Proof. Let

,

and

be as in the statement of the theorem. Assume that

. Thus, by the corollary above, there exists an unramified Galois extension
field 
of

of degree

. Notice that the fact that

does not divide the degree of the extension
implies that

. In particular, the
compositum 
is a Galois extension of

and
Thus, the extension

is of degree

, Galois, and therefore
abelian. By the corollary above, in
order to prove the theorem it suffices to show that the extension

is unramified. Suppose for a
contradiction that

is a
prime ideal which ramifies in the extension

. Let

be a prime lying above

and let

be a prime of

such that

lies above it. Similarly, let

be a prime of

lying above

and such that the prime

lies above

. For an arbitrary extension

, the ramification index of a prime

is denoted by

. Then, by the multiplicativity of the ramification index in towers, we have:
Since we assumed that

is ramified in

, and the degree of the extension is

, we must have

. Therefore, by the
equality above,

divides

. Notice that the extension

is everywhere unramified, therefore

. Also,
![$ [EF:E]=[F:K]$ $ [EF:E]=[F:K]$](http://images.planetmath.org:8080/cache/objects/6792/l2h/img60.png)
which, by
hypothesis, is
relatively prime to

. Thus

is also relatively prime to

, and so,

is not a
divisor of

, which leads to the desired contradiction, finishing the proof of the theorem.
