Tutorial 6 : Cyclotomic Extensions


1. Determine $ [\mathbb{Q}(\zeta _7,\zeta _3):\mathbb{Q}(\zeta _3)].$



2. Determine a primitive element of a subfield $ K$ of $ E=\mathbb{Q}(\zeta _{13})$ so that $ [K:\mathbb{Q}]=3.$




3. Put $ \zeta =\zeta _7.$ Determine the degrees of $ \zeta +\zeta ^{5}$ and $ \zeta +\zeta ^5+\zeta ^8$ over $ \mathbb{Q}.$




4. Put $ \zeta =\zeta _{11}$ and $ \alpha=\zeta +\zeta ^3+\zeta ^4+\zeta ^5 +\zeta ^9.$ Show that $ [\mathbb{Q}(\alpha):\mathbb{Q}]=2.$



5. Let $ \nu$ be a primitive element modulo $ p$ where $ p$ is a prime. Thus $ \mathbb{F}_p^{\times}=(\nu).$ Let $ \zeta=\zeta _p.$ Using the list $ \{ \zeta ^{\nu^0}, \zeta ^{\nu^1}, \zeta ^{\nu^2}, \ldots, \zeta ^{\nu^{p-2}}\},$ show how to find the sum $ \beta$ of powers of $ \zeta$ which determines a subfield $ \mathbb{Q}(\beta)$ of $ \mathbb{Q}(\zeta)$ so that $ [\mathbb{Q}(\beta):\mathbb{Q}]=d$ where $ d \,\vert\, (p-1).$



6. Suppose $ A \in \mathbb{C}^{n \times n}$ and $ A^k=I$ for some integer $ k \in \mathbb{N}.$ Show that $ A$ can be diagonalized. Prove that the matrix $ A= \left[\begin{array}{ll}
1 & \alpha \\ 0 & 1 \end{array}\right]$ where $ \alpha \in K$ and $ K$ is a field of chracteristic $ p$ satisfies $ A^p=I$ and cannot be diagonalized if $ \alpha \neq 0.$



7. Show that $ n=\sum_{d\vert n} \phi(d)$ and deduce that $ \phi(n)=\sum_{d\vert n} \mu(n/d)d.$



8. Show that % latex2html id marker 35501
$ \Phi_n(x)=\prod_{d\mid n}(x^{n/d}-1)^{\mu(d)}$.



9. Show that $ \Phi_n(x)=x^{\phi(n)}\Phi_n(1/x)$ and deduce that the coefficients of $ \Phi_n(x)$ satisfy $ a_k=a_{\phi(n)-k}$ for all $ 0 \leq k \leq \phi(n).$



10. Let $ p$ be a prime number and $ (p,n)=1.$ Show that
$\displaystyle \Phi_p(x)$ $\displaystyle =$ $\displaystyle x^{p-1}+x^{p-2}+\cdots+x^2+x+1$  
$\displaystyle \Phi_{p^r}(x)$ $\displaystyle =$ $\displaystyle \Phi_p(x^{p^{r-1}})$  
$\displaystyle \Phi_{2n}(x)$ $\displaystyle =$ $\displaystyle \Phi_n(-x),$    if n is odd   
$\displaystyle \Phi_{np}(x)$ $\displaystyle =$ $\displaystyle \Phi_n(x^p)/\Phi_n(x)$  
$\displaystyle \Phi_{p_1^{r_1} p_2^{r_2}\ldots p_m^{r_m}}(x)$ $\displaystyle =$ $\displaystyle \Phi_{p_1\ldots p_m}
(x^{p_1^{r_1-1}p_2^{r_2-1}\ldots p_m^{r_m-1}}).$