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\subsection{Separability}

Over $\RR$ or $\CC$ we know from calculus that a
polynomial $f$ has a repeated root $\alpha$ if and
only if
\[ f(\alpha) = f'(\alpha) = 0 \]
To work over arbitrary \glspl{field} we proceed
purely algebraically (no calculus!).

\glsadjdefn{simple_root}{simple}{root}
Note that we call a root \emph{simple} if it is
not a repeated root.

\begin{flashcard}[formal-deriv-defn]
\begin{definition*}[Formal derivative]
  \glsnoundefn{form_deriv}{formal
  derivative}{formal derivatives}
  \glssymboldefn{form_deriv}{'}{'}
  \cloze{
  The \emph{formal derivative} of $f = \sum_{i =
  0}^d c_i X^i \in \radj K[X]$ is
  \[ f' = \sum_{i = 1}^d i c_i X^{i - 1} .\]
  }
\end{definition*}
\end{flashcard}

\textbf{Exercise:} Check with this definition that
\[ \begin{cases}
  (f + g)\fd = f\fd + g\fd \\
  (fg)\fd = fg\fd + f\fd g
\end{cases} \]

\begin{flashcard}[simple-root-if-fd-lemma]
\begin{lemma}
  % Lemma 5.2
  \label{simple_root_if_fd_lemma}
  Let $f \in \radj K[X]$ and $\alpha \in K$ a root
  of $f$. Then $\alpha$ is a \gls{simple_root} root if
  and only if $f\fd(\alpha) \neq 0$.
\end{lemma}

\begin{proof}
  \cloze{
  Write $f(X) = (X - \alpha) g(X)$ for some $g \in
  \radj K[X]$. Then
  \begin{align*}
    \text{$\alpha$ is a \gls{simple_root} root of
    $f$}
    &\iff \text{$X - \alpha$ is not a factor of
    $g$} \\
    &\iff g(\alpha) \neq 0
  \end{align*}
  But $f\fd(X) = (X - \alpha) g\fd(X) + g(X)$, so
  $f\fd(\alpha) = g(\alpha)$.
  }
\end{proof}
\end{flashcard}

By the GCD of polynomials $f, g \in \radj K[X]$
not both zero, we mean the unique monic polynomial
$\gcd(f, g)$ which generates the ideal $(f, g)
\subset \radj K[X]$.

This is the unique monic polynomial which divides
both $f$ and $g$ and can be written as $af + bg$
for some $a, b \in \radj K[X]$.

We can compute $\gcd(f, g)$, together with $a, b$,
using Euclid's algorithm.

\begin{flashcard}[gcd-same-in-extension-lemma]
\begin{lemma}
  \label{gcd_same_in_extension_lemma}
  Let $f, g \in \radj K[X]$ and let $L \extendover
  K$ be any \gls{field_ex}. Then $\gcd(f, g)$ is
  the same computed in $\radj K[X]$ and in $\radj L[X]$.
\end{lemma}

\begin{proof}
  \cloze{
  Running Euclid's algorithm on $f, g \in \radj
  K[X]$ gives the same answer whether we work in
  $\radj K[X]$ or $\radj L[X]$.
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[separable-poly-defn]
\begin{definition*}[Separable]
  \glsadjdefn{sep_poly}{separable}{polynomial}
  \glsadjdefn[sep_poly]{insep_poly}{inseparable}{polynomial}
  \cloze{
  An irreducible polynomial $f \in \radj K[X]$ is
  \emph{separable} if it splits into distinct
  linear factors in a \gls{splitting_field}.

  The convention in this course is that we use the
  same definition for any $0 \neq f \in \radj
  K[X]$. Anything which is not \gls{sep_poly} is
  called \emph{inseparable}.
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[separable-iff-lemma]
\begin{lemma}
  Let $0 \neq f \in \radj K[X]$. Then
  \[ \text{$f$ is \gls{sep_poly}} \iff
  \cloze{\gcd(f, f\fd) = 1.} \]
\end{lemma}

\begin{proof}
  \cloze{
  Let $L$ be a \gls{splitting_field} of $f$. Then
  \begin{align*}
    \text{$f$ \gls{sep_poly}}
    &\iff \text{$f$ and $f\fd$ have no common
    roots in $L$} \\
    &\iff \text{$\gcd(f, f\fd) = 1$ in $\radj
    L[X]$} \\
    &\iff \text{$\gcd(f, f\fd) = 1$ in $\radj
    K[X]$}
  \end{align*}
}
\end{proof}
\end{flashcard}

\begin{flashcard}[irred-implies-separable]
\begin{theorem}
  \label{irred_implies_sep}
  Let $f \in \radj K[X]$ \cloze{irreducible}. Then
  $f$ is \gls{sep_poly} \cloze{unless
  $\characteristic(K) = p > 0$ and $f(X) = g(X^p)$
  for some $g \in \radj K[X]$.}
\end{theorem}

\begin{proof}
  \cloze{
  Assume $f$ is monic. Since $f$ is irreducible,
  $\gcd(f, f\fd) = 1$ or $f$. If $f\fd \neq 0$
  then since $\deg f\fd < \deg f$ we have $\gcd(f,
  f\fd) \neq f$, so $\gcd(f, f\fd) = 1$, and $f$
  is \gls{sep_poly}. Now suppose that $f\fd = 0$.
  If $f = \sum_{i = 0}^d c_i X^i$ then $f\fd =
  \sum_{i = 1}^d ic_i X^{i - 1}$. So $f\fd = 0
  \implies i c_i = 0 ~\forall 1 \le i \le d$.

  If $\characteristic(K) = 0$, then this implies
  that $c_i = 0$ for all $1 \le i \le d$, so $f$
  is constant (hence not irreducible). If
  $\characteristic(K) = p > 0$ we still get $c_i =
  0$ for all $i$ with $p \nmid i$. Therefore $f(X)
  = g(X^p)$ for some $g \in \radj K[X]$.
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[separable-ex-el-defn]
\begin{definition*}[Separable element / extension]
  \glsadjdefn{sep_el}{separable}{element}
  \glsadjdefn{sep_ex}{separable}{extension}
  Let $L \extendover K$ be a \gls{field_ex}. We
  define:
  \begin{enumerate}[(i)]
    \item $\alpha \in L$ is \emph{separable} over
      $K$ if it is \gls{alg_ex} over $K$ and its
      \gls{min_poly} if \gls{sep_poly}.
    \item $L \extendover K$ is \emph{separable} if
      every $\alpha \in L$ is \gls{sep_el} over
      $K$ (in particular $L \extendover K$ is
      \gls{alg_ex}).
  \end{enumerate}
\end{definition*}
\end{flashcard}

\begin{flashcard}[primitive-el-thm]
\begin{theorem}[Theorem of the Primitive Element]
  \label{prim_el_thm}
  \cloze{
  If $L \extendover K$ is \gls{finite_ex} and
  \gls{sep_ex} then $L \extendover K$ is
  \gls{simple_ex} (that is, $L = \fadj K(\theta)$ for some
  $\theta \in L$).
  }
\end{theorem}

\begin{proof}
  \cloze{
  Write $L = \fadj L(\alpha_1, \ldots, \alpha_n)$
  for some $\alpha_i \in L$. We must show that $L
  = \fadj K(\theta)$ for some $\theta \in L$. It
  suffices to prove the case $n = 2$, since the
  general case follows by induction on $n$.

  Write $L = \fadj K(\alpha, \beta)$, and let $f$
  and $g$ be the \glspl{min_poly} of $\alpha$ and
  $\beta$ over $K$. Let $M$ be a
  \gls{splitting_field} for $fg$ over $L$.
  Write
  \[ f(X) = \prod_{i = 1}^r (X - \alpha_i), \qquad
  \alpha_i \in M, \alpha = \alpha_1 \]
  \[ g(X) = \prod_{i = 1}^s (X - \beta_i), \qquad
  \beta_i \in M, \beta = \beta_1 \]
  Now $L \extendover K$ \gls{sep_ex} implies
  $\beta$ \gls{sep_el} over $K$, which implies
  $\beta_1, \ldots, \beta_s$ are distinct. We pick
  some $c \in K$ and let $\theta = \alpha +
  c\beta$. Let $F(X) = f(\theta - cX) \in
  \radj\fadj K(\theta)[X]$. Then $F(\beta) =
  f(\theta - c\beta) = f(\alpha) = 0$, and
  $g(\beta) = 0$.

  \textbf{If} $\beta_2, \ldots, \beta_s$ are not
  roots of $F$ then
  \begin{align*}
    \text{$\gcd(F, g) = X - \beta$ in $\radj M[X]$}
    &\implies \text{$\gcd(F, g) = X - \beta$ in
    $\radj\fadj K(\theta)[X]$} \\
    &\implies \beta \in \fadj K(\theta)
  \end{align*}
  But then $\alpha = \theta - c\beta \in \fadj
  K(\theta)$ so $\fadj K(\alpha, \beta) \subset
  \fadj K(\theta)$. But clearly $\fadj K(\theta)
  \subset \fadj K(\alpha, \beta)$, and hence $L =
  \fadj K(\alpha, \beta) = \fadj K(\theta)$.

  We are done unless $\fadj F(\beta_j) = 0$ for
  some $2 \le j \le s$. In this case, we have
  $f(\theta - c\beta_j) = 0$ for some $2 \le j \le
  s$, and so $\alpha + c\beta = \alpha_i +
  c\beta_j$ for some $1 \le i \le r$, $2 \le j \le
  s$. If $|K| = \infty$ then since $\beta \not\in
  \{\beta_2, \ldots, \beta_s\}$ we can pick $c \in
  K$ such that this never happens. If $|K| <
  \infty$, then $|L| < \infty$ and by
  \cref{cyclic_mult_subgroup}, $\mult{L}$ is
  cyclic and generated by $\theta$ (say). Then
  $L = \fadj K(\theta)$.
  }
\end{proof}
\end{flashcard}