i'm confused with some notation that involves reduction of polynomyals on $\mathbb{Z}[x]$ to $\mathbb{F}_p[x]$. It's part of the proof that Cyclotomic polynomials are irreducible over $\mathbb{Q}[x]$.

I have the polynomial $f(x)=x^n-1$ that I assuming that is factorized in $f=gh$, with $g,h \in \mathbb{Z}[x]$ monic polynomials and such that $g$ is the minimal pol. of $\zeta_{n}$. Let $p$ be an integer such that $gcd(n,p)=1$, and we assume that $\zeta_{n}^p$ is a root of $h$, so $h(x^p)$ has $\zeta_{n}$ as root and then $h=fz$, where by the Gauss Lema $z$ has intenger coefieficients.

Here is where I getting problem.

We reduce all the polynomial $\mathbb{Z}[x]$ to $\mathbb{F}_{p}[x]$ and the by the Little Teo. of Fermat:

$\bar{h(x^p)}=\bar{g}\bar{z}$ and then $\bar{h(\zeta_n)}=\bar{0}$.

So the text affirm that $x^n-\bar{1}$ has multiple roots and did exctaly this:

"let $\alpha \in \mathbb{Z}$ be the multiple root, then $\alpha^n=1$ and $n\alpha^{n-1}=0$ so $n=0$ and since p doest not divide n this is absurd."

This made me very confuse because we are still in $\mathbb{F}_p[x]$ since the arguments of the absurd depends on the field characteristic.

In my way of view the sencence should be:

" So $\zeta_n$ is a multiple root of $x^n-\bar{1}$ and then $\bar{\zeta_n}^n=\bar{1}$ and then $\bar{n}\bar{\zeta_n}^{n-1}=\bar{0}$. Since $p$ does not divide $n$ this is an absurd!"

My doubts are:

Is there a convention on how to write ordinary polynomials in $\mathbb{F}_p[x]$?

Why did the text take $\alpha \in \mathbb{Z}$ since we just prove that the multiple root is $\zeta_n$?

  • $\begingroup$ What is mdc? I assume it is $\gcd$ (greatest common divisor), but I wonder what language this is from.. $\endgroup$ – Hagen von Eitzen Nov 21 '18 at 21:19
  • $\begingroup$ you right, thats was the usual notation on portuguese, I've edited. $\endgroup$ – Eduardo Silva Nov 21 '18 at 21:22
  • 1
    $\begingroup$ Error: when $h(\zeta_n^p) = 0$, so $\zeta_n$ is a root of $h(x^p)$, this means $h(x^p)$ is divisible in $\mathbf Z[x]$ by $g(x)$, not $f(x) = x^n - 1$. For example, $i$ has minimal polynomial $x^2 + 1$ and $i$ is a root of $x^4 + 3x^2 + 2$, but this does not mean $x^4 + 3x^2 + 2$ is divisible by $x^4 - 1$; it is divisible by $x^2 + 1$. You can find a treatment of this proof in Theorem 2.5 of math.uconn.edu/~kconrad/blurbs/galoistheory/cyclotomic.pdf. $\endgroup$ – KCd Nov 21 '18 at 21:56
  • $\begingroup$ Oh, sorry, that should be $\bar{h(x^p)}=\bar{g}\bar{z}$ instead $\endgroup$ – Eduardo Silva Nov 22 '18 at 2:02

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