If $u$ is an $n$-th root of unity, $p$ is a prime number such that $p \nmid n$, then $P(u^p)=0$ where $P$ is the minimal polynomial of $u$

Question

This is a step in a guided proof that the cyclotomic polynomial $\Phi_n$ is the minimal polynomial of $u$. I already know that $\Phi_n(0)=0$ so $P$ divides $\Phi_n$, I need to show the converse. Any hints?

Answer 1

Here’s the elementary argument.

If $\Phi_n$ isn’t irreducible, it is a product of two monic polynomials $A$ and $B$ with integer coefficients and positive degree. Write $X^n-1=ABC$, then $X(X^n-1)’-n(X^n-1)=n$, so that $n \in (A,B)$.

We can assume that $A$ is irreducible and that it has a root $\omega$ and a prime $p$ not dividing $n$ such that $A(\omega^p) \neq 0$.

Then $\omega$ is a root of $B(X^p)$, so that $A|B(X^p)$. Write $a,b$ the reductions mod $p$ of $A$ and $B$. The above divisibility means that, because $p|B(X^p)-B(X)^p$, $a|b^p$. But $n \in (a,b)$, so that $a$ and $b$ are coprime. So $a=1$ and we get a contradiction.

Answer 2

Suppose $P$ has roots $\zeta_1,\zeta_2,...,\zeta_k$. All of them are $n$-th root of unity. Without the loss of generality, let $P$ be monic. Thus,

$$P(X)=(X-\zeta_1)(X-\zeta_2)...(X-\zeta_k)$$

$$P(X^p)=(X-\zeta_1^p)(X-\zeta_2^p)...(X-\zeta_k^p)$$

You can try to expand these two polynomials. Some knowledge in elementary symmetric polynomials and multinomial theorem will helpful.

When I learnt about cyclotomic polynomials, I proved they are minimal using Galois theory, so I don't really have an elementary proof like this in my mind. I found this proof here https://www.lehigh.edu/~shw2/c-poly/several_proofs.pdf.