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%\title{Math 772 -- Field Theory}
\title{Field Theory}

\author{Math 552}

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\makeindex


\begin{document}

\maketitle

\tableofcontents

\section{Algebraic Extensions}

\subsection{Finite and Algebraic Extensions}

\begin{defn}
  Let $1_F$ be the multiplicative unity of the field $F$.  
  \begin{enumerate}
  \item If $\sum_{i=1}^n 1_F \not= 0$ for any positive integer $n$, we
    say that $F$ has \midx{characteristic $0$}.

  \item Otherwise, if $p$ is the smallest positive integer such that
    $\sum_{i=1}^p 1_F = 0$, then $F$ has \midx{characteristic $p$}.
    (In this case, $p$ is necessarily prime.)

  \item We denote the characteristic of the field by $\chr(F)$.

  \item The \midx{prime field} of $F$ is the smallest subfield of $F$.
    (Thus, if $\chr(F)=p>0$, then the prime field of $F$ is $\F_p
    \idef \Z/p\Z$ (the filed with $p$ elements) and if $\chr(F)=0$,
    then the prime field of $F$ is $\Q$.)

  \item If $F$ and $K$ are fields with $F \subseteq K$, we say that
    $K$ is an \midx{extension} of $F$ and we write $K/F$.  $F$ is
    called the \midx{base field}.

  \item The \midx{degree} of $K/F$, denoted by $\dege{K}{F} \idef
    \dim_F K$, i.e., the dimension of $K$ as a vector space over $F$.
    We say that $K/F$ is a \midx{finite extension} (resp.,
    \midx{infinite extension}) if the degree is finite (resp.,
    infinite).

  \item $\alpha$ is \midx{algebraic} over $F$ if there exists a
    polynomial $f \in F[X]-\{0\}$ such that $f(\alpha)=0$.

  \end{enumerate}

\end{defn}\xsep



\begin{defn}
  If $F$ is a field, then
  $$
  F(\alpha) \idef \chav{ \frac{f(\alpha)}{g(\alpha)} \st f,g \in F[X]
    \text{ and } g(\alpha) \not=0 },
  $$
  is the smallest extension of $F$ containing $\alpha$.  (Hence
  $\alpha$ is algebraic over $F$ if, and only if,
  $F[\alpha]=F(\alpha)$.)

  In the same
  way,
  \begin{align*}
    F(\alpha_1, \dots , \alpha_n) &\idef \chav{ \frac{f(\alpha_1, \dots ,
        \alpha_n)}{g(\alpha_1, \dots , \alpha_n)} \st f,g \in
      F[X_1,\dots ,X_n]
      \text{ and } g(\alpha_1, \dots , \alpha_n) \not=0 } \\
    &= F(\alpha_1, \dots ,\alpha_{n-1})(\alpha_n)
  \end{align*}
  is the smallest extension of $F$ containing $\{\alpha_1, \dots, \alpha_n\}$.
\end{defn}\xsep


\begin{defn}
  If $K/F$ is a finite extension and $K=F[\alpha]$, then $\alpha$ is
  called a \midx{primitive element} of $K/F$.
\end{defn}\xsep

\begin{prop}
  For any $f \in F[X]-\{0\}$ there exists an extension $K/F$ such that
  $f$ has a root in $K$.  (E.g., $K \idef F[X]/(g)$, where $g$ is an
  irreducible factor of $f$.)
\end{prop}\xsep

\begin{theo}
  If $p(X) \in F[X]$ is irreducible of degree $n$, $K \idef
  F[X]/(p(X))$ and $\theta$ is the class of $X$ in $K$, then $\theta$
  is a root of $p(X)$ in $K$, $\dege{K}{F}=n$ and
  $\{1,\theta,\theta^2,\dots,\theta^{n-1}\}$ is an $F$-basis of $K$.
\end{theo}\xsep


\begin{rem}
  Observe that $F[\theta]$ (polynomials over $F$ evaluated at
  $\theta$), where $\theta$ is a root of an irreducible polynomial
  $p(X)$, is then a \emph{field}.  Observe that $1/\theta$ can be
  obtained with the \emph{extended Euclidean algorithm}: if $d(X)$ is
  the $\gcd(X,p(X))$ and $d(X)=a(X) \cdot X + b(X) \cdot p(X)$, the
  $1/\theta = a(\theta)$.
\end{rem}\xsep


\begin{defn}
  If $\alpha$ is algebraic over $F$, then there is a \emph{unique}
  monic irreducible over $F$ that has $\alpha$ as a root, called the
  \midx{irreducible polynomial} (or \midx{minimal polynomial}) of
  $\alpha$ over $F$, and we shall denote it $\irrp{\alpha}{F}(X)$.
  [\textbf{Note:} $(\irrp{\alpha}{F}(X)) = \ker \phi$, where $\phi :
  F[X] \rarrow F[\alpha]$ is the evaluation map.]
\end{defn}\xsep


\begin{cor}
  If $\alpha$ is algebraic over $F$, then $F(\alpha)=F[\alpha] \cong
  F[x]/(\irrp{\alpha}{F})$, and $\dege{F[\alpha]}{F} = \deg
  \irrp{\alpha}{F}$.  
\end{cor}\xsep


\begin{prop}
  If $K$ is a finite extension of $F$ and $\alpha$ is algebraic over
  $K$, then $\alpha$ is algebraic over $F$ and $\irrp{\alpha}{K}(X)
  \mid \irrp{\alpha}{F}(X)$. 
\end{prop}\xsep


\begin{defn}
  Let $\phi : R \rarrow S$ be a ring homomorphism.  If $f(X) = a_n X^n
  + \cdots + a_1 X + a_0$, then $f^{\phi} \idef \phi(a_n) X^n + \cdots
  + \phi(a_1) X + \phi(a_0) \in S[X]$.  [Note that $f \mapsto
  f^{\phi}$ is a ring homomorphism.]
\end{defn}\xsep

\begin{theo}
  Let $\phi : F \rarrow F'$ be an isomorphism, and $f \in F[X]$ be an
  irreducible polynomial.  If $\alpha$ is a root of $f$ in some
  extension of $F$ and $\alpha'$ is a root of $f^{\phi}$ in some
  extension of $F'$, then there exists an isomorphism $\Phi :
  F[\alpha] \rarrow F'[\alpha']$ such that $\Phi(\alpha)=\alpha'$ and
  $\restr{\Phi}{F} = \phi$.
\end{theo}\xsep


\begin{defn}
  $K/F$ is an \midx{algebraic extension} if every $\alpha \in K$ is
  algebraic over $F$.
\end{defn}\xsep


\begin{prop}
  If $\dege{K}{F}< \infty$, then $K/F$ is algebraic.
\end{prop}\xsep

\begin{rem}
  The converse is false.  E.g., $\alc{\Q} \idef \{ \alpha \in \C \st
  \alpha \text{ is algebraic over } \Q\}$ is an infinite algebraic
  extension of $\Q$.
\end{rem}


\begin{prop}
  If $L$ is a finite extension $K$ and $K$ is a finite extension of
  $F$, then
  $$
  \dege{L}{F}=\dege{L}{K}\cdot\dege{K}{F}.
  $$
  Moreover, if $\{\alpha_1,\dots,\alpha_n\}$ is an $F$-basis of $K$
  and $\{\beta_1,\dots,\beta_m\}$ is a $K$-basis of $L$, then $\{
  \alpha_i \cdot \beta_j : i \in \{1,\dots,n\} \text{ and } j \in
  \{1,\dots,m\}\}$ is an $F$-basis of $L$.
\end{prop}\xsep

\begin{defn}
  $\{\alpha_1, \dots ,\alpha_n\}$ \midx{generates} $K/F$ if
  $K=F(\alpha_1, \dots ,\alpha_n)$ and $K/F$ is \midx{finitely
    generated}.  (Not necessarily algebraic!)
\end{defn}\xsep


\begin{prop}
  $\dege{K}{F} < \infty$ if, and only if, $K$ is finitely generated over $F$
  by algebraic elements.
\end{prop}\xsep


\begin{cor}
  Let $K/F$ be an arbitrary extension, then
  $$
  E \idef \chav{\alpha \in K \st \alpha \text{ is a algebraic over } F},
  $$
  is a subfield of $K$ containing $F$.
\end{cor}\xsep



\begin{defn}
  If $F$ and $K$ are fields contained in the field $\bfd$, then the
  \midx{composite} (or \midx{compositum}) of $F$ and $K$ is the
  smallest subfield of $\bfd$ containing $F$ and $K$, and is denoted
  by $F \, K$.
\end{defn}\xsep

\begin{prop}
  \begin{enumerate}

  \item In general, we have:
    $$
    F \, K = \chav{ \frac{\alpha_1 \beta_1 + \cdots + \alpha_m
        \beta_m}{\gamma_1 \delta_1 + \cdots + \gamma_n \delta_n} \st
      \alpha_i, \gamma_i \in F; \; \beta_j, \delta_j \in K; \; \gamma_1
      \delta_1 + \cdots + \gamma_n \delta_n \neq 0}
    $$

    \item If $K_1/F$ and $K_2/F$ are finite extensions, with $K_1 =
      F[\alpha_1, \ldots, \alpha_m]$ and $K_2 = F[\beta_1, \ldots ,
      \beta_n]$, then $K_1 \, K_2 = F[\alpha_1, \ldots , \alpha_m,
      \beta_1 , \ldots , \beta_n]$, and $\dege{K_1 \, K_2}{F} \leq
      \dege{K_1}{F} \cdot \dege{K_2}{F}$.

  \end{enumerate}
\end{prop}\xsep


\begin{defn}
  Let ${\mcal C}$ be a class of field extensions.  We say that ${\mcal
    C}$ is \midx{distinguished} if the following three conditions are
  satisfied:
  \begin{enumerate}

  \item Let $F \subseteq K \subseteq L$. Then, $L/F$ is in ${\mcal C}$
    if, and only if, $L/K$ and $K/F$ are in ${\mcal C}$.

  \item If $K_1$ and $K_2$ are extensions of $F$, both contained in
    $\bfd$, then if $K_1/F$ is in ${\mcal C}$, then $K_1\, K_2/K_2$ is
    also in ${\mcal C}$.

  \item If $K_1$ and $K_2$ are extensions of $F$, both contained in
    $\bfd$, then if $K_1/F$ and $K_2/F$ are in ${\mcal C}$, then
    $K_1\, K_2/F$ is also in ${\mcal C}$.  [Note that this follows
    from the previous two.]

  \end{enumerate}
\end{defn}\xsep


\begin{defn}\label{defn:qd}
  Let ${\mcal C}$ be a class of field extensions.  We say that ${\mcal
    C}$ is \midx{quasi-distinguished} if the following three
  conditions are satisfied:
  \begin{enumerate}

  \item[($1'$)] Let $F \subseteq K \subseteq L$. Then, if $L/F$ is in
    ${\mcal C}$ then $L/K$ in ${\mcal C}$.

  \item[(2)] Same as (2) of distinguished.

  \item[(3')] Same as (3) of distinguished \emph{and} $(K_1 \cap
    K_2)/F$ also in $\mcal{C}$.

  \end{enumerate}
\end{defn}\xsep

\begin{rem}
  The above definition is \emph{not} standard.
\end{rem}


\begin{prop}
  The classes of algebraic extensions and finite extensions are
  distinguished.
\end{prop}\xsep


\subsection{Algebraic Closure}


\begin{defn}
  Let $K$ and $L$ be extensions of $F$.  
  \begin{enumerate}
  \item An \midx{embedding} (i.e., an injective homomorphism) $\phi: K
    \rarrow L$ is \midx{over} $F$ if $\rst{\phi}{F}=\idt_F$.

  \item If $E/K$ and $\psi:E \rarrow L$ is also an embedding, we say
    that $\psi$ is \midx{over} $\phi$, or is an \midx{extension} of
    $\phi$, if $\rst{\psi}{K}=\phi$.
  \end{enumerate}
\end{defn}\xsep


\begin{rem}
  Remember that if $\phi:F \rarrow F'$ is field homomorphism, then
  $\phi$ is either injective or $\phi \equiv 0$.
\end{rem}\xsep


\begin{defn}
  An \midx{algebraic closure} of $F$ is an algebraic extension $K$ in
  which any polynomial in $F[X]$ \midx{splits} [i.e., can be written as
  a product of linear factors] in $K[X]$.  We say that $F$ is
  \midx{algebraically closed} if it is an algebraic closure of itself.
\end{defn}\xsep


\begin{lemma}
  Let $K/F$ be algebraic.  If $\phi:K \rarrow K$ is an embedding over
  $F$, then $\phi$ is an automorphism.
\end{lemma}\xsep


\begin{lemma}
  Let $F$ and $K$ be subfields of $\bfd$ and $\phi: \bfd \rarrow L$ be
  an embedding into some field $L$.  Then $\phi(F \, K) = \phi(F) \,
  \phi(K)$.
\end{lemma}\xsep


\begin{theo}
  \begin{enumerate}

  \item For any field $F$, there exists an algebraic closure of $F$.

  \item An algebraic closure of $F$ is algebraically closed.

  \end{enumerate}
\end{theo}\xsep


\begin{defn}
  If
  $$
  f(X) = \sum_{i=0}^n a_i \, X^i \in F[X],
  $$
  then the \midx{formal derivative} of $f$ is
  $$
  \fder{f}(X) = \sum_{i=0}^n i \, a_i \, X^{i-1}.
  $$
\end{defn}\xsep


\begin{rem}
  The same formulas from calculus still hold (product rule, chain
  rule, etc.).
\end{rem}\xsep



\begin{lemma}
  Let $f \in F[X]$ and $\alpha$ a root of $f$.  Then $\alpha$ is a
  multiple root if, and only if, $\fder{f}(\alpha)=0$.
\end{lemma}\xsep

\begin{lemma}
  Let $\phi : F \rarrow F'$ be an embedding, $c, a_1 , \ldots, a_k \in
  F$, and $f \idef c(X-a_1) \cdots (X-a_k) \in F[X]$.  Then,
  $f^{\phi}(X) = \phi(c)(X-\phi(a_1)) \cdots (X-\phi(a_k))$.
\end{lemma}\xsep


\begin{theo}
  Let $f \in F[X]$ be an irreducible polynomial.  If $f$ splits in $K$
  as $f=c(X-\alpha_1)^{n_1} \cdots (X-\alpha_k)^{n_k}$, with the
  $\alpha_i$'s distinct, then $n_1= \cdots =n_k$.  [So, $f$ is a
  $n_1$-th power of a polynomial with simple roots.]  Moreover, if
  $K'$ is any other field where $f$ splits, and $n$ is the common
  exponent above [e.g, $n=n_1$], we must have $f = c(X-\alpha_1')^{n}
  \cdots (X-\alpha_k')^{n}$ in $K'[X]$.  [I.e., the number of distinct
  roots $k$ and the exponent $n$ are the same.]
\end{theo}\xsep


\begin{cor}
  If $f \in F[x]$ is irreducible and $\chr(F)=0$ [or $\fder{f} \neq
  0$], then $f$ has only simple roots [in any extension of $F$].
\end{cor}\xsep




\begin{theo}
  \begin{enumerate}

  \item If $\phi:F \rarrow K$ is an embedding of $F$, $K$ is
    algebraically closed and $\alpha$ is algebraic over $F$, then the
    number of extensions of $\phi$ to $F[\alpha]$ is equal to the
    number of distinct roots of $\irrp{\alpha}{F}(X)$.

  \item If $K/F$ is an algebraic extension, $\phi:F \rarrow L$, with
    $L$ algebraically closed, then there exists an extension $\psi:K
    \rarrow L$ of $\phi$.  Moreover, if $K$ is also algebraically
    closed and $L/\phi(F)$ is algebraic, then $\psi$ is an
    isomorphism.  [Hence the algebraic closure of a field is unique up
    to isomorphism, and we denote the algebraic closure of $F$ by
    $\alc{F}$.]

  \item If $K/F$ is an algebraic extension and $\alc{K}$ is an
    algebraic closure of $K$, then it is also an algebraic closure of
    $F$.  Conversely, if $\alc{F}$ is an algebraic closure of $F$ and
    $K'$ is the image of the embedding of $K$ into $\alc{F}$, then
    $\alc{F}$ is an algebraic closure of $K'$.

  \end{enumerate}
\end{theo}\xsep



\subsection{Splitting Fields}

\begin{defn}
  $K$ is a \midx{splitting field} of $f \in F[X]$ if $f(X)$ splits in
  $K$, but not in any proper subfield of $K$.  In particular if $f$
  splits in an extension of $F$ as $f = c(X-\alpha_1) \cdots (X -
  \alpha_n)$, then $F[\alpha_1, \ldots ,\alpha_n]$ is a splitting
  field of $f$.
\end{defn}\xsep


\begin{theo}
  If $K_1/F$ and $K_2/F$ are two splitting fields of $f \in F[X]$ [or
  of the same \emph{families} of polynomials] in different algebraic
  closure [so that they are distinct], then there exists an
  isomorphism between $K_1$ and $K_2$ over $F$ [induced by the
  isomorphism of the algebraic closures].
\end{theo}\xsep


\begin{rem}
  If $\alc{F}$ is an algebraic closure of $F$ and $\alpha_1, \dots ,
  \alpha_n \in \alc{F}$ are \emph{all} the roots of $f(X)$, then the
  splitting field of $F$ is $F[\alpha_1, \dots, \alpha_n]$.
\end{rem}\xsep


\begin{defn}
  $K$ is \midx{normal extension} of $F$ is it is algebraic over $F$
  and any embedding $\phi:K \rarrow \alc{K} = \alc{F}$ over $F$ is an
  automorphism of $K$.
\end{defn}\xsep


\begin{theo}
  Let $F \subseteq K \subseteq \alc{F}$.  The following are
  equivalent:
  \begin{enumerate}
  \item $K$ is normal.

  \item $K$ is a splitting field of a family of polynomials.

  \item Every polynomials in $F[X]$ that has a root in $K$, splits in
    $K[X]$.  
    
  \end{enumerate}
\end{theo}\xsep


\begin{theo}
  The class of normal extensions is \emph{quasi}-distinguished [but
  not distinguished].  Also, if $K_1/F$ and $K_2/F$ are normal, then
  so is $K_1 \cap K_2/F$.
\end{theo}\xsep

\begin{prop}
  If $\dege{K}{F}=2$, then $K/F$ is normal.
\end{prop}


\begin{rem}
  \begin{enumerate}
  \item $\Q(\sqrt{2})/\Q$ and $\Q(\sqrt[4]{2})/\Q(\sqrt{2})$ are
    normal extensions, but $\Q(\sqrt[4]{2})/\Q$ is \emph{not} normal.

  \item $\Q(\zeta_3,\sqrt[3]{2})/\Q$, where $\zeta_3 = \me^{2 \pi
      \mi/3}$, is normal, and $\Q \subseteq \Q(\sqrt[3]{2}) \subseteq
    \Q(\zeta_3,\sqrt[3]{2})$, but $\Q(\sqrt[3]{2})/\Q$ is \emph{not}
    normal.
  \end{enumerate}
\end{rem}\xsep



\subsection{Separable Extensions}

\begin{lemma}
  Let $\sigma : F \rarrow L$ and $\tau : F \rarrow L'$ be embeddings
  of $F$ into algebraically closed fields, and let $K/F$ be an
  algebraic extension.  Then, the number [or cardinality] of
  extensions of $\sigma$ to $K$ is the same as the number of
  extensions of $\tau$ to $K$.
\end{lemma}\xsep



\begin{defn}
  \begin{enumerate}
    
  \item Let $K/F$ be a finite extension and $\alc{F}$ be an algebraic
    closure of $F$.  Then, the \midx{separable degree} of $K/F$ is
    $$
    \degs{K}{F} \idef \text{number of embeddings } \phi:K \rarrow
    \alc{F} \text{ over } F.
    $$

  \item A polynomial $f \in F[X]$ is a \midx{separable polynomial} if
    it has no multiple roots.

  \item Let $\alpha$ be algebraic over $F$.  Then $\alpha$ is
    \midx{separable} over $F$ if $\irrp{\alpha}{F}(X)$ is separable.

  \item $K/F$ is a \midx{separable extension} if every element of $K$
    is separable over $F$.

  \end{enumerate}
\end{defn}\xsep


\begin{rem}
  If $\phi: F \rarrow L$ is embedding of $F$ and $L$ is
  algebraic closed, then
  $$
  \degs{K}{F} = \text{number of extensions } \psi:K \rarrow
  L \text{ of } \phi.
  $$
\end{rem}\xsep


\begin{theo}
  If $L/K$ and $K/F$ are algebraic extensions, then
  $$
  \degs{L}{F} = \degs{L}{K} \cdot \degs{K}{F}.
  $$
  Moreover, if $\dege{L}{F}< \infty$, then 
  $$
  \degs{L}{F} \leq \dege{L}{F},
  $$
  and $K/F$ is separable if, and only if, $\degs{L}{F} = \dege{L}{F}$.
\end{theo}\xsep


\begin{theo}
  If $K=F[\{ \alpha_i \st i \in I \}]$, where $I$ is a set of indices
  and $\alpha_i$ is separable over $F$ for all $i \in I$, then $K/F$
  is separable.
\end{theo}\xsep



\begin{theo}
  The class of separable extensions is distinguished.
\end{theo}\xsep


\begin{prop}
  Let $K$ be a finite extension of $F$ inside $\alc{F}$.  Then the
  smallest extension of $K$ which is normal over $F$ is $L \idef
  \phi_1(K)\dots \phi_n(K)$, where $\{\phi_1, \dots , \phi_n\}$ are
  \emph{all} the embeddings of $K$ into $\alc{F}$ over $F$.  (The
  $\phi_i(K)$'s are called the \midx{conjugates} of $K$.)  Moreover,
  if $K/F$ is separable, then $L$ is also separable over $F$.
\end{prop}\xsep


\begin{defn}
  \begin{enumerate}
    
  \item The field $L$ in the proposition above is called the
    \midx{normal closure} of $K/F$. 

    
  \item Let
    $$
    \sepc{F} \idef \text{compositum of all separable extensions of }F.
    $$
    $\sepc{F}$ is called the \midx{separable closure} of all $F$.

  \item If $K=F[\alpha]$, then $K$ is said to be a \midx{simple
      extension} of $F$.

  \end{enumerate}
\end{defn}\xsep




\begin{theo}[Primitive Element Theorem]\index{Primitive Element Theorem}
  If $\dege{F}{F} < \infty$, then $K/F$ has a primitive element if,
  and only if, there are finitely many \midx{intermediate fields}
  (i.e., fields $L$ such that $F \subseteq L \subseteq K$).  Moreover,
  if $K/F$ is (finite and) separable, then $K/F$ has a primitive
  element. 
\end{theo}\xsep


\begin{lemma}
  If $f \in F[X]$ is \emph{irreducible}, then $f$ has distinct roots
  if, and only if, $\fder{f}(X)$ is a non-zero polynomial.
\end{lemma}\xsep


\begin{prop}
  \begin{enumerate}

  \item $\alpha$ is separable over $F$ if, and only if,
    $\fder{(\irrp{\alpha}{F})} \not\equiv 0$.

  \item If $\chr(F)=0$, then any extension of $F$ is separable.

  \item Let $\chr(F)=p>0$. Then $\alpha$ is inseparable over $F$ if,
    and only if, $\irrp{\alpha}{F} \in F[X^p]$.  (And thus,
    $\irrp{\alpha}{F}$ is a $p$-power in $\alc{F}[X]$.)
  \end{enumerate}
\end{prop}\xsep



\subsection{Inseparable Extensions}


\begin{defn}
  An algebraic extension $K/F$ is \midx{inseparable} if it is not
  separable.  (Note that if $K/F$ is inseparable, then $\chr(F)=p>0$.)
\end{defn}\xsep


\begin{prop}
  If $F[\alpha]/F$ is finite and inseparable, then
  $\irrp{\alpha}{F}(X)=f(X^{p^k})$, where $p=\chr(F)$ [necessarily
  positive], for some positive integer $k$ and \emph{separable and
    irreducible} polynomial $f \in F[X]$.  Moreover,
  $\degs{F[\alpha]}{F} = \deg f$, $\dege{F[\alpha]}{F} = p^k \cdot
  \deg f$, and $\alpha^{p^k}$ is separable over $F$.
\end{prop}\xsep

\begin{cor}
  If $K/F$ is finite, then $\degs{K}{F} \mid \dege{K}{F}$.  If
  $\chr(F)=0$, then the quotient is $1$, and if $\chr(F)=p>0$, then
  the quotient is a power of $p$.
\end{cor}\xsep

\begin{defn}
  Let $K/F$ be a finite algebraic extension.  The inseparable degree
  of $K/F$ is
  $$
  \degi{K}{F} \idef \frac{\dege{K}{F}}{\degs{K}{F}}.
  $$
\end{defn}\xsep


\begin{prop}
  Let $K/F$ be a finite algebraic extension.  Then:
  \begin{enumerate}
  \item $K/F$ is separable if, and only if, $\degi{K}{F}=1$;

  \item if $E$ is an intermediate field, then
    $\degi{K}{F}=\degi{K}{E} \cdot \degi{E}{F}$.
  \end{enumerate}
\end{prop}\xsep

\begin{defn}
  \begin{enumerate}

  \item Let $\alpha$ be algebraic over $F$, with $\chr(F)=p$.  We say
    that $\alpha$ is \midx{purely inseparable} over $F$ if
    $\alpha^{p^n} \in F$ for some positive integer $n$.  [Thus,
    $\irrp{\alpha}{F} \mid X^{p^n}-\alpha^{p^n} = (X -
    \alpha)^{p^n}$.]

  \item An algebraic [maybe infinite] extension $K/F$ is a
    \midx{purely inseparable extension} if $\degs{K}{F}=1$.

  \end{enumerate}
\end{defn}\xsep


\begin{prop}
  An element $\alpha$ is purely inseparable if, and only if,
  $\irrp{\alpha}{F}(X) = X^{p^n} - a$ for some positive integer $n$
  and $a \in F$.  [Observe that $a=\alpha^{p^n}$.]
\end{prop}\xsep


\begin{prop}
  Let $K/F$ be an algebraic extension.  The following are equivalent:
  \begin{enumerate}
  \item $K/F$ is purely inseparable [i.e., $\degs{K}{F}=1$].

  \item All elements of $K$ are purely inseparable over $F$.

  \item $K=F[ \alpha_i \st i \in I ]$, for some set of indices $I$,
    with $\alpha_i$ purely inseparable over $F$.
  \end{enumerate}
\end{prop}\xsep


\begin{prop}
  The class of purely inseparable extensions is distinguished.
\end{prop}\xsep

\begin{defn} \label{def:gal}
  \begin{enumerate}

  \item Let $F$ be a field and $G$ be a subgroup of $\aut(F)$.  Then:
    $$
    \ffd{F}{G} \idef \{ \alpha \in F \st \phi(\alpha)=\alpha, \, \forall
    \phi \in G \},
    $$
    is the \midx{fixed field} of $G$.  (\textbf{Note:} it is a field.)

  \item The extension $K/F$ is a \midx{Galois extension} if it is
    normal and separable.  In this case, the \midx{Galois group} of
    $K/F$, denoted by $\gal(K/F)$ is the group of automorphisms of $K$
    over $F$ [i.e., automorphisms of $K$ which fix $F$].

  \end{enumerate}
\end{defn}\xsep


\begin{rem}
  If $K/F$ is Galois, then $\gal(K/F)$ is equal to the set of
  embeddings of $K$ into $\alc{K}$.  Also, if $K/F$ is finite, then
  $K/F$ is Galois if, and only if, $\card{\aut_F(K)} = \dege{K}{F}$,
  and so $\card{\gal(K/F)} = \dege{K}{F}$.
\end{rem}\xsep



\begin{rem}
  Note that for any field extension $K/F$ we have a group of
  automorphisms over $F$, which we denote by $\aut_F(K)$.  But,
  usually, the notation $\gal(K/F)$ is reserved for Galois extensions
  only.  [A few authors do use $\gal(K/F)$ for $\aut_F(K)$, though.]
\end{rem}\xsep


\begin{prop}
  Let $K/F$ be an algebraic extension.  Then
  $$
  K' \idef \{ x \in K \st x \text{ is separable over } F\}
  $$
  is a field [equal to the compositum of all separable extensions of
  $F$ that are contained in $K$].  [So, it is clearly the
  \emph{maximal} separable extension of $F$ contained in $K$.]  Then,
  $K'/F$ is separable and $K/K'$ is purely inseparable.
\end{prop}\xsep


\begin{cor}
  \begin{enumerate}
  \item $K/F$ is separable and purely inseparable, then $K=F$.

  \item If $\alpha$ is separable and purely inseparable over $F$, then
    $\alpha \in F$.
  \end{enumerate}
\end{cor}\xsep


\begin{cor}
  If $K/F$ is normal, then the maximal separable extension of $F$
  contained in $K$ [i.e., the $K'$ in the proposition above] is normal
  over $F$.  [Hence, $K'/F$ is Galois.]
\end{cor}\xsep


\begin{cor}
  If $F/E$ and $K/E$ are finite, with $F,K \subseteq \bfd$, with $F/E$
  separable and $K/E$ purely inseparable, then
  \begin{align*}
    \dege{F \, K}{K} &= \dege{F}{E} = \degs{F\, K}{E}, \\
    \dege{F \, K}{F} &= \dege{K}{E} = \degi{F\, K}{E}.
  \end{align*}
\end{cor}\xsep


\begin{defn}
  Let $F$ be a field [or a ring] of characteristic $p$, with $p$
  prime.  The \midx{Frobenius morphism} of $F$ is the map
  $$
  \begin{array}{rcl}
    \frob : F & \rarrow & F \\
            x & \mapsto & x^p.
  \end{array}
  $$
\end{defn}\xsep


\begin{cor}
  Let $K/F$ be a \emph{finite} extension in characteristic $p>0$ and
  $\frob$ be the Frobenius.
  \begin{enumerate}
  \item If $K^{\frob} F = K$, then $K/F$ is separable, where
    $$
    K^{\frob} = \frob(K) = \{ \frob(x) \st x \in K \}.
    $$

  \item If $K/F$ is separable, then $K^{\frob^n} F=K$ for any
    positive integer $n$.
  \end{enumerate}
\end{cor}\xsep


\begin{rem}
  \begin{enumerate}
  \item If $K=F[\alpha_1, \dots , \alpha_m ]$, then $K^{\frob^n} F =
    F[\alpha_1^{p^n}, \dots , \alpha_m^{p^n} ]$.

  \item Notice that if $K/F$ is an algebraic extension, we can always
    have an intermediate field $K'$ such that $K'/F$ is separable and
    $K/K'$ is purely inseparable, but not always we can have a $K''$
    such that $K''/F$ is purely inseparable and $K/K''$ is separable.
    [For example, take $F=\F_p(s,t)$, with $p>2$, and $K=F[\alpha]$,
    where $\alpha$ is a root of $X^p - \beta$ and $\beta$ is a root of
    $X^2 -s\, X + t$.]

    The next proposition states that if $K/F$ is normal, then there is
    such  a $K''$.
  \end{enumerate}
\end{rem}\xsep


\begin{prop}
  Let $K/F$ be \emph{normal} and $G \idef \aut_F(K)$ [where
  $\aut_F(K)$ is the set of automorphisms of $K$ over $F$] and
  $\ffd{K}{G}$ be the fixed field of $G$ [as in Definition
  \ref{def:gal}].  Then $\ffd{K}{G}/F$ is purely inseparable and
  $K/\ffd{K}{G}$ is separable.  [Hence, $K/\ffd{K}{G}$ is Galois.]

  Moreover, if $K'$ is the maximal separable extension of $F$
  contained in $K$, then $K=K' \, K^G$ and $K' \cap K^G=F$.
\end{prop}\xsep


\begin{defn}
  A field $F$ is a \midx{perfect field} if either $\chr(F)=0$ or
  $\chr(F)=p>0$ and the Frobenius $\frob:F \rarrow F$ is onto [or
  equivalently, every element of $F$ has a $p$-th root].  [Note that
  $\frob$ is always injective, so $\frob$ is, in in this case, an
  automorphism of $F$.]
\end{defn}\xsep


\begin{prop}
  Every algebraic extension of a perfect field $F$ is both perfect and
  separable over $F$.
\end{prop}\xsep



\subsection{Finite Fields}


\begin{theo}
  If $F$ is a field with $q$ [finite] elements, then:
  \begin{enumerate}
  \item $\chr(F)=p>0$ and so $\F_p \subseteq F$;

  \item $q=p^n$ for some positive integer $n$;

  \item $F$ is the splitting field of $X^q-X$ (over $\F_p$);

  \item any other field with $q$ elements is isomorphic to $F$, and
    in a fixed algebraic closure of $\F_p$, there exists only one
    field with $q$ elements, usually denoted by $\F_q$;

  \item there exists $\xi \in F$, such that $F^{\times}=\langle \xi
    \rangle$;

  \item for any positive integer $r$, there is a unique field with
    $p^r$ elements in a fixed algebraic closure $\alc{\F}_p$ of
    $\F_p$, which is the unique extension of $\F_p$ of degree $r$ in
    $\alc{\F}_p$.
  \end{enumerate}
\end{theo}\xsep


\begin{prop}
  Any algebraic extension of a finite field Galois [i.e., it is both
  normal and separable].
\end{prop}\xsep



\begin{prop}
  The set of automorphisms of $\F_{p^r}$ is $\{\idt, \frob , \frob^2,
  \dots , \frob^{r-1}\}$, where $\frob$ is the Frobenius map.  [Note
  that these are \emph{all} automorphisms, and they are automorphisms
  over $\F_p$.]
\end{prop}\xsep


\begin{prop}
  $\F_{p^s}$ is an extension of $\F_{p^r}$ if, and only if, $r \mid
  s$.  In this case, the set of embeddings of $\F_{p^s}$ into
  $\alc{\F}_p$ over $\F_{p^r}$ [or equivalently, since normal, the set
  of automorphisms of $\F_{p^s}$ over $\F_{p^r}$] is $\{\idt, \frob^r,
  \frob^{2r}, \dots ,\frob^{s-r}\}$, where $\frob$ is the Frobenius
  map.  [In other words, $\gal(\F_{p^s}/\F_{p^r}) = \cyclg{\frob^r}$.]
\end{prop}\xsep


\begin{prop}
  The algebraic closure $\alc{\F}_p$ is $\displaystyle \bigcup_{r > 0}
  \F_{p^r}$.  [Note that any finite union is contained in a single
  finite field.]
\end{prop}\xsep



\section{Galois Theory}

\subsection{Galois Extensions}

\begin{prop}
  Galois extensions form a quasi-distinguished class, and if $K_1/F$
  and $K_2/F$ are Galois, then so is $K_1 \cap K_2/F$.
\end{prop}\xsep


\begin{theo}
  Let $K/F$ be a Galois extension and $G \idef \gal(K/F)$.  Then
  \begin{enumerate}
  \item $\ffd{K}{G} = F$;

  \item if $E$ is an intermediate field ($F \subseteq E \subseteq K$),
    then $K/E$ is also Galois;

  \item the map $E \mapsto \gal(K/E)$ is injective.
  \end{enumerate}
\end{theo}\xsep

\begin{cor}
  Let $K/F$ be a Galois extension and $G \idef \gal(K/F)$.  If $E_i$
  is an intermediate field and $H_i \idef \gal(K/E_i)$, for $i=1,2$,
  then:

  \begin{enumerate}
  \item  $H_1 \cap H_2 = \gal(K/E_1 \, E_2)$;

  \item if $H = \cyclg{H_1, H_2}$ [i.e., $H$ is the smallest subgroup
    of $G$ containing $H_1$ and $H_2$], then $\ffd{K}{H} = E_1 \cap
    E_2$.
  \end{enumerate}
\end{cor}\xsep



\begin{cor}
  Let $K/F$ be separable and \textbf{finite}, and $L$ be the normal
  closure of $K/F$ [i.e., the smallest normal extension of $F$
  containing $K$].  Then $L/F$ is finite and Galois.
\end{cor}\xsep

\begin{lemma}
  Let $K/F$ be a separable extension such that for all $\alpha \in K$,
  $\dege{F[\alpha]}{F} \leq n$, for some fixed $n$.  Then $\dege{K}{F}
  \leq n$.
\end{lemma}\xsep


\begin{theo}[Artin]
  Let $K$ be a field, $G$ be a subgroup of $\aut(K)$ with $\card{G}=n
  < \infty$, and $F \idef \ffd{K}{G}$.  Then $K/F$ is Galois and
  $G=\gal(K/F)$ (and $\dege{K}{F}=n$).
\end{theo}\xsep


\begin{cor}
  Let $K/F$ be Galois and \textbf{finite} and $G \idef \gal(K/F)$.
  Then, for any subgroup $H$ of $G$, $H=\gal(K/\ffd{K}{H})$.
\end{cor}\xsep

\begin{rem}
  The above corollary is not true if the extension is infinite!  The
  map $H \mapsto \ffd{K}{H}$ is not injective!  For example,
  $\alc{\F}_p/\F_p$ is Galois, the cyclic group $H$ generated by the
  Frobenius is not the Galois group, and yet $\ffd{K}{H}=\F_p$.
\end{rem}\xsep


\begin{lemma}
  Let $K_1$ and $K_2$ be two extensions of $F$ with $\phi:K_1 \rarrow
  K_2$ an \emph{isomorphism over $F$}.  Then $\aut_F(K_2) = \phi \circ
  \aut_F(K_1) \circ \phi^{-1}$.
\end{lemma}\xsep

\begin{theo}
  Let $K/F$ be a Galois extension and $G \idef \gal(K/F)$.  If $E$ is
  an intermediate extension, then $E/F$ is normal [and thus Galois]
  if, and only if, $H\idef \gal(K/E)$ is a normal subgroup of $G$.  In
  this case, $\phi \mapsto \rst{\phi}{E}$ induces an isomorphism
  between $G/H$ and $\gal(E/F)$.
\end{theo}\xsep

\begin{defn}
  An extension $K/F$ is an \midx{Abelian extension} (resp., a
  \midx{cyclic extension}) if it is Galois and $\gal(K/F)$ is Abelian
  (resp., cyclic).
\end{defn}\xsep


\begin{cor}
  If $K/F$ is Abelian (resp., cyclic), then for any intermediate field
  $E$, $K/E$ and $E/F$ are Abelian (resp., cyclic).
\end{cor}\xsep

\begin{theo}[Fundamental Theorem of Galois Theory]\index{Fundamental Theorem of Galois Theory}
  Let $K/F$ be \textbf{finite} and Galois, with $G \idef \gal(K/F)$.
  The results above gives: the map
  $$
  \begin{array}{rcl}
    \{ \text{subgroups of } G \} & \longrightarrow & \{
    \text{intermediate fields of } K/F \} \\
    H & \longmapsto & \ffd{K}{H}
  \end{array}
  $$
  is a bijection with inverse
  $$
  \begin{array}{rcl}
    \{
    \text{intermediate fields of } K/F \} & \longrightarrow 
    & \{ \text{subgroups of } G \} \\
    E & \longmapsto & \gal(K/E).
  \end{array}
  $$
  Moreover an intermediate field $E$ is Galois if, and only if,
  $H \idef \gal(K/E)$ is normal in $G$, and $\gal(E/F) \cong G/H$,
  induced by $\phi \mapsto \rst{\phi}{E}$.
  
\end{theo}\xsep

\begin{rem}
  Note that the maps $H \mapsto \ffd{K}{H}$ and $E \mapsto \gal(K/E)$
  are \emph{inclusion reversing}, i.e., $H_1 \leq H_2$ implies
  $\ffd{K}{H_1} \supseteq \ffd{K}{H_2}$, and if $E_1 \subseteq E_2$,
  then $\gal(K/E_1) \geq \gal(K/E_2)$.
\end{rem}\xsep


\begin{theo}[Natural Irrationalities]\index{Natural Irrationalities}
  Let $K/F$ be a Galois extension and $L/F$ be an arbitrary extension,
  with $K,L \subseteq \bfd$ [so that we can consider the compositum
  $L\, K$].  Then $K\, L$ is Galois over $L$ and $K$ is Galois over $K
  \cap L$.  Moreover, if $G \idef \gal(K/F)$ and $H \idef \gal(K\,
    L/L)$, then for any $\phi \in H$, $\rst{\phi}{K} \in G$ and $\phi
  \mapsto \rst{\phi}{K}$ is an isomorphism between $H$ and $\gal(K/K
    \cap L)$.
\end{theo}\xsep

\begin{cor}
  If $K/F$ is finite and Galois and $L/F$ is an arbitrary extension,
  then $\dege{K\,L}{L} \mid \dege{K}{F}$.
\end{cor}\xsep

\begin{rem}
  The above theorem does not hold for if $K/F$ is not Galois.  For
  example, $F \idef \Q$, $K \idef \Q(\sqrt[3]{2})$ and $L \idef
  \Q(\zeta_3  \sqrt[3]{2})$, where $\zeta_3=\me^{2 \pi \mi/3}$.
\end{rem}\xsep


\begin{theo}
  Let $K_1/F$ and $K_2/F$ be Galois extensions with $K_1,K_2 \in
  \bfd$.  Then $K_1 \, K_2/F$ is Galois.  Moreover, if $G \idef
  \gal(K_1 \, K_2/F)$, $G_1 \idef \gal(K_1/F)$, $G_2 \idef
  \gal(K_2/F)$ and
  $$
  \begin{array}{rcl}
    \Phi : G & \rarrow & G_1 \times G_2 \\
    \phi & \mapsto & (\rst{\phi}{K_1},\rst{\phi}{K_2}),\\[0.8em]
  \end{array}
  $$
  then $\Phi$ is injective and if $K_1 \cap K_2 = F$, then $\Phi$ is
  an isomorphism.
\end{theo}\xsep

\begin{cor}
  If $K_i/F$ is Galois and $G_i \idef \gal(K_i/F)$ for $i=1, \dots ,
  n$ and $K_{i+1} \cap (K_1 \dots K_i)=F$ for $i=1,\dots,(n-1)$, then
  $\gal(K_1 \dots K_n/F)=G_1 \times \dots \times G_n$.
\end{cor}\xsep

\begin{cor}
  Let $K/F$ be finite and Galois, with $G \idef \gal(K/F) = G_1 \times
  \dots \times G_n$, $H_i \idef G_1 \times \dots \times G_{i-1} \times
  1 \times G_{i+1} \times \dots \times G_n$ and $K_i \idef
  \ffd{K}{H_i}$.  Then $K_i/F$ is Galois with $\gal(K_i/F) \cong G_i$,
  $K_{i+1} \cap (K_1 \dots K_i)=F$ and $K = K_1 \dots K_n$.
\end{cor}\xsep

\begin{cor}
  Abelian extensions are \emph{quasi}-distinguished [see Definition
  \ref{defn:qd}].  Moreover, if $K$ is an Abelian extension of $F$ and
  $E$ is an intermediate field, then $E/F$ is also Abelian.  [Hence,
  intersections of Abelian extensions are also Abelian.]
%   Let $F$, $K$ and $L$ be subfields of $\bfd$.
%   \begin{enumerate}
%   \item If $K$ and $L$ are Abelian extensions of $F$, then $K\, L/F$
%     is also Abelian.

%   \item If $K/F$ is Abelian and $L/F$ be any extension, then $K\, L/L$
%     is also Abelian.

%   \item If $K/F$ is Abelian and $E$ is an intermediate field, then
%     $K/E$ and $E/F$ are Abelian.
%   \end{enumerate}
\end{cor}\xsep

\begin{rem}
  Observe that, as with Galois extensions [and Abelian extensions are
  Galois by definition], we do \emph{not} always have that if $K/E$
  and $E/F$ are Abelian, then $K/F$ is Abelian.  For example,
  $\Q(\sqrt[4]{2})/\Q(\sqrt{2})$ and $\Q(\sqrt{2})/\Q$ are Abelian
  (since they are degree two extensions), but $\Q(\sqrt[4]{2})/\Q$ is
  not even Galois [since $X^4-2$ does not split in $\Q(\sqrt[4]{2})$].
\end{rem}\xsep



\subsection{Examples and Applications}

\begin{defn}
  The \midx{Galois group of a separable polynomial} $f \in F[X]$ is
  the Galois group of the splitting field of $f$ over $F$.  We will
  denote it by $\galp{f}$ or $\galp{f,F}$.
\end{defn}\xsep


\begin{prop}
  \begin{enumerate}

  \item Let $f \in F[X]$ be a [not necessarily separable or
    irreducible] polynomial, $K$ be its splitting field, and $n$ be
    the number of \emph{distinct} roots of $f$ [in $K$].  Then, $G
    \idef \aut_F(K)$ is a subgroup of the symmetric group $\symg{n}$,
    seen as permutations of the roots of $f$.  [In particular, any
    $\sigma \in G$ is determined by its values on the roots of $f$,
    and hence, if $\sigma \in G$ fixes all roots of $f$, then $\sigma
    = \idt_K$.]

  \item If $f \in F[X]$ is \emph{irreducible} [but not necessarily
    separable] and $K$, $n$, and $G$ are as above, then $G$ is a
    \midx{transitive subgroup} of $\symg{n}$ [i.e., for all $i,j \in
    \{1 , \ldots, n\}$, there is $\sigma \in G$ such that $\sigma(i) =
    j$.]

  \item Let $K/F$ be Galois [and hence \emph{separable}] with $G \idef
    \gal(K/F)$, $\alpha \in K$,
    $$
    \mcal{O} \idef \{ \sigma(\alpha) \st \sigma \in G \}
    $$
    be the \midx{orbit} of $\alpha$ by the action of $G$ in $K$.
    Then, $\mcal{O}$ is finite, say, $\mcal{O} = \{\alpha_1, \ldots,
      \alpha_k\}$, and 
    $$
    \irrp{\alpha}{F}= (x-\alpha_1) \cdots (x-\alpha_k).
    $$
    Note that $\card{\mcal{O}} \mid \dege{K}{F} = \card{G}$.

  \item Let $K/F$ be \emph{finite} and Galois with $G \idef \gal(K/F)$,
    and let $\alpha \in K$.  Then, $K=F[\alpha]$ if, and only if, the
    orbit of $\alpha$ by $G$ has exactly $[K:F]$ elements.

  \end{enumerate}
  
\end{prop}\xsep



\begin{prop}[Quadratic Extensions]
  \strut
  \begin{enumerate}

  \item If $\chr(F) \not=2$ and $[K:F]=2$, then there exists an $a \in
    F$ such that $K=F[\alpha]$, with $\irrp{\alpha}{F}=X^2 -a$.  Also,
    $\gal(K/F) \cong \Z/2\Z$ and the non-identity element is such that
    $\phi(\alpha)=-\alpha$.

  \item If $f \in F[X]$ is a quadratic separable polynomial, then the
    splitting field of $F$ has degree two over $F$, $\galp{f} \cong
    \Z/2\Z$ and the non-zero element of $\galp{f}$ is takes a root of
    $f$ to the other root.

%   \item If $\chr(F)=2$ and $[K:F]=2$, then for any $\alpha \in K-F$,
%     $K=F[\alpha]$.  If $K/F$ is \emph{separable}, then
%     $\gal(K/F)=\galp{\irrp{\alpha}{F}}$ [as above].

  \end{enumerate}
\end{prop}\xsep


\begin{defn}
  Let $f \in F[X]$, such that
  $$
  f(X) = \prod_{i=1}^n (X - \alpha_i).
  $$
  Then the \midx{discriminant} of $f$ is defined as
  $$
  \disc_f = \disc \idef \prod_{i < j} (\alpha_i - \alpha_j)^2.
  $$
\end{defn}\xsep


\begin{prop}
  For any $f \in F[X]$, $\disc_f \in F$.  In particular if $f=aX^2 +
  bX + c$, then $\disc_f = b^2 -4ac$ and if $f=X^3 +aX +b$, then
  $\disc_f = -4a^3 - 27 b^2$.
\end{prop}\xsep

\begin{prop}[Cubic Extensions and Polynomials]
  \strut
  \begin{enumerate}

  \item If $[K:F]=3$, then for any $\alpha \in K-F$, we have
    $K=F[\alpha]$.

  \item If $\chr(F)\not=3$ and $f \in F[X]$ is irreducible of degree
    $3$, say $f(X) = X^3 + a X^2 + bX +c$, then the splitting field of
    $f$ is the same as the splitting field of the polynomial
    $\tilde{f}(X) \idef f(X-a/3) = X^3 + \tilde{a}X + \tilde{b}$.
    [Hence $\galp{f} = \galp{\tilde{f}}$.]

  \item If the splitting field of a separable $f \in F[X]$ is of
    degree $3$, then $\galp{f} \cong \Z/3\Z$ and if $\alpha_1,
    \alpha_2, \alpha_3$ are the [distinct] roots of $f$, then
    $\galp{f} = \langle \phi \rangle$, where $\phi(\alpha_1)=\alpha_2$
    and $\phi(\alpha_2)=\alpha_3$ and $\phi(\alpha_3)=\alpha_1$. Note
    that in this case, $\galp{f} \cong \altg{3}$, where $\altg{n}$ is
    the \emph{alternating subgroup} of $\symg{n}$ [i.e., the subgroup
    of \emph{even} permutations].

  \item If the splitting field of a separable $f \in F[X]$ is
    \emph{not }of degree $3$, then $\galp{f} \cong \symg{3}$ [and
    hence $\galp{f}$ can permute the roots of $f$ in all possible
    ways].

  \item Let $f = \prod_{i=1}^3 (X-\alpha_i) \in F[X]$ and
    $$
    \delta \idef (\alpha_1 - \alpha_2)(\alpha_1 - \alpha_3)(\alpha_2
    - \alpha_3).
    $$
    [Thus, $\delta^2 = \disc_f$.]  If $f$ is irreducible in $F[X]$,
    $\disc_f \not=0$ [i.e., $f$ is separable] and $\chr(F) \not=2$,
    then $\galp{f} \cong \symg{3}$ if, and only if, $\delta \not\in F$
    [or equivalently, $\disc_f$ is not a square in $F$.]  [Note that
    if $\delta \not\in F$, then $F[\delta]/F$ is a degree two
    extension contained in the splitting field of $f$.]

  \end{enumerate}
\end{prop}\xsep


\begin{exs}
  From the above, we can deduce:

  \begin{enumerate}

  \item If $f \idef X^3 - X + 1 \in \Q[X]$, then $\disc_f = -23$, and
    hence $\galp{f} = \symg{3}$.

  \item If $f \idef X^3 - 3X + 1 \in \Q[X]$, then $\disc_f = 81$, and
    hence $\galp{f} = \Z/3\Z$.

  \end{enumerate}

\end{exs}\xsep


\begin{ex}
  If $f = X^4 - 2 \in \Q[X]$, then $\galp{f} \cong \mathrm{D}_8$, the
  dihedral group of $8$ elements.  More precisely, if $\phi \in
  \gal(Q[\sqrt[4]{2},\mi]/\Q[\mi])$ such that $\phi(\sqrt[4]{2}) =
  \sqrt[4]{2} \, \mi$ and $\psi \in \gal(Q[\sqrt[4]{2},\mi]/\Q[\sqrt[4]{2}])$
  such that $\psi(\mi) = - \mi$ [i.e., $\psi$ is the complex
  conjugation], then
  \begin{align*}
    \galp{f} &= \left\langle \phi, \psi \st \phi^4 = \idt, \; \psi^2 =
    \idt, \;
    \psi \circ \phi = \phi^3 \circ \psi \right\rangle \\
    &= \chav{ \idt, \; \phi, \; \phi^2, \; \phi^3, \; \psi, \; \phi
      \circ \psi, \; \phi^2 \circ \psi , \; \phi^3 \circ \psi }.
  \end{align*}
\end{ex}\xsep

\begin{prop}
  Let $E$ be a field, $t_1, \dots , t_n$ be algebraically independent
  variables over $E$, $s_1, \dots , s_n$ be their elementary symmetric
  functions, $F \idef E(s_1, \dots , s_n)$ and $K \idef E(t_1, \dots ,
  t_n)$.  Then $\irrp{t_i}{F} = \prod_{i=1}^n (X-t_i)$ and $\gal(K/F)
  \cong \symg{n}$.
\end{prop}\xsep


\begin{theo}[Fundamental Theorem of Algebra]\index{Fundamental Theorem
    of Algebra}
  $\C$ is the algebraic closure of $\R$.  
\end{theo}\xsep

\begin{lemma} \label{lemma:transpcy}
  If $G \subseteq \symg{p}$, with $p$ prime, and $G$ contains a
  transposition and a $p$-cycle, then $G=\symg{p}$.
\end{lemma}\xsep


\begin{prop}
  If $f \in \Q[X]$ is irreducible, $\deg f = p$, with $p$ prime, and if
  $f$ has exactly two complex roots, then $\galp{f} \cong \symg{p}$.
\end{prop}\xsep


\begin{ex}
  As an application of the proposition above, let $f \idef X^5 - 4X +2
  \in \Q[X]$.  Then $\galp{f} \cong \symg{5}$.  In fact, one can use
  the above proposition to prove that for every prime $p$ there is a
  polynomial $f_p \in \Q[X]$ such that $\galp{f_p,\Q} = \symg{p}$.
  [One can get all $\symg{n}$, in fact, but it is harder.]
\end{ex}\xsep


% \begin{prop}\label{prop:ffcycle}
%   Let $f \in \F_q[X]$ be a separable polynomial and $f = f_1 \dots
%   f_k$, with $f_i$ irreducible of degree $n_i$.  Then $\galp{f}$ is
%   cyclic generated by $\frob^r$, where $\frob$ is the Frobenius and
%   $q=p^r$, with $p=\chr(\F_q)$. Moreover, as a permutation of the
%   roots, $\frob^r$ is a product of $k$ disjoint cycles of orders
%   $n_1, n_2, \dots, n_k$, where the $n_i$-cycle is a permutation of
%   the roots of $f_i$.
% \end{prop}\xsep


\begin{theo}
  Let $f \in \Z[X]$ be a monic separable polynomial, $p$ be a prime
  that does not divide the discriminant of $f$, and $\bar{f} \in
  \Z/p\Z[X]$ be the reduction modulo $p$ of $f$ [i.e., obtained by
  reducing the coefficients].  Then, there is a bijection between the
  roots of $f$ and the roots of $\bar{f}$, denoted by $\alpha \mapsto
  \bar{\alpha}$, and an injection $i : \galp{\bar{f}} \rarrow
  \galp{f}$, such that, if $\phi \in \galp{\bar{f}}$ and
  $\bar{\alpha}_i$ and $\bar{\alpha}_j$ are roots of $\bar{f}$, with
  $\phi(\bar{\alpha}_i) = \bar{\alpha}_j$, then
  $i(\phi)(\alpha_i)=\alpha_j$.

  In particular, if $\phi \in \galp{\bar{f}}$, then $\galp{f}$ has an
  element [namely $i(\phi)$] that has the same cycle structure [seen
  as a permutation] as $\phi$ itself.  [E.g., if $\phi$ as a
  permutation is a product of a two-cycle, a 4-cycle and a 7-cycle
  [all disjoint], then $i(\phi)$ is also a product of a two-cycle, a
  4-cycle and a 7-cycle [all disjoint] in $\galp{f}$.]
\end{theo}\xsep


% \begin{prop}
%   Let $f \in \F_p[X]$.  Then $f$ has an irreducible factor of degree
%   $n$ if, and only if, for every divisor $d$ of $n$ different from $n$
%   itself, $\gcd(f,g_d) \not=1$, where
%   $$
%   g_d \idef \frac{X^{p^n-1}-1}{X^{p^d-1}-1}.
%   $$
%   (Observe that $g_d \in \F_p[X]$.)
% \end{prop}\xsep

\begin{ex}
  As an application of the theorem above, one can prove that $f \idef
  X^5 - X - 1 \in \Z[X]$ is such that $\galp{f}=\symg{5}$, by reducing
  $f$ modulo $5$ and modulo $2$.
\end{ex}\xsep


% \begin{theo}[Schur]
%   \strut
%   \begin{enumerate}

%   \item Let 
%     $$
%     f \idef \sum_{i=0}^n \frac{X^i}{i!}.
%     $$
%     Then, if $4 \nmid n$, then $\galp{f}=\symg{n}$ and if $4 \mid n$,
%     then $\galp{f}=\altg{n}$.

%   \item Let 
%     $$
%     H_n(X) \idef (-1)^n \, \me^{X^2/2} \,
%     \frac{\md^n}{\md X^n}\parent{\me^{-X^2/2}}
%     $$
%     be the $n$-th \midx{Hermitian polynomial} (note that $H_m \in
%     \Z[X]$) and let $K_n^{(0)}, K_n^{(1)} \in \Z[X]$ be defined as
%     $$
%     H_{2n}(X)=K_n^{(0)}(X^2), \quad  H_{2n+1}(X)=X \, K_n^{(1)}(X^2).
%     $$
%     Then, $\galp{K_n^{(i)}}=\symg{n}$ for $i=0,1$ and $n>12$.
    
%   \end{enumerate}
% \end{theo}\xsep


% \begin{lemma}
%   If $G \subseteq \symg{n}$ and $G$ is transitive, contains a
%   transposition and an $(n-1)$-cycle, then $G=\symg{n}$.
% \end{lemma}\xsep


% \begin{theo}
%   For any $n>0$, there exists $f \in \Q[X]$ such that $\galp{f} =
%   \symg{n}$.
% \end{theo}\xsep



\subsection{Roots of Unity}


\begin{defn}
  \strut
  \begin{enumerate}

  \item A $n$-th \midx{root of unity} in a field $F$ is a root of $X^n
    -1$ in $F$.  A \emph{root of unity} [with no $n$ specified] is a
    root of unit for some $n$.

  \item The set of all roots of unity form an Abelian group, denoted
    by $\gru{}(F)$ or simply $\gru{}$.

  \item The set of $n$-th roots of unity in $F$ is a \emph{cyclic}
    group denoted by $\gru{n}(F)$ or simply $\gru{n}$.

    \item If $\chr(F) \nmid n$, then $\card{\gru{n}}=n$ and a
      generator of $\gru{n}$ is called a \midx{primitive $n$-th root
        of unity}.

  \end{enumerate}
\end{defn}\xsep


\begin{prop}
  \begin{enumerate}

  \item If $\chr(F) = p>0$, $n=p^rm$, and $p \nmid m$, then
    $\gru{n}(F) = \gru{m}(F)$ [and so $\card{\gru{n}(F)}=m$].
    
  \item If $\gcd(n,m)=1$, then $\gru{n}\times\gru{m} \cong \gru{n}
    \cdot \gru{m} = \gru{n\, m}$ and the isomorphism is given by
    $(\zeta,\zeta') \mapsto \zeta \, \zeta'$.  [In particular, if
    $\zeta_n$ and $\zeta_m$ are primitive $n$-th and $m$-th roots of
    unity, then $\zeta_n \zeta_m$ is a primitive $nm$-th root of
    unity.]

  \end{enumerate}
\end{prop}\xsep

\begin{prop} \label{prop:cyclo1}
  Let $F$ be a field such that $\chr(F) \nmid n$, and $\zeta_n$ a
  primitive $n$-th root of unity.  Then $F[\zeta_n]/F$ is Galois.  If
  $\phi \in \gal(F[\zeta_n]/F)$, then
  $\phi(\zeta_n)=\zeta_n^{i(\phi)}$, for some $i(\phi) \in (\Z/n
  \Z)^{\times}$ and this map $i : \gal(F[\zeta_n]/F) \rarrow
  (\Z/n\Z)^{\times}$ is injective.  Thus, $\gal(F[\zeta_n]/F)$
  is Abelian.
\end{prop}\xsep


\begin{rem}
  Note that $\gal(F[\zeta_n]/F)$ is not necessarily cyclic.  For
  example, $\gal(\Q[\zeta_8]/\Q) \cong (\Z/2\Z)\times(\Z/2\Z)$.
\end{rem}\xsep


\begin{defn}
  We say that $K/F$ is a \midx{cyclotomic extension} if there exists a
  root of unity $\zeta$ over $F$ such that $K=F[\zeta]$.
  [\emph{Careful}: in Lang, an extension is cyclotomic if there exists
  a root of unity $\zeta$ over $F$ such that $K \subseteq F[\zeta]$!]
\end{defn}\xsep


\begin{defn}
  Let $\ephi : \Z \rarrow \Z$ denote the \midx{Euler phi-function},
  which is defined as
  $$
  \ephi(n) \idef \card{\{ m \in \Z \st 0<m<n \text{ and } \gcd(m,n)=1\}}.
  $$
\end{defn}\xsep


\begin{theo}
  If $\zeta_n$ is a primitive $n$-th root of unity in $\Q$, then
  $[\Q[\zeta_n]:\Q] = \ephi(n)$ and the map $i : \gal(F[\zeta_n]/F)
  \rarrow (\Z/n\Z)^{\times}$ [as in Proposition \ref{prop:cyclo1}] is
  an isomorphism.
\end{theo}\xsep


\begin{cor}
  If $\zeta_m$ and $\zeta_n$ are a primitive $m$-th root of unity and
  primitive $n$-th root of unity, respectively, with
  $\gcd(m,n)=1$,then $\Q[\zeta_m] \cap \Q[\zeta_n]=\Q$,
\end{cor}\xsep


\begin{rem}
  If $m = \lcm(n_1,\dots,n_r)$, and $\zeta_{n_i}$ is a primitive
  $n_i$-th root of unity for $i=1,\dots,r$, then $\Q[\zeta_{n_1}]
  \cdots
  \Q[\zeta_{n_r}] = \Q[\zeta_m]$.
\end{rem}\xsep


\begin{defn}
  Let $n$ be a positive integer not divisible by $\chr(F)$.  The
  polynomial
  $$
  \cyclop{n}(X) \idef \prod_{\substack{\zeta \text{ prim.\ $n$-th} \\
      \text{ root of $1$ in $F$}}} (X - \zeta)
  $$
  is called the $n$-th \midx{cyclotomic polynomial} [over $F$].
\end{defn}\xsep

\begin{prop}
  \strut
  \begin{enumerate}

  \item $\deg \cyclop{n} = \ephi(n)$.

  \item If $\zeta_n$ is a primitive $n$-th root of unity, then
    $\cyclop{n}(X) = \irrp{\zeta_n}{\Q}(X)$.

  \item If $\zeta_n$ is a primitive $n$-th root of unity, then
    $$
    \cyclop{n}(X) = \prod_{\phi \in \gal(\Q[\zeta_n]/\Q)} (X -
    \phi(\zeta_n))
     $$

   \item $\displaystyle X^n -1 = \prod_{d \mid n} \cyclop{d}(X)$.\vspace{1.8ex}

   \item If $\chr(F)=0$, then $\cyclop{n} \in \Z[X]$ for all $n$.  If
     $\chr(F)=p>0$, then $\cyclop{n} \in \F_p[X]$ for all $n$ [not
     divisible by $p$].

  \end{enumerate}
\end{prop}\xsep



\begin{prop}
  \strut
  \begin{enumerate}

  \item If $p$ is prime, then $\cyclop{p}(X)=X^{p-1} + X^{p-2} +
    \cdots + X + 1$.

  \item If $p$ is prime, then
    $\cyclop{p^r}(X)=\cyclop{p}(X^{p^{r-1}})$.

  \item If $n=p_1^{r_1} \cdots p_s^{r_s}$, with $p_i$'s distinct
    primes, then $\cyclop{n}(X) = \cyclop{p_1 \cdots
      p_s}(X^{p_1^{r_1-1} \cdots p_s^{r_s-1}})$.

  \item If $n>1$ is odd, then $\cyclop{2n}(X)=\cyclop{n}(-X)$.\vspace{1.2ex}

  \item If $p \nmid n$, with $p$ an odd prime, then $\displaystyle
    \cyclop{p\,n}(X)=\frac{\cyclop{n}(X^p)}{\cyclop{n}(X)}$.\vspace{1.8ex}

  \item If $p \mid n$, with $p$ prime, then $\cyclop{p \, n}(X) =
    \cyclop{n}(X^p)$.

  \end{enumerate}
\end{prop}\xsep


\begin{rem}
  It is \emph{not} true that for all $n$, the coefficients of
  $\cyclop{n}(X)$ are either $0$, $1$ or $-1$.  The first $n$ for
  which this fails is $105 = 3 \cdot 5 \cdot 7$.
\end{rem}\xsep


% \begin{defn}
%   Let $p$ be a prime.  The \midx{Legendre symbol} modulo $p$ is
%   defined as
%   $$
%   \legs{k}{p}\idef
%   \begin{cases}
%     0, & \text{if } \; k \equiv 0 \pmod{p}; \\
%     1, & \text{if } \; k \equiv x^2 \pmod{p} \; \text{ for some } x
%     \not\equiv 0
%     \pmod{p}; \\
%     -1, & \text{if } \; k \not\equiv x^2 \pmod{p} \; \text{ for any }
%     x \in \Z.
%   \end{cases}
%   $$
% \end{defn}\xsep


% \begin{theo}
%   Let $p$ be an odd prime and $\zeta_p$ a primitive $p$-th root of
%   unity.  If
%   $$
%   S \idef \sum_{k=1}^{p-1} \legs{k}{p} \cdot \zeta_p^k.
%   $$
%   Then $\displaystyle S^2=\legs{-1}{p} \cdot p$.
% \end{theo}\xsep


% \begin{cor}
%   Every quadratic extension of $\Q$  is contained in a cyclotomic extension.
% \end{cor}\xsep


% \begin{prop}
%   If $\Q \subseteq F \subseteq \Q[\zeta_p]$, where $p$ is prime and
%   $\zeta_p$ is a primitive $p$-th root of unity, then $F=\Q[\alpha]$,
%   where
%   $$
%   \alpha \idef \sum_{\phi \in \gal(\Q[\zeta_p]/F)} \phi(\zeta_p).
%   $$
% \end{prop}\xsep


\begin{theo}[Dirichlet's Theorem of Primes in Arithmetic
  Progression]\index{Dirichlet's Theorem of Primes in Arithmetic
    Progression}
  If $\gcd(a,r)=1$, there are infinitely many primes in the arithmetic
  progression
  $$
  a, \; a + r, \; a + 2r , \; a + 3r, \dots.
  $$
\end{theo}\xsep

\begin{theo}
  Given a finite Abelian group $G$, there exists an extension $F/\Q$
  such that $\gal(F/\Q)=G$.
\end{theo}\xsep


\begin{theo}[Kronecker-Weber]\index{Kronecker-Weber Theorem}
  If $F/\Q$ is finite and Abelian, then there exists a cyclotomic
  extension $\Q[\zeta]/\Q$ such that $F \subseteq \Q[\zeta]$.
\end{theo}\xsep



\subsection{Linear Independence of Characters}

\begin{defn}
  Let $G$ be a \midx{monoid} [i.e., a ``group'' which might not have
  inverses] and $F$ be a field.  A \midx{character} of $G$ in $F$ is a
  homomorphism $\chi: G \rarrow F^{\times}$.  The \midx{trivial
    character} is the map constant equal to $1$.

  Let $f_i:G \rarrow F$ for $i=1, \dots ,n$.  We say that the $f_i$'s
  are \midx{linearly independent} if
  $$
  \alpha_1 \, f_1 + \dots \alpha_n \, f_n =0, \quad \alpha_i \in F,
  $$
  then $\alpha_i=0$ for all $i$.
\end{defn}\xsep


\begin{rems}
  \begin{enumerate}
  \item If $K/F$ is a field extension and $\{ \phi_1, \dots ,
    \phi_n\}$ are the embedding of $K$ over $F$, then we can think of
    $\restr{\phi}{K^{\times}}$ as characters of $K^{\times}$ in $K$.

  \item If one says only a character in $G$ (without mention of the
    field), one usually means a character from $G$ in $\C^{\times}$ or
    even in
    $$
    S^1 \idef \{ \zeta \in \C \st \abs{\alpha}=1 \}.
    $$
  \end{enumerate}
\end{rems}\xsep


\begin{theo}[Artin]
  If $\chi_1, \dots, \chi_n$ \emph{distinct} characters of $G$ in $F$,
  then they are linearly independent.
\end{theo}\xsep


\begin{cor}
  Let $\alpha_1, \dots , \alpha_n$ be \emph{distinct}
  elements of a field $F^{\times}$.  If $a_1, \dots , a_n \in F$ such
  that for all positive integer $r$ we have
  $$
  a_1 \, \alpha_1^r + \dots + a_n \, \alpha_n^r = 0,
  $$
  then $a_i=0$ for all $i$.
\end{cor}\xsep


\begin{cor}
  For any extension $K/F$, the set $\emb{K/F}$ is linearly
  independent over $K$.
\end{cor}\xsep

\subsection{Norm and Trace}

\begin{defn}
  Let $K/F$ be a finite extension, with $\degs{K}{F}=r$ and
  $\degi{K}{F}=p^{\mu}$.  [So, $\chr(F)=p$ or $\degi{K}{F}=1$.]  Let
  $\emb{K/F} = \{ \phi_1, \dots , \phi_n \}$ and $\alpha \in K$:
  \begin{enumerate}

  \item The \midx{norm} of $\alpha$ from $K$ to $F$ is
    $$
    \norm{K}{F}(\alpha) \idef \prod_{i=1}^n \phi_i(\alpha^{p^{\mu}}) =
    \parent{ \prod_{i=1}^n \phi_i(\alpha) }^{\degi{K}{F}}.
    $$

  \item The \midx{trace} of $\alpha$ from $K$ to $F$ is
    $$
    \trace{K}{F}(\alpha) \idef \degi{K}{F} \cdot \sum_{i=1}^n
    \phi_i(\alpha).
    $$

  \end{enumerate}
\end{defn}\xsep


\begin{rem}
  Note that if $K/F$ is inseparable, then $\trace{K}{F}(\alpha)=0$.
\end{rem}\xsep


\begin{lemma}
  \strut

  \begin{enumerate}

  \item Let $K/F$ be a finite extension, and $\emb{K/F} = \{ \phi_1,
    \dots , \phi_n \}$ be the set of embeddings of $K$ over $F$.  If
    $L/K$ is an algebraic extension and $\psi : L \rarrow \alc{F}$ is
    an embedding over $F$, then
    $$
    \{ \psi \circ \phi_1 , \dots , \psi\circ \phi_n \}=\emb{K/F}.
    $$

  \item Let $F \subseteq K \subseteq L$ be field extensions.  Let
    $$
    \emb{K/F} = \{ \phi_1 , \dots \phi_r \},
    $$
    and
    $$
    \emb{L/K} = \{ \psi_1 , \dots \psi_s \}.
    $$
    If $\tilde{\phi}_i : \alc{F} \rarrow \alc{F}$ is an extension of
    $\phi_i$ to $\alc{F}$ (which exists since $\alc{F}/F$ is
    algebraic), then
    $$
    \emb{L/F} = \{ \tilde{\phi}_i \circ \psi_j \st i \in \{1, \dots , r
    \} \text{ and } j \in \{1, \dots , s\} \}.
    $$
    
  \item Let $K/F$ be a \emph{separable} extension.  If $\alpha \in K$
    is such that $\phi(\alpha)=\alpha$ for all embeddings $\phi \in
    \emb{K/F}$, then $\alpha \in F$.

  \end{enumerate}
\end{lemma}\xsep


\begin{theo}
  Let $L/F$ be a finite extension.

  \begin{enumerate}

  \item For all $\alpha \in K$, $\norm{K}{F}(\alpha),
    \trace{K}{F}(\alpha) \in F$.

  \item If $\dege{K}{F}=n$ and $\alpha \in F$,
    $\norm{K}{F}(\alpha)=\alpha^n$ and $\trace{K}{F}(\alpha)=n \cdot
    \alpha$.

  \item $\restr{\norm{K}{F}}{K^{\times}} : K^{\times} \rarrow
    F^{\times}$ is a [multiplicative] group homomorphism and
    $\trace{K}{F}: K \rarrow F$ is an [additive] group homomorphism.

  \item If $K$ is an intermediate field, then 
    \begin{align*}
      \norm{L}{F} &=\norm{K}{F} \circ \norm{L}{K}, \\
      \trace{L}{F} &=\norm{K}{F} \circ \trace{L}{K}.
    \end{align*}

  \item If $L=F(\alpha)$, where $\irrp_{\alpha,F}(X) = X^{n} + a_{n-1}
    \, X^{n-1} + \dots + a_1 \, X + a_0$, then
    $$
    \norm{L}{F}(\alpha) = (-1)^n \,  a_0,  \qquad
    \trace{L}{F}(\alpha)=-a_{n-1}. 
    $$

  \end{enumerate}
\end{theo}\xsep


\begin{cor}
  If $F \subseteq F(\alpha) \subseteq K$, with $\dege{K}{F}=n$,
  $\irrp_{\alpha,F}(X)=X^d + a_{d-1} \, X^{d-1} + \dots + a_1 \, X +
  a_0$, and $\dege{L}{F(\alpha)}=e$, then
  $$
  \norm{L}{F}(\alpha) = (-1)^n \, a_0^e, \qquad 
  \trace{L}{F}(\alpha) = (-a_{d-1})^e. \qquad 
  $$
\end{cor}\xsep

\begin{rem}
  $\trace{K}{F}: K \rarrow F$ is an $F$-linear map.
\end{rem}\xsep


% \begin{defn}
%   Let $V$ be a vector space over $F$.  Then the \midx{dual space} of
%   $V$ is
%   $$
%   \dual{V} \idef \chav{ T : V \rarrow F \st T \text{ is linear over }
%     F }.
%   $$
%   If $\basis{B}=\{w_1, \dots , w_n\}$ is an $F$-basis of $V$, then
%   the \midx{dual basis} of $\basis{B}$ is an $F$-basis of $\dual{V}$,
%   say $\dual{\basis{B}}=\{T_1, \dots , T_n\}$, such that
%   $T_i(w_j)=\krd_{i\, j}$, where $\krd_{i \, j}$ is the Kronecker
%   delta-function:
%   $$
%   \krd_{i\, j} =\begin{cases}
%     1, & \text{if }i=j; \\
%     0, & \text{if }i \not= j.
%   \end{cases}
%   $$
% \end{defn}\xsep


% \begin{theo}
%   Let $K/F$ be finite and separable.  Then, $\trace{K}{F}$ is a
%   non-zero map, $(x,y) \mapsto \trace{K}{F}(x \, y)$ is a bilinear
%   map, and identify $K$ (as an $F$-vector space) with its dual space.
% \end{theo}\xsep


% \begin{cor}
%   Let $\{w_1, \dots, w_n\}$ be an $F$-basis of $K$.  Then there exists
%   another $F$-basis $\{w_1', \dots, w_n'\}$ of $K$ such that
%   $\trace{K}{F}(w_i \, w_j')=\krd_{i\, j}$.
% \end{cor}\xsep


% \begin{cor}
%   Let $K/F$ is finite, separable, $\{ w_1, \dots , w_n\}$ is an
%   $F$-basis of $K$ and $\emb{K/F}=\{ \phi_1, \dots , \phi_n \}$.
%   Then, the vectors
%   \begin{align*}
%     \xi_1 &= (\phi_1(w_1) , \dots , \phi_1(w_n)), \\
%           &\vdots \\
%     \xi_n &= (\phi_n(w_1), \dots , \phi_n(w_n)),
%   \end{align*}
%   are linearly independent over $K$.
% \end{cor}\xsep

% \begin{prop}
%   Let $K=F(\alpha)/F$ be a separable extension, $f \idef
%   \irrp_{\alpha,F}$, 
%   $$
%   \frac{f(X)}{X-\alpha} = \sum_{i=0}^{n-1} \beta_i \, X^i, \quad
%   \beta_i \in K.
%   $$
%   Then
%   $$
%   \trace{K}{F} \parent{\alpha^i \, \frac{\beta_j}{\fder{f}(\alpha)}}=
%   \krd_{i\, j}, \quad i,j \in \{0, \dots , (n-1)\}.
%   $$
%   (Remember that $\fder{f}$ denotes the formal derivative of $f$.)
% \end{prop}\xsep


% \begin{defn}
%   Let $K/F$ be a field extension and $\alpha \in K$.  Then, the map
%   $\mult{\alpha} : K \rarrow K$, defined by
%   $\mult{\alpha}(x)=\alpha \, x$, is an $F$-linear map.
% \end{defn}\xsep


% \begin{prop}
%   Let $K/F$ be a field extension and $\alpha \in K$.  Then,
%   $$
%   \norm{K}{F}(\alpha)=\det(\mult{\alpha}), \quad \text{ and } \quad
%   \trace{K}{F}=\tr(\mult{\alpha}).
%   $$
% \end{prop}\xsep


% \begin{defn}
%   Let $K/F$ be a finite separable extension, $\emb{K/F}=\{\phi_1,
%   \dots , \phi_n\}$ be the embeddings of $K$ into $\alc{F}$ over $F$
%   and $\basis{B}=\{w_1, \dots , w_n\}$.  Then the \midx{discriminant}
%   of $K/F$ with respect to $\basis{B}$ is defined as
%   $$
%   \disc_{K/F}(\basis{B}) \idef \colch{\det\parent{\phi_j(w_i) }}^2 =
%   \abs{
%     \begin{array}{cccc}
%       \phi_1(w_1) & \phi_2(w_1) & \dots & \phi_n(w_1) \\
%       \phi_1(w_2) & \phi_2(w_2) & \dots & \phi_n(w_2) \\
%       \vdots      & \vdots     & \vdots & \vdots \\
%       \phi_1(w_n) & \phi_2(w_n) & \dots & \phi_n(w_n)
%     \end{array}}^2
%   $$
%   (One could define the discriminant of an inseparable extension as
%   zero.) 
% \end{defn}\xsep


% \begin{rem}
%   If $K=F(\alpha)$, where $\alpha$ is algebraic and separable over
%   $F$, with $f \idef \irrp{\alpha}{F}$ and $[F(\alpha):F]=n$, and
%   $\basis{B}\idef \{1, \alpha, \alpha^2, \dots, \alpha^{n-1}\}$, then
%   $\disc_{K/F}(\basis{B}) = \disc_{f}$.
% \end{rem}\xsep


% \begin{defn}
%   Let $V$ be a vector space over $F$ and $\basis{B}=\{v_1, \dots ,
%   v_n\}$ and $\basis{C}=\{w_1, \dots , w_n\}$ be two $F$-bases of $V$.
%   Then a \midx{change of bases matrix} (or map) from $\basis{B}$ to
%   $\basis{C}$ is a matrix $(a_{i \, j})$ such that
%   $$
%   \parent{
%     \begin{array}{cccc}
%       a_{1 \, 1} & a_{1 \, 2} & \dots & a_{1 \, n} \\
%       a_{2 \, 1} & a_{2 \, 2} & \dots & a_{2 \, n} \\
%       \vdots    & \vdots    & \vdots & \vdots \\
%       a_{n \, 1} & a_{n \, 2} & \dots & a_{n \, n} \\
%     \end{array}
%   } \cdot
%   \parent{
%     \begin{array}{c}
%       v_1 \\
%       v_2 \\
%       \vdots \\
%       v_n
%     \end{array}} =
%   \parent{
%     \begin{array}{c}
%       w_1 \\
%       w_2 \\
%       \vdots \\
%       w_n
%     \end{array}}
%   $$
% \end{defn}\xsep


% \begin{prop}
%   Let $K/F$ be a finite separable extension.
%   \begin{enumerate}

%   \item If $\basis{B}$ is any $F$-bases of $K$, then
%     $\disc_{K/F}(\basis{B}) \in F$.

%   \item If $\basis{B}$ and $\basis{C}$ are two of $F$-bases of $K$ and
%     $C$ is a change of bases matrix from $\basis{B}$ to $\basis{C}$,
%     then $\disc_{K/F}(\basis{C}) = \colch{\det(C)}^2 \cdot
%     \disc_{K/F}(\basis{B})$.

%   \item Let $K=F(\alpha)$, $f \idef \irrp{\alpha}{F}$, and $\alpha_1,
%     \dots , \alpha_n$ be the roots of $f$.  Then,
%     $$
%     \disc_f =(-1)^{n(n-1)/2} \cdot \norm{K}{F}(\fder{f}(\alpha)).
%     $$

%     \item If $\basis{B}=\{w_1, \dots , w_n\}$ is an $F$-basis of $K$,
%         then $\disc_{K/F}(\basis{B}) = \det \parent{ \trace{K}{F}(w_i
%           \cdot w_j) }$.

%   \end{enumerate}
% \end{prop}\xsep


% \begin{defn}
%   If $F=\Q$ and $K/\Q$ is finite, then there exists an \midx{integral
%     basis} for $K/\Q$.  Basically, an integral basis in a $\Q$-basis
%   of $K$, say $\{w_1, \dots , w_n\}$, such that $\Z[w_1, \dots , w_n]$
%   generates all the \midx{algebraic integers} of $K$, i.e., the
%   elements of $K$ that are roots of \emph{monic} polynomials in
%   $\Z[X]$.  The discriminants of $K/\Q$ with respect to all integral
%   bases are equal, and the discriminant of $K/\Q$ (without a basis
%   specified) is the the discriminant with respect to an integral
%   basis.
% \end{defn}\xsep



\subsection{Cyclic Extensions}

\begin{theo}[Hilbert's Theorem 90 -- multiplicative
  form]\index{Hilbert's Theorem 90 -- multiplicative form} Let $K/F$ be
  a cyclic extension of degree $n$ and $\gal(K/F)=\cyclg{\sigma}$.
  Then, $\beta \in K$ is such that $\norm{K}{F}(\beta)=1$ if, and only
  if, there exists $\alpha \in K^{\times}$ such that
  $\beta=\alpha/\sigma(\alpha)$.
\end{theo}\xsep

\begin{theo}
  Let $F$ be a field such that $F$ contains a \emph{primitive} $n$-th
  root of unity for some fixed $n$ not divisible by $\chr(F)$.
  \begin{enumerate}

  \item If $K/F$ is cyclic of degree $n$, then $K=F[\alpha]$ where
    $\alpha$ is a root of $X^n-a$, for some $a \in F$.  [In
    particular, $\irrp{\alpha}{F}=X^n-a$.]

  \item Conversely, if $a \in F$ and $\alpha$ is a root of $X^n -a$,
    then $F[\alpha]/F$ is cyclic, its degree, say $d$, is a
    divisor of $n$, and $\alpha^d \in F$. 

  \end{enumerate}
\end{theo}\xsep

\begin{rem}
  Note that, by linear independence of characters, if $K/F$ is
  separable, then $\trace{K}{F}$ is not constant equal to zero.
\end{rem}\xsep



\begin{theo}[Hilbert's Theorem 90 -- additive form]\index{Hilbert's
    Theorem 90 -- additive form} Let $K/F$ be a
  cyclic extension of degree $n$ and $\gal(K/F)=\cyclg{\sigma}$.
  Then, $\beta \in K$ is such that $\trace{K}{F}(\beta)=0$ if, and
  only if, there exists $\alpha \in K^{\times}$ such that
  $\beta=\alpha - \sigma(\alpha)$.
\end{theo}\xsep

\begin{theo}[Artin-Schreier]\index{Artin-Schreier Theorem}
  Let $F$ be a field of characteristic $p>0$.
  \begin{enumerate}

  \item If $K/F$ is cyclic of degree $p$, then $K=F[\alpha]$ where
    $\alpha$ is a root of $X^p-X-a$, for some $a \in F$.  [In
    particular, $\irrp{\alpha}{F}=X^p-X-a$.]

  \item Conversely, if $a \in F$ and $f \idef X^p-X -a$, then either
    $f$ splits completely in $F$ or is irreducible over $F$.  In the
    latter case, if $\alpha$ is a root of $f$, then $F[\alpha]/F$ is
    cyclic of degree $p$.

  \end{enumerate}
\end{theo}\xsep


\subsection{Solvable and Radical Extensions}

\begin{defn}
  A \emph{finite} extension $K/F$ is a \midx{solvable extension} if it
  is separable and the normal closure $L$ of $K/F$ [which is then
  finite Galois over $F$] is such that $\gal(L/F)$ is a solvable group.
\end{defn}\xsep

\begin{rem}
  Note that for a finite separable extension $K/F$ to be solvable, it
  suffices that there exists some finite Galois extension of $F$
  containing $K$ with its Galois group solvable.
\end{rem}\xsep


\begin{prop}
  The class of solvable extensions is distinguished.
\end{prop}\xsep


\begin{defn}
  \begin{enumerate}

  \item A finite extension $K/F$ is a \midx{repeated radical
      extension} if there is a tower:
    $$
    F=F_0 \subseteq F_1 \subseteq F_2 \subseteq \cdots \subseteq F_r = K,
    $$
    such that $F_i=F_{i-1}[\alpha_i]$, where $\alpha_i$ is either a
    root of a polynomial $X^n-a$, for some $a \in F_{i-1}$ and with
    $\chr(F) \nmid n$, or a root of $X^p-X-a$, for some $a \in
    F_{i-1}$, where $p=\chr(F)$.  [Note that $\alpha_i$ might then be
    a root of unity.]

  \item A finite extension $K/F$ is a \midx{radical extension} if
    there is $L \supseteq K$ such that $L/F$ is repeated radical.

  \end{enumerate}
\end{defn}\xsep


\begin{rem}
  Note that, by definition, if $K$ is the splitting field of a
  separable polynomial $f \in F[X]$, then the roots of $f$ are given
  by radicals [i.e., $f$ is \midx{solvable by radicals}] if, and only
  if, $K$ is radical.
\end{rem}\xsep


\begin{prop}
  The class of radical extensions is distinguished.
\end{prop}\xsep

\begin{theo}
  Let $K/F$ be \emph{separable}.  Then, $K/F$ is solvable if, and only
  if, it is radical.
\end{theo}\xsep

\begin{rem}
  This allows us to determine when a polynomial can be solved by
  radicals simply by looking at its Galois group!
\end{rem}\xsep

\begin{theo}
  For $n=2,3,4$ [and $\chr(F) \neq 2, 3$] there are formulas for
  solving [general] polynomial equations of degree $n$ by means of
  radicals.  For $n \geq 5$, there aren't.
\end{theo}\xsep

\begin{theo}
  Suppose that $f \in \Q[X]$ is irreducible and splits completely in
  $\R$.  If any root of $f$ lies in a \emph{real} repeated radical
  extension of $\Q$, then $\deg f = 2^r$ for some non-negative integer
  $r$.
\end{theo}\xsep

\begin{rem}
  Note that the above theorem tells us that we cannot replace
  \emph{radical} by \emph{repeated radical} in trying to express all
  roots of a polynomials in terms of radicals.  For example, the
  polynomial $f=X^3-4X+2$ splits completely in $\R$ and is solvable.
  So, we can write its roots in terms of radicals [since its radical],
  but we \emph{must} have \emph{complex numbers} to write them in
  terms of radicals [since is not repeated radical by the theorem
  above].  More precisely, if
  $$
  \alpha \idef \sqrt[3]{\frac{\sqrt{111}}{9}-1}, \qquad \text{and}
    \qquad \zeta_3 \idef \frac{\sqrt{3}}{2} \mi - \frac{1}{2},
  $$
  then the [all real] roots of $f$ are 
  $$
  \alpha + \frac{4}{3\alpha}, \quad \alpha\, \zeta_3 +
  \frac{4}{3\alpha \, \zeta_3}, \quad \alpha\, \zeta_3^2 +
  \frac{4}{3\alpha \, \zeta_3^2}.
  $$
  [We \emph{cannot} rewrite the above roots only using radicals of
  real numbers!]
\end{rem}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\printindex



\end{document}



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% LocalWords:  automorphisms embeddings Artin Frobenius Abelian resp compositum
% LocalWords:  Finotti automorphism monic rcl Schur cyclotomic monoid bX ac aX
% LocalWords:  cccc subfield subfields th bijection Hermitian nm bilinear
% LocalWords:  Schreier
