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\begin{flashcard}[algebraically-closed-defn]
\begin{definition*}[algebraically closed field]
  \glsdefn{alg_closed}{algebraically closed}{N/A}
  \cloze{
  A \gls{field} $K$ is \emph{algebraically closed} if
  every nonconstant polynomial in $\ringadjoin K[X]$ has a
  root in $K$.

  Equivalently, if every irreducible polynomial in
  $\ringadjoin K[X]$ is linear.
  }
\end{definition*}
\end{flashcard}

\begin{example*}
  $\CC$, by the fundamental theorem of algebra.
\end{example*}

\begin{flashcard}[algebraically-closed-tfae-lemma]
\begin{lemma}
  \label{alg_closed_tfae}
  Let $K$ be a \gls{field}. Then the following are equivalent:
  \begin{enumerate}[(i)]
    \item $K$ is \gls{alg_closed}.
    \item \cloze{If $L \extendover K$ is a
      \gls{field_ex} and $\alpha \in L$ is
      \gls{alg_el} over $K$ then $\alpha \in K$.}
    \item \cloze{If $L \extendover K$ is
      \gls{alg_ex} then $L = K$.}
    \item \cloze{If $L \extendover K$ is
      \gls{finite_ex} then $L =
      K$.}
  \end{enumerate}
\end{lemma}

\begin{proof}
  \cloze{
  (ii) $\implies$ (iii) $\implies$ (iv) are all
  clear.

  For (iv) $\implies$ (i), let $f \in \ringadjoin K[X]$ be an
  irreducible polynomial. Then $L = \ringadjoin K[X] / (f)$ is
  a finite extension of $K$ with $\exdeg[L : K] =
  \deg f$. By (iv) we have $L = K$. Therefore $f$
  is linear.
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[algebraic-closure-defn]
\begin{definition*}[algebraic closure]
  \glsdefn{alg_closure}{algebraic
  closure}{algebraic closures}
  \cloze{
  If $L \extendover K$ is \gls{alg_ex} and $L$ is
  \gls{alg_closed} then we say that $L$ is an
  \emph{algebraic closure} of $K$.
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[algebraically-closed-extension-lemma]
\begin{lemma}
  % Lemma 3.9
  \label{alg_closed_extension_lemma}
  Let $L \extendover K$ be an \gls{alg_ex} \gls{ex}
  such that every polynomial in $\ringadjoin K[X]$ splits into
  linear factors over $L$. Then $L$ is
  \gls{alg_closed} (and hence an \gls{alg_closure}
  of $K$).
\end{lemma}

\begin{proof}
  \cloze{
  If $L$ is not \gls{alg_closed} then by
  \cref{alg_closed_tfae} there exists $M
  \extendover L$ \gls{alg_ex} with $\degover{M}{L}
  > 1$. Both $M \extendover L$ and $L \extendover
  K$ are \gls{alg_ex}. So by \cref{alg_tower_prop},
  $M \extendover K$ is \gls{alg_ex}.

  Pick any $\alpha \in M$. Let $f$ be the
  \gls{min_poly} for $\alpha$ over $K$. By our
  assumption, $f$ splits over $L$, so $\alpha \in
  L$. So $M = L$.}
\end{proof}
\end{flashcard}

Later we will show that every \gls{field} $K$ has an
\gls{alg_closure}. Here are two easy cases:

\begin{flashcard}[algebraic-closure-in-C-or-finite]
\begin{theorem}
  Suppose that (i) $K \subset \CC$ or (ii) $K$ is
  countable. Then $K$ has an \gls{alg_closure}.
\end{theorem}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item \cloze{If $K \subset \CC$ then let
      \[ L = \{\alpha \in \CC \mid \text{$\alpha$
      is \gls{alg_el} over $K$}\} .\]
      $L$ is a \gls{field} by
      \cref{algebraic_is_subfield_coro}, $L
      \extendover K$ is \gls{alg_ex}. If $f \in
      \ringadjoin K[X]$
      then write $f(X) = \prod_{i = 1}^n (X -
      \alpha_i)$ for some $\alpha_i \in \CC$. The
      definition of $L$ implies that all the
      $\alpha_i \in L$, i.e. $f$ splits into
      linear factors over $L$. Then
      \cref{alg_closed_extension_lemma} gives that
      $L$ is \gls{alg_closed}. Therefore $L$ is
      the \gls{alg_closure} of $K$.}
    \item \cloze{If $K$ is countable then so
      $\ringadjoin K[X]$.
      Enumerate the monic irreducible polynomials
      $f_1, f_2, f_3, \ldots$. Let $L_0 = k$ and
      for each $i \ge 1$ let $L_i$ be a
      \gls{splitting_field} for $f_i$ over $L_{i - 1}$. Then
      \[ L_0 \subset L_1 \subset L_2 \subset
      \cdots \]
      Then $L = \bigcup_{n \ge 0} L_n$ is a
      \gls{field},
      $L \extendover K$ is \gls{alg_ex}, and every
      polynomial in $\ringadjoin K[X]$ splits over $L$. Then
      \cref{alg_closed_extension_lemma} implies
      that $L$ is \gls{alg_closed}. Therefore $L$
      is an \gls{alg_closure} of $K$.}
      \qedhere
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{remark*}
  Taking $K = \QQ$ in the proof of (i), we see
  that $\algclos{\QQ} = \{\text{\gls{alg_Q} numbers}\}
  \subset \CC$ is \gls{alg_closed}.
\end{remark*}

\newpage
\section{Symmetric Polynomials}
\label{sec_4}

Suppose we wish to find the roots of a cubic
polynomial $f(X) = X^3 + a X^2 + b X + c \in
\ringadjoin \QQ[X]$. After substituting $X - \frac{a}{3}$ for
$X$ we may assume $a = 0$. Writing
\[ f(X) = (X - \alpha)(X - \beta)(X - \gamma) \]
and comparing coefficients gives
\begin{align*}
  \alpha + \beta + \gamma
  &= -a = 0 \\
  \alpha \beta + \beta \gamma + \gamma \alpha
  &= b \\
  \alpha \beta \gamma
  &= -c
\end{align*}
Let $\omega = e^{2\pi i / 3}$. Write
\[ \alpha = \frac{1}{3} [\ub{(\alpha + \beta +
\gamma)}_{=0}
+ \ub{(\alpha + \omega \beta + \omega^2
\gamma)}_{=u} +
\ub{(\alpha + \omega^2 \beta + \omega \gamma)}_{=v}] \]
Then $u^3 + v^3$ and $uv$ are unchanged under
permuting $\alpha, \beta, \gamma$. After some
calculation we find (remembering $a = 0$) that
\[ u^3 + v^3 = -27c \qquad \text{and} \qquad uv =
-3b \]
Therefore $u^3$ and $v^3$ are the roots of
\[ X^2 + 27cX - 27b^3 = 0 \]
\glsdefn{cardano_formula}{Cardano's formula}{N/A}
Solving this quadratic and taking cube roots gives
a formula for the roots of a cube, usually called
Cardano's formula.

Let $S_n$ be the symmetric group on $n$ letters.

\begin{flashcard}[symmetric-polynomial-defn]
\begin{definition*}[symmetric polynomial]
  \glsadjdefn{symmetric}{symmetric}{polynomial}
  \glsdefn{symm_poly}{symmetric
  polynomial}{symmetric polynomials}
  \cloze{
  Let $R$ be a ring. A polynomial $f \in
  \ringadjoin R[X_1,
  \ldots, X_n]$ is \emph{symmetric} if
  \[ f(X_{\sigma(1)}, \ldots, X_{\sigma(n)}) =
  f(X_1, \ldots, X_n) \qquad \forall \sigma \in
  S_n \]
  If $f$ and $g$ are \gls{symmetric} then so are $f + g$
  and $fg$. Therefore the \glspl{symm_poly}
  form a subring of $\ringadjoin R[X_1, \ldots, X_n]$.
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[elementary-symmetric-functions-defn]
\begin{definition*}[elementary symmetric functions]
  \glsdefn{elem_symm_func}{elementary
  symmetric function}{elementary symmetric
  functions}
  \glssymboldefn{symm_poly}{elementary
  symmetric function notation}{$s_i$}
  \cloze{
  The \emph{elementary symmetric functions} are
  the polynomials $\symmpoly_1, \ldots,
  \symmpoly_n$ in $\ringadjoin \ZZ[X_1,
  \ldots, X_n]$ such that
  \[ \prod_{i = 1}^n (T + X_i) = T^n +
  \symmpoly_1 T^{n -
  1} + \cdots + \symmpoly_{n - 1} T + \symmpoly_n \]
  }
\end{definition*}
\end{flashcard}

\begin{example*}
  When $n = 3$,
  \begin{align*}
    \symmpoly_1
    &= X_1 + X_2 + X_3 \\
    \symmpoly_2
    &= X_1 X_2 + X_1 X_3 + X_2 X_3 \\
    \symmpoly_3
    &= X_1 X_2 X_3
  \end{align*}
  In general,
  \[ \symmpoly_r = \sum_{1 \le i_1 < \cdots < i_r \le n}
  X_{i_1} X_{i_2} \cdots X_{i_r} \]
\end{example*}

\begin{flashcard}[symm-func-thm]
\begin{theorem}[Symmetric Function Theorem]
  \label{symm_func_thm}
  \cloze{
  \phantom{}
  \begin{enumerate}[(i)]
    \item Every \gls{symm_poly} over $R$ can
      be expressed as a polynomial in the
      \glspl{elem_symm_func}, with
      coefficients in $R$.
    \item There are no non-trivial relations
      between the $\symmpoly_i$.
  \end{enumerate}
  }
\end{theorem}
\end{flashcard}

\begin{remark*}
  Consider the ring homomorphism
  \begin{align*}
    R[\tau_1, \ldots, \tau_n]
    &\stackrel{\theta}{\to}
    \ringadjoin R[X_1, \ldots, X_n] \\
    \tau_i
    &\mapsto \symmpoly_i
  \end{align*}
  Then part (i) of \nameref{symm_func_thm} says
  that
  \[ \Im(\theta) = \{\text{\glspl{symm_poly}
  in $\ringadjoin R[X_1, \ldots, X_n]$}\} ,\]
  and part (ii) says $\theta$ is injective.
\end{remark*}