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\begin{center}
{\bf \Large \underline{Honors Algebra 4, MATH 371 Winter 2010}}\\ 
	{\large Assignment 6}\\Due Wednesday, March 24 at 08:35
\end{center}


\begin{enumerate}

	\item Let $K/F$ be a degree $2$ extension of fields.  
	\begin{enumerate}
		\item If the characteristic of $F$ is not 2, prove that $K=F(a)$ for some $a\in K\setminus F$ 
		with $a^2\in F$.\label{one}
		
		\item Give a counterexample to (\ref{one}) if $F$ has characteristic 2. 
	
		\item Fix $F$ of characteristic not 2 and let $K_1,K_2$ be quadratic extensions of $F$
		with $K_1 = F(a_1)$ and $K_2=F(a_2)$ where $a_i^2=b_i\in F$.  Prove that $K_1\simeq K_2$
		as extensions of $F$ (i.e. that there exists an isomorphism of fields $K_1\simeq K_2$
		restricting to the identity on $F$) if and only if $b_1/b_2\in (F^{\times})^2$ is a square.\label{three}
		Conclude that the isomorphism classes of quadratic extensions of $F$ are in bijection with
		the group $F^{\times}/(F^{\times})^2$.
		
		\item Using (\ref{three}), give a complete list (without repetition) of all isomorphism classes of quadratic
		extensions of $\Q$.
		
	\end{enumerate}
	
\noindent{\bf Solution}	:
\begin{enumerate}
	\item Fix $b\in K\setminus F$.  Then $\{1,b\}$ is an $F$-basis of $K$, so $b$ satisfies a degree 2 polynomial
	$b^2+ ub +v=0$ with $u,v\in F$.  Since the characteristic of $F$ is not 2, $2\in F^{times}$ so $u/2$ makes
	sense and we have $(b+u/2)^2 =u^2/4-v$ by completing the square.  Thus, $a:=b+u/2\in K\setminus F$
	has $a^2\in F$ and clearly $K=F(a)$.

	\item The extension $\F_2[X]/(X^2+X+1)$ of $\F_2$ gives a counterexample, since
	$(a+bX)^2 = (a^2+b^2)+ b^2X$ lies in $F$ if and only if $b=0$.
	
	\item If $K_1\simeq K_2$ as extensions of $F$, then $b_1$ must be a square in $K_2$,
	say $b_1= (u+va_2)^2$.  This gives $b_1=u^2+v^2b_2 + 2uva_2$ from which it follows
	(as $2\neq 0$ in $F$) that either $u$ or $v$ must be zero.  The second case cannot occur
	as otherwise $b_1$ would be a square in $F$ and $K_1=F$.  Thus ,$b_1=v^2b_2$ for some $v\in F$.
	Conversely, If $b_1=v^2b_2$ then $b_1=(va_2)^2$ is a square in $K_2$, and the map
	$F[X]\rightarrow K_2$ sending $X$ to $va_2$ is surjective and yields an isomorphism
	$F[X]/(X^2-b_1)\simeq K_2$.  As the source of this isomorphism is isomorphic to $K_1$,
	we get $K_1\simeq K_2$ as extensions of $F$.
	
	\item These are parameterized by $\Q^{\times}/(\Q^{\times})^2 = \{2,3,5,6,7,10,11,13,14,15,\ldots\}$
	(the positive square-free integers).
\end{enumerate}
	

	\item For $a\in \F_p$, set
	$$f_a(x):=X^p-X-a\in \F_p[X].$$
	\begin{enumerate}
		\item If $a=0$, show that $f_a(X)=\prod_{u\in \F_p} (X-u)$.
		\item Suppose that $a\neq 0$ and let $E_a$ be a splitting field of $f_a(X)$.
		If $r_1,r_2\in E_a$ are roots of $f_a$, prove that $r_1-r_2\in \F_p$.
		\item Show that $f_a(X)$ is irreducible for all $a\in \F_p^{\times}$.
		\item Prove that $f_b(X)$ splits completely over $E_a$ for each fixed $a\in \F_p^{\times}$
		and all $b\in \F_p^{\times}$.  Conclude that $E_a$ is independent of $a$.
	\end{enumerate}

\noindent{Solution:} 
\begin{enumerate}
	\item Every $u\in \F_p$ satisfies $X^p-X=0$ and this gives $p$ roots of the degree $p$ polynomial
	$X^p-X$ in the Euclidean domain $F[X]$, so we get the claimed factorization.
	
	\item Observe that $(r_1-r_2)^p-(r_1-r_2) = (r_1^p-r_1)-(r_2^p-r_2) = 0$
	so $r_1-r_2$ is a root of $X^p-X$ and hence an element of $\F_p$ by the first part.

	\item Certainly $f_a$ has no root in $\F_p$ for $a\in \F_p^{\times}$, by part 1.  
	Over $E_a$, we have the factorization 
	$$f_a=\prod_{0\le i <p} (X-(r+i))$$
	for a fixed root $b$ of $f_a$ in $E_a$ (by the previous part).  
	If $f_a=gh$ in $\F_p[X]$ then $g=X^d+\alpha X^{d-1}$ for some $0< d < p$.
	But $g$ is a product over certain integers $i$ of $(X-(r+i))$ in $E_a[X]$
	so we must have $-\alpha = dr + u$ for some $u\in \F_p$.  As $d\in \F_p^{\times}$,
	this gives $r\in \F_p$ (as $\alpha\in \F_p$), a contradiction as $f_a$ has no roots
	in $\F_p$.  Hence $f_a$ is irreducible.
	
	\item If $r$ is any root of $f_a$ in $E_a$, then $(vr)^p - (vr) + va=0$ for any $v\in \F_p^{\times}$.
	Thus, the roots of $f_{va}$ are precisely $vr,vr+1,\ldots,vr+p-1\in E_a$ so $f_{va}$
	splits completely over $E_a$.  This shows that $E_a$ contains $E_{va}$ for all $a\in \F_p^{\times}$
	and hence that $E_a$ is independent of $a$.
\end{enumerate}


	\item Find the minimal polynomials of $2\cos(2\pi/5)$ and $2\cos(2\pi/7)$ over $\Q$.

\noindent{\bf Solution} We treat the case of $2\cos(2\pi /7)$ as it is the harder of the
two.  Let $\zeta=e^{2\pi i/7}$ and set $K=\Q(\zeta)$, $G=\Gal(K/\Q)$.
Put $\eta:=\zeta+\zeta^{-1}=2\cos(2\pi/7)$.  We know that $G\simeq (\Z/7\Z)^{\times}$
is cyclic of order 6, generated by the automorphism $\sigma:\zeta\mapsto \zeta^3$
(since $3\in (\Z/7\Z)^{\times}$ is a generator of this cyclic group).  The conjugates of
$\eta$ are 
$$\eta,\quad \sigma\eta = \zeta^{3}+\zeta^{-3},\quad \sigma^2\eta = \zeta^2 + \zeta^{-2}.$$
Using the binomial theorem, we compute
$$
 \sigma^2\eta = \eta^2 - 2,\quad
 \sigma\eta = \eta^3 -3\eta
$$
and hence we find that
$$
 \eta + \sigma\eta + \sigma^2\eta = \eta^3 + \eta^2 -2\eta -2.
$$
Using the fact that the minimal polynomial of $\zeta$ is $X^6+X^5+\cdots + X+ 1$,
the left hand side above is $-1$ whence $\eta$ is a root of the degree 3 polynomial
$$X^3+X^2-2X-1$$
which must therefore be the minimal polynomial of $\eta$ since $\eta$ has 3 distinct
conjugates.

	\item For each of the following extensions, determine $[K:F]$ and an $F$-basis of $K$.
	\begin{enumerate}
		\item $F=\Q$, $L=\Q(a,b)$ with $a^2=6$ and $b^3=2$.
		
		\item $F=\C(T)$ and $L$ is the splitting field of $X^n-T$ over $F$, with $n$ a positive integer.
		
		\item $F=\F_p(T)$ and $L$ is the splitting field of $X^p-T$ over $F$, with $p$ a prime.
	\end{enumerate}
	
\noindent{\bf Solution:} 
\begin{enumerate}
	\item An $F$-basis is $\{1,b,b^2,a,ab,ab^2\}$
	\item Let $r$ be a root of $X^n-T$ in $L$.  An $F$-basis is $\{1,r,r^2,\ldots,r^{n-1}\}$
	(note that this polynomial is irreducible by Eisenstein's criterion, so $F(r)$ is a degree $n$
	extension and since $\C$ contains all $n$-th roots of unity, $X^n-T$ splits completely
	over $F(r)$).
	\item Again, Eisenstein's criterion gives irreducibility.  If $r$ the unique(!) root of 
	$X^p-T$ in $L$, then an $F$-basis of $L$ is $1,r^2,\ldots, r^{p-1}$.
\end{enumerate}	
	
	
	\item Let $K/F$ be a finite extension of fields and let $\alpha\in K$.  Then $\alpha$ induces
	an $F$-linear map of finite-dimensional $F$-vector spaces
	$$m_{\alpha}:K\rightarrow K.$$
	\begin{enumerate}
		\item Prove that $\alpha$ is a root of the characteristic polynomial of the linear map $m_{\alpha}$.
		Hint: select a suitable $F$-basis of $F(\alpha)$. \label{charp}
		
		\item Use (\ref{charp}) to find a monic degree 3 polynomial with $\Q$-coefficients 
		satisfied by $1+\sqrt[3]{2}+\sqrt[3]{4}$.
		
		\item Prove that if $K=F(\alpha)$, then the characteristic polynomial of $m_{\alpha}$
		as a linear map $K\rightarrow K$ is in fact the minimal polynomial of $\alpha$ over $F$.
	\end{enumerate} 
	
\noindent{\bf Solution:} 
\begin{enumerate}
	\item Let $m(x):=x^d+a_{d-1}\alpha^{d-1}+\cdots+a_0$ be the minimal polynomial of $\alpha$ over $F$.
	Then $1,\alpha,\ldots, \alpha^{d-1}$ is an $F$-basis of $F(\alpha)$.  
	Let $\{b_1,\ldots,b_e\}$ be an $F(\alpha)$-basis of $K$ so $\{\alpha^i b_j\}$ is an $F$-basis
	of $K$.  Then the matrix of multiplication by $\alpha$ on $K$ with respect to this basis
	is a block diagonal matrix, with blocks given by the matrix of multiplication 
	by $\alpha$ on $F(\alpha)$, which is easily seen to be
	$$
	\left(
		\begin{array}{ccccc}
			0 & 1 & 0 & \cdots & 0\\
			0 & 0 & 1 & \cdots & 0\\
			\vdots & & & \vdots & 0\\
			-a_0 & -a_1 & -a_2 & \cdots & -a_{d-1}
		\end{array}
		\right)
	$$
	This matrix has characteristic polynomial $m(x)$ so the characteristic polynomial of $m_{\alpha}$
	is a power of $m(x)$.  This also handles part c).
	
	\item Let $\alpha=\sqrt[3]{2}$ and $\beta:=1+\alpha+\alpha^2$.  The matrix of 
	multiplication by $\beta$ on $\Q(\alpha)$ with respect to the basis $1,\alpha,\alpha^2$
	is
	$$
	\left(
		\begin{array}{ccc}
			1 & 1 & 1\\
			2 & 1 & 1\\
			2 & 2 & 1
		\end{array}
		\right)
	$$
	and this has characteristic polynomial $X^3-3X^2-3X-1$.
\end{enumerate}	
	
	
	\item For each of the following algebraic elements $\alpha$ of the given field extension 
	$K/\Q$, express $1/\alpha$ and $1/(\alpha+1)$ as polynomials in $\alpha$ with $\Q$-coefficients.
	\begin{enumerate}
		\item $K$ is the splitting field of $f=X^3-3X+1$ and $\alpha$ is a root of $f$.
		
		\item $K$ is the splitting field of $f=X^4+X^3+X^2+X+1$ and $\alpha$ is a root of $f$.
	
		\item $K$ is the splitting field of $f=X^5-3X+3$ and $\alpha$ is a root of $f$.
	\end{enumerate}
	
\noindent{\bf Solution:}
\begin{enumerate}
	\item $1/\alpha = 3-\alpha^2$ and $1/(\alpha+1) = \frac{1}{3}(-a^2+a+2)$.
	
	\item $1/\alpha = -a^3-a^2-a-1$ and $1/(\alpha+1)=-a^3-a$
	
	\item $1/\alpha = \frac{1}{3}(-a^4 + 3)$ and $1/(\alpha+1)=\frac{1}{5}(-a^4 + a^3 - a^2 + a + 2)$
\end{enumerate}	
	
	
	\item Prove that $X^4-5$ is irreducible over $\Q$ and has splitting field $K$ of degree $8$ over $\Q$.
	Describe this splitting field explicitly as $\Q(a,b)$ where $a$ is a root of $X^4-5$ and $b^2\in \Q$.
	In terms of $a$ and $b$, write down a $\Q$-basis for $K$.
	
\noindent{\bf Solution:} Use Eisenstein with $p=5$.  The splitting field is easily seen to be
$\Q(a,b)$ where $a:=\sqrt[4]{5}$ and $b:=i$, which has degree 8 since $i$ is not in $\Q(a)$
as $\Q(a)$ is a subfield of $\R$.  A $\Q$-basis for $K$ is $\{1,a,a^2,a^3,b,ba,ba^2,ba^3\}$.	
	
	\item Describe the splitting fields of $f:=X^3-5$ over $\F_{11}$ and $\F_7$ and factor $f$
	into linear factors over each extension.
	
\noindent{\bf Solution:} Over $\F_{11}$, the given polynomial has a root $X=3$ and factors
as $(X - 3)(X^2 + 3X -2)$ with irreducible quadratic.  The splitting field is therefore
degree 2, and is obtained by adjoining the square root of any nonsquare in $\F_{11}$.
Explicitly, the splitting field is $\F_{11}(a)$ where $a^2=-1$ and then $X^3-5$
factors over $\F_{11}(a)$ as 
$$X^3-5 = (X -3)(X + 2a -4)(X - 2a - 4).$$
The case of $\F_7$ is similar and is left to the reader.

	




\end{enumerate}
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