Tensor product of fields

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In mathematics, the theory of fields in abstract algebra lacks a direct product: the direct product of two fields, considered as a 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 subfields of a larger field M, or when K and L are both field extensions of a smaller field N (for example a prime field).

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).

If K and L do not have isomorphic prime fields, or in other words they have different characteristics, they have no possibility of being common subfields of a field M. Correspondingly their tensor product will in that case be the zero ring (collapse of the construction to nothing of interest).

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[edit] Compositum of fields

Firstly, in field theory, the compositum of subfields K and L of a field M is defined, without a problem, as the smallest subfield of M containing both K and L. It can be written

K.L.

In many cases we 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 we adjoin to the rational field Q √2 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)

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

as a vector space over Q. (This type of result can be verified, in general, by using the 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

K\otimes{}_NL

to K.L is injective. Naturally enough this isn't always the case, for example when K = L. When the degrees are finite, injective is equivalent here to bijective.

A significant case in the theory of cyclotomic fields is that for the n-th roots of unity, for n a composite number, the subfields generated by the pkth roots of unity for prime powers dividing n are linearly disjoint for distinct p.

[edit] The tensor product as ring

To get a general theory, we need to consider a ring structure on K\otimes{}_{\mathrm N}L. We can define (a\otimes b)\star(c\otimes d) = ac \otimes bd. This formula is multilinear over N in each variable; and so makes sense as a candidate for a ring structure on the tensor product. One can check that \star in fact makes K\otimes{}_{\mathrm N}L into a commutative N-algebra. This is the tensor product of fields.

[edit] 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. Note for this that the construction assumes the common subfield N; but does not assume a priori that K and L are subfields of some field M. Whenever we embed 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 K\otimes{}_{\mathrm N}L into M defined by:

\gamma(a\otimes b) = (\alpha(a)\otimes1)\star(1\otimes\beta(b)) = \alpha(a).\beta(b).

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.

In this way one can analyse the structure of K\otimes{}_{\mathrm N}L: there may in principle be a non-zero Jacobson radical (intersection of all prime ideals) - and after taking the quotient by that we 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). We can then say that if R is the radical we have (K\otimes{}_{\mathrm N}\mathrm L)/\mathrm R 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 of M. When K is a number field, this result can be combined with Dirichlet's unit theorem to yield the rank of the group of units of K.

[edit] Examples

For example, if K is generated over Q by the cube root of 2, then K\otimes{}_{\mathbb Q}K is the product of (a copy) of K, and a splitting field of

X3 − 2,

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.

An example leading to a non-zero nilpotent: let

P(X) = Xp − T

with K the field of rational functions 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(T1/p) (the splitting field of P) then L/K is an example of a purely inseparable field extension. In L\otimes{}_{\mathrm K}L the element

T^{1/p}\otimes1-1\otimes T^{1/p}

is nilpotent: by taking its pth power one gets 0 by using K-linearity.

[edit] 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 Q of finite degree n, K\otimes{}_{\mathbb Q}\mathbb R is always a product of fields isomorphic to R or C. The totally real number fields are those for which only real fields occur: in general there are r1 real and r2 complex fields, with r1 + 2r2 = 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 K\otimes{}_{\mathbb Q}\mathbb Q_p, where Qp is the field of p-adic numbers. This is a product of finite extensions of Qp, in 1-1 correspondence with the completions of K for extensions of the p-adic metric on Q.

[edit] 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 extensions the radical is always {0}; therefore the Galois theory case is the semisimple one, of products of fields alone.

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