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{{Short description|Ring produced from two fields}}{{Other uses|Tensor product (disambiguation)}}{{distinguish|Tensor field}}
{{textbook|date=August 2021}}

In [[mathematics]], the '''tensor product''' of two [[field (mathematics)|fields]]  is their [[tensor product of algebras|tensor product]] as [[algebra over a field|algebras]] over a common [[subfield (mathematics)|subfield]]. If no subfield is explicitly specified, the two fields must have the same [[characteristic (algebra)|characteristic]] and the common subfield is their [[prime field|prime subfield]].

The tensor product of two fields is sometimes a field, and often a [[product ring|direct product]] of fields; In some cases, it can contain non-zero [[nilpotent element]]s.

The tensor product of two fields expresses in a single structure the different way to embed the two fields in a common [[extension field]].

==Compositum of fields==

First, 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 ''K.L'', is defined to be <math> K.L = k(K \cup L) </math> where the right-hand side denotes the extension generated by ''K'' and ''L''. This assumes ''some'' field containing both ''K'' and ''L''. Either one starts in a situation where an ambient field is easy to identify (for example if ''K'' and ''L'' are both subfields of the [[complex number]]s), or one proves a result that allows one to place both ''K'' and ''L'' (as [[isomorphic]] copies) in some large enough field.

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 numbers|rational field]] <math>\mathbb{Q}</math> to get ''K'', and √3 to get ''L'', it is true that the field ''M'' obtained as ''K''.''L'' inside the complex numbers <math>\mathbb{C}</math> is ([[up to]] isomorphism)

:<math>K\otimes_{\mathbb Q}L</math>

as a vector space over <math>\mathbb{Q}</math>. (This type of result can be verified, in general, by using the [[ramification (mathematics)|ramification]] theory of [[algebraic number theory]].)

Subfields ''K'' and ''L'' of ''M'' are [[linearly disjoint]] (over a subfield ''N'') when in this way the natural ''N''-linear map of

:<math>K\otimes_NL</math>

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, injectivity is equivalent here to [[bijectivity]]. Hence, when ''K'' and ''L'' are linearly disjoint finite-degree extension fields over ''N'', <math>K.L \cong K \otimes_N L</math>, as with the aforementioned extensions of the rationals.

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>&hairsp;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>

==The tensor product as ring==

To get a general theory, one needs to consider a [[ring (mathematics)|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> (see [[Tensor product of algebras]]). 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 (structure)|commutative ''N''-algebra]], called the '''tensor product of fields'''.

==Analysis of the ring structure==

The structure of the ring can be analysed by considering all ways of embedding both ''K'' and ''L'' in some field extension of ''N''. 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:

:<math>\gamma(a\otimes b) = (\alpha(a)\otimes1)\star(1\otimes\beta(b)) = \alpha(a).\beta(b).</math>

The [[kernel (algebra)|kernel]] of γ will be a [[prime ideal]] of the tensor product; and [[converse (logic)|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''.

In this way one can analyse the structure of <math>K \otimes_N L</math>: there may in principle be a non-zero [[Nilradical of a ring|nilradical]] (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''.

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''.

==Examples==

To give an explicit example consider the fields <math>K=\mathbb{Q}[x]/(x^2-2)</math> and <math>L=\mathbb{Q}[y]/(y^2-2)</math>. Clearly <math>\mathbb{Q}(\sqrt{2})\cong K\cong L\neq K</math> are isomorphic but technically unequal fields with their (set theoretic) intersection being the prime field <math>N=\mathbb{Q}</math>. Their tensor product

: <math>K\otimes L=K\otimes_{N} L=\mathbb{Q}[x]/(x^2-2)\otimes_{\mathbb{Q}}\mathbb{Q}[y]/(y^2-2) \cong \mathbb{Q}(\sqrt{2})[z]/(z^2-2)\cong \mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{2})</math> 

is not a field, but a 4-dimensional <math>\mathbb{Q}</math>-algebra. Furthermore this algebra is isomorphic to a direct sum of fields

: <math>K\otimes L\cong \mathbb{Q}(\sqrt{2})[z]/(z^2-2)\cong \mathbb{Q}(\sqrt{2})\oplus\mathbb{Q}(\sqrt{2})</math> 

via the map induced by <math>1\mapsto (1,1), z\mapsto (\sqrt{2},-\sqrt{2})</math>.
Morally <math>\tilde{N}=\mathbb{Q}(\sqrt{2})</math> should be considered the ''largest common subfield up to isomorphism'' of ''K'' and ''L'' via the isomorphisms <math>\tilde{N}=\mathbb{Q}(\sqrt{2})\cong K\cong L</math>. When one performs the tensor product over this better candidate for the largest common subfield we actually get a (rather trivial) field

: <math>K\otimes_{\tilde{N}} L=\mathbb{Q}[x]/(x^2-2)\otimes_{\mathbb{Q}(\sqrt{2})}\mathbb{Q}[y]/(y^2-2) \cong \mathbb{Q}(\sqrt{2})=\tilde{N}\cong K\cong L</math>.

For another example, if ''K'' is generated over <math>\mathbb{Q}</math> by the [[cube root]] of 2, then <math>K \otimes_{\mathbb Q} K</math> is the sum of (a copy of) ''K'', and a [[splitting field]] of

:''X''<sup>&hairsp;&hairsp;3</sup> − 2,

of degree 6 over <math>\mathbb{Q}</math>. One can prove this by calculating the dimension of the tensor product over <math>\mathbb{Q}</math> 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.

An example leading to a non-zero nilpotent: let

:''P''(''X'') = ''X''<sup>&hairsp;&hairsp;''p''</sup> − ''T''

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>&thinsp;1/''p''</sup>) (the [[splitting field]] of ''P'') then ''L''/''K'' is an example of a [[purely inseparable field extension]]. In <math>L \otimes_K L</math> the element

:<math>T^{1/p}\otimes1-1\otimes T^{1/p}</math>

is nilpotent: by taking its ''p''th power one gets 0 by using ''K''-linearity.

==Classical theory of real and complex embeddings==

In [[algebraic number theory]], tensor products of fields are (implicitly, often) a basic tool. If ''K'' is an extension of <math>\mathbb{Q}</math> of finite degree ''n'', <math>K\otimes_{\mathbb Q}\mathbb R</math> is always a product of fields isomorphic to <math>\mathbb{R}</math> or <math>\mathbb{C}</math>. The [[totally real number field]]s are those for which only [[real number|real]] fields occur: in general there are ''r''<sub>1</sub> real and ''r''<sub>2</sub> complex fields, with ''r''<sub>1</sub>&thinsp;+&nbsp;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.

This idea applies also to <math>K\otimes_{\mathbb Q}\mathbb Q_p,</math> where <math>\mathbb{Q}</math><sub>''p''</sub> is the field of [[p-adic number|''p''-adic numbers]]. This is a product of finite extensions of <math>\mathbb{Q}</math><sub>''p''</sub>, in 1–1 correspondence with the completions of ''K'' for extensions of the ''p''-adic metric on <math>\mathbb{Q}</math>.

==Consequences for Galois theory==

This gives a general picture, and indeed a way of developing [[Galois theory]]
(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.

==See also==
*[[Extension of scalars]]—tensor product of a field extension and a vector space over that field

==Notes==
{{Reflist}}

==References==
*{{SpringerEOM|title=Compositum}}
*{{cite book |first=George R. |last=Kempf |title=Algebraic Structures |chapter=9.2 Decomposition of Tensor Products of Fields |chapter-url=https://books.google.com/books?id=B3T0BwAAQBAJ&pg=PA85 |date=2012 |publisher=Springer |isbn=978-3-322-80278-1 |pages=85–87 |orig-year=1995}}
*{{cite book |url=http://www.jmilne.org/math/CourseNotes/ANT.pdf |title=Algebraic Number Theory |first=J.S. |last=Milne |page=17 |id=3.07 |date=18 March 2017 }}
*{{cite web |title=A Brief Introduction to Classical and Adelic Algebraic Number Theory |first=William |last=Stein |year=2004 |pages=140–2 |url=http://abel.math.harvard.edu/archive/129_spring_04/ant/ant.pdf }}
*{{Cite book | last1=Zariski | first1=Oscar | author1-link=Oscar Zariski | last2=Samuel | first2=Pierre | author2-link=Pierre Samuel | title=Commutative algebra I | orig-year=1958 | publisher=[[Springer-Verlag]]  | series=Graduate Texts in Mathematics | isbn=978-0-387-90089-6   | mr=0090581 | year=1975 | volume=28}}

==External links==
*[http://mathoverflow.net/questions/8324/what-does-linearly-disjoint-mean-for-abstract-field-extensions MathOverflow thread on the definition of linear disjointness]

[[Category:Field (mathematics)]]