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 non-zero nilpotents (see radical of a ring).


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>QL 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 N-linear map of K<math> \otimes <math>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.

The tensor product as ring

To get a general theory, we need to consider a ring structure on K<math> \otimes <math>NL. 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>NL into a commutative N-algebra. 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>NL 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 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<math> \otimes <math>NL: 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 N-algebra (and thus an Artinian ring). We can then say that if R is the radical we have K<math> \otimes <math>NL/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.


For example, if K is generated over Q by the cube root of 2, then K<math> \otimes <math>QK 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<math> \otimes <math>KL the element T1/p<math> \otimes <math>1 - 1<math> \otimes <math>T1/p is nilpotent: by taking its pth power one gets 0 by using K-linearity.

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>QR 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 1-1 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>QQp, 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.

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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