# Notation and interpretation of Polynomials in $\mathbb{F}_{p}[x]$

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 $$\zeta_n$$?

• What is mdc? I assume it is $\gcd$ (greatest common divisor), but I wonder what language this is from.. – Hagen von Eitzen Nov 21 '18 at 21:19
• you right, thats was the usual notation on portuguese, I've edited. – Eduardo Silva Nov 21 '18 at 21:22
• 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. – KCd Nov 21 '18 at 21:56
• Oh, sorry, that should be $\bar{h(x^p)}=\bar{g}\bar{z}$ instead – Eduardo Silva Nov 22 '18 at 2:02