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\begin{remark*}
  \glsadjdefn{hom_poly}{homogeneous}{polynomial}
  We can write any $f \in \radj R[X_1, \ldots,
  X_n]$ as $f = \sum_d f_d$ where $f_d$ is
  \emph{homogeneous} of degree $d$ (i.e. each monomial
  has total degree $d$).

  Clearly $f$ \gls{symmetric} implies all $f_d$
  are \gls{symmetric}. So for the proof of
  \nameref{symm_func_thm}(i), it suffices to
  consider $f \in \radj R[X_1, \ldots, X_n]$ which
  is \gls{symmetric} and homogeneous of degree $d$.
\end{remark*}

\begin{flashcard}[lexicographic-ordering-defn]
\begin{definition*}[lexicocraphic ordering]
  \glsnoundefn{lex_ord}{lexicographic
  ordering}{N/A}
  \cloze{
  Define the \emph{lexicographic ordering} of monomials
  such that
  \[ X_1^{i_1} X_2^{i_2} \cdots X_n^{i_n} >
  X_1^{k_1} X^{j_2} \cdots X_n^{j_n} \]
  if $i_1 = j_1, i_2 = j_2, \ldots, i_{r - 1} =
  j_{r - 1}, i_r > j_r$ for some $1 \le r \le n$.

  This is a total ordering.
  } 
\end{definition*}
\end{flashcard}

\begin{flashcard}[symm-func-thm-proof]
\prompt{Proof of \nameref{symm_func_thm}?}
\begin{proof}[Proof of \nameref{symm_func_thm}]
  \phantom{}
  \begin{enumerate}[(i)]
    \item \cloze{
      By previous remark, we may split $f$ into a sum
      of homogeneous polynomials, and just prove that
      each term in the sum can be written as a sum of
      \gls{elem_symm_func}.

      Let $X_1^{i_1} X_2^{i_2} \cdots X_n^{i_n}$ be
      the largest monomial (with respect to
      \gls{lex_ord}) to appear in $f$ with non-zero
      coefficient ($c$ say). Since
      $X_{\sigma(1)}^{i_1} X_{\sigma(2)}^{i_2} \cdots
      X_{\sigma(n)}^{i_n}$ also appears in $f$ for all
      $\sigma \in S_n$, we must have $i_1 \ge i_2 \ge
      i_3 \ge \cdots i_n$.

      Write
      \[ X_1^{i_1} X_2^{i_2} \cdots X_n^{i_n}
      = X_1^{i_1 - i_2} (X_1 X_2)^{i_2 - i_3} \cdots
      (X_1 X_2 \cdots X_n)^{i_{n - 1} - i_n} \]
      Let $g = \symmpoly_1^{i_1 - i_2} \symmpoly_2^{i_2 - i_3} \cdots
      \symmpoly_n^{i_n}$. Then $f$ and $g$ are both
      \gls{hom_poly} of degree $d$ and have the same
      \glsref[lex_ord]{largest} monomial. So $f - cg$
      is either $0$, or it is a \gls{symm_poly}
      of degree $d$, whose leading monomial is
      \glsref[lex_ord]{smaller} than that of $f$.

      Since there are only finitely many monomials of
      degree $d$ in $\radj R[X_1, \ldots, X_n]$, the
      process stops after finitely many steps.
      Therefore we can write $f$ as a polynomial in
      $\symmpoly_1, \ldots, \symmpoly_n$.
      }
    \item \cloze{Write $\symmpoly_{r, n}$ instead of
      $s_r$ to indicate the number of variables
      involved. Suppose $G \in \radj R[Y_1,
      \ldots, Y_n]$ with $G(\symmpoly_{1, n},
      \symmpoly_{2, n}, \ldots, \symmpoly_{n, n})
      = 0$. We must prove that $G = 0$. The proof
      is by induction on $n$. The case $n = 1$ is
      clear.

      Write $G = Y_n^k H$ with $Y_n \nmid H$, $k
      \ge 0$. Since $\symmpoly_{n, n} = X_1 X_2
      \cdots X_n$, it is not a zero divisor in
      $\radj R[X_1, \ldots, X_n]$, so we have
      \[ H(\symmpoly_{1, n}, \symmpoly_{2, n},
      \ldots, \symmpoly_{n, n}) = 0 \]
      So we may assume that $G$, if non-zero, is
      not divisible by $Y_n$. Replacing $X_n$ by
      $0$ gives
      \[ \symmpoly_{r, n} (X_1, \ldots, X_{n - 1},
      0) = \begin{cases}
        \symmpoly_{r, n - 1}(X_1, \ldots, X_{n - 1}) & r <
        n \\
        0 & r = n
      \end{cases} \]
      Therefore
      \[ G(\symmpoly_{1, n - 1}, \symmpoly_{2, n -
      1}, \ldots, \symmpoly_{n - 1, n - 1}, 0) = 0 \]
      By induction hypothesis, $G(Y_1, \ldots,
      Y_{n - 1}, 0) = 0$, hence $Y_n \mid G$. So
      by the above $G$ must be zero.} \qedhere
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{example}
  Let $f = \sum_{i \neq j} X_i^2 X_j$. The leading
  term (in \gls{lex_ord}) is $X_1^2 X_2 = X_1(X_1
  X_2)$. Calculate:
  \begin{align*}
    \symmpoly_1 \symmpoly_2
    &= \sum_i \sum_{j < k} X_i X_j X_k \\
    &= \ub{\sum_{i \neq j} X_i^2
    X_j}_{=f} + 3 \sum_{i < j <
    k} X_i X_j X_k
  \end{align*}
  So $f = \symmpoly_1 \symmpoly_2 - 3
  \symmpoly_3$.
\end{example}

\begin{flashcard}[]
\begin{example}
  \glsnoundefn{disc_f}{discriminant}{discriminants}
  \glssymboldefn{disc_f}{Disc}{Disc}
  Let $f(X) = \prod_{i = 1}^n (X - \alpha_i)$ be a
  monis polynomial with roots $\alpha_1, \ldots,
  \alpha_n$. The \emph{discriminant of $f$} is
  \[ \Disc(f) = \cloze{\prod_{i < j} (\alpha_i -
  \alpha_j)^2} \]
  \cloze{By the \nameref{symm_func_thm}, we can
  write $\Disc(f)$ as a polynomial in the
  coefficients of $f$.}
  \fcscrap{
  
  $n = 2$, $f(X) = X^2 + bX + c = (X - \alpha_1)(X
  - \alpha_2)$. Then
  \begin{align*}
    \Disc(f) 
    &= (\alpha_1 - \alpha_2)^2 \\
    &= (\alpha_1 + \alpha_2)^2 - 4\alpha_1
    \alpha_2 \\
    &= b^2 - 4ac
  \end{align*}
  It is clear from the definition that
  \[ \Disc(f) = 0 \iff \text{$f$ has repeated
  roots} \]
  }
\end{example}
\end{flashcard}

\newpage
\section{Normal and separable extensions}

\begin{flashcard}[normal-extension-defn]
\begin{definition*}[normal extension]
  \glsadjdefn{norm_ex}{normal}{extension}
  \cloze{
  An \gls{ex} $L \extendover K$ is \emph{normal}
  if it is \gls{alg_ex} and the \gls{min_poly} of
  every $\alpha \in L$ splits into linear factors
  over $L$.

  Equivalently if $f \in \radj K[X]$ is
  irreducible and has a root in $L$, then it
  splits into linear factors over $L$.
  \fcscrap{(slogan: ``one out -- all out'').}
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[splitting-fields-are-normal-thm]
\begin{theorem}[Splitting fields are normal]
  \cloze{Let $\exdeg [L : K] < \infty$. Then
  \[ \text{$L \extendover K$ is \gls{norm_ex}} \iff \text{$L$ is a
  \gls{splitting_field} for some $f \in \radj
  K[X]$} \]
  }
\end{theorem}

\begin{proof}
  \phantom{}
  \begin{enumerate}[$\Rightarrow$]
    \item[$\Rightarrow$] \cloze{Write $L = \fadj
      K(\alpha_1, \ldots, \alpha_n)$. Let $f_i$ be
      the \gls{min_poly} of $\alpha_i$ over $K$.
      Then
      \begin{align*}
        \text{$L \extendover K$ \gls{norm_ex}}
        &\implies \text{$f_i$ splits into linear
        factors over $L$} \\
        &\implies \text{$L$ is a
        \gls{splitting_field} for $\prod_i f_i$}
      \end{align*}
      }
    \item \cloze{Let $L$ be the \gls{splitting_field} of
      $f \in \radj K[X]$ over $K$. Let $\alpha \in
      L$ have \gls{min_poly} $g$ over $K$. Let $M
      \extendover L$ be a \gls{splitting_field}
      for $g$. We must show that whenever $\beta$
      is a root of $g$ then in fact $\beta \in L$.
      Then $L = \fadj L(\alpha)$ is a
      \gls{splitting_field} for $f$ over $\fadj
      K(\alpha)$, and $\fadj L(\beta)$ is a
      \gls{splitting_field} for $f$ over $\fadj
      K(\beta)$. $\alpha$ and $\beta$ have the
      same \gls{min_poly} $g$ over $K$, so $\fadj
      K(\alpha)$ and $\fadj K(\beta)$ are
      \Kisomic{K}. By
      \nameref{uniqueness_of_splitting_fields},
      $\fadj L(\alpha)$ and $\fadj L(\beta)$ are
      \Kisomic{K}. Therefore $\exdeg[L : K] =
      \exdeg[\fadj L(\beta) : K]$. So by
      \nameref{tower_law}, $[\fadj L(\beta) : L] =
      1$, so $\fadj L(\beta) = L$, so $\beta \in
      L$.
      }\qedhere
  \end{enumerate}
\end{proof}
\end{flashcard}