0

I have some confusion regarding Dummit and foote Book Algebra, page number $554$

Theorem $41$.The cyclotomic polynomial $\Phi_n(x)$ is an irreducible monic polynomialin $\mathbb{Z}[x]$ of degree $\varphi(n)$

Proof : we must show that $\Phi_n(x)$ is irreducible. If not then we have a factorization $$\Phi_n(x) = f(x)g(x)$$ with $f(x),g(x)$ monic polynomials in $\mathbb{Z}[x]$ where we take $f(x)$ to be an irreducible factor of $\Phi_n(x)$. let $\zeta$ to be a primitive $n^{th}$ root of $1$ which is a root of $f(x)$ (so then $f(x)$ is the minimal polynomial for $\zeta$ over $\mathbb{Q}$) and let $p$ denote any prime not dividing $n$ .Then $\zeta^p$ is again a primitive $n^{th}$ root of $1$, hence is a root of either $f(x)$ or $g(x)$.

My confusion : Im not getting why $\zeta^p$ is again a primitive $n^{th}$ root of $1?$

My thinking : It is given that $\zeta$ is a primitive $n^{th}$ root of $1$

That means only the power $n $ will give $1$ and the remaining power (not equal to $n$) will not give unity

So i think $\zeta^p \neq 1$ because $p \neq n$

i,e $\zeta^p$ is not again a primitive $n^{th}$ root of $1$

jasmine
  • 14,457
  • 2
    If $\zeta$ is a primitive $n$-th root of $1$ then it's multiplicative order is $n$ and hence $\text {ord} (\zeta^p) = \frac {n} {\text {gcd} (p,n)} = n$ (since $p$ doesn't divide $n$). – Anil Bagchi. Jul 31 '21 at 20:19
  • 1
    You can observe that $(\zeta^p) ^n=(\zeta^n) ^p=1$ so $\zeta^p$ is an $n$th root of unity. It is primitive because $p\nmid n$ implies that $\zeta ^p=\zeta^k$ where $k$ is remainder when $p$ is divided by $n$ and $k$ is coprime to $n$. – Paramanand Singh Aug 01 '21 at 09:22
  • 1
    You may have a look at this answer : https://math.stackexchange.com/a/532977/72031 – Paramanand Singh Aug 01 '21 at 09:24

0 Answers0