Let
be a primitive
-th root of unity, and let
. I am going to outline a proof that
, based on several homework problems from one of my recent courses. There are probably many other proofs of this, but I particularly like this one because it’s easy to follow, touches on a wide range of topics, and I worked hard through it!
First, let
be a prime power. The minimal polynomial of
is the
-th cyclotomic polynomial
.
Let
.
Exercise 1. Following the above notation, show that
satisfies the conditions of Eisenstein’s criterion for the prime
.
Consider the discriminant
, where
. If
is a prime factor of
, then
must have a multiple root modulo
. Hence
will also have a multiple root modulo
. But
in
unless
. Thus
is a power of
. In particular,
is a power of
.
Exercise 2. Suppose that
satisfies the conditions of Eisenstein’s criterion for a prime number
. Let
be a root of
and let
. Prove that there is exactly one prime ideal
that contains
, and that the local ring
is a DVR with uniformiser
.
Now if
is another prime ideal, then
. So
.
Exercise 3. Let
be a number field and
a subring of finite index
. If
is a prime ideal not containing
, show that
is a DVR.
I’ll include this solution because I love it!
Solution. Let
. By going up, there is a prime ideal
with
. We’ll show that
, so the result will follow.
Let
, so that
,
. If
, then
, a contradiction. Hence
, i.e.
. Thus
.
Now let
, so that
,
. Then
,
, and
. Note that
, and
by assumption. So
since
is a prime ideal. Thus
, i.e.
. 
So
localised at any prime ideal is a DVR, implying that
is a Dedekind domain. Thus
. This completes the proof for the prime power case.
Now we proceed by induction on the number of distinct prime factors of
. The base case has already been taken care of. So suppose that
, where
are coprime integers. Let
and
. By the induction hypothesis,
and
.
Exercise 4. Let
and
be number fields with discriminants
and
respectively. Let
and
be integral bases for
and
respectively. Let
be the composite field of
and
. Suppose that
and that
. Show that
is an integral basis for
.
So our main result will follow from exercise 4 once we have checked that all the hypotheses are satisfied.
(i) Checking
. Firstly,
contains
.
Exercise 5. If
, show that
.
So writing
shows that
and
. If
is the multiplicative inverse of
, then
. Thus
.
Again, since
and
, we have
. Thus
.
(ii) Checking
. We have
.
(iii) Checking
. This is slightly harder. We need a few more facts.
Exercise 6. Let
be a number field, and let
be the minimal polynomial of
. Show that
.
Solution. Let
. Then
.
Using Leibniz’s rule,
.
Hence
. 
Exercise 7. Using the previous exercise, show that
.
Hint. Write
, and use Leibniz’s rule to get
.
So
divides some power of
, and
divides some power of
. Since
, we conclude that
.