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\def\al{\alpha}
\def\aut{{\rm aut}}
\def\be{\beta}
\def\ccc{\mathcal{C}}
\def\cc{{\mathbb C}}
\def\comb(#1,#2){\left(\begin{array}{c} #1 \\ #2  \end{array} \right) }
\def\de{\delta}
\def\div{\slash}
\def\fix{{\rm fix}}
\def\ga{\gamma}
\def\ii{{\cal F}}
\def\isom{\simeq}
\def\lam{\lambda}
\def\la{\langle}
\def\mod{\mbox{ mod }}
\def\nn{{\mathbb N}}
\def\normal{\triangleleft}
\def\pf{{\bf Proof:}\par\noindent}
\def\pr{^\prime}
\def\qed{\par\noindent {\bf :foorP}}
\def\qq{{\mathbb Q}}
\def\ra{\rangle}
\def\reals{\mathbb{R}}
\def\rr{{\mathbb R}}
\def\si{\sigma}
\def\sm{\setminus}
\def\span{{ \rm span }}
\def\ss{{\cal G}}
\def\st{\;:\;}
\def\su{\subseteq}
\def\zz{{\mathbb Z}}

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\newtheorem{theorem}{Theorem}[section]
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\newtheorem{thm}[theorem]{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{define}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{prop}[theorem]{Proposition}
\newtheorem{ques}[theorem]{Question}
\newtheorem{quest}[theorem]{Question}
\newtheorem{cor}[theorem]{Corollary}
\newtheorem{examp}[theorem]{Example}

%endthms

\newtheorem{prob}{Problem}

\newenvironment{enum}
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{\end{list}}

\begin{document}

\noindent  Last revised: \today

\begin{flushright}
  A.Miller $\;\;\;\;$  M542 \\
  www.math.wisc.edu/$\sim$miller/\\
\end{flushright}

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%thmcutbegin   line 78 in "up" head -78 alg.tex > prob.tex
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\bigskip
Each Problem is due one week from the date it is assigned.
Do not hand them in early.
Please put them on the desk in front of the room at the
beginning or end of class. Include the statement of the problem
as part of your solution.

The date the problem was assigned in class is in parentheses.

\bigskip

%thmcutend

\section{Finite abelian groups}

\begin{theorem}
(Chinese remainder theorem) Given $n,m$ relatively prime integers for
every $i,j\in\zz$ there is an $x\in\zz$ such that
$x=i$ mod  $n$ and $x=j$ mod  $m$.
\end{theorem}


\begin{prob} (Fri Jan 24) (a) Find an integer $x$ such that
$x= 6 \mod 10$ and $x=15 \mod 21$ and $0\leq x\leq 210$.
(b) Find the smallest positive integer $y$ such that 
$y= 6 \mod 10$ and $y=15 \mod 21$ and $y= 8 \mod 11$.
\end{prob}

\begin{prob}(Fri Jan 24)
(a) Find integers $i,j$ such that there is no integer $x$ with
$x= i \mod 6$ and $x= j \mod 15$.   (b) Find all pairs $i,j$
with $i=0,1,\dots 5$ and $j=0,1,\ldots, 14$ such that there
is an integer $x$ with $x= i \mod 6$ and $x= j \mod 15$.
\end{prob}


\begin{theorem}
$\zz_n\times\zz_m\isom \zz_{nm}$ iff $n,m$ are relatively prime.
\end{theorem}

\begin{lemma}
Suppose $n,m$ are relatively prime, $G$ is a finite abelian group such that
$x^{nm}=e$ for every $x\in G$.  Let $G_n=\{x\in G\st x^n=e\}$ and
$G_m=\{x\in G\st x^m=e\}$.  Then
\begin{itemize}
\item $G_n$ and $G_m$ are subgroups of $G$,
\item $G_n\cap G_m=\{e\}$,
\item $G_nG_m=G$, and therefore
\item $G\isom G_n\times G_m$
\end{itemize}
\end{lemma}

\begin{cor} (Decomposition into $p$-groups)
Suppose $G$ is an abelian group and $|G|=p_1^{i_1}\cdot p_2^{i_2}\cdots p_n^{i_n}$
where $p_1<p_2<\cdots <p_n$ are primes.  Then
$$G\isom G_1\times G_2\times\cdots\times G_n$$ 
where for each $j$ if $x\in G_j$ then $x^{n_j}=e$ where $n_j=p_j^{i_j}$.
\end{cor}

\begin{prob}
(Mon Jan 27) Prove that for any $n$ there is only one abelian group (up to 
isomorphism) of size $n$ iff $n$ is square-free.  Square-free
mean that no $p^2$ divides $n$ for $p$ a prime.
\end{prob}

\begin{lemma}
Suppose $G$ is a finite abelian $p$-group and $a\in G$ has maximum
order, then there exists a subgroup $K\su G$ such that
\begin{itemize}
\item $\la a\ra\cdot K=G$ and 
\item $\la a\ra\cap K=\{e\}$.
\end{itemize}
\end{lemma}

The proof given in class is like the one in Gallian or Judson.

\begin{theorem}
Any finite abelian group is isomorphic to a product of cyclic
groups each of which has prime-power order.
\end{theorem}

\begin{prob}(Wed Jan 29)
Let $G$ be a finite abelian group.  Prove that the following are equivalent
\begin{enumerate}
\item For every subgroup $H$ of $G$ there is a subgroup $K$ of $G$ with
$HK=G$ and $H\cap K=\{e\}$. 
\item Every  element of $G$ has square-free order.
\end{enumerate}
\end{prob}

%thmcutbegin

Hint: Polya's Dictum: ``If you can't do a problem, then there is an
easier problem you can't do.  Find it.''

Lets call property (1) the Complementation Property for $G$
or CP for short.  Here are some easier problems:
\begin{enum}
\item Prove that $C_{p^2}$ fails to have CP.
\item Prove that $C_p\times C_p$ has CP.
\item Let $|G|$ and $|H|$ be
relatively prime.  Prove that $G\times H$ has CP iff both
$G$ and $H$ have CP.
\end{enum}

%thmcutend

\begin{theorem}(Uniqueness)
Suppose 
$$C_{p^{n_1}}\times C_{p^{n_1}}\times\cdots\times C_{p^{n_k}}\isom
C_{p^{m_1}}\times C_{p^{m_1}}\times\cdots\times C_{p^{m_l}}$$
where $n_1\geq n_2\geq \cdots n_k\geq 1$ and 
$m_1\geq m_2\geq \cdots m_l\geq 1$.  Then $k=l$ and $n_i=m_i$ for
all $i$.
\end{theorem}

\begin{prob}(Fri Jan 31)
How many abelian groups of order 144 are there up to isomorphism?
Explain.
\end{prob}

\begin{prob}
(Mon Feb 3) Suppose $G_1,G_2,H_1,H_2$ are finite abelian groups,
$G_1\times G_2\isom H_1\times H_2$ and $G_1\isom H_1$.
Prove that $G_2\isom H_2$.
\par Give a counterexample if the word finite is dropped, i.e.,
$G_1\times G_2\isom H_1\times H_2$ and $G_1\isom H_1$ but $G_2$
is not isomorphic to $H_2$.   
\end{prob}

%thmcutbegin

\noindent 
I do not know if problem 6 is true or false for finite non-abelian groups.

%thmcutend

\section{Group Actions and Sylow Theorems}

\bigskip
For the group $G$ acting on the set $X$
the orbit of $a\in X$ is
$$orb(a)=^{def}\{ga\st g\in G\}\su X.$$

\begin{prop} Orbits are either disjoint or the same.
\end{prop}

\begin{prob}
(Wed Feb 5) Prove or disprove:
\par For any finite abelian groups $G_1$ and $G_2$ with
subgroups, $H_1\su G_1$ and $H_2\su G_2$ such that $H_1\isom H_2$,
if $G_1/H_1\not\isom G_2/H_2$ then $G_1\not\isom G_2$.
\end{prob}

For a given group action of group $G$ on set
$X$, define $Stab(a)=\{g\in G\st ga=a\}$ for each $a\in X$.
Called stabilizer or fixed subgroup.

\begin{prop}
$Stab(a)$ is a subgroup of $G$.
\end{prop}

\begin{prob}
(Wed Feb 5) Prove that $Stab(ga)=g\,Stab(a)\,g^{-1}$.
\end{prob}

For $H\su G$ a subgroup the index of $H$, $[G:H]$ is the
number of $H$-cosets, $|\{gH\st g\in G\}|$.  Lagrange's
Theorem says $|G|=[G:H]\cdot |H|$.

\begin{prop}
(Orbit-stabilizer formula) $|orb(a)|=[G:Stab(a)]$.
\end{prop}

The conjugacy action of $G$ on $G$ is given by $(g,h) \to ghg^{-1}$.
Under this action the orbits are called the conjugacy classes.
$Z(G)$ the center of $G$ is the subgroup of all elements of
$G$ which commute with every other element of $g$.  Equivalently
it is the set of elements of $G$ with orbits (conjugacy classes) 
of size one. $C(g)=Stab(g)$ is called the centralizer subgroup of $g$.

\begin{theorem}
(Class formula) If $conj(g_1),\;\cdots, \; conj(g_n)$ are the
conjugacy classes of size greater
than one, then
$$|G|=|Z(G)| +\sum_{k=1}^n [G:C(g_k)]$$
\end{theorem}

\begin{theorem}
(Cauchy) If $p$ is a prime which divides $|G|$, then $G$ has
an element of order $p$.
\end{theorem}

\begin{prob}
This is due in lecture on valentines day.  It will be graded
in class so do not hand-in.

\par\noindent (a) Suppose $G$ is a finite abelian group which contains an
element which has non-square-free order. Prove that for some prime
$p$ it has an element of order $p^2$.
\par\noindent (b) Suppose $a$ is an element of a finite abelian group $G$ with
order $p^2$ let $b=a^p$, let $H=<b>$ be the subgroup generated
by $b$ and suppose $K$ is a subgroup of $G$ with $K\cap H=\{e\}$.
Prove that $a$ is not an element of $HK$.
\par\noindent (c) Suppose $G_1,G_2$ are finite abelian groups with $|G_1|$ and
$|G_2|$ relatively prime.  Show that for any subgroup $H\su G_1\times G_2$
there are subgroups $H_1\su G_1$ and $H_2\su G_2$ such that
$H=H_1\times H_2$.   (Warning: the relatively prime hypothesis is
necessary.)
\par\noindent (d) Suppose $G_1,G_2$ are finite abelian groups with $|G_1|$ and
$|G_2|$ relatively prime.  Show that if $G_1$ and $G_2$ both have
the CP then $G_1\times G_2$ has CP.\footnote{CP is defined after Problem 4.}
\par\noindent (e) Prove that $C_p\times C_p\times \cdots \times C_p$ has the CP.
\par\noindent (f) Prove Problem 4.

\end{prob}

\begin{cor}
Groups of order $p^2$ are abelian.
\end{cor}

\begin{thm}
(Sylow 1) If $G$ is a finite group and $p^n$ divides $|G|$,
then there exists a subgroup $H\su G$ with $|H|=p^n$.
\end{thm}

\begin{prop}
Any two n-cycles in $S_N$ are conjugates.  If $\tau=c_1c_2\cdots c_n$
and $\rho=c_1\pr c_2\pr \cdots c_n\pr$ are
disjoint cycle decomposition with $|c_i|=|c_i\pr|$ all $i$, then
$\tau$ and $\rho$ are conjugates.  Similarly for the converse.
\end{prop}

\begin{prob}
(Mon Feb 10) Prove for any $n\geq 3$ that
$Z(S_n)=\{id\}$.
\end{prob}

\begin{define}
$H\su G$ is a $p$-subgroup iff its order is a power of $p$. $P\su G$
is a $p$-Sylow subgroup of $G$ iff $|P|=p^n$ where $|G|=p^nm$ and
$p$ does not divide $m$.
\end{define}

\begin{lemma}
Suppose $P$ is a $p$-Sylow subgroup of $G$, $g\in G$ has
order a power of $p$, and $gPg^{-1}=P$.  Then $g\in P$.
\end{lemma}

\begin{thm}
(Sylow 2) If $G$ is a finite group, $H$ a p-subgroup of $G$, and $P$
a $p$-Sylow subgroup of $G$, then there exists $g\in G$ such that
$H\su gPg^{-1}$.
\end{thm}

\begin{cor} Let $G$ be a finite group such that $p$ divides $|G|$.
\par (a) Any $p$-subgroup of $G$ is contained in a $p$-Sylow subgroup of $G$.
\par (b) Any two $p$-Sylow subgroups of $G$ are conjugates.
\par (c) Any two $p$-Sylow subgroups of $G$ are isomorphic.
\par (d) A $p$-Sylow subgroup of $G$ is normal iff it is the only
$p$-Sylow subgroup of $G$.
\end{cor}

\begin{thm}
(Sylow 3) If $|G|=p^nm$ where $p$ does not divide $m$ and
$n(p)$ is the number of $p$-Sylow subgroups of $G$, then:
\par (a) $n(p)=[G:N(P)]$ for any $P$ a $p$-Sylow subgroup of $G$,
\par (b) $n(p)$ divides $m$, and
\par (c) $n(p)=1$ mod $p$
\end{thm}

\begin{prob}(Wed Feb 12)
\par (a) Prove that there are no simple groups of order
either $575$ or $272$.
\par (b) For any prime $p$ prove there are no simple groups
of order $p(p-1)$ or $p(p+2)$.
\end{prob}

\begin{thm}
If $p<q$ are primes and $q$ is not $1$ mod $p$, then every
group of order $pq$ is abelian.
\end{thm}

\begin{prob}
(Fri Feb 14) Question (August J.) Suppose every subgroup
of finite group $G$ is a normal subgroup.  Must $G$ be abelian?
\end{prob}

\begin{prob}
(Fri Feb 14)
\par (a) Suppose $P$ is a $p$-Sylow subgroup of $G$ and
$H$ a subgroup such that $P\normal H$ and $H\normal G$.
Prove that $P\normal G$.
\par (b) If $K\normal H$ and $H\normal G$, does it follow
that $K\normal G$?  Show that the answer is No.
Consider $G=S_4$, $K=\{id,\si\}$ where $\si=(12)(34)$ and
$H=\{id,\si,\tau,\rho\}$ where $\tau$ and $\rho$ are conjugates
of $\si$.  Determine what $\tau$ and $\rho$ are and show
that $K\normal H$ and $H\normal G$, but $K$ is not a normal subgroup
of  $G$.
\end{prob}

\begin{thm}
$aut(\zz_p,+_p)$ is isomorphic to 
$(\zz_p\sm\{0\},\times_p)$ the multiplicative group
of nonzero elements.
\end{thm}

\begin{examp}
If $p<q$ are primes and $q=1$ mod $p$, then there is a twisted product of
$\zz_p$ and $\zz_q$ which has order $pq$ and is not abelian.
\end{examp}

\begin{prob}
(Mon Feb 17) Suppose for every $x\in G$ that $x^2=e$.  Prove
that $G$ is abelian.
\end{prob}

\begin{prob}
(Mon Feb 17) Suppose $H\su G$ is subgroup of index 2, i.e., 
$[G:H]=2$.  Prove that it is a normal subgroup of $G$.
\end{prob}

\begin{thm}
If $p<q$ are primes and $q=1$ mod $p$, then up to isomorphism
there is a unique nonabelian group of order $pq$.
\end{thm}


\section{Polynomials and finite field extensions}

\begin{thm}
Suppose that $p(x)$ is a polynomial over the field $F$ and for some $\al\in F$
$\;p(\al)=0$.  Then $p(x)=(x-\al)q(x)$ for
some polynomial $q(x)$.
\end{thm}

\begin{cor}
Any polynomial $p\in F[x]$ of degree $\leq n$ with more than $n$ roots
must be identically zero.
\end{cor}

\begin{thm}
Let the exponent of $G$ be the least $n$ such that
$x^n=e$ for every $x\in G$.  If $G$ is finite abelian group then 
$G$ is cyclic iff $exp(G)=|G|$.
\end{thm}

\begin{cor}
The multiplicative group of a finite field is cyclic.
\end{cor}

\begin{prob}(Wed Feb 19)
For $F$ a finite field call $a\in F$ a generator of $F$ iff
every nonzero element of $F$ is a power of $a$.
\par (a) Find a generator of $\zz_7$.
\par (b) How many generators does $\zz_{17}$ have?
\par (c) How many generators does $\zz_{31}$ have?
\end{prob}



\section{Vector spaces over an abstract field}

Before taking up finite field extensions we review some
elementary results on vector spaces. See:

\url{http://www.math.wisc.edu/~miller/old/m542-00/vector.pdf}


\begin{lemma}
(Exchange Lemma) Suppose $\span(A\cup B)=V$ and $a$ is not in $\span(A)$.
Then there exists $b\in B$ such that $\span(A\cup\{a\}\cup(B\sm\{b\}))=V$.
\end{lemma}

\begin{thm}
Every vector space has a basis.  Any two bases have the
same cardinality.  Any set of $n+1$ vectors in a vector
space of dimension $n$ is linearly dependent.
\end{thm}

\begin{cor}
Any finite field $F$ of characteristic $p$ has cardinality
$p^n$ for some integer $n$.
\end{cor}

\begin{prob}
(Fri Feb 21) Prove that $v_1,v_2,\ldots,v_n$ are linearly dependent
iff $v_1=\vec{0}$ or $v_{i+1}\in\span\{v_1,v_2,\ldots,v_i\}$
for some $i$ with $1\leq i<n$.
\end{prob}

\section{Extension fields}

\begin{thm}
(Kronecker) If $p(x)\in F[x]$ is a non-constant polynomial, then
there exists a field $E\supseteq F$ and $\al\in E$ with
$p(\al)=0$.
\end{thm}


\begin{cor}(Kronecker)
If $p(x)\in F[x]$ is a polynomial of degree $n$, then
there exists a field $E\supseteq F$ and $\al_i\in E$ such that
$$p(x)=a(x-\al_1)(x-\al_2)\cdots (x-\al_n)$$
\end{cor}

\begin{thm}
If $p(x)\in F[x]$ is irreducible and $\al,\be$ are roots in
some extension fields of $F$ then $F(\al)$ and $F(\be)$ are
isomorphic via an isomorphism which fixes $F$.
\end{thm}

\begin{cor}
If  $p(x)\in F[x]$ is irreducible and splits in an extension field $E$
of $F$ then the multiplicity of each root of $p$ is the same. 
\end{cor}

\begin{prob}
(Mon Feb 24) Let $R$ be a commutative ring with 1.  Let $I$ be
a maximal ideal in $R$.  Suppose $ab=0$. Prove that $a\in I$ or 
$b\in I$.
\end{prob}

\begin{prob}
(Mon Feb 24) Consider $p(x)=x^3+x+1$ as a polynomial in $\zz_2[x]$.
Suppose $p$ has a root $\al$ is in some field extension.  Construct
the multiplication table for 
$$\zz_2[\al]=^{def}\{a+b\al+c\al^2 \st a,b,c\in \zz_2\}$$
\end{prob}

\begin{prob}
(Wed Feb 26) Let $\al$ be transcendental over $\zz_2$.  Let
$F=\zz_2(\al)$ and let $p(x)=x^2-\al$.  
\par (a) Prove that $p$ is irreducible over $F$.
\par (b) Prove that if $\be$ is a root of $p$ in some
some extension field, then $p(x)=(x-\be)^2$.
\par (c) Suppose that $F$ is a finite field of characteristic $2$.
Prove that for every $a\in F$ there is a $b\in F$ such that $b^2=a$.
\par (d)  Suppose that $F$ is a finite field of odd characteristic.
Prove that there exists $a\in F$ for every $b\in F$ such that $b^2\neq a$.
\par (e) Find a field $F$ and an irreducible polynomial $p(x)$ of
degree three such that in any extension field in which $p$ splits
there exist a $\be$ such that $p(x)=(x-\be)^3$.
\end{prob}

\begin{thm}
The formal derivative for an abstract polynomial $f(x)\in F[x]$ satisfies
the usual derivative laws:
\begin{enum}
\item If $a\in F$ and $f\in F[x]$, then $(af)\pr=af\pr$.
\item If $f,g\in F[x]$, then $(f+g)\pr=f\pr+g\pr$.
\item If $f,g\in F[x]$, then $(fg)\pr=f\pr g+fg\pr$.
\end{enum}
\end{thm}

\begin{prob}
(Fri Feb 28) Prove that the formal derivative for polynomials
in $F[x]$ satisfies 
\begin{enum}
\item The power rule: $(f^n)\pr =n(f^{n-1})f\pr$
\item The chain rule: $f(g(x))\pr=f\pr(g(x))g\pr(x)$
\end{enum}
\end{prob}

\begin{thm}
For any $\al\in F$ and $f\in F[x]$ $\;\;\;\;$
\par $\al$ is repeated root of $f$ iff it is a root of $f\pr$.
\end{thm}

\begin{cor}
The roots of an irreducible polynomial in a field of characteristic
zero, are always distinct.
\end{cor}

\begin{lemma}
If $E$ is any field of characteristic $p$, then for any $\al,\be\in E$
$$(\al+\be)^{p^n}=\al^{p^n}+\be^{p^n}$$
\end{lemma}

\begin{thm}
For any $p^n$ and there is a 
field $F$ with $|F|=p^n$.
\end{thm}

\begin{prob}
(Mon Mar 3) Prove for any prime $p$ and positive integer $n$
that $p$ divides $\comb(p^n,k)$ for any $k$ with $0<k<p^n$.
\end{prob}

\begin{define}
For fields $F\su E$ define $[E:F]$ to be the dimension
of $E$ viewed as a vector space over $F$.
\end{define}

\begin{thm}
For fields $F\su K\su E$
$$[E:F]=[E:K]\cdot [K:F]$$
\end{thm}

\begin{thm}
For $p(x)\in F[x]$ irreducible and $\al$ a root of $p$ in
some extension field, $[F[\al]:F]$ is the degree of $p$.
\end{thm}

\begin{thm}
If $E\supseteq F$ is the splitting field of some polynomial in $F[x]$,
then $[E:F]$ is finite.
\end{thm}

\begin{thm}
If $[E:F]$ is finite and $\al\in E$, then there is an irreducible
polynomial $p\in F[x]$ with $p(\al)=0$.
\end{thm}


\begin{prob}
(Wed Mar 5) $p$ is a prime and $n$ a positive integer.
Prove:
\par (a) If $F$ is a field such that $|F|=p^n$ and $m$ is a
positive integer then there is a field $E$ with $F\su E$
and $E=p^{nm}$.
\par (b) If $F\su E$ are fields, $|F|=p^n$
and $|E|=p^N$, then $n$ divides $N$.
\end{prob}

\section{Algebraic closure}

\begin{define}
$\al$ is algebraic over $F$ iff it is the root of a nontrivial polynomial
in $F[x]$.  A field $K$ is algebraically closed iff every nonconstant 
polynomial $f\in K[x]$ has a root in $K$.
\end{define}

\begin{thm}
If $F\su E$ are fields define
$$K=\{\al\in E\st \al \mbox{ is algebraic over } F\}$$
Then $K$ is a field and $F\su K\su F$.
\end{thm}

Steinitz proved that every field $F$ is a subfield of an algebraically
closed field $K$.  This requires the Axiom of Choice.

\begin{thm}
Suppose $F\su K$ and $K$ is algebraically closed.  Let
$E$ be the elements of $K$ which are algebraic over $F$.  Then
$E$ is algebraically closed.
\end{thm}

\section{Compass and straight-edge}

\begin{thm}(Wantzel 1837)
Let $\ccc\su\reals\times\reals$ be the smallest set containing
$(0,0)$ and $(1,0)$ and closed under constructions using straight
edge and compass. Then $\ccc=F_c\times F_c$ where $F_c$ is the
smallest subfield of $\reals$ which closed under square roots.
\end{thm}

\begin{lemma} For any $\al$
\par $\al\in F_c$ iff for some $n$ there are fields $F_k$ for $k=0,1,\ldots,n$
with $\al\in F_n$ and such that $F_0=\qq$ and for each $k<n\;\;$
$F_{k+1}=F_k[\sqrt{a_k}]$ for some $a_k\in F_k$.
\end{lemma}

\begin{thm} For any $\al\in F_c$
$$[\qq[\al]:\qq]=2^n\;\;\;\;
\mbox{ for some integer $n$.} $$
\end{thm}

\begin{cor}
$^3\sqrt{2}\notin F_c$ so it is impossible to ``double the cube''.
\end{cor}

\begin{cor}
$\cos(20^\circ)\notin F_c$ so it is impossible to trisect every angle.
\end{cor}

\begin{cor} Since $\pi$ is transcendental and every element of
$F_c$ is algebraic, it is impossible to ``square the circle''.
\end{cor}

\begin{prob}(Wed Mar 26)
Prove or disprove:
\par Using a straight edge and compass it is possible to construct
an equilateral triangle with area 1.
\end{prob}

\section{Irreducibility criterion}

\begin{lemma}
(Gauss's Lemma) Suppose $f\in\zz[x]$,  then 
\par $f$ is irreducible in $\qq[x]$ iff $f$ is irreducible in $\zz[x]$.
\end{lemma}

\begin{lemma}
(Eisenstein's Criterion) Suppose $f\in \zz[x]$ has degree $n$
$$f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n$$
and for some prime $p$
\par (a) $p$ does not divide $a_n$, 
\par (b) $p$ divides $a_k$ for all $k=0,1,\ldots,n-1$, and
\par (c) $p^2$ does not divide $a_0$.
\par\noindent Then $f$ is irreducible in $\zz[x]$.
\end{lemma}

\begin{thm}
For any prime $p$ the polynomial $f(x)=1+x+x^2+\cdots+x^{p-1}$ is irreducible
in $\qq[x]$.
\end{thm}

\begin{prop}
If $2^m+1$ is prime, then $m$ is a power of $2$.
\end{prop}

\begin{thm}(Gauss) If the regular $p$-gon is constructible with
straight edge and compass, then $p=2^{2^n}+1$ for some integer $n$.
\end{thm}

\begin{prob}(Fri Mar 28)
Prove that $[F_c:\qq]$ is infinite. $F_c$ is the field of
constructible reals (straight edge and compass).
\end{prob}

\begin{prob}(Fri Mar 28)
Prove that if $2^m-1$ is prime, then $m$ is prime.
\end{prob}

\section{Solvability by radicals}

\par\noindent For Tartaglia method of solving a cubic polynomial see
\par\url{https://www.math.wisc.edu/~miller/old/m542-00/galois.pdf}

\par\noindent For a brief history see:
\par\url{www.dwick.org/pages/cubicquartic.pdf}

\begin{prob}(Mon Mar 31) Find the roots
of $$x^3+3x^2+6x+5=0$$
using addition, subtraction, multiplication, division,
and extraction of roots, i.e., solvability by radicals.
\end{prob}

\begin{thm}
(Steinitz 1910) Suppose $F\su E$ are fields of characteristic 0 and
$[E:F]$ is finite.  Then there exists $\al\in E$ such that
$E=F[\al]$.  The same is true if $E$ is a finite field.
\end{thm}

\begin{examp}
There is a field $F$ and $\al,\be$ with $[F[\al,\be]:F]$ finite
but there is no $\ga$ with $F[\al,\be]=F[\ga]$.
\end{examp}

See
\par\url{http://www.math.wisc.edu/~miller/old/m542-00/examp.pdf}

\begin{prob}(Wed Apr 2)
Suppose  $[F[\al]:F]=n$,  $[F[\be]:F]=m$, and $gcd(n,m)=1$.
Prove that $[F[\al,\be]:F]=nm$.
\end{prob}

\section{Galois Theory}

Proofs and definitions can be found in  galois.pdf see:
\par \url{https://www.math.wisc.edu/~miller/old/m542-00/galois.pdf}

\begin{prop}(2.4 galois.pdf)
$\aut(E|F)$ is a group.  Furthermore, if $F\su E\su K$
are fields, then  $\aut(K|E)$ is a subgroup of $\aut(K|F)$.
\end{prop}

\begin{lemma} \label{roots}(2.5 galois.pdf)
Suppose $\si,\rho\in \aut(F(\al)|F)$.  Then
$\si=\rho$ iff $\si(\al)=\rho(\al)$.
Similarly, if  $\si,\rho\in \aut(F(\al_1,\al_2,\ldots,\al_n)|F)$
then $\si=\rho$ iff $\si(\al_k)=\rho(\al_k)$ for all
$k=1,2,\ldots,n$.
\end{lemma}

\begin{thm}(2.6 galois.pdf)
Suppose that $K$ is the splitting field of a polynomial in
$F[x]$ of degree $n$.  Then $\aut(K|F)$ is isomorphic to
a subgroup of $S_n$.
\end{thm}

\begin{prob}
(Fri Apr 4) Prove the following:
\par (a) Suppose $\al+\be$ is algebraic over $F$, then
$\al$ is algebraic over $F[\be]$.
\par (b) Suppose $\al+\be$ and $\al\be$ are both algebraic over $F$, then
$\al$ is algebraic over $F$.
\end{prob}

\begin{define}
For fields $F\su K$ we say that $K$ is a splitting field over $F$ iff
$K$ is the splitting field of some polynomial in $F[x]$.
\end{define}

Every polynomial in $\qq[x]$ splits in $\cc$ but $\cc$ is not a
splitting field over $\qq$.

\begin{lemma}
(Extension Lemma 2.9 galois.pdf) Suppose that $F\su F_1\su K$ and 
$F\su F_2\su K$
are fields, $K$ is a splitting field over $F$, and $\si:F_1\to F_2$
is an isomorphism which fixes $F$.  Then there exists $\rho:K\to K$
an automorphism which extends $\si$.
\end{lemma}


\begin{thm}(3.1 galois.pdf)
Suppose $F\su K$, $K$ is a splitting field over $F$, 
$p\in F[x]$ is irreducible, and there is $\al\in K$ such 
that $p(\al)=0$.   Then $p$ splits in $K$.
\end{thm}

\begin{prob}
(Mon Apr 7) Suppose that $F\su K_1\su L$ and $F\su K_2\su L$
and $K_1$ and $K_2$ are splitting fields over $F$.  Prove
that $K_1\cap K_2$ is a splitting field over $F$.
\end{prob}

\begin{thm}(2.8 galois.pdf)
Suppose $F\su K$, $K$ is a splitting field over $F$, and these
fields have characteristic zero.  Then $|\aut(K,F)|=[K:F]$.
\end{thm}

\begin{thm}(2.10 galois.pdf)
Suppose $F\su K\su E$, $K$ and $E$ are splitting fields over $F$.
Then $\aut(E|K)\normal\aut(E|F)$ and
$$\frac{\aut(E|F)}{\aut(E|K)}\isom \aut(K|F)$$
\end{thm}

\begin{prop}
Suppose $F\su K\su E$, $E$ is a splitting fields over $F$,
and $\aut(E|K)\normal\aut(E|F)$.  Then $K$ is a splitting field over $F$.
\end{prop}

\begin{prob}
(Wed Apr 9) Suppose that $E$ is a splitting field over $F$ and
$p\in F[x]$ splits in $E$ as 
$p(x)=(x-\al_1)(x-\al_2)\cdots(x-\al_n)$ where
$\al_i\neq \al_j$ whenever $i\neq j$. 
Prove that the following are equivalent:
\par (1) $p$ is irreducible.
\par (2) for any $i,j$ there is $\si\in\aut(E|F)$ such $\si(\al_i)=\al_j$.
\end{prob}

\begin{thm}(5.3 galois.pdf)
Suppose $F\su E$ is a radical Galois extension, then
$\aut(E|F)$ is a solvable group.
\end{thm}

\begin{examp} If $2$ generates the multiplicative group of
$\zz_p$, then $$f(x)=1+x+x^2+\cdots + x^{p-1}$$ is irreducible
over $\zz_2$.
\end{examp}

See Gurrier 1968
\par\noindent \url{http://www.jstor.org/stable/2315109}

See also Artin's conjecture on primitive roots
\par\noindent\url{http://en.wikipedia.org/wiki/Artin_conjecture}


\begin{thm}(5.4 galois.pdf)
Subgroups of solvable groups are solvable and homomorphic images
of solvable groups are solvable.
\end{thm}

\begin{prob}(Mon Apr 14)
For each of the following polynomials compute its Galois
group, i.e., $\aut(K|\qq)$ where $K$ is the splitting field
of the polynomial.
\par (a) $x^5-1$
\par (b) $x^4-2$
\par (c) $x^4-2x^2-2$
\end{prob}

\begin{thm}
Suppose $K$ is the splitting field of a polynomial in $F_1[x]$
and $F_1\su F_2\su\cdots \su F_m$ satisfies $F_1\su K\su F_m$ and
$F_{k+1}$ is a radical Galois extension of $F_k$ for each $k<m$.
Then $\aut(K|F_1)$ is a solvable group.
\end{thm}

\begin{lemma}. \label{generate}
\par(a) $\{(i,i+1):1\leq i<n\}$ generates $S_n$. (adjacent swaps)
\par(b) $\{(1,2),(1,2,3,\ldots,n)\}$ generates $S_n$. 
\par(c) $\{(1,i),(1,2,3,\ldots,n)\}$ generates $S_n$ if $n$ is prime.
\par(d) If $n$ is prime, then any subgroup of $S_n$ which contains
an $n$-cycle and at least one transposition must be $S_n$.
\end{lemma}

\begin{prob}
(Wed Apr 16) In the Lemma \ref{generate} (d) must $n$ be prime?
\par Prove or disprove: $\{(1,2,3,4),(1,3)\}$ generates $S_4$.
\end{prob}

\begin{thm}
Suppose $f(x)\in \qq[x]$ is an irreducible polynomial of prime degree
$p$ such that $f$ has exactly $p-2$ real roots.  If $K$ is the
splitting field of $f$, then $\aut(K,\qq)$ is isomorphic to $S_p$.
\end{thm}

\begin{examp}
If $f(x)=x^5-5x +\frac52$ then $f$ is irreducible and has exactly
three real roots.
\end{examp}

\begin{thm}
The alternating group $A_5$ is simple. Hence $S_5$ is not solvable.
\end{thm}

\begin{cor}
There is polynomial in $\qq[x]$ of degree 5 which cannot be solved
by radicals.
\end{cor}

\begin{prob}
(Mon Apr 21) $\si,\tau\in A_5$ are conjugate in $A_5$
iff there is $\rho\in A_5$ such that $\si=\rho^{-1}\tau\rho$.  Prove
that every element of $A_5$ except the identity is conjugate to 
exactly one of the following:
\par (a) $(1,2,3)$
\par (b) $(1,2)(3,4)$
\par (c) $(1,2,3,4,5)$
\par (d) $(2,1,3,4,5)$
\par\noindent In particular (c) and (d) are not conjugates.
\end{prob}

\begin{thm}
For any $n$ there are fields $E\su K$ such that $K$ is the splitting 
field of a polynomial in $E[x]$ and $\aut(K|E)$ is isomorphic to $S_n$.
\end{thm}

\begin{thm}
(char 0) Suppose $F\su K$ and $K$ is the splitting 
field of a polynomial in $F[x]$ and $H\su \aut(K|F)$ is a subgroup.
Then there exists a field $E$ with $F\su E\su K$ and $\aut(K|E)=H$.
\end{thm}

\begin{cor}
Every finite group is a Galois group.
\end{cor}

\noindent Proof of the fundamental theorem of algebra using Galois theory:
\par\url{http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra}

\begin{prob}
(Wed Apr 23) Let $K\supseteq\qq$ be the splitting field
of $$f(x)=(x^2+1)\cdot(x^2-2)=x^4-x^2-2$$
\par (a) Prove that $\aut(K|\qq)\isom \zz_2\times \zz_2$
\par (b) Find an irreducible polynomial $p(x)$ whose
splitting field is $K$.
\end{prob}

\begin{define}
A polynomial $f(x)\in\qq[x]$ is solvable by real radicals iff
its roots are in the smallest subfield $S\su\rr$ which is closed
under taking real roots, i.e., if $a\in S$, $a>0$ and $n\in \nn$
then $^n\sqrt{a}\in S$.
\end{define}

\begin{lemma}
Suppose $F\su\cc$ is a subfield, $p$ a prime, and $a\in F$.
Then $f(x)=x^p-a$ is reducible in $F$ iff it has a root in $F$.
\end{lemma}

\begin{thm}
Suppose $f(x)\in\qq[x]$ is an irreducible cubic with three real
roots.  Then $f(x)$ is not solvable by real radicals.
\end{thm}


\section{Similar Matrices}

For this material see 
\par\url{http://www.math.wisc.edu/~miller/old/m542-00/similar.pdf}


\begin{prob}(Mon Apr 28)
Suppose $A$ and $B$ are  matrices with real
entries and there exists a matrix $P$ with complex entries
such that $A=PBP^{-1}$.  Prove
there exists a matrix $P$ with real entries
such that $A=PBP^{-1}$.
\par Hint: Show $\{Q:AQ=QB\}$ is a subspace.
\end{prob}

\begin{thm}
Suppose $F$ is an infinite field, $A$ and $B$ are $F$-matrices, and
for some field extension $E\supseteq F$ there is an $E$-matrix
$P$ such that $A=PBP^{-1}$.  Then there is an $F$-matrix
$P$ such that $A=PBP^{-1}$.
\end{thm}

For algebraically closed fields $A$ and $B$ are similar iff
they have the same Jordan Normal forms up to a permutation
of the Jordan blocks.  So without loss we may as well assume
that $E$ is the algebraic closure of $F$.  By adding one new
element at a time it suffices to prove the Theorem for
$E=F[\al]$ with $[E:F]$ finite.  Let $p(x)\in F[x]$ be the
minimal polynomial for $\al$.  Consider the vector space
$$\mathcal{M}=\{P\;:\; AP=PB\}$$
where the $P$ are $E$-matrix.  Note that any such $P$ has entries
which are a polnomial in $\al$.  So we can write
$$P=P_0+\al P_1+\ldots+\al^nP_n$$
where the $P_i$ are $F$-matrices.  Let
$f(x)\in F[x]$ be the determinate of
$$P_0+xP_1+\ldots+x^nP_n$$
Since $f(\al)\neq 0$ and $F$ is an infinite field there is
an $\be\in F$ such that $f(\be)\neq 0$.  The $F$-matrix
$$P^\prime=P_0+\be P_1+\ldots+\be^nP_n$$
is invertible and witnesses the similarity of $A$ and $B$.



\end{document}
