# Tensor Products

We need only a special case of the tensor product construction. Let  and  be commutative rings containing a field  and suppose that  is of finite dimension  over , say, with basis

Then  is determined up to isomorphism as a ring over  by the multiplication table defined by

We define a new ring  containing  whose elements are the set of all expressions

where the have the same multiplication rule

as the .

There are injective ring homomorphisms

(note that $w_1=1$)

and

Moreover  is defined, up to isomorphism, by  and  and is independent of the particular choice of basis of  (i.e., a change of basis of induces a canonical isomorphism of the defined by the first basis to the defined by the second basis). We write

since  is, in fact, a special case of the ring tensor product.

Let us now suppose, further, that  is a topological ring, i.e., has a topology with respect to which addition and multiplication are continuous. Then the map

defines a bijection between  and the product of  copies of  (considered as sets). We give  the product topology. It is readily verified that this topology is independent of the choice of basis and that multiplication and addition on  are continuous, so  is a topological ring. We call this topology on  the tensor product topology.

Now drop our assumption that  and  have a topology, but suppose that  and  are not merely rings but fields. Recall that a finite extension of fields is separable if the number of embeddings that fix  equals the degree of  over , where is an algebraic closure of . The primitive element theorem from Galois theory asserts that any such extension is generated by a single element, i.e., for some .

Lemma 16.2.1   Let  and  be fields containing the field  and suppose that  is a separable extension of finite degree . Then is the direct sum of a finite number of fields , each containing an isomorphic image of  and an isomorphic image of .

Proof. By the primitive element theorem, we have , where  is a root of some separable irreducible polynomial of degree . Then is a basis for  over , so

where are linearly independent over  and satisfies .

Although the polynomial is irreducible as an element of , it need not be irreducible in . Since  is a field, we have a factorization

where is irreducible. The are distinct because is separable (i.e., has distinct roots in any algebraic closure).

For each , let be a root of , where is a fixed algebraic closure of the field . Let . Then the map

 (16.1)

given by sending any polynomial in (where ) to is a ring homomorphism, because the image of  satisfies the polynomial , and .

By the Chinese Remainder Theorem, the maps from (16.2.1) combine to define a ring isomorphism

Each is of the form , so contains an isomorphic image of . It thus remains to show that the ring homomorphisms

are injections. Since and are both fields, is either the 0 map or injective. However, is not the 0 map since .

Example 16.2.2   If and are finite extensions of , then is an algebra of degree . For example, suppose is generated by a root of and is generated by a root of . We can view as either or . The polynomial is irreducible over , and if it factored over the cubic field , then there would be a root of in , i.e., the quadratic field would be a subfield of the cubic field , which is impossible. Thus is irreducible over , so is a degree extension of . Notice that contains a copy  and a copy of . By the primitive element theorem the composite field can be generated by the root of a single polynomial. For example, the minimal polynomial of is , hence .

Example 16.2.3   The case is even more exciting. For example, suppose . Using the Chinese Remainder Theorem we have that

since and are coprime. The last isomorphism sends , with , to . Since has zero divisors, the tensor product must also have zero divisors. For example, and is a zero divisor pair on the right hand side, and we can trace back to the elements of the tensor product that they define. First, by solving the system

and

we see that corresponds to and , i.e., to the element

This element in turn corresponds to

Similarly the other element corresponds to

As a double check, observe that

Clearing the denominator of and writing , we have , so is a root of the polynomimal , and is not , so has more than roots.

In general, to understand explicitly is the same as factoring either the defining polynomial of  over the field , or factoring the defining polynomial of  over .

Corollary 16.2.4   Let be any element and let be the characteristic polynomials of over and let (for ) be the characteristic polynomials of the images of  under over , respectively. Then

 (16.2)

Proof. We show that both sides of (16.2.2) are the characteristic polynomial of the image of in over . That follows at once by computing the characteristic polynomial in terms of a basis of , where is a basis for over (this is because the matrix of left multiplication by on is exactly the same as the matrix of left multiplication on , so the characteristic polynomial doesn't change). To see that , compute the action of the image of  in with respect to a basis of

 (16.3)

composed of basis of the individual extensions of . The resulting matrix will be a block direct sum of submatrices, each of whose characteristic polynomials is one of the . Taking the product gives the claimed identity (16.2.2).

Corollary 16.2.5   For we have

and

Proof. This follows from Corollary 16.2.4. First, the norm is the constant term of the characteristic polynomial, and the constant term of the product of polynomials is the product of the constant terms (and one sees that the sign matches up correctly). Second, the trace is minus the second coefficient of the characteristic polynomial, and second coefficients add when one multiplies polynomials:

One could also see both the statements by considering a matrix of left multiplication by first with respect to the basis of and second with respect to the basis coming from the left side of (16.2.3).

William Stein 2012-09-24