4
votes
2answers
43 views

Irreducibility of $X^n-a$

Let ${\mathbb K}$ be a subfield of ${\mathbb C}$. Let $a\in{\mathbb K}$ such that $X^d-a$ has no root in ${\mathbb K}$, for any divisor $d>1$ of $n$. Does it follow that $X^n-a$ is irreducible ...
8
votes
1answer
118 views

Cyclotomic polynomial - coefficient

For a polynomial $f=X^n+a_1X^{n-1}+\ldots+a_n \in \mathbb{Q}[X]$ we define $\varphi(f):=a_1 \in \mathbb{Q}$. Now I want to show that for the $n$th cyclotomic polynomial $\Phi_n$ it holds that ...
3
votes
3answers
70 views

Roots of unity modulo $p$

Let $f(X)$ be the minimal polynomial of something like $\zeta + \frac{1}{\zeta}$, where $\zeta$ is a primitive $p$-th root of unity for some prime $p > 2$. I'd like to show that $f(X) \equiv ...
2
votes
0answers
41 views

Properties of cyclotomic polynomial

Assume first that $p$ a prime divides $n$. I have to show that $\Phi_{np}(X)=\Phi_n(X^p)$. Here is what I tried: Suppose $\eta_i$ are roots of $\Phi_{np}(X)$ so $\eta_i=\text{exp}(\frac{2\pi i ...
1
vote
1answer
30 views

Proving that $|\Phi_n(x)| > x-1$

Let $\Phi_n$ be the n-th cyclotomic polynomial. I'd like to prove that $$\forall n \geq 2, \forall x \in [2, \infty[, |\Phi_n(x)| > x-1$$ The result is clear when $n$ is prime, but I'm struggling ...
0
votes
1answer
36 views

Does the succesion of two radical extensions yield a radical extension only in the obvious case?

Answering that recent stackoverflow question, I encountered the following related problem : Let $n,m,p\geq 2$ be integers, and let $K$ be a subfield of $\mathbb C$ containing all $nmp$-th roots of ...
1
vote
0answers
38 views

Divisors and cyclotomic polynomials

Let $n \in \mathbb{N}^{\ast}$ and $\Phi_{n}(X)$ be the $n$-th cyclotomic polynomial defined by : $$ \Phi_{n}(X) = \prod \limits_{\substack{1 \leq k \leq n-1 \\ \gcd(k,n)=1}} \Big( X - \exp \big( ...
4
votes
2answers
303 views

When is a cyclotomic polynomial over a finite field a minimal polynomial?

When is the cyclotomic polynomial $f(x)$ over a finite field $\mathrm{F}_q$ also the minimal polynomial of some element $\alpha \in \mathrm{F}_q$?
8
votes
4answers
656 views

showing that $n$th cyclotomic polynomial $\Phi_n(x)$ is irreducible over $\mathbb{Q}$

I studied the cyclotomic extension using Fraleigh's text. To prove that Galois group of the $n$th cyclotomic extension has order $\phi(n)$( $\phi$ is the Euler's phi function.), the writer assumed, ...
2
votes
0answers
96 views

Cyclic linear codes and idempotents

Got this assignment from coding class and would be very thankful for checking if my solutions are correct. a) Find all idempotents modulo $1 + x^{17}$ of degree at most $15$ So first i find $r$ from ...
11
votes
0answers
217 views

Is $\bigl(X(X-a)(X-b)\bigr)^{2^n} +1$ an irreducible polynomial over $\mathbb{Q}[X]$?

Let $a, b \in \mathbb{Q}$, with $a\neq b$ and $ab\neq 0$, and $n$ a positive integer. Is the polynomial $\bigl(X(X-a)(X-b)\bigr)^{2^n} +1$ irreducible over $\mathbb{Q}[X]$? I know that ...
0
votes
1answer
38 views

Formula for the $nm$th cyclotomic polynomial when $(n,m) = 1$

Let $n,m$ be coprime. I want to find a formulae for $\Phi_{n\cdot m, \mathbb Q}$. I conjecture that because $$d \mid nm \implies d \mid n \lor d \mid m,$$ that $$ \Phi_{n\cdot m, \mathbb Q} = ...
5
votes
1answer
174 views

Cyclotomic Polynomials and GCD

Since Cyclotomic polynomials are irreducible over $\mathbb{Q}$, $\phi_n(x)$, $\phi_m(x)$ are coprime as polynomials in $\mathbb{Z}[x]$. Working over $\mathbb{Q}$, $(\phi_n(x)$, $\phi_m(x))=(1)$. ...
3
votes
3answers
149 views

Cyclotomic Polynomial of a Prime

I have this question on a homework sheet: Claim:$$\Phi_{p}(x)=1+x+x^2+...+x^{p-1}\space$$ for $p$ prime. which was followed by the claim that $\Phi_{p^n}(x)=\Phi_p(x^{p^{n-1}})$ which I have ...
5
votes
2answers
211 views

Given a prime $p\in\mathbb{N}$, is $A=\frac{x^{p^{2}}-1}{x^{p}-1}$ irreducible in $\mathbb{Q}[x]$?

If $p \in \mathbb{N}$ is a prime, is $\displaystyle A=\frac{x^{p^{2}}-1}{x^{p}-1}$ irreducible in $\mathbb{Q}[x]$? I don't think it is. If somebody sees a contradiction, I would be glad to see it. ...