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\newpage
\section{Algebraic Closure}

\begin{lemma*}[Zorn's Lemma]
  \label{zorn}
  Let $S$ be a nonempty partially ordered set.
  Assume that every chain in $S$ has an upper
  bound. Then $S$ has a maximal element.
\end{lemma*}

\begin{definition*}
  A relation $\le $ on a set $S$ is a partial
  order if for all $x, y, z \in S$:
  \begin{enumerate}[(i)]
    \item $x \le x$.
    \item If $x \le y$ and $y \le z$ then $x \le
      z$.
    \item If $x \le y \le z$ and $y \le x$ then $x
      = y$.
  \end{enumerate}
  \glsnoundefn{poset}{partially ordered
  set}{partially ordered sets}
  \glsnoundefn{porder}{partial order}{N/A}
  $(S, \le)$ is called a \emph{partially ordered
  set} (or
  poset).

  \glsadjdefn{tot_orded}{totally
  ordered}{\gls{poset}}%
  It is \emph{totally ordered} if moreover for each $x, y
  \in S$
  \begin{enumerate}[(i)]
    \setcounter{enumi}{3}
    \item Either $x \le y$ or $y \le x$.
  \end{enumerate}
  Let $T \subset S$ be a subset.
  \begin{itemize}
    \item \glsnoundefn{chain}{chain}{chains}%
      $T$ is a \emph{chain} if it is totally
      ordered by $\le$.
    \item \glsnoundefn{upp_bd}{upper bound}{upper
      bounds}%
      $x \in S$ is an \emph{upper bound} for $T$ if
      $t \le x$ for all $t \in T$.
    \item \glsadjdefn{max_el}{maximal}{element}%
      $x \in S$ is \emph{maximal} if $\nexists
      y \in S$ with $x \le y$ and $x \neq y$.
  \end{itemize}
\end{definition*}

\begin{example*}
  Let $V$ be a vector space and $(S, \le)$ be the
  set of linearly independent subsets of $V$
  ordered by inclusion. If $T \subset S$ is a
  \gls{chain} then let $Y = \bigcup_{X \in T} X$. It may
  be checked that $Y$ is linearly independent,
  hence an upper bound for $T$. \nameref{zorn} shows
  that $S$ has a \gls{max_el} $B$. Then:
  \begin{enumerate}[(i)]
    \item $B$ is linearly independent.
    \item $B \cup \{v\}$ is not linearly
      independent for any $v \in V \setminus B$.
  \end{enumerate}
  (i) and (ii) $\implies$ $B$ spans $V$. Therefore
  $B$ is a basis for $V$.
\end{example*}

\begin{example*}[Maximal ideal]
  \refstepcounter{customlink}
  \label{max_ideal_example}
  Let $R$ be a nonzero ring. Let $(S, \le)$ be the
  set of all proper (ie $\neq R$) ideals of $R$
  ordered by inclusion. $R$ nonzero implies $\{0\}
  \in S$ so $S$ is nonempty. If $T \subset S$ is a
  \gls{chain} then let $J = \bigcup_{I \in T} I$.
  If $x, y \in J$ then $x \in I_1$ and $y \in I_2$
  for some $I_1, I_2 \in T$. Since $T$ is
  \gls{tot_orded}, either $I_1 \subset I_2$ or
  $I_2 \subset I_1$. Therefore $x + y \in J$.
  Also, $r \in R$, $x \in J$ implies $rx \in J$.
  So $J$ is an ideal in $R$. It is a proper ideal
  since $1 \notin J$. Therefore $J \in S$ is an
  upper bound for $T$. \nameref{zorn} shows that
  $S$ has a \gls{max_el}, hence $R$ has a maximal
  ideal.
\end{example*}

\begin{flashcard}[algebraic-closure-existence-thm]
\begin{theorem}[Existence of algebraic closure]
  % Theorem 11.1
  \cloze{
  Let $K$ be a \gls{field}. Then
  \begin{enumerate}[(i)]
    \item There is an \gls{alg_ex} \gls{ex} $L
      \extendover K$ such that every nonconstant
      $f \in \radj K[X]$ has a root in $L$.
    \item $K$ has an \gls{alg_closure}
      $\algclos{K}$.
  \end{enumerate}
  }
\end{theorem}

\begin{proof}
  \phantom{}
  \cloze{
  \begin{enumerate}[(i)]
    \item Let $S = \{\text{monic constant
      polynomials in $\radj K[X]$}\}$. Rough idea:
      $L = \fadj K(\alpha_f : f \in S)$ where
      $\alpha_f$ is a root of $f$.

      In detail: Let $R = \radj K[X_f : f \in S]$
      be the polynomial ring in indeterminates
      $\{X_f : f \in S\}$. So elements of $R$ are
      finite $K$-linear combinations of monomials
      of the form $X_{f_1}^{d_1} X_{f_2}^{d_2}
      \cdots X_{f_r}^{d_r}$ where $f_i \in S$ and
      $d_i \in \NN$. Let $I \subset R$ be the
      ideal generated by $\{f(X_f) : f \in S\}$.

      \textbf{Claim:} $I \neq R$.

      \textit{Proof of claim:} If not then $1 \in
      I$, i.e.
      \[ 1 = \sum_{f \in T} g_f f(X_f) \tag{$*$}
      \label{lec22_l125_eq} \]
      for some finite subset $T \subset S$ and
      polynomials $g_f \in R$. Let $L \extendover
      K$ be a \gls{splitting_field} for $\prod_{f
      \in T} f$ and for each $f \in T$, let
      $\alpha_f \in L$ be a root of $f$. We define
      a ring homomorphism
      \begin{align*}
        \phi : R
        &\to \radj L[X_f : f \in S \setminus T] \\
        X_f
        &\mapsto \begin{cases}
          \alpha_f & f \in T \\
          X_f & f \notin T
        \end{cases} \\
        c
        &\mapsto c \qquad \forall c \in K
      \end{align*}
      Applying $\phi$ to \eqref{lec22_l125_eq}
      gives
      \[ 1 = \sum_{f \in T} \phi(g_f)
      \ub{f(\alpha_f)}_{= 0} = 0 \]
      which gives a contradiction. Hence $I \neq
      R$, which proves the claim.

      Since $I \neq R$,
      by the earlier example
      (\nameref{max_ideal_example}), we get that
      $R / I$ has a maximal ideal, so equivalently
      $R$ has a maximal ideal $J$ containing $I$.
      Let $L = R / J$ and $\alpha_j = X_j + J \in
      L$. Then $f(\alpha_f) = 0$ (since $f(X_f)
      \in I \subset J$). Since
      \[ L = \bigcup_{\substack{T \subset S
      \text{ finite} \\ \text{subsets}}} \fadj
      K(\alpha_f : f \in T) \]
      it follows that $L \extendover K$ is an
      \gls{alg_ex} \gls{ex}.
    \item Repeating the construction in (i) gives
      \[ K \subset K_1 \subset K_2 \subset \cdots \]
      Each nonconstant polynomial in $\radj
      K_n[X]$ has a root in $K_{n + 1}$. If $f \in
      \radj K[X]$ has degree $n \ge 1$ then is has
      a root $\alpha_1$ in $K_1$. Then
      $\frac{f(X)}{X - \alpha_1}$ is either a
      constant, or a nonconstant polynomial, hence
      has a root $\alpha_2$ in $K_2$, and so on,
      so that $f$ splits into linear factors in
      $K_n$. Let
      \[ \algclos{K} = \bigcup_{n \ge 1} K_n .\]
      This is a field since it is a union of
      fields \gls{tot_orded} by inclusion. Then
      every polynomial in $\radj K[X]$ splits into
      linear factors over $\algclos{K}$, and each
      element of $\algclos{K}$ belongs to some
      $K_n$, so is \gls{alg_ex} over $K$. Now
      apply \cref{alg_closed_extension_lemma}.
      \qedhere
  \end{enumerate}
  }
\end{proof}
\end{flashcard}
