Question

Explain that with details thanks

(3) Let ơ (1, 2, ,n) E S,, be the cycle of length n. Let C, be the n x n matrix over an algebraically closed field k corresponding to σ, so Co (e) et+1 for i 1,..,n -1 and Ca(en)-e1. Show that and hence that C, is diagonalizible, similar to a diagonal matrix Dơ with diagonal entries 1,f, ξ2..-5n-1, where ξ is a primitive n-th root of unity (so ξ e2m/n if k C the field of complex numbers.)

Topic: bilinear map and Tensor product

Answer 1

a) Let $f(x)=x^2+bx+c$ ; then

$\begin{align*}f_2(x)&=f(x^2+bx+c)\\ &=(x^2+bx+c)^2+b(x^2+bx+c)+c\end{align*}$

shows that $\begin{align*}\deg f_2(x)&=2^2\end{align*}$ . We claim that $\begin{align*}\deg f_n(x)&=2^n\end{align*}$ .

Proof of claim: We already have checked the formula for $\begin{align*}n=1,2\end{align*}$ . Suppose $\begin{align*}n\geq 2\end{align*}$ and $\begin{align*}\deg f_n(x)&=2^n\end{align*}$ . Since $\begin{align*}f_{n+1}(x)=f(f_n(x))\end{align*}$ , we have

$\begin{align*}\deg f_{n+1}(x)&=\deg f(f_n(x))\\ &=2\deg f_n(x)\\ &=2^{n+1}\end{align*}$

By induction, we have $\begin{align*}\deg f_n(x)&=2^n\end{align*}$ for all $\begin{align*}n\geq 1\end{align*}$ .

Let $\begin{align*}\alpha_1,\cdots,\alpha_{2^n}\in F^a\end{align*}$ be all the roots of $\begin{align*}f_n(x)\in F[x]\end{align*}$ . Then $\begin{align*}K_n=F(\alpha_1,\cdots,\alpha_{2^n})\end{align*}$ . Thus, to show that $\begin{align*}K_n\subseteq K_{n+1}\end{align*}$ it suffices to show that $\begin{align*}\alpha_i\in K_{n+1}\end{align*}$ for all $\begin{align*}1\leq i\leq 2^n\end{align*}$ . Let $\begin{align*}\gamma\in F^a\end{align*}$ be a zero of the quadratic $\begin{align*}K_n[x]\end{align*}$ -polynomial $\begin{align*}p(x):=f(x)-\alpha_i\end{align*}$ ; then $\begin{align*}f(\gamma)=\alpha_i\end{align*}$ , so that

$\begin{align*}f_{n+1}(\gamma)&=f_n(f(\gamma))\\ &=f_n(\alpha_i)\\ &=0\end{align*}$

Thus, $\begin{align*}\gamma\in K_{n+1}\end{align*}$ . Now, quadratic formula implies

$\begin{align*}\gamma\in \begin{Bmatrix}{\frac{-b+\sqrt{b^2-4(c-\alpha_i)}}2}, {\frac{-b-\sqrt{b^2-4(c-\alpha_i)}}2}\end{Bmatrix}\end{align*}$

Therefore,

$\begin{align*}\alpha_i=c+b\gamma+\gamma^2\end{align*}$

which makes $\begin{align*}\alpha_i\in K_{n+1}\end{align*}$ .

b) Let $\begin{align*}\sigma: K_{n+1}\rightarrow F^a\end{align*}$ be any embedding over $\begin{align*}K_n\end{align*}$ . If $\begin{align*}\gamma\in K_{n+1}\end{align*}$ is a zero of $\begin{align*}f_{n+1}(x)\end{align*}$ then

$\begin{align*}f_{n+1}(\sigma (\gamma))&=\sigma(f_{n+1})(\gamma)\\ &=f_{n+1}(\gamma)\\ &=0\end{align*}$

Thus, $\begin{align*}\sigma: K_{n+1}\rightarrow F^a\end{align*}$ satisfies $\begin{align*}\sigma (K_{n+1})\subseteq K_{n+1}\end{align*}$ , showing that $\begin{align*}\sigma\in\mbox{Gal}(K_{n+1}/K_n)\end{align*}$ . Since $\begin{align*}|\mbox{Gal}(K_{n+1}/K_n)|=[K_{n+1}:K_n]\leq 2\end{align*}$ , we conclude that the order of $\begin{align*}\sigma\in\mbox{Gal}(K_{n+1}/K_n)\end{align*}$ is $\begin{align*}1\end{align*}$ or $\begin{align*}2\end{align*}$ .