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| In [[abstract algebra]], the theory of [[field (mathematics)|fields]] lacks a [[direct product]]: the direct product of two fields, considered as a [[ring (mathematics)|ring]] is ''never'' itself a field. On the other hand it is often required to 'join' two fields ''K'' and ''L'', either in cases where ''K'' and ''L'' are given as [[subfield (mathematics)|subfields]] of a larger field ''M'', or when ''K'' and ''L'' are both [[field extension]]s of a smaller field ''N'' (for example a [[prime field]]).
| | I'm Bobbie and was born on 22 July 1971. My hobbies are Stone collecting and Gardening.<br><br>Here is my page [http://gjycorp.com/Gcentre_Advisors/43509?ckattempt=2 Belinda Broido] |
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| The '''tensor product of fields''' is the best available construction on fields with which to discuss all the phenomena arising. As a ring, it is sometimes a field, and often a direct product of fields; it can, though, contain non-zero nilpotents (see [[radical of a ring]]).
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| If ''K'' and ''L'' do not have isomorphic prime fields, or in other words they have different [[characteristic (field)|characteristic]]s, they have no possibility of being common subfields of a field ''M''. Correspondingly their tensor product will in that case be the [[trivial ring]] (collapse of the construction to nothing of interest).
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| ==Compositum of fields==
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| Firstly, one defines the notion of the compositum of fields. This construction occurs frequently in [[field theory (mathematics)|field theory]]. The idea behind the compositum is to make the smallest field containing two other fields. In order to formally define the compositum, one must first specify a [[tower of fields]]. Let ''k'' be a field and ''L'' and ''K'' be two extensions of ''k''. The compositum, denoted ''KL'' is defined to be <math> KL = k(K \cup L) </math> where the right-hand side denotes the extension generated by ''K'' and ''L''. Note that this assumes ''some'' field containing both ''K'' and ''L''. Either one starts in a situation where such a common over-field is easy to identify (for example if ''K'' and ''L'' are both subfields of the complex numbers); or one proves a result that allows one to place both ''K'' and ''L'' (as isomorphic copies) in some large enough field.
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| In many cases one can identify ''K''.''L'' as a [[vector space]] [[tensor product]], taken over the field ''N'' that is the intersection of ''K'' and ''L''. For example if one adjoins √2 to the rational field ℚ to get ''K'', and √3 to get ''L'', it is true that the field ''M'' obtained as ''K''.''L'' inside the complex numbers ℂ is ([[up to]] isomorphism)
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| :<math>K\otimes_{\mathbb Q}L</math>
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| as a vector space over ℚ. (This type of result can be verified, in general, by using the [[ramification]] theory of [[algebraic number theory]].)
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| Subfields ''K'' and ''L'' of ''M'' are [[linearly disjoint]] (over a subfield ''N'') when in this way the natural ''N''-linear map of
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| :<math>K\otimes_NL</math>
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| to ''K''.''L'' is [[injective]].<ref>{{springer|id=L/l059560|title=Linearly-disjoint extensions}}</ref> Naturally enough this isn't always the case, for example when ''K'' = ''L''. When the degrees are finite, injective is equivalent here to [[bijective]].
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| A significant case in the theory of [[cyclotomic field]]s is that for the ''n''th [[roots of unity]], for ''n'' a composite number, the subfields generated by the ''p''<sup>''k''</sup>th roots of unity for [[prime power]]s dividing ''n'' are linearly disjoint for distinct ''p''.<ref>{{springer|id=c/c027570|title=Cyclotomic field}}</ref>
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| ==The tensor product as ring==
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| To get a general theory, one needs to consider a ring structure on <math>K \otimes_N L</math>. One can define the product <math>(a\otimes b)(c\otimes d)</math> to be <math> ac \otimes bd</math>. This formula is multilinear over ''N'' in each variable; and so defines a ring structure on the tensor product, making <math>K \otimes_N L</math> into a commutative [[algebra over a field|''N''-algebra]], called the '''tensor product of fields'''.
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| ==Analysis of the ring structure==
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| The structure of the ring can be analysed by considering all ways of embedding both ''K'' and ''L'' in some field extension of ''N''. Note that the construction here assumes the common subfield ''N''; but does not assume ''[[A priori and a posteriori|a priori]]'' that ''K'' and ''L'' are subfields of some field ''M'' (thus getting round the caveats about constructing a compositum field). Whenever one embeds ''K'' and ''L'' in such a field ''M'', say using embeddings α of ''K'' and β of ''L'', there results a ring homomorphism γ from <math>K \otimes_N L</math> into ''M'' defined by:
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| :<math>\gamma(a\otimes b) = (\alpha(a)\otimes1)\star(1\otimes\beta(b)) = \alpha(a).\beta(b).</math>
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| The kernel of γ will be a [[prime ideal]] of the tensor product; and conversely any prime ideal of the tensor product will give a homomorphism of ''N''-algebras to an [[integral domain]] (inside a [[field of fractions]]) and so provides embeddings of ''K'' and ''L'' in some field as extensions of (a copy of) ''N''.
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| In this way one can analyse the structure of <math>K \otimes_N L</math>: there may in principle be a non-zero [[Jacobson radical]] (intersection of all prime ideals) - and after taking the quotient by that one can speak of the product of all embeddings of ''K'' and ''L'' in various ''M'', ''over'' ''N''.
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| In case ''K'' and ''L'' are finite extensions of N, the situation is particularly simple since the tensor product is of finite dimension as an ''N''-algebra (and thus an [[Artinian ring]]). One can then say that if ''R'' is the radical, one has <math>(K \otimes_N L) / R</math> as a direct product of finitely many fields. Each such field is a representative of an equivalence class of (essentially distinct) field embeddings for ''K'' and ''L'' in some extension ''M''.
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| ==Examples==
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| For example, if ''K'' is generated over ℚ by the cube root of 2, then <math>K\otimes_{\mathbb Q}K</math> is the product of (a copy of) ''K'', and a [[splitting field]] of
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| :''X''<sup>3</sup> − 2,
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| of degree 6 over ℚ. One can prove this by calculating the dimension of the tensor product over ℚ as 9, and observing that the splitting field does contain two (indeed three) copies of ''K'', and is the compositum of two of them. That incidentally shows that ''R'' = {0} in this case.
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| An example leading to a non-zero nilpotent: let
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| :''P''(''X'') = ''X''<sup>''p''</sup> − ''T''
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| with K the field of [[rational function]]s in the indeterminate ''T'' over the finite field with ''p'' elements. (See [[separable polynomial]]: the point here is that ''P'' is ''not'' separable). If L is the field extension ''K''(''T''<sup>1/''p''</sup>) (the [[splitting field]] of ''P'') then ''L''/''K'' is an example of a [[purely inseparable field extension]]. In <math>L\otimes_KL</math> the element
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| :<math>T^{1/p}\otimes1-1\otimes T^{1/p}</math>
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| is nilpotent: by taking its ''p''th power one gets 0 by using ''K''-linearity.
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| ==Classical theory of real and complex embeddings==
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| In [[algebraic number theory]], tensor products of fields are (implicitly, often) a basic tool. If ''K'' is an extension of ℚ of finite degree ''n'', <math>K\otimes_{\mathbb Q}\mathbb R</math> is always a product of fields isomorphic to ℝ or ℂ. The [[totally real number field]]s are those for which only real fields occur: in general there are ''r''<sub>1</sub> real and ''r''<sub>2</sub> complex fields, with ''r''<sub>1</sub> + 2''r''<sub>2</sub> = ''n'' as one sees by counting dimensions. The field factors are in 1–1 correspondence with the ''real embeddings'', and ''pairs of complex conjugate embeddings'', described in the classical literature.
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| This idea applies also to <math>K\otimes_{\mathbb Q}\mathbb Q_p,</math> where ℚ<sub>''p''</sub> is the field of [[p-adic number|''p''-adic numbers]]. This is a product of finite extensions of ℚ<sub>''p''</sub>, in 1–1 correspondence with the completions of K for extensions of the ''p''-adic metric on ℚ.
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| ==Consequences for Galois theory==
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| This gives a general picture, and indeed a way of developing [[Galois theory]]
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| (along lines exploited in [[Grothendieck's Galois theory]]). It can be shown that for [[separable extension]]s the radical is always {0}; therefore the Galois theory case is the ''semisimple'' one, of products of fields alone.
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| ==See also==
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| *[[Extension of scalars]]—tensor product of a field extension and a vector space over that field
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| ==Notes==
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| {{Reflist}}
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| ==References==
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| *{{springer|id=C/c024310|title=Compositum of field extensions}}
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| *George Kempf (1995) ''Algebraic Structures'', pp. 85–87.
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| *[http://www.jmilne.org/math/CourseNotes/ANT.pdf ''Algebraic Number Theory'', J. S. Milne Notes (PDF)] at p. 17.
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| *[http://boxen.math.washington.edu/papers/ant/ant.pdf ''A Brief Introduction to Classical and Adelic Algebraic Number Theory'', William Stein (PDF)] pp. 140–142.
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| *{{Citation | last1=Zariski | first1=Oscar | author1-link=Oscar Zariski | last2=Samuel | first2=Pierre | author2-link=Pierre Samuel | title=Commutative algebra I | origyear=1958 | publisher=[[Springer-Verlag]] | location=Berlin, New York | series=Graduate Texts in Mathematics | isbn=978-0-387-90089-6 | mr=0090581 | year=1975 | volume=28}}
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| ==External links==
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| *[http://mathoverflow.net/questions/8324/what-does-linearly-disjoint-mean-for-abstract-field-extensions MathOverflow thread on the definition of linear disjointness]
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| [[Category:Field theory]]
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I'm Bobbie and was born on 22 July 1971. My hobbies are Stone collecting and Gardening.
Here is my page Belinda Broido