proof every cyclotomic polynomial is irreducible

Question

I would like some help on a proof.

I’m pretty sure my professor on the last lesson of his course got a little confused and tried to prove that the polynomial $\xi_n = \sum_{i=0}^{n-1}x^i=\frac{x^n-1}{x-1}$ was irreducible (in general), which is obviously not. It has happened a few times during the course so I wouldn’t be surprised.

I believe that’s because initially he said “when $n$ is prime $\xi_n$ is the n-th cyclotomic polynomial” so I suppose he meant to prove that every cyclotomic polynomial is irreducible over $\mathbb Q$: what I’m trying to do now is to try follow roughly what he said during the proof, which is similar to the classical one but really messy, and adapt it to what I think he meant to prove. I know it would be easier to look up another proof, and I did, but I’m not sure how would he take it… i surely will prove it another way if I don’t get any improvement. The attempted adaptation of the proof is as follows.

Thesis: every cyclotomic polynomial $\Phi_n$ is irreducible over $\mathbb Q$.

Suppose instead they weren’t, say $\Phi_n$ is reducible over $\mathbb Q$, therefore by Gauss’s lemma it is reducible over $\mathbb Z$ i.e. there exists two polynomials f and g over $\mathbb Z$ of positive degree such that $\Phi_n$ = fg. wlog f is irreducible.

Now, the roots of $\Phi_n$ are the primitive n-th roots of unity by definition, so take a $z $ root of f: if we show that any $z^p$ with p prime not dividing n can’t be a root of g, inductively we can prove any root of $\Phi_n$ is a root of f, so they have the same degree and g has 0-degree, which is a contradiction and the statement is proven.

What my professor did (and I saw it also elsewhere but in a different way) is reducing all the polynomials mod p i.e. considering all the polynomials with integer coefficients as coefficients mod p, so he’s considering all the polynomials over $\mathbb Z_p = \mathbb Z /p \mathbb Z$.

Now he proceeded proving the following things and I don’t even know how they’re related neither to each other (ok maybe a bit) nor to concluding the main claim, (remember though he wanted to show that “$\Phi_n=\xi_n$” is irreducible):

1. Monic polynomials over $\mathbb Z$ have the same degree over $\mathbb Z_p$ (it’s trivial), and thus $\Phi_n$ = fg over $\mathbb Z_p$ still is a non trivial factorization, i.e. $\Phi_n$ is still reducible.

2. If $z$ is a root of f mod p (over an adequate splitting extension of $\mathbb Z_p$), then $z^p$ is a root of f mod p as well and can’t be a root of g mod p: since $\Phi_n$ divides $x^n - 1$, it also divides it mod p (they’re monic) and since $D(x^n - 1)=nx^{n-1}$ and p doesn’t divide n all of its roots are singular, thus if $z$ or $z^p$ are roots or f they can’t be roots of g; $z^p$ is a root of f because, if $a_i$ are its coefficients, $a_i=a_i^p \mod p$ so $f(z^p) = \sum_{i=0}^m a_i(z^p)^i= \sum_{i=0}^m a_i^p (z^i)^p = \big( \sum_{i=0}^m a_iz^i \big )^p= (f(z))^p =0$

3. The function $ a \rightarrow a^p $ in the group of n-th roots of unity over $\mathbb Z_p$ is bijective it is injective since p doesn’t divide n and the group is finite

4. If $z^{p^u} = z^{p^v} $(say u>v), then $n $divides $p^u - p^v= p^v(p^{u-v}-1)$, since p doesn’t divide n we get that $n$ divides $p^{u-v}-1$ and therefore u-v>n (???).

Now this should have led to the fact that all the (distinct by (4) ) elements $z, z^p, z^{p^2},…, z^{p^{n-1}}$ are roots of f mod p and thus mod p f has degree n-1 and g has degree 0, which is false because g should have kept the same degree when reduced mod p “$\square$“.

I have no idea what that means, but it’s a pretty idea (?).

So what I’m trying to adapt in the right proof of “any $z^p$ with p prime not dividing n can’t be a root of g” is the use of the reduction mod p:

First of all, the reduced polynomial $\Phi_n$ has as roots all primitive the n-th roots of unity over $\mathbb Z_p$, which is proven by induction: if n=1 it’s obvious, and supposed true for any d<n $x^n-1= \prod^n_{d \vert n} \Phi_d$ since as we’ve seen all the roots of $x^n-1$ are singular mod p the ones of order $d$ can only be the roots of $\Phi_d$.

So taken a primitive root $z$ of f over $\mathbb Z_p$, $z^p$ is also one of its roots and can’t be a root of g, as in (2).

Now the group of roots of unity over $\mathbb C$ and the ones over $\mathbb Z_p$ are isomorphic (take a primitive root $z$ over the complex and $\alpha$ over $\mathbb Z_p$ and let $z^k \rightarrow \alpha ^k$ ), and i would like to show that if $z$ is a complex primitive root of f and $z^p$ were a root of g, then I’d get a contradiction because i can “inject this information” over $\mathbb Z_p$ where $\alpha $ is a root of f and $\alpha^p$ of g, which is absurd.

I can’t formalize this injection and I would like some help <3 or clarifications. thanks in advance.