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\begin{lemma*}
  Let $M$ be a finitely generated $A$ module for
  $A$ a commutative ring $I \subseteq A$, $\phi :
  M \to M$ an $A$-module homomorphism such that
  \[ \phi(M) \subseteq I \cdot M = \langle a \cdot
  m \st a \in I, \in M \rangle ,\]
  where $\langle \cdots \rangle$ represents the
  submodule of $M$ generated by those elements.
  Then there exists an equation
  \[ \phi^n + a_i \phi^{n - 1} + \cdots + a_n
  \equiv 0 \]
  with $a_i \in I$. Interpretation: $a_i$
  represents the homomorphism $m \mapsto a_i m$.
\end{lemma*}

\begin{proof}
  Let $x_1, \ldots, x_n \in M$ be a set of
  generators for $M$. Then each $\phi(x_i) \in I
  \cdot M$, so can write
  \[ \phi(x_i) = \sum_{j = 1}^n a_{ij} \cdot x_j \]
  with $a_{ij} \in I$, i.e.
  \[ \sum_{j = 1}^n (\delta_{ij} \phi - a_{ij})
  x_j = 0 \]
  So
  \[ \begin{pmatrix}
    \phi - a_{11} & -a_{12} & \cdots \\
    -a_{21} & \phi - a_{22} & \cdots \\
    \vdots & \vdots & \ddots
  \end{pmatrix}
  \begin{pmatrix}
    x_1 \\
    \vdots \\
    x_n
  \end{pmatrix} = 0 \]
  Multiplying by the adjoing matrix, we get
  \[ \det((\delta_{ij} \phi - a_{ij})) x_j = 0
  \qquad \forall j \]
  But $\det(\delta_{ij} \phi - a_{ij})$ is a
  degree $n$ polynomial in $\phi$ annihiliting
  each $x_j$, hence annihilating every element in
  $M$. The leading term in $\phi$ is $\phi^n$ and
  all other coefficients involve $a_{ij}$'s, hence
  lie in $I$.
\end{proof}

\subsubsection*{Integrality}

\begin{flashcard}[integral-element-defn]
\begin{definition*}[Integral element]
  \glsadjdefn{intel}{integral}{element}
  \cloze{Let $A \subseteq B$ be integral domains. An
  element $b \in B$ is \emph{integral} over $A$ if
  $f(b) = 0$ for a \emph{monic} polynomial $f(X)
  \in A[X]$. (recall that monic means that the leading
  coefficient is $1$).}
\end{definition*}
\end{flashcard}

\begin{flashcard}[intel-iff-subring-containing-Ab-prop]
\begin{proposition*}
  $b \in B$ \gls{intel} over $A$ if and only if
  there is a subring $C \subseteq B$ containing
  $A[b]$ with $C$ a finitely generated $A$-module.
\end{proposition*}

\begin{proof}
  \phantom{}
  \begin{enumerate}
    \item[$\Rightarrow$] \cloze{Suppose $b^n + a_1 b^{n -
      1} + \cdots + a_n = 0$. Then since $A[b]$ is
      generated as an $A$-module by $1, b, b^2, b^2,
      \ldots$. It is also generated by $1, \ldots,
      b^{n - 1}$. So $A[b]$ is ginitely generated,
      and can take $C = A[b]$.}
    \item[$\Leftarrow$] \cloze{If $C$ is finitely
      generated, let $\phi : C \to C$ be given by
      $\phi(x) = b \cdot x$. Apply the above Lemma to
      the finitely generated $A$-module $C$ with
      $I = A$. We get $\phi^n + a_1 \phi^{n - 1}
      \cdots + a_n \equiv 0$ or $b^n + a_1 b^{n -
      1} + \cdots + a_n$, acting by
      multiplication on $C$, is the zero map.
      Since $C$ is an integral domain, we have
      \[ b^n + a_1 b^{n - 1} + \cdots + a_n = 0
      \qedhere \]}
  \end{enumerate}
\end{proof}
\end{flashcard}