A'' and ''B'' are two unital algebras, then an algebra homomorphism~~ <~~math~~>~~F:A\rightarrow B~~<~~/math~~> is ~~said to be ''unital'' if it maps the unity~~ of ~~''A''~~ to ~~the unity of ''B''~~. ~~Often the words "algebra homomorphism" are actually used in the meaning of "unital algebra homomorphism", so non-unital algebra homomorphisms are excluded.~~

	~~==Examples==~~
	~~Let ''A'' = ''K''[''x''] be the set of all polynomials over~~ a ~~field ''K'' and ''B'' be the set of all polynomial functions over ''K''. Both ''A'' and ''B''~~ are ~~algebras over ''K'' given by~~ the ~~standard multiplication and addition of polynomials and functions, respectively~~. ~~We can map each <math>f\~~,~~</math> in ''A''~~ to ~~<math>\hat{f}\,</math> in ''B'' by the rule <math>\hat{f}(t) = f(t) \, </math>~~. ~~A routine check shows that the mapping~~ <~~math~~>~~f \mapsto \hat{f}\,~~<~~/math~~> is a ~~homomorphism~~ of the ~~algebras ''A''~~ and ~~''B''~~. ~~This homomorphism~~ is an ~~isomorphism if and only if ''K'' is an infinite field~~.

	''~~Proof~~.~~'' If ''K'' is~~ a ~~finite field then let~~

	~~:<math>p(x) = \prod\limits_{t \in K} (x-t).\,</math>~~

	~~''p'' is~~ a ~~nonzero polynomial in ''K''[''x''], however <math>p(t) = 0\,</math>~~ for ~~all ''t'' in ''K'', so~~ <~~math~~>~~\hat{p} = 0\,~~<~~/math~~> ~~is the zero function and~~ our ~~homomorphism is~~ not ~~an isomorphism (and, actually~~, the ~~algebras are not isomorphic, since~~ the ~~algebra~~ of ~~polynomials is infinite while that of polynomial functions is finite).~~

	~~If ''K'' is infinite then choose a polynomial ''f'' such that <math>\hat{f} = 0\,</math>. We want~~ to ~~show this implies that <math>f = 0\~~,~~</math>. Let <math>\deg f = n\~~,~~</math> and let <math>t_0~~,~~t_1~~,~~\dots~~,~~t_n\,</math> be ''n'' + 1 distinct elements of ''K''. Then <math>f(t_i) = 0\~~,~~</math> for <math>0 \le i \le n</math> and by [[Lagrange interpolation]] we have <math>f = 0\~~,~~</math>~~. ~~Hence the mapping~~ <~~math~~>~~f \mapsto \hat{f}\,~~<~~/math~~> ~~is injective. Since this mapping is clearly surjective~~, it ~~is bijective and thus an algebra isomorphism~~ of ~~''A'' and ''B''~~.

	If ~~''A'' is a [[subalgebra]] of ''B'', then for every [[group of units\|invertible]] ''b'' in ''B'' the function that takes every ''~~a'' in ~~''A''~~ to ~~''b''<sup>−1</sup> ''~~a~~'' ''b'' is an algebra homomorphism (in case~~ <~~math~~>~~A=B~~<~~/math~~>, ~~this is called an inner automorphism of ''B''). If ''A'' is also [[simple algebra\|simple]]~~ and ~~''B'' is~~ a ~~[[central simple algebra]], then every homomorphism from ''A'' to ''B'' is given in this way by some ''b'' in ''B''; this is the [[Skolem-Noether theorem]]~~.

	~~==References==~~
	~~{{reflist}}~~

	~~{{DEFAULTSORT:Algebra Homomorphism}}~~
	~~[[Category:Algebras]]~~
	~~[[Category:Ring theory]]~~
	~~[[Category:Morphisms]]~~

67.174.255.201: /* See also */

2014-02-01T19:57:35Z

See also

← Older revision		Revision as of 21:57, 1 February 2014
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	~~Anybody who wrote the text~~ is ~~called Eusebio~~. ~~His friends say it~~'~~s a bad one~~ for ~~him but what he loves doing typically~~ is ~~acting~~ and he'~~s ever been doing~~ it ~~for quite~~ a ~~while~~. ~~Filing has been his profession for~~ a ~~short time~~. ~~Massachusetts has always been his~~ [~~http:~~//~~www~~.~~lifestyle~~.~~org~~/ ~~lifestyle~~] ~~place~~ and ~~his family~~ group ~~loves it. Go~~ to ~~his website to identify~~ a ~~out more: http://prometeu.net~~<br><br>~~Have~~ a ~~look at my web~~-~~site~~ :: ~~hack clash of clans~~ [[~~http~~:~~//prometeu.net stay with me~~]]		A '''homomorphism''' between two algebras, ''A'' and ''B'', [[Algebra over a field\|over a field]] (or [[Algebra (ring theory)\|ring]]) ''K'', is a [[Function (mathematics)\|map]] <math>F:A\rightarrow B</math> such that for all ''k'' in ''K'' and ''x'',''y'' in ''A'',

			* ''F''(''kx'') = ''kF''(''x'')

			* ''F''(''x'' + ''y'') = ''F''(''x'') + ''F''(''y'')

			* ''F''(''xy'') = ''F''(''x'')''F''(''y'')<ref>{{cite book \| last1=Dummit \| first1=David S. \| last2=Foote \| first2=Richard M. \| title=Abstract Algebra \| publisher=[[John Wiley & Sons]] \| year=2004 \| edition=3rd \| isbn=0-471-43334-9}}</ref><ref>{{cite book \| last=Lang \| first=Serge \| authorlink=Serge Lang \| title=Algebra \| publisher=[[Springer Science+Business Media\|Springer]] \| series=[[Graduate Texts in Mathematics]] \| year=2002 \| isbn=0-387-95385-X}}</ref>

			If ''F'' is [[bijective]] then ''F'' is said to be an '''isomorphism''' between ''A'' and ''B''.

			A common abbreviation for "homomorphism between algebras" is "algebra homomorphism" or "algebra map". Every algebra homomorphism is a homomorphism of ''K''-modules.

			== Unital algebra homomorphisms ==

			If ''A'' and ''B'' are two unital algebras, then an algebra homomorphism <math>F:A\rightarrow B</math> is said to be ''unital'' if it maps the unity of ''A'' to the unity of ''B''. Often the words "algebra homomorphism" are actually used in the meaning of "unital algebra homomorphism", so non-unital algebra homomorphisms are excluded.

			==Examples==
			Let ''A'' = ''K''[''x''] be the set of all polynomials over a field ''K'' and ''B'' be the set of all polynomial functions over ''K''. Both ''A'' and ''B'' are algebras over ''K'' given by the standard multiplication and addition of polynomials and functions, respectively. We can map each <math>f\,</math> in ''A'' to <math>\hat{f}\,</math> in ''B'' by the rule <math>\hat{f}(t) = f(t) \, </math>. A routine check shows that the mapping <math>f \mapsto \hat{f}\,</math> is a homomorphism of the algebras ''A'' and ''B''. This homomorphism is an isomorphism if and only if ''K'' is an infinite field.

			''Proof.'' If ''K'' is a finite field then let

			:<math>p(x) = \prod\limits_{t \in K} (x-t).\,</math>

			''p'' is a nonzero polynomial in ''K''[''x''], however <math>p(t) = 0\,</math> for all ''t'' in ''K'', so <math>\hat{p} = 0\,</math> is the zero function and our homomorphism is not an isomorphism (and, actually, the algebras are not isomorphic, since the algebra of polynomials is infinite while that of polynomial functions is finite).

			If ''K'' is infinite then choose a polynomial ''f'' such that <math>\hat{f} = 0\,</math>. We want to show this implies that <math>f = 0\,</math>. Let <math>\deg f = n\,</math> and let <math>t_0,t_1,\dots,t_n\,</math> be ''n'' + 1 distinct elements of ''K''. Then <math>f(t_i) = 0\,</math> for <math>0 \le i \le n</math> and by [[Lagrange interpolation]] we have <math>f = 0\,</math>. Hence the mapping <math>f \mapsto \hat{f}\,</math> is injective. Since this mapping is clearly surjective, it is bijective and thus an algebra isomorphism of ''A'' and ''B''.

			If ''A'' is a [[subalgebra]] of ''B'', then for every [[group of units\|invertible]] ''b'' in ''B'' the function that takes every ''a'' in ''A'' to ''b''<sup>−1</sup> ''a'' ''b'' is an algebra homomorphism (in case <math>A=B</math>, this is called an inner automorphism of ''B''). If ''A'' is also [[simple algebra\|simple]] and ''B'' is a [[central simple algebra]], then every homomorphism from ''A'' to ''B'' is given in this way by some ''b'' in ''B''; this is the [[Skolem-Noether theorem]].

			==References==
			{{reflist}}

			{{DEFAULTSORT:Algebra Homomorphism}}
			[[Category:Algebras]]
			[[Category:Ring theory]]
			[[Category:Morphisms]]

70.247.162.84: link

2012-08-31T12:50:45Z

link

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