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% 11/03/2023 12AM

\begin{flashcard}[primary-decomposition-theorem]
\begin{theorem*}[Primary decomposition theorem]
    \cloze{
    Let $R$ be a Euclidean Domain and $M$ a
    finitely generated $R$-module. Then
    \[ M \cong R / (p_1^{n_1}) \oplus \cdots \oplus
    R / (p_k^{n_k}) \oplus R^m \]
    (as $R$-modules) where $p_1, \ldots, p_k$ are
    primes (not necessarily distinct) and $m \ge
    0$.
}
\end{theorem*}

\begin{proof}
    \cloze{
    By the structure theorem
    \[ M \cong R / (d_1) \oplus \cdots \oplus R /
    (d_t) \oplus R^m \]
    So it suffices to consider $M \cong R /
    (d_i)$, $d_i = u p_1^{a_1} \cdots p_r^{a_r}$
    where $u$ is a unit and $p_1, \ldots, p_r$ are
    distinct (non-associate) primes. Lemma 16.6
    implies
    \[ R / (d_i) \cong R / (p_1^{a_1}) \oplus
    \cdots \oplus R / (p_r^{a_r}) \qedhere \]
}
\end{proof}
\end{flashcard}

\myskip
Let $V$ be a vector space over a field $F$. Let
$\alpha : V \to V$ be a linear map and let
$V_\alpha$ denote the $F[X]$-module $V$ where
$F[X] \times V \to V$ is given by $(f(X), v)
\mapsto f(\alpha)(v)$.

\begin{lemma}
    If $V$ finite dimensional, then $V_\alpha$ is
    a finitely generated $F[X]$-module.
\end{lemma}

\begin{proof}
    If $v_1, \ldots, v_n$ generate $V$ as an
    $F$-vector space, then they generate
    $V_\alpha$ as an $F[X]$-module since $F \le
    F[X]$.
\end{proof}

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item Suppose $V_\alpha \cong F[X] / (X^n)$ as
        $F[X]$-module. Then $1, X, X^2, \ldots,
        X^{n - 1}$ is a basis for $F[X] / (X^n)$
        as an $F$-vector space, and with respect
        to this basis $\alpha$ has matrix
        \[ (*) =
        \begin{pmatrix}
            0 & 0 & 0 & \cdots & 0 & 0 \\
            1 & 0 & 0 & \cdots & 0 & 0 \\
            0 & 1 & 0 & \cdots & 0 & 0 \\
            \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
            0 & 0 & 0 & \cdots & 0 & 0 \\
            0 & 0 & 0 & \cdots & 1 & 0
        \end{pmatrix}
        \]
    \item Suppose $V_\alpha \cong F[X] / (X -
        \lambda)^n$ as $F[X]$-modules. Then with
        respect to basis $1, (X - \lambda), (X -
        \lambda)^2, \ldots, (X - \lambda)^{n -
        1}$, $\alpha - \lambda \id$ has matrix
        ($*$), thus $\alpha$ has matrix
        \[
        \begin{pmatrix}
            \lambda & 0 & 0 & \cdots & 0 & 0 \\
            1 & \lambda & 0 & \cdots & 0 & 0 \\
            0 & 1 & \lambda & \cdots & 0 & 0 \\
            \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
            0 & 0 & 0 & \cdots & \lambda & 0 \\
            0 & 0 & 0 & \cdots & 1 & \lambda
        \end{pmatrix}
        \]
    \item
        \begin{flashcard}[companion-matrix]
        Suppose $V_\alpha \cong F[X] / (f(X))$
        where $f(X) = X^n + a_{n - 1} X^{n - 1} +
        \cdots + a_0$, then with respect to basis
        $1, X, X^2, \ldots, X^{n - 1}$, $\alpha$
        has matrix
        \[
        \cloze{
        \begin{pmatrix}
            0 & 0 & 0 & \cdots & 0 & -a_0 \\
            1 & 0 & 0 & \cdots & 0 & -a_1 \\
            0 & 1 & 0 & \cdots & 0 & -a_2 \\
            \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
            0 & 0 & 0 & \cdots & 0 & -a_{n - 2} \\
            0 & 0 & 0 & \cdots & 1 & -a_{n - 1}
        \end{pmatrix}
        }
        \]
        This is called the
        \cloze{\fcemph{companion matrix}} $C(f)$
        of the monic polynomial $f$.
        \end{flashcard}
\end{enumerate}

\begin{flashcard}[rational-canonical-form]
\begin{theorem}[Rational canonical form]
    \cloze{
    Let $\alpha : V \to V$ be an endomorphism of a
    finite dimensional $F$-vector space, where $F$
    is a field. Then $F[X]$-module $V_\alpha$
    decomposes as
    \[ V_\alpha \cong F[X] / (f_1) \oplus \cdots
    \oplus F[X] / (f_t) \]
    where $f_i \in F[X]$ monic and $f_1 \mid f_2
    \mid \cdots \mid f_t$. Moreover, with respect
    to a suitable basis for $V$ (as an $F$ vector
    space), $\alpha$ has matrix
    \[
    \begin{pmatrix}
        C(f_1) & 0 & \cdots & 0 \\
        0 & C(f_2) & \cdots & 0 \\
        \vdots & \vdots & \ddots & \vdots \\
        0 & 0 & \cdots & C(f_t)
    \end{pmatrix}
    \tag{$**$} \]
}
\end{theorem}

\begin{proof}
    \cloze{
    By Lemma 16.7, $V_\alpha$ is a finitely
    generated $F[X]$-module. Since $F[X]$ is a
    Euclidean Domain, structure theorem implies
    \[ V_\alpha \cong F[X] / (f_1) \oplus \cdots
    \oplus F[X] / (f_t) \oplus F[X]^m \]
    with $f_1 \mid f_2 \mid \cdots \mid f_t$.
    Since $V$ is finite dimensional as an $F$
    vector space, $m = 0$. Upon multiplying $f_i$
    by a unit we may assume $f_i$ is monic.}
\end{proof}
\end{flashcard}

\begin{remark*}
    \begin{enumerate}[(i)]
        \item If $\alpha$ is represented by an $n
            \times n$ matrix $A$, then the theorem
            says that $A$ is similar to ($**$).
        \item The minimal polynomial of $\alpha$
            is $f_t$.
        \item The characteristic polynomial of
            $\alpha$ is $\prod_{i = 1}^t f_i$.
    \end{enumerate}
    The last two properties show that the minimal
    polynomial divides the characteristic
    polynomial, which is the Cayley-Hamilton
    Theorem.
\end{remark*}

\begin{example*}
    If $\dim V = 2$, then $\sum \deg f_i = 2$. So
    \[ V_\alpha = F[X] / (X - \lambda) \oplus F[X]
    / (X - \lambda) \]
    or
    \[ V_\alpha \cong F[X] / (f) \]
    where $f$ is the characteristic polynomial of
    $\alpha$.
\end{example*}

\begin{corollary}
    Let $A, B \in \GL_2(F)$ non-scalar. Then
    \[ \text{$A$ and $B$ are similar ($=$
    conjugate)} \iff \text{they have the same
    characteristic polynomial} \]
\end{corollary}

\begin{proof}
    \begin{enumerate}
        \item[$\Rightarrow$] Linear algebra.
        \item[$\Leftarrow$] By the last example,
            $A$ and $B$ are similar to $C(f)$.
    \end{enumerate}
\end{proof}

\begin{flashcard}[annihilator-module-defn]
\begin{definition*}
    The \emph{annihilator} of an $R$ module $M$ is
    \[ \Ann_R(M) = \cloze{\{r \in R \mid rm = 0 \forall m
    \in M\} \normalsub R} \]
\end{definition*}
\end{flashcard}

\begin{example*}
    \begin{enumerate}[(i)]
        \item $I \normalsub R$, then $\Ann_R(R /
            I) = I$.
        \item If $A$ is a finite abelian group,
            then $\Ann_\ZZ(A) = (e)$ where $e$ is
            the exponent of $A$.
        \item If $V_\alpha$ as above, then
            $\Ann_{F[X]}(V_\alpha)$ is the ideal
            generated by the minimal polynomial of
            $\alpha$.
    \end{enumerate}
\end{example*}


