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\title{MAT 250B HW10}
\author{[add your name here]}
\date{Due Friday, 3/15/24 at 11:59 pm on Gradescope}

\begin{document}

\maketitle


\begin{comment}
%%%%%%%%%%%%%%%%%%%%%%
\exercise
\bea

\item Let $\BF$ be a field of characteristic $p > 0$. Prove that $|\BF| = p^n$ for some $n \in \N$.

\item Prove that $\BF_{p^n}^\times$ is a cyclic group.
\ee
\end{comment}


%%%%%%%%%%%%%%%%%%%%%%
\exercise
Let $\sigma_p$ denote the Frobenius map $a \mapsto a^p$ on the finite field $\BF_{p^n}$. Verify that $\sigma_p$ is an automorphism of $\BF_{p^n}$, and that the order of $\sigma_p$ is $n$. 

%%%%%%%%%%%%%%%%%%%%%%
\exercise
Let $\mu_n \subset \C$ denote the set of $n$th roots of unity.
The $n$-th cyclotomic polynomial is 
\[
    \Phi_n(x) = \prod_{\text{primitive } \zeta \in \mu_n} (x-\zeta).
\]
\paragraph{Fact}
$\Phi_n(x)$ is an irreducible monic polynomial in $\Z[x]$. Hence $[\Q(\zeta_n): \Q] = \phi(n)$.

\textit{Unfortunately, we don't have time to talk about the proof of this in class. You can find the proof in various textbooks, and already have the tools to understand the proof.}

\paragraph{Observations}

\begin{enumerate}
    \item $\displaystyle
        x^n - 1 = \prod_{\zeta \in \mu_n} (x-\zeta) 
        = \prod_{d \divides n} \prod_{\text{primitive }\zeta \in \mu_d} (x-\zeta)
        = \prod_{d \divides n} \Phi_d(x)$
    \item $\deg \Phi_n(x) = \phi(n)$, where $\phi$ is \href{https://en.wikipedia.org/wiki/Euler%27s_totient_function}{Euler's totient function}.
\end{enumerate}

\paragraph{Over $\BF_p$}
Let $p$ be a prime. The splitting field of $x^n-1$ contains all the $n$-th roots of unity $\mu_n \subset \overline{\BF}_p$. The observations above still hold, since we are just taking the coefficients of polynomials mod $p$.

If $a \in \BF_{p^n}^\times$ and $|a|= m$, then we still have $\Phi_m(a) = 0$. But also, for all $d < m$, $\Phi_d(a) \neq 0$ since $a$ is not a $d$th root of $1 \in \BF_p$. So $\Phi_m(x) = m_{a,\BF_p}(x)$ still holds.





\bea

\item Determine $\Phi_p(x) \in \Z[x]$. Then, for $p$ prime, show that $\Phi_p(x) \equiv (x-1)^{p-1} \mod p$. \textit{This should be a fairly short explanation.} 

\item Prove that if $d \divides (p^n-1) = |\BF_{p^n}^\times|$, then $\Phi_d(x) \in \BF_p[x]$ has exactly $\phi(d)$ roots in $\BF_{p^n}^\times$.

\textit{Hint: These roots are precisely the primitive $d$th roots of unity over $\BF_p$. Use the fact that $|\BF_{p^n}^\times| = p^n -1 =  \sum_{d \divides p^n-1} \phi(d)$.}

\item
Prove that $n$ divides $\phi(p^n-1)$. 
\textit{Hint: Think about $\Aut(\BF_{p^n}^\times)$.}


\ee





%%%%%%%%%%%%%%%%%%%%%%
\exercise
Let $d,n \in \N$. 
\bea

\item Prove that $d \divides n$ if and only if $x^d -1$ divides $x^n-1$. 

\emph{Hint:} If $n= qd +r$, then $x^n-1 = (x^{qd +r} - x^r) + (x^r-1)$. 

\item Prove that for any $a \in \N$, 
\[
    d \divides n 
    \qquad
    \text{if and only if}
    \qquad
    a^d-1 \divides a^n-1.
\]

\item 
Conclude that $\BF_{p^d} \subseteq \BF_{p^n}$ if and only if $d \divides n$. 

\ee




\begin{comment}
%%%%%%%%%%%%%%%%%%%%%%
\exercise
Prove that one of 2, 3, or 6 is a square in $\BF_p$ for every prime $p$. 
Conclude that the polynomial 
\[
    f(x) = (x^2-2)(x^2-3)(x^2-6)
\]
has a root  mod $p$ for every prime $p$, but has no root in $\Z$.
\end{comment}


%%%%%%%%%%%%%%%%%%%%%%
\exercise
Compute the Galois groups of the following polynomials over the given fields.
\bea
\item $x^8-x$ over $\Q$
\item $x^8-x$ over $\F_2$
\item $x^4-1$ over $\F_7$
\ee

%%%%%%%%%%%%%%%%%%%%%%
\exercise
Let $p$ be a prime. Determine the elements of the Galois group of $x^p-2 \in \Q[x]$. 




\end{document}
