Irreducibility of prime degree cyclotomic polynomials I came across this proof that the cyclotomic polynomials of prime degree are irreducible over the rationals.  I was wondering if anyone has come across this particular proof before.
Let $p$ be prime, and $\phi(x) = 1 + x + \cdots + x^{p-1}$ be the cyclotomic polynomial of degree $p$.  Let $\phi(x) = g_1(x) g_2(x) \cdots g_k(x)$ be its decomposition into irreducible factors.  Without loss of generality, we may suppose that the factors have integer coefficients.
We first show that $|g_i(1)| = |g_j(1)|$ for $1 \le i,j \le k$.  Let $u$ be  root of $g_i(x)$, and $v$ be a root of $g_j(x)$.  Then $v = u^s$ for some positive natural number $s$.  Hence $u$ is a root of $g_j(x^s)$.  Hence $g_i(x)$ divides into $g_j(x^s)$.  Hence $g_i(1)$ divides into $g_j(1)$.  Similarly $g_j(1)$ divides into $g_i(1)$.
So $p = \phi(1) = g_1(1) g_2(1) \cdots g_k(1) = |g_1(1)|^k$.  Therefore $|g_1(1)| = p$ and $k=1$.
 A: I think your proof deserve to be known. It is exactly the same as the one we use to show irreducibility of $\Phi_{p^m}(x)$ over $\Bbb{Q}_p$, showing that $v_p(\zeta_{p^m}^a-1)$ doesn't depend on $a$ implies $v_p(\zeta_{p^m}-1) = 1/(p^m-p^{m-1})$ so that we need all the roots to obtain a polynomial with integer valuations, except that your proof doesn't even need $p$-adics or valuations.

Assumption : idea of minimal polynomials and that $\overline{\Bbb{Z}}$ is a ring
Let $$\Phi_{p^m}(x) =\sum_{l=0}^{p-1} (x^{p^{m-1}})^l =\prod_{a\bmod p^m,\ p\, \nmid\, a} (x-\zeta_{p^m}^a) = \prod_{j=1}^k g_j(x)$$
where  $g_j(x) = \prod_{r=1}^{R_j} (x-\zeta_{p^m}^{a_{j,r}}) \in \Bbb{Q}[x]$  is the minimal polynomial of $\zeta_{p^m}^{a_{j,1}}$. Since the latter is an algebraic integer the coefficients of $g_j$ are algebraic integers ie. $g_j(x) \in \overline{\Bbb{Z}}[x]_{monic} \cap \Bbb{Q}[x] = \Bbb{Z}[x]_{monic}$.
$g_1(x)$ is the minimal polynomial of $\zeta_{p^m}$.
$\zeta_{p^m}$ is a root of $g_j(x^{a_{j,1}})$ thus $g_1(x)$ divides $g_j(x^{a_{j,1}})$ (divides in $\Bbb{Z}[x]_{monic}$ from the same minimal-polynomial-of-algebraic-integer argument as before) and $g_1(1)\, |\, g_j(1)$.
$\zeta_{p^m}^{a_{j,1}}$ is a root of $g_1(x^{s_j})$ where $s_j$ is an inverse of $a_{j,1}\bmod p^m$ thus $g_j(x)$ divides $g_1(x^{s_j})$ and $g_j(1)\, |\, g_1(1)$. 
In other words $|g_j(1)| = |g_1(1)|$ and $$p=|\Phi_{p^m}(1)| = \prod_{j=1}^k |g_j(1)|= |g_1(1)|^k$$ implies $|g_1(1)|=p,k=1$ and $\Phi_{p^m}(x)$ is irreducible.
