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\pagestyle{fancy} \lhead{{\sf Homework 8 $|$ Due March 17 (Friday)
}} \rhead{\thepage} \cfoot{{\sf Math 4130 $|$ Abstract Algebra $|$
    Spring 2023 $|$ M.~Macauley}}

\begin{document}

$\;$ \vspace{-6mm}

\begin{enumerate}



  
\item The splitting field of the polynomial $f(x)=x^4-2$ is
  \[
  K=\Q(\sqrt[4]{2},i)=\big\{a+b\sqrt[4]{2}+c\sqrt{2}+d\sqrt[4]{8}
  +ei+f\sqrt[4]{2}i+g\sqrt{2}i+h\sqrt[4]{8}i\mid a,\dots,h\in\Q\big\},
  \]
  which has degree $[\Q(\sqrt[4]{2},i):\Q]=8$ over $\Q$. 
  \begin{enumerate}
  \item There are eight non-trivial proper subfields of $\Q(\sqrt[4]{2},i)$:
    \begin{itemize}
    \item Five of degree $4$ over the rationals: $\Q(\sqrt[4]{2})$,
      $\Q(i\sqrt[4]{2})$, $\Q(\sqrt{2},i)$, $\Q((1+i)\sqrt[4]{2})$,
      and $\Q((1-i)\sqrt[4]{2})$.
    \item Three of degree $2$ over the rationals: $\Q(\sqrt{2})$,
      $\Q(i)$, and $\Q(i\sqrt{2})$.
    \end{itemize}
    For each subfield $F$, find a polynomial $f(x)$ of minimal degree that has a root $r$ such that $F=\Q(r)$. Then determine how many of the roots of $f(x)$ lie in $K$.
  \item Arrange the 10 subfields in a lattice, with $\Q(\sqrt[4]{2},i)$ on top, and $\Q$ on the bottom. 
\item The field $\Q(\sqrt[4]{2},i)$ can be generated by just $\sqrt[4]{2}+i$, which is called a \emph{primitive element}. For each of the ten subfields $F\subseteq \Q(\sqrt[4]{2},i)=\Q(\sqrt[4]{2}+i)$, find a primitive element $\alpha$ over $F$, and its minimal polynomial. The degree of the polynomial will be equal to the degree of the extension $[F:\Q]$. \\
  \end{enumerate}

  
\item Consider the polynomial $f(x)=x^5-2$.
  \begin{enumerate}
  \item Show that the splitting field of $f(x)=x^5-2$ is $K=\Q(\sqrt[5]{2},\zeta)$, where $\zeta=e^{2\pi i/5}$ is a primitive $5^{\rm th}$ root of unity.
  \item In the complex plane, sketch all five roots of $f(x)$, and all five fifth roots of unity. 
  \item Find the subfields of $K$, and arrange them in a subfield lattice, whose structure is shown below.
    \end{enumerate}
  \[
  \begin{tikzpicture}[shorten >= -0pt, shorten <= -0pt,scale=1.45]
  \tikzstyle{every node}=[font=\normalsize]
  \tikzstyle{f} = [faded]
    \begin{scope}[shift={(0,0)},scale=1.5,xscale=1.3]
      \node (Fr20) at (0,-4) {$\Q$};
      \node(s2-t) at (-.15,-2.9) {\color{white}$\<s^2,\!t\>$};
      \node (t) at (-1.4,-2.3) {\color{white}$\<t\>$};
      \node (s) at (-.7,-2) {\color{white}$\<s\>$};
      \node (st) at (.25,-2) {\color{white}$\<st\>$};
      \node (s3t) at (1.15,-2) {\color{white}$\<s^3t\>$};
      \node (ts) at (1.8,-2) {\color{white}$\<ts\>$};
      \node (ts3) at (2.5,-2) {\color{white}$\<ts^3\>$};
      \node(s2) at (-.45,-1.1) {\color{white}$\<s^2\>$}; 
      \node(ts2) at (.25,-1.1) {\color{white}$\<ts^2\>$}; 
      \node(sts) at (1.15,-1.1) {\color{white}$\<sts\>$}; 
      \node(s2t2) at (1.8,-1.1) {\color{white}$\<s^2t^2\>$};
      \node(s2t) at (2.5,-1.1) {\color{white}$\<s^2t\>$};
      \node (1) at (0,0) {$\Q(\sqrt[5]{2},\zeta)$};
      \draw[bend left=5,f] (Fr20) to (s); 
      \draw[f] (Fr20) to (st); 
      \draw[f] (Fr20) to (s3t);
      \draw[f] (Fr20) to (ts); 
      \draw[f] (Fr20) to (ts3); 
      \draw[f] (Fr20) to (s2-t);
      %%
      \draw[f](s2-t) to (t); \draw[f](t) to (1);
      \draw[f](s2-t) to (s2); \draw[f](s2-t) to (ts2);
      \draw[f](s2-t) to (sts);
      \draw[f](s2-t) to (s2t2); 
      \draw[bend right=10,f] (s2-t) to (s2t);
      %%
      \draw[faded] (s) to (s2); \draw[faded] (st) to (ts2);
      \draw[faded] (s3t) to (sts); \draw[faded] (ts) to (s2t2);
      \draw[faded] (ts3) to (s2t); 
      %%
      \draw[f](1) to (s2); \draw[f](1) to (ts2); \draw[f](1) to (sts);
      \draw[f](1) to (s2t2); \draw[f](1) to (s2t);
  \end{scope}
\end{tikzpicture}
\]
 
\item Consider the function 
  \[
  \phi\colon\Q(\sqrt{2})\longto\Q(\sqrt{2})\,,\qquad
  \phi(a+b\sqrt{2})=a-b\sqrt{2}\,.
  \]
  Show that $\phi$ is a field automorphism. \\
  


\item Let $\alpha$ be a root of a polynomial $f(x)$ that irreducible
  over $\Q$, which means that it generates a maximal ideal in $\Q[x]$.
  \begin{enumerate}
  \item Show that
    \[
    \phi\colon\Q[x]\longto\Q(\alpha),\qquad\phi\colon f(x)\longmapsto f(\alpha)
    \]
    is a ring homomorphism.
  \item Show that $\Q[x]/(f(x))\cong\Q(a)$.
  \item Show that if $\beta$ is another root of $f(x)$, then there is a field isomorphism $\Q(\alpha)\to\Q(\beta)$ that fixes $\Q$, elementwise, and sends $\alpha\mapsto\beta$. 
  \end{enumerate}
  
\end{enumerate}

\end{document}
