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 .
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
20120924