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!
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. .
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,
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 .