I'm trying to solve solve the next problem:
- Find all complex roots of the polynomial $f(x)=x^{101}-1$
- Decompose the polynomial $f(x)=\sum\limits_{n=0}^{100}x^n$ as product of irreducible polynomials over each one of the next fields: (i)$\mathbb{C}$, (ii)$\mathbb{R}$, (iii)$\mathbb{Q}$.
This is my attempt:
- The roots of $f(x)$ are the $101-$roots of unity, which are $x_k=e^\frac{2ki\pi}{101}$.
- Given that $101$ is prime, $f$ is the $101$th cyclotomic polynomial, so over $\mathbb{C}$ it factorize as $f(x)=(x-x_1)(x-x_2)\cdots(x-x_{101})$ when $x_k$'s are the $101-$roots of unity.
That's what i've got, but i don't know how to complete the solution.
You already have the factorization over $\mathbb C$, except you only have $100$ of the $101$th roots of unity (excluding $1$). Over $\mathbb Q$, $x^{100}+\cdots+x+1$ is irreducible because $101$ is prime. To get a factorization over $\mathbb R$, you can pair together roots of unity that are conjugates, and thus inverses, of each other. If $\zeta$ is a $101$th root of unit not equal to $1$, you get $(x-\zeta)(x-\zeta^{-1})=x^2-(\zeta+\zeta^{-1})x+1$. If $\zeta=e^{i\theta}$, then $\zeta+\zeta^{-1}=2\cos(\theta)$. Thus, you can factor $x^{100}+\cdots+x+1=\prod_{k=1}^{50} (x^2-2\cos(2\pi k/101)+1)$ over $\mathbb R$.