biology daily - the biology and biochemistry encyclopedia
biology daily articles and research Encyclopedia Dictionary Forums biology research links Weblinks Pictures Articles Blogs Newsletter

Tensor product of fields

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).

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\otimesQL 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\otimesNL 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\otimesNL. We can define a\otimesb.c\otimesd = ab\otimescd. 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\otimesNL 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\otimesNL into M defined by

γ(a\otimesb) = 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\otimesNL: 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\otimesNL/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\otimesQK 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\otimesKL the element T1/p\otimes1 - 1\otimesT1/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\otimesQR 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\otimesQQp, 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.



08-19-2006 15:59:36
The contents of this article are licensed from Wikipedia.org under the GNU Free Documentation License. How to see transparent copy
BiologyDaily.com 2005. Legal info