Articles of roots of unity

$p^a\mid f(v) \implies p^a\mid f(w)$ in $\mathbb Z$

Let $p$ be a prime, $n$ a positive integer and $q=p^n$ a $p$- power. Let $w,v$ is a root of the $q,p$-th cyclotomic polynomial $Q_q,Q_p$, respectively. Let $f$ be a polynomial with coefficient in $\mathbb Z$. Prove or disprove: “$p^a\mid f(v) \implies p^a\mid f(w)$” in $\mathbb Z[w]$. Edit: This question was an “iff” question, now […]

Finding the Roots of Unity

I have the following equation, $$z^5 = -16 + (16\sqrt 3)i$$I am asked to write down the 5th roots of unity and find all the roots for the above equation expressing each root in the form $re^{i\theta}$. I am just wondering if my solutions are correct. Here are my solutions, 5th roots of unity, $$z […]

Discriminant of $x^n-1$

Question is to find discriminant of polynomial $x^n-1$ I consider $f(x)=x^n-1=(x-a_1)(x-a_2)(x-a_3)\cdots(x-a_n)$ Now, $$f'(x)=[(x-a_2)(x-a_3)\cdots(x-a_n)]+\cdots+[(x-a_1)(x-a_2)\cdots(x-a_{n-1})]$$ $f'(a_1)=(a_1-a_2)(a_1-a_3)\cdots(a_1-a_n)$ $f'(a_2)=(a_2-a_1)(a_2-a_3)\cdots(a_2-a_n)$ $f'(a_3)=(a_3-a_1)(a_3-a_2)\cdots (a_3-a_n)$ and so on.. Now i need to know how many sign changes do i need to get something which looks like discriminant I would write this in a matrix form to get some idea… $$\begin{bmatrix}12&13&14&15&\cdots&1n\\21&23&24&25&\cdots&2n\\31&32&34&35&\cdots&3n\\\\n1&n2&n3&n4&\cdots&n(n-1) \end{bmatrix}$$ See that […]

Why do I get an imaginary result for the cube root of a negative number?

I have a function that includes the phrase $(-x)^{1/3}$. It seems like this should always evaluate to $-(x^{1/3})$. For example, $-1 \cdot -1 \cdot -1 = -1$, so it seems that $(-1)^{1/3}$ should equal $-1$. When I plug $(-1)^{(1/3)}$ into something like Mathematica, I get: 0.5 + 0.866025i Cubing this answer does in fact compute […]

Prove that $(x^2-x^3)(x^4-x) = \sqrt{5}$, where $x= \cos(2\pi/5)+i\sin(2\pi/5)$

Prove $(x^2-x^3)(x^4-x) = \sqrt{5}$ if $x= \cos(2\pi/5)+i\sin(2\pi/5)$. I have tried it by substituting $x = \exp(2i\pi/5)$ but it is getting complicated.

Radical extension

Let $K=\mathbb{Q}(\sqrt[n]a)$ where $a\in\mathbb{Q}$, $a>0$ and suppose $[K:\mathbb{Q}]=n$. Let $E$ be any subfield of $K$ and let $[E:\mathbb{Q}]=d.$ Prove that $E=\mathbb{Q}(\sqrt[d]a)$. It’s quite clear that $\mathbb{Q}(\sqrt[d]a)$ must have $x^d-a$ as its minimal polynomial in order to have degree $d$. And also $d|n$. We also have $x^n-a$ is irreduicible. How does it follow that $x^d-a$ must […]

How can we conclude $n\xi_j^{n-1}=\prod_{i \ne j}(\xi_{i}-\xi_{j})$ from $x^n-1=\prod_{i}(x-\xi_{i})$

I need to conclude $n\xi_j^{n-1}=\prod_{i \ne j}(\xi_{i}-\xi_{j})$ from $x^n-1=\prod_{i}(x-\xi_{i})$ where $\xi_i$ is the $i$th root of the unit. And another thing I do not quite understand is that why $\prod_{i=1}^n\xi_{i}^{-1}=(-1)^{n-1}$ where $\xi_i$ is the $i$th root of the unit. I think maybe it can be done by differentiation but how can I ceal with the […]

Simplifying this (perhaps) real expression containing roots of unity

Let $k\in\mathbb{N}$ be odd and $N\in\mathbb{N}$. You may assume that $N>k^2/4$ although I don’t think that is relevant. Let $\zeta:=\exp(2\pi i/k)$ and $\alpha_v:=\zeta^v+\zeta^{-v}+\zeta^{-1}$. As it comes from the trace of a positive matrix I know that the following is real: $$\sum_{v=1}^{\frac{k-1}{2}}\sec^2\left(\frac{2\pi v}{k}\right)\frac{(\overline{\alpha_v}^N+\alpha_v^N\zeta^{2N})(\zeta^{2v}-1)^2}{\zeta^N\zeta^{2v}}.$$ I am guessing, and numerical evidence suggests, that in fact $$\frac{(\overline{\alpha_v}^N+\alpha_v^N\zeta^{2N})(\zeta^{2v}-1)^2}{\zeta^N\zeta^{2v}}$$ is real […]

Problem based on sum of reciprocal of $n^{th}$ roots of unity

Let $1,x_{1},x_{2},x_{3},\ldots,x_{n-1}$ be the $\bf{n^{th}}$ roots of unity. Find: $$\frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+……+\frac{1}{1-x_{n-1}}$$ $\bf{My\; Try::}$ Given $x=(1)^{\frac{1}{n}}\Rightarrow x^n=1\Rightarrow x^n-1=0$ Now Put $\displaystyle y = \frac{1}{1-x}\Rightarrow 1-x=\frac{1}{y}\Rightarrow x=\frac{y-1}{y}$ Put that value into $\displaystyle x^n-1=0\;,$ We get $\displaystyle \left(\frac{y-1}{y}\right)^n-1=0$ So we get $(y-1)^n-y^n=0\Rightarrow \displaystyle \left\{y^n-\binom{n}{1}y^{n-1}+\binom{n}{2}y^2-……\right\}-y^n=0$ So $\displaystyle \binom{n}{1}y^{n-1}-\binom{n}{2}y^{n-1}+…+(-1)^n=0$ has roots $y=y_{1}\;,y_{2}\;,y_{3}\;,…….,y_{n-1}$ So $$y_{1}+y_{2}+y_{3}+…..+y_{n-1} = \binom{n}{2} =\frac{n(n-1)}{2}\;,$$ Where $\displaystyle y_{i} = […]

More on primes $p=u^2+27v^2$ and roots of unity

Given, $$p=u^2+27v^2=6m+1\tag1$$ and the cubic, $$x^3+x^2-2mx+N=0\tag2$$ with its constant expressed in terms of $(1)$ as, $$N = \frac{1}{27}(1-3p\pm2pu)\tag3$$ and the sign $\pm u$ chosen appropriately. (The discriminant is $D=-108p^2v^2$, so all real roots.) Starting with Ramanujan’s general cubic identity, they are a special case, yielding the simple, $$(a+b\,x_1)^{1/3}+(a+b\,x_2)^{1/3}+(a+b\,x_3)^{1/3}=\big(c+\sqrt[3]{dp}\big)^{1/3}\tag4$$ for some rational $a,b,c,d$. Question: Is it […]