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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 $k=1$.

Servaes
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Stephen Montgomery-Smith
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  • I do not see how you conclude that "Hence $g_i(x)$ divides into $g_j(x^s)$.". – Servaes Jul 12 '19 at 21:36
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    @Servaes If $u$ is a root of $h(x) \in \mathbb Q[x]$, then the minimal polynomial of $u$ divides into $h(x)$. – Stephen Montgomery-Smith Jul 12 '19 at 21:38
  • Then this seems like a fine proof, assuming that you know $$1+x+\cdots+x^{p-1}=\prod_{i=1}^{p-1}(x-\zeta_p^i),$$ or something similar to ensure that $v=u^s$ for some positive natural number $s$. As for your question: I myself haven't seen such a proof before. – Servaes Jul 12 '19 at 21:41
  • For this proof we need results from Galois theory, and also of the sort of Gauss' lemma (factors live already over the integers, which is not far away from the usual proof involving the Eisenstein criterion). Maybe one has to consider all roots $u$ of the fixed $g_i$ polynomial, for each associate the $s(u)$ as above and then set $$s=\prod_{u}s(u)\ .$$ (For me, it is hard to distinguish now which part of the Galois theory of the cyclotomic field $\Bbb Q(\zeta_p)$ over $\Bbb Q$ should i forget, since all is based on the irreducibility of $\Phi_p$ in my mind.) – dan_fulea Jul 12 '19 at 22:10
  • @dan_fulea This proof definitely uses Gauss's Lemma. But I don't see where it uses any Galois Theory. I do agree that Eisenstein's proof is simpler. I don't understand why you take that product of the $s(u)$. – Stephen Montgomery-Smith Jul 12 '19 at 22:15
  • @Servaes The existence of $s$ follows because the primitive $p$th roots are of the form $\exp(2\pi i t/p)$ where $t$ is coprime to $p$. So you just need to show there is a positive natural number $s$ such that $s t_1 \equiv t_2 \pmod p$ for any $t_1, t_2$ coprime to $p$. – Stephen Montgomery-Smith Jul 12 '19 at 22:19
  • @reuns I must admit that my knowledge of $p$-adics is insufficient to understand your proof. – Stephen Montgomery-Smith Jul 12 '19 at 22:29
  • @Stephen Montgomery-Smith We are starting with one factor, $g_1$ say. It has many roots, $u,v,\dots,w$. Now let us take an other factor, if there is one other. (Else we are done.) Then we construct some $s$ so that $u, v,\dots, w$ are also roots of $g_2(x^s)$. (We use Galois theory if we say now it is enough to do this for only one root, $u$ say.) To have this $s$, we construct as in the OP the many powers $s(u)$, $s(v)$, ... , $s(w)$ and take $s$ to be their product. (All involved powers are prime to $p$, of course.) – dan_fulea Jul 12 '19 at 22:39

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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.

reuns
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  • I'm really struggling to understand this. What is $\overline{\mathbb Z}$? And with respect to which field is $g_i(x)$ irreducible? – Stephen Montgomery-Smith Jul 13 '19 at 21:41
  • I just rephrased your proof, $\overline{\Bbb{Z}}$ is the algebraic integers and the $g_j(x)$ are the monic $\Bbb{Q}$-minimal polynomials of their roots, their coefficients are algebraic integers thus integers – reuns Jul 13 '19 at 21:43
  • Can't you just use Gauss's lemma to show the coefficients of $g_j(x)$ are integers? – Stephen Montgomery-Smith Jul 13 '19 at 21:47
  • Sure you can I just prefer not to do so. Someone asked me if it works for $\Phi_n(x)$, I think the point is to understand what is $\Phi_n(1)$ – reuns Jul 13 '19 at 21:50
  • I can only make it work for $n$ a power of a prime. If $n$ is divisible by two distinct primes, $\Phi_n(1) = 1$. – Stephen Montgomery-Smith Jul 13 '19 at 21:54