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Inmathematics, thetensor product of twofields is theirtensor product asalgebras over a commonsubfield. If no subfield is explicitly specified, the two fields must have the samecharacteristic and the common subfield is theirprime subfield.

The tensor product of two fields is sometimes a field, and often adirect product of fields; In some cases, it can contain non-zeronilpotent elements.

The tensor product of two fields expresses in a single structure the different way to embed the two fields in a commonextension field.

First, one defines the notion of the compositum of fields. This construction occurs frequently infield 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 atower of fields. Letk be a field andL andK be two extensions ofk. The compositum, denotedK.L, is defined to be${\displaystyle K.L=k(K\cup L)}$ where the right-hand side denotes the extension generated byK andL. This assumessome field containing bothK andL. Either one starts in a situation where an ambient field is easy to identify (for example ifK andL are both subfields of thecomplex numbers), or one proves a result that allows one to place bothK andL (asisomorphic copies) in some large enough field.

In many cases one can identifyK.L as avector spacetensor product, taken over the fieldN that is the intersection ofK andL. For example, if one adjoins √2 to therational field${\displaystyle \mathbb{Q}}$ to getK, and √3 to getL, it is true that the fieldM obtained asK.L inside the complex numbers${\displaystyle \mathbb{C}}$ is (up to isomorphism)

${\displaystyle K{\otimes }_{\mathbb{Q}}L}$

as a vector space over${\displaystyle \mathbb{Q}}$. (This type of result can be verified, in general, by using theramification theory ofalgebraic number theory.)

SubfieldsK andL ofM arelinearly disjoint (over a subfieldN) when in this way the naturalN-linear map of

${\displaystyle K{\otimes }_{N}L}$

toK.L isinjective.^[1] Naturally enough this isn't always the case, for example whenK =L. When the degrees are finite, injectivity is equivalent here tobijectivity. Hence, whenK andL are linearly disjoint finite-degree extension fields overN,${\displaystyle K.L\cong K{\otimes }_{N}L}$, as with the aforementioned extensions of the rationals.

A significant case in the theory ofcyclotomic fields is that for thenthroots of unity, forn acomposite number, the subfields generated by thep^k th roots of unity forprime powers dividingn are linearly disjoint for distinctp.^[2]

To get a general theory, one needs to consider aring structure on${\displaystyle K{\otimes }_{N}L}$. One can define the product${\displaystyle (a\otimes b)(c\otimes d)}$ to be${\displaystyle ac\otimes bd}$ (seeTensor product of algebras). This formula is multilinear overN in each variable; and so defines a ring structure on the tensor product, making${\displaystyle K{\otimes }_{N}L}$ into acommutativeN-algebra, called thetensor product of fields.

The structure of the ring can be analysed by considering all ways of embedding bothK andL in some field extension ofN. The construction here assumes the common subfieldN; but does not assumea priori thatK andL are subfields of some fieldM (thus getting round the caveats about constructing a compositum field). Whenever one embedsK andL in such a fieldM, say using embeddings α ofK and β ofL, there results aring homomorphism γ from${\displaystyle K{\otimes }_{N}L}$ intoM defined by:

${\displaystyle \gamma (a\otimes b)=(\alpha (a)\otimes 1)\star (1\otimes \beta (b))=\alpha (a).\beta (b).}$

Thekernel of γ will be aprime ideal of the tensor product; andconversely any prime ideal of the tensor product will give a homomorphism ofN-algebras to anintegral domain (inside afield of fractions) and so provides embeddings ofK andL in some field as extensions of (a copy of)N.

In this way one can analyse the structure of${\displaystyle K{\otimes }_{N}L}$: there may in principle be a non-zeronilradical (intersection of all prime ideals) – and after taking the quotient by that one can speak of the product of all embeddings ofK andL in variousM,overN.

In caseK andL are finite extensions ofN, the situation is particularly simple since the tensor product is of finite dimension as anN-algebra (and thus anArtinian ring). One can then say that ifR is the radical, one has${\displaystyle (K{\otimes }_{N}L)/R}$ as a direct product of finitely many fields. Each such field is a representative of anequivalence class of (essentially distinct) field embeddings forK andL in some extensionM.

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

${\displaystyle K\otimes L=K{\otimes }_{N}L=\mathbb{Q}[x]/(x^2-2){\otimes }_{\mathbb{Q}}\mathbb{Q}[y]/(y^2-2)\cong \mathbb{Q}(\sqrt{2})[z]/(z^2-2)\cong \mathbb{Q}(\sqrt{2}){\otimes }_{\mathbb{Q}}\mathbb{Q}(\sqrt{2})}$

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

${\displaystyle K\otimes L\cong \mathbb{Q}(\sqrt{2})[z]/(z^2-2)\cong \mathbb{Q}(\sqrt{2})\oplus \mathbb{Q}(\sqrt{2})}$

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

${\displaystyle K{\otimes }_{\tilde{N}}L=\mathbb{Q}[x]/(x^2-2){\otimes }_{\mathbb{Q}(\sqrt{2})}\mathbb{Q}[y]/(y^2-2)\cong \mathbb{Q}(\sqrt{2})=\tilde{N}\cong K\cong L}$.

For another example, ifK is generated over${\displaystyle \mathbb{Q}}$ by thecube root of 2, then${\displaystyle K{\otimes }_{\mathbb{Q}}K}$ is the sum of (a copy of)K, and asplitting field of

X³ − 2,

of degree 6 over${\displaystyle \mathbb{Q}}$. One can prove this by calculating the dimension of the tensor product over${\displaystyle \mathbb{Q}}$ as 9, and observing that the splitting field does contain two (indeed three) copies ofK, and is the compositum of two of them. That incidentally shows thatR = {0} in this case.

An example leading to a non-zero nilpotent: let

P(X) =X^p −T

withK the field ofrational functions in the indeterminateT over thefinite field withp elements (seeSeparable polynomial: the point here is thatP isnot separable). IfL is the field extensionK(T^1/p) (thesplitting field ofP) thenL/K is an example of apurely inseparable field extension. In${\displaystyle L{\otimes }_{K}L}$ the element

${\displaystyle T^{1/p}\otimes 1-1\otimes T^{1/p}}$

is nilpotent: by taking itspth power one gets 0 by usingK-linearity.

Inalgebraic number theory, tensor products of fields are (implicitly, often) a basic tool. IfK is an extension of${\displaystyle \mathbb{Q}}$ of finite degreen,${\displaystyle K{\otimes }_{\mathbb{Q}}\mathbb{R}}$ is always a product of fields isomorphic to${\displaystyle \mathbb{R}}$ or${\displaystyle \mathbb{C}}$. Thetotally real number fields are those for which onlyreal fields occur: in general there arer₁ real andr₂ complex fields, withr₁ + 2r₂ =n as one sees by counting dimensions. The field factors are in 1–1 correspondence with thereal embeddings, andpairs of complex conjugate embeddings, described in the classical literature.

This idea applies also to${\displaystyle K{\otimes }_{\mathbb{Q}}{\mathbb{Q}}_p,}$ where${\displaystyle \mathbb{Q}}$_p is the field ofp-adic numbers. This is a product of finite extensions of${\displaystyle \mathbb{Q}}$_p, in 1–1 correspondence with the completions ofK for extensions of thep-adic metric on${\displaystyle \mathbb{Q}}$.

This gives a general picture, and indeed a way of developingGalois theory(along lines exploited inGrothendieck's Galois theory). It can be shown that forseparable extensions the radical is always {0}; therefore the Galois theory case is thesemisimple one, of products of fields alone.

• Extension of scalars—tensor product of a field extension and a vector space over that field

• "Compositum",Encyclopedia of Mathematics,EMS Press, 2001 [1994]
• Kempf, George R. (2012) [1995]."9.2 Decomposition of Tensor Products of Fields".Algebraic Structures. Springer. pp. 85–87.ISBN 978-3-322-80278-1.
• Milne, J.S. (18 March 2017).Algebraic Number Theory(PDF). p. 17. 3.07.
• Stein, William (2004)."A Brief Introduction to Classical and Adelic Algebraic Number Theory"(PDF). pp. 140–2.
• Zariski, Oscar;Samuel, Pierre (1975) [1958].Commutative algebra I. Graduate Texts in Mathematics. Vol. 28.Springer-Verlag.ISBN 978-0-387-90089-6.MR 0090581.

• MathOverflow thread on the definition of linear disjointness

• Field (mathematics)

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