Tensor product of fields
From Academic Kids

In mathematics, the theory of fields in abstract algebra lacks a direct product: the direct product of two fields, considered as ring (mathematics) 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.
The tensor product of fields is the best available operation on fields with which to discuss the phenomena. As a ring, it is sometimes a field, and often a direct product of fields; it can though contain nonzero nilpotents (see radical of a ring).
Contents 
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 the square root of 2 to get K, and the square root of 3 to get L, it is true that the field M obtained as K.L inside the complex numbers C is K<math> \otimes <math>_{Q}L as a vector space over Q. This kind of result can be proved in general using the ramification theory of algebraic number theory. We say that subfields K and L of M are linearly disjoint (over a subfield N) when in this way the natural Nlinear map of K<math> \otimes <math>_{N}L 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.
The tensor product as ring
To get a general theory, we need to consider a ring structure on K<math> \otimes <math>_{N}L. We can define a<math> \otimes <math>b.c<math> \otimes <math>d = ab<math> \otimes <math>cd. 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 this in fact makes K<math> \otimes <math>_{N}L into a commutative Nalgebra. This is 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. 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<math> \otimes <math>_{N}L into M defined by
γ(a<math> \otimes <math>b) = α(a).β(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 Nalgebras 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<math> \otimes <math>_{N}L: there may in principle be a 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 Nalgebra (and thus an Artinian ring). We can then say that if R is the radical we have K<math> \otimes <math>_{N}L/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.
Examples
For example, if K is generated over Q by the cube root of 2, then K<math> \otimes <math>_{Q}K is the product of (a copy) of K, and a splitting field of X^{3}  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 nonzero nilpotent: let P(X) = X^{p}  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(T^{1/p}) (the splitting field of P) then L/K is an example of a purely inseparable field extension. In L<math> \otimes <math>_{K}L the element T^{1/p}<math> \otimes <math>1  1<math> \otimes <math>T^{1/p} is nilpotent: by taking its pth power one gets 0 by using Klinearity.
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<math> \otimes <math>_{Q}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 r real and s complex fields, with r + 2s = n as one sees by counting dimensions. The field factors are in 11 correspondence with the real embeddings, and pairs of complex conjugate embeddings, described in the classical literature.
This idea applies also to K<math> \otimes <math>_{Q}Q_{p}, where Q_{p} is the field of padic numbers. This is a product of finite extensions of Q_{p}, in 11 correspondence with the completions of K for extensions of the padic metric on Q.
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.