Properties of Cyclotomic Polynomials

Background in Basic Field Theory

Let be a field (in this post we mostly assume that ) and an integer which is not divisible by the characteristic of . Then the polynomial

is separable because its derivative is . Hence in the algebraic closure , the polynomial has distinct roots, which forms a group , and is cyclic. In fact, as an exercise, one can show that, for a field , any subgroup of the multiplicative group is a cyclic group.

The generator of is called the primitive -th root of unity. Let and be the smallest extension that contains all elements of , then we have . As a matter of fact, is a Galois extension (to be shown later), and the cyclotomic polynomial is the irreducible polynomial of over . We first need to find the degree .

Proposition 1. Notation being above, is Galois, the Galois group (the group of units in ) and .

Let's first elaborate the fact that . Let be representatives of . An element in is a unit if and only if there exists such that , which is to say, . Notice that if and only if for some , if and only if . Therefore is proved.

The proof can be produced by two lemmas, the first of which is independent to the characteristic of the field.

Lemma 1. Let be a field and be not divisible by the characteristic . Let be a primitive -th root of unity in , then and therefore . Besides, is a normal abelian extension.

Proof. Let be an embedding of in over , then

so that is also an -th root of unity also. Hence for some , uniquely determined modulo . It follows that maps into itself. This is to say, is normal over . Let be another automorphism of over then

It follows that and are prime to (otherwise, would have a period smaller than , implying that the period of is smaller than , which is absurd). Therefore for each , can be embedded into , thus proving our theorem.

It is easy to find an example with strict inclusion. One only needs to look at or .

Lemma 2. Let be a primitive -th root of polynomial over , then for any , is also a primitive -th root of unity.

Proof. Let be the irreducible polynomial of over , then by definition. As a result we can write where has leading coefficient . By Gauss's lemma, both and have integral coefficients.

Suppose is not a root of . Since , it follows that is a root of , and is a root of . As a result, divides and we write

Again by Gauss's lemma, has integral coefficients.

Next we reduce these equations in . We firstly have

By Fermat's little theorem for all , we also have

Therefore

which implies that . Hence and must have a common factor. As a result, has multiple roots in , which is impossible because of our choice of .

Now we are ready for Proposition 1.

Proof of Proposition 1. Since is a perfect field, is automatically separable. This extension is Galois because of lemma 1. By lemma 1, it suffices to show that .

Recall in elementary group theory, if is a finite cyclic group of order and is a generator of , then the set of generators consists elements of the form where . In this occasion, if generates , then also generates because . It follows that every primitive -th root of unity can be obtained by raising to a succession of prime numbers that do not divide (as a result we obtain exactly such primitive roots). By lemma 2, all these numbers are roots of in the proof of lemma 2. Therefore . Hence the proposition is proved.

We will show that in the proof lemma 2 is actually the cyclotomic polynomial you are looking for. The following procedure works for all fields where the characteristic does not divide , but we assume characteristic to be for simplicity.

We have

where the product is taken over all -th roots of unity. Collecting all roots with the same period (i.e., those such that ), we put

It follows that and

This presentation makes our computation much easier. But to understand , we still should keep in mind that the -th cyclotomic polynomial is defined to be

whose roots are all primitive -th roots of unity. As stated in the proof of proposition 1, there are primitive -th roots of unity, and therefore . Besides, . Since both have the same degree, these two polynomials equal. It also follows that .

Proposition 2. The cyclotomic polynomial is irreducible and is the irreducible polynomial of over , where is a primitive -th root of unity.

We end this section by a problem in number fields, making use of what we have studied above.

Problem 0. A number field only contains finitely many roots of unity.

Solution. Let be a root of unity with period . Then and therefore has degree . Since is also a subfield of , we also have . Since

is certainly a finite set, the number of roots of unity lie in is finite.

Technical Computations

We will do some dirty computation in this section.

Problem 1. If is prime, then , and for an integer , .

Solution. The only integer that divides is and we can only have

For the second statement, we use induction on . When we have nothing to prove. Suppose now

is proved, then and therefore

Problem 2. Let be a prime number. If , then

Solution. Assume first. It holds clearly for . Suppose now the statement holds for all integers that are prime to . We see

Problem 3. If is an odd number , then .

Solution. By problem 2, . To show the identity it suffices to show that

For we see

Now suppose it holds for all odd numbers , then

The following problem would not be very easy without the Möbius inversion formula so we will use it anyway. Problems above can also be deduced from this formula. Let be a function and , then the Möbius inversion formula states that

Putting , we see

Now we proceed.

Problem 4. If , then .

Solution. By the Möbius inversion formula, we see

because all that divides but not must be divisible by . Problem 2 can also follow from here.

Problem 5. Let , then

Solution. This problem can be solved by induction on the number of primes. For it is problem 1. Suppose it has been proved for primes, then for

and a prime , we have

On the other hand,

if we put . When it comes to higher degree of , it's merely problem 2. Therefore we have shown what we want.

Computing the Norm

Let be a primitive -th root of unity, put and the Galois group.. We will compute the norm of with respect to the extension . Since this extension is separable, we have

Since acts on the set of primitive roots transitively, is exactly the set of primitive roots of unity, which are roots of . It follows that

If , then . On the other hand, if