# 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).

## 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$\displaystyle \otimes$ QL

as a vector space over Q. (This kind of result can be proved, 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$\displaystyle \otimes$ 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.

## The tensor product as ring

To get a general theory, we need to consider a ring structure on K$\displaystyle \otimes$ NL. We can define (a$\displaystyle \otimes$ b)$\displaystyle \star$ (c$\displaystyle \otimes$ d) = ab$\displaystyle \otimes$ 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 $\displaystyle \star$ in fact makes K$\displaystyle \otimes$ 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$\displaystyle \otimes$ NL into M defined by:

γ(a$\displaystyle \otimes$ b) = (α(a)$\displaystyle \otimes$ 1)$\displaystyle \star$ (1$\displaystyle \otimes$ β(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$\displaystyle \otimes$ NL: 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$\displaystyle \otimes$ 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. 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.

## Examples

For example, if K is generated over Q by the cube root of 2, then K$\displaystyle \otimes$ 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$\displaystyle \otimes$ KL the element

T1/p$\displaystyle \otimes$ 1 - 1$\displaystyle \otimes$ T1/p

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

## 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$\displaystyle \otimes$ 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$\displaystyle \otimes$ 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.