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\noindent
\parbox{1.95in}{\bf Math 4010}
\hfill {\Large \bf Homework \#6} \hfill
\parbox{1.95in}{\bf \hfill Due: Wed., Oct.\! 22${}^\mathrm{\bf nd}$\!, 2025}

\vspace{0.3in}

\noindent
\fbox{\bf Important Note:} In problems \#2 and \#3, you are asked to ``find'' Galois groups.\\ \force\hfill I will clarify what I mean by this (and help with these problems) in class.

\vspace{0.3in}


%\noindent {\bf Name:} \  \underline{\qquad My Name Goes Here \qquad}

\noindent
{\sc Background Theorems:} Let $\varphi:\mathbb{F} \to \widehat{\mathbb{F}}$ be an isomorphism of fields. This induces an isomorphism $\widetilde{\varphi}:\mathbb{F}[x] \to \widehat{\mathbb{F}}[x]$ of the corresponding polynomial rings (extending $\varphi$ and sending $x$ to $x$). Let $f(x) \in \mathbb{F}[x]$ and $\widetilde{\varphi}(f(x))=\widehat{f}(x) \in \widehat{\mathbb{F}}[x]$.
 
\begin{description}
\item[Theorem \#1:] Suppose $f(x)$ is irreducible (over $\mathbb{F}$). If $\alpha$ is any root of $f(x)$ and $\beta$ is any root of $\widehat{f}(x)$,\\ \force\hfill then there exists an isomorphism $\widehat{\varphi}:\mathbb{F}[\alpha] \to \widehat{\mathbb{F}}[\beta]$ extending $\varphi:\mathbb{F} \to \widehat{\mathbb{F}}$ and sending $\alpha$ to $\beta$.

\item[Theorem \#2:] $\mathbb{E}$ be a splitting field for $f(x)$ (over $\mathbb{F}$) and $\widehat{\mathbb{E}}$ is a splitting field for $\widehat{f}(x)$ (over $\widehat{\mathbb{F}}$),\\ \force \hfill then there exists an isomorphism $\widehat{\varphi}:\mathbb{E} \to \widehat{\mathbb{E}}$ extending $\varphi:\mathbb{F} \to \widehat{\mathbb{F}}$.

\item[Theorem \#3:] If $\widehat{\varphi}:\mathbb{E} \to \widehat{\mathbb{E}}$ is an isomorphism of fields extending $\varphi:\mathbb{F} \to \widehat{\mathbb{F}}$,\\ \force\hfill then for any $\alpha \in \mathbb{E}$ such that $f(\alpha)=0$ we also have $\widehat{f}(\widehat{\varphi}(\alpha))=0$ (i.e., roots map to roots).
\end{description}

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\begin{description}
\item[\#1 Rootin' Around:] Let $\mathbb{F}$ be a field, $f(x) \in \mathbb{F}[x]$ (not a constant polynomial), and $\mathbb{E}$ be a splitting field for $f(x)$ over $\mathbb{F}$. In addition, let $G=\mathrm{Gal}(\mathbb{E}/\mathbb{F})$ (we say $G$ is the Galois group of the polynomial $f(x)$).
   \begin{enumerate}[(a)]
   \item Suppose that $f(x)$ is irreducible. Show that $G$ acts {\it transitively} on the roots of $f(x)$ in $\mathbb{E}$. 
   
   In particular, given any two roots for $f(x)$, say $\alpha, \beta \in \mathbb{E}$, there exists some $\sigma \in G$ such that $\sigma(\alpha)=\beta$. 
   
   \item Suppose $G$ acts transitively on the roots of $f(x)$ in $\mathbb{E}$. In addition, suppose $f(x)$ has no repeated roots.\\ \force\hfill Show that $f(x)$ is irreducible (over $\mathbb{F}$). 
   
   {\it Hint:} Suppose that $f(x)$ factors. Explain why $G$ must send roots of a factor to other roots of that same factor. Transitivity of the action will force distinct factors to share a root (why?). This is a problem (why?).
   \end{enumerate}
   
{\it Note:} In summary, for separable polynomials, the Galois group of a a polynomial acts transitively\\ \force\hfill on the polynomial's roots if and only if it is an irreducible polynomial (over the base field).

\item[\#2 Finding Galois:] Let $f(x)=x^4-10x^2+1 \in \mathbb{Q}[x]$. Find the Galois group of $f(x)$ (over $\mathbb{Q}$).

{\it Note:} $f(x)=(x-\sqrt{2}-\sqrt{3})(x+\sqrt{2}-\sqrt{3})(x-\sqrt{2}+\sqrt{3})(x+\sqrt{2}+\sqrt{3})$.\\ \phantom{{\it Note:} }You may find Rotman's Example 20 on page 54 helpful. 
   
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{\it Claim:} $f(x)$ is irreducible in $\mathbb{Q}[x]$. 

\bproof By the Rational Root Theorem, the only possible rational roots are $\pm 1$, since $f(\pm 1)=-8\not=0$, $f(x)$ has no roots in $\mathbb{Q}$. Thus the only way $f(x)$ could be reducible is if it factors into a product of (irreducible) quadratics. Given $f(x)$ is monic, without loss of generality we can assume the factors are monic. Suppose $f(x)=(x^2+ax+b)(x^2+cx+d)$. Noting the coefficient of $x^3$ in $f(x)$ is zero, we must have $c=-a$. Finally, noting that the constant term of $f(x)$ is 1, we must either have $b=d=1$ or $b=d=-1$. 
Case 1: $f(x)=(x^2+ax+1)(x^2-ax+1) = x^4+(2-a^2)x^2+1$ so that $2-a^2=-10$ and so $a^2=12$ and thus $a \not\in \mathbb{Q}$. Case 2: $f(x)=(x^2+ax-1)(x^2-ax-1) = x^4+(-2-a^2)+1$ so that $-2-a^2=-10$ and so $a^2=8$ and thus $a \not\in \mathbb{Q}$. Therefore, $f(x)$ cannot factor into quadratics in $\mathbb{Q}[x]$. Thus $f(x)$ is irreducible in $\mathbb{Q}[x]$.
\eproof

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{\it Suggestive Calculation:} $(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}$, $\sqrt{6}(\sqrt{2}+\sqrt{3})=3\sqrt{2}+2\sqrt{3}$, and $3(\sqrt{2}+\sqrt{3})-(3\sqrt{2}+2\sqrt{3})=\sqrt{3}$.


\item[\#3 Roots of Unity:] Just remember $\sqrt{1}=1$. 
    \begin{enumerate}[(a)]
    \item Find a primitive $n^{\mbox{th}}$-root of unity when $n=1,2,3,4,5,$ and $6$. Your final formulas should not involve sines and cosines. For example: $e^{2\pi i/3} = \cos(2\pi/3)+i\sin(2\pi/3)$ is a primitive third root of unity, but an unacceptable (final) answer. You may find Wolfram Alpha helpful as you seek out what numbers like $\cos(2\pi/5)$ are (unless you have more special triangles memorized than I do).
    
    %\begin{tabular}{c||c|c|c|c|c|c|}
    %$n=$  & 1 & 2 & 3 & 4 & 5 & 6 {\large\strut} \\ \hline
    %Root & & & & & & {\Huge\strut} \\ \hline
    %\end{tabular}
    
    \item Find the Galois  groups of $x^n-1$ over $\mathbb{Q}$ when $n=1,2,3,4,5,6$ and $7$. 
    
    %\begin{itemize}
    %\item $\mathrm{Gal}(\mathbb{Q}[\zeta]/\mathbb{Q}) \cong$ ??? when $\zeta = e^{2\pi i/1}=1$ because ???
    %\item $\mathrm{Gal}(\mathbb{Q}[\zeta]/\mathbb{Q}) \cong$ ??? when $\zeta = e^{2\pi i/2}=-1$ because ???
    %\end{itemize}
   
    \end{enumerate}
\end{description}

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