2
votes
1answer
55 views

Why are two statements about a polynomial equivalent?

I am reading a claim that the following two statements are equivalent. One of the roots of a polynomial $v(t)$ is a $2^j$-th root of unity, for some $j$. The polynomial $v(t)$ is divisible either by ...
4
votes
2answers
54 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 ...
9
votes
1answer
148 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
92 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
56 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
33 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
37 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
40 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
453 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 $\alpha \in \mathrm{F}_q$?
9
votes
4answers
974 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
126 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
240 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
39 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
189 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
157 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
236 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. ...