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% 14/03/2023 12AM

\begin{lemma}
    The primes in $\CC[X]$ (up to associates) are
    the polynomials $X - \lambda$, for some
    $\lambda \in \CC$.
\end{lemma}

\begin{proof}
    By the fundamental theorem of algebra, any
    non-constant polynomial in $\CC[X]$ has a root
    in $\CC$, so a factor $X - \lambda$. Hence, the
    irreducibles have degree 1.
\end{proof}

\begin{flashcard}[jordan-normal-form]
\begin{theorem}[Jordan Normal form]
    \cloze{
    Let $\alpha : V \to V$ be an endomorphism of a
    finite dimensional $\CC$-vector space. Let
    $V_\alpha$ be $V$ regarded as a
    $\CC[X]$-module with $X$ acting as $\alpha$.
    There is an isomorphism of $\CC[X]$-modules
    \[ V_\alpha \cong \CC[X] / ((X -
    \lambda_1)^{n_1}) \oplus \cdots \oplus \CC[X]
    / ((X - \lambda_t)^{n_t}) \]
    where $\lambda_1, \ldots, \lambda_t \in \CC$
    (not necessarily distinct). In particular
    there exists a basis for $V$ such that
    $\alpha$ has matrix
    \[
    \begin{pmatrix}
        J_{n_1}(\lambda_1) & 0 & \cdots & 0 \\
        0 & J_{n_2}(\lambda_2) & \cdots & 0 \\
        \vdots & \vdots & \ddots & \vdots \\
        0 & 0 & \cdots & 0
    \end{pmatrix}
    \]
    where
    \[ J_n(\lambda) =
    \begin{pmatrix}
        \lambda & 0 & 0 & \cdots & 0 & 0 \\
        1 & \lambda & 0 & \cdots & 0 & 0 \\
        0 & 1 & \lambda & \cdots & 0 & 0 \\
        \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
        0 & 0 & 0 & \cdots & \lambda & 0 \\
        0 & 0 & 0 & \cdots & 1 & \lambda
    \end{pmatrix}
    \]
}
\end{theorem}

\begin{proof}
    \cloze{
    $\CC[X]$ is a Euclidean Domain and $V_\alpha$
    is finitely generated by Lemma 16.7. We apply
    the \fcemph{primary decomposition}, noting that the
    primes in $\CC[X]$ are as in Lemma 16.10. $V$
    finite dimensional implies we get no copies of
    $\CC[X]$. $J_n(\lambda)$ represents
    multiplying by $X$ on $\CC[X] / (X -
    \lambda)^n$ with respect to the basis $1, X -
    \lambda, (X - \lambda^2, \ldots, (X -
    \lambda)^{n - 1}$.
}
\end{proof}
\end{flashcard}

\begin{remark*}
    \begin{enumerate}[(i)]
        \item If $\alpha$ represented by matrix
            $A$, then the theorem says that $A$ is
            similar to a matrix in JNF.
        \item The Jordan blocks are uniquely
            determined up to reordering. Can be
            proved by considering the dimensions
            of the generalised eigenspace
            $\Ker((\alpha - \lambda \id)^m)$, $m =
            1, 2, 3, \ldots$ (omitted).
        \item The minimal polynomial of $\alpha$
            is $\prod_\lambda (X -
            \lambda)^{c_\lambda}$ where
            $c_\lambda$ is the size of the largest
            $\lambda$-block.
        \item The \emph{characteristic polynomial}
            of $\alpha$ is $\prod_\lambda (X -
            \lambda)^{a_\lambda}$ where
            $a_\lambda$ is the sum of the sizes of
            $\lambda$-blocks.
        \item The number of $\lambda$ blocks is
            the dimension of the
            $\lambda$-eigenspace.
    \end{enumerate}
\end{remark*}

\newpage
\section{Modules over PID (non-examinable)}

The \emph{structure theorem} holds for PID's. We
illustrate some ideas which go into the proof.

\begin{theorem}
    Let $R$ be a PID. Then any finitely generated
    torsion-free $R$-module is free. (For $R$ a
    Euclidean Domain, this is Corollary 16.5).
\end{theorem}

\begin{lemma}
    Let $R$ be a PID and $M$ an $R$-module. Let
    $r_1, r_2 \in R$ not both zero and let $d =
    \gcd(r_1, r_2)$.
    \begin{enumerate}[(i)]
        \item There exists $A \in \SL_2(R)$ such
            that
            \[ A
            \begin{pmatrix}
                r_1 \\
                r_2
            \end{pmatrix}
            =
            \begin{pmatrix}
                \alpha \\
                0
            \end{pmatrix}
            \]
        \item If $x_1, x_2 \in M$ then there
            exists $x_1', x_2 \in M$ such that $R
            x_1 + R x_2 = R x_1' + x_2'$ and $r_1
            x_1 + r_2 x_2 = d x_1' + 0
            x_2'$.
    \end{enumerate}
\end{lemma}

\begin{proof}
    $R$ a PID implies $(r_1, r_2) = (d)$, hence
    there exists $\alpha, \beta \in R$ such that
    $\alpha r_1 + \beta r_2 = d$. Write $r_1 = s_1
    d$, $r_2 = s_2 d$ for some $s_1, s_2 \in R$.
    Then $\alpha s_1 + \beta s_2 = 1$.
    \begin{enumerate}[(i)]
        \item \[ \ub{
            \begin{pmatrix}
                \alpha & \beta \\
                -s_2 & s_1
            \end{pmatrix}
            }_{\det = \alpha s_1 + \beta s_2 = 1}
            \begin{pmatrix}
                r_1 \\
                r_2
            \end{pmatrix}
            =
            \begin{pmatrix}
                d \\
                0
            \end{pmatrix}
            \]
        \item  Let $x_1' = s_1 x_1 + s_2 x_2$,
            $x_2' = -\beta x_1 + \alpha x_2$. Then
            $R x_1' + R x_2' \subseteq R x_1 + R
            x_2$. To prove the reverse inclusion
            we solve for $x_1$ and $x_2$ in terms
            of $x_1'$ and $x_2'$. This is possible
            since
            \[ \det
            \begin{pmatrix}
                s_1 & s_2 \\
                -\beta & \alpha
            \end{pmatrix}
            = \alpha s_1 + \beta s_2 = 1 \]
            Finally
            \begin{align*}
                r_1 x_1 + r_2 x_2
                &= d(s_1 x_1 + s_2 x_2) \\
                &= d x_1' \qedhere
            \end{align*}
    \end{enumerate}
\end{proof}

\begin{proof}[Proof of Theorem 17.1]
    Let $M = R x_1 + R x_n$ with $n$ as small as
    possible. If $x_1, \ldots, x_n$ are
    independent then $M$ is free, and we're done.
    Otherwise, $\exists r_1, \ldots, r_n \in R$
    not all zero with $\sum_{r = 1}^n r_i x_i =
    0$. WLOG $r_1 \neq 0$. Lemma 17.2 (ii) shows
    that after replacing $x_1$ and $x_2$ by
    suitable $x_1'$ and $x_2'$, we may assume $r_1
    \neq 0$ and $r_2 = 0$. Repeating this process
    (changing $x_1$ and $x_3$, then $x_1$ and
    $x_4$ and so on), we may assume $r_1 \neq 0$,
    $r_2 = 0, \ldots, r_n = 0$. Now $r_1 x_1 = 0
    \implies x_1 = 0$ (since $M$ is torsion free).
    Thus, $M = R x_2 + \cdots + R x_n$, which
    contradicts our choice of $n$.
\end{proof}