% vim: tw=50
% 04/03/2023 12AM

\newpage
\section{Direct Sums and Free Module}

\begin{definition*}
    Let $M_1, \ldots, M_n$ be $R$-modules. The direct sum
    \[ M_1 \oplus M_2 \oplus \cdots \oplus M_n \]
    is the set $M_1 \times \cdots \times M_n$ with
    operations
    \begin{align*}
        (m_1, \ldots, m_n) + (m_1', \ldots, m_n')
        &= (m_1 + m_1', \ldots, m_n + m_n') \\
        r(m_1, \ldots, m_n)
        &= (rm_1, \ldots, rm_n)
        &&(r \in R)
    \end{align*}
\end{definition*}

\begin{example*}
    $R^n = R \oplus \cdots \oplus R$.
\end{example*}

\begin{flashcard}[direct-sum-quotient-lemma]
\begin{lemma}
    If $M = \bigoplus_{i = 1}^n M_i$ and $N_i \le
    M_i$ for all $i$, then setting $N =
    \bigoplus_{i = 1}^n N_i \le M$, we have
    \[ \cloze{M / N \cong \bigoplus_{i = 1}^n M_i
    / N_i} \]
\end{lemma}

\begin{proof}
    \cloze{
    Apply first isomorphism theorem to the
    surjective $R$-module homomorphism
    \begin{align*}
        M
        &\to \bigoplus_{i = 1}^n M_i / N_i \\
        (m_1, \ldots, m_n)
        &\mapsto (m_1 + N_1, \ldots, m_n + N_n)
    \end{align*}
    with kernel $N = \bigoplus_{i = 1}^n N_i$.
}
\end{proof}
\end{flashcard}

\begin{flashcard}[module-independent-set-defn]
\begin{definition*}
    Let $m_1, \ldots, m_n \in M$. The set $\{m_1,
    \ldots, m_n\}$ is independent if
    \[ \cloze{\sum_{i =
    1}^n r_i m_i = 0 \implies r_1 = r_2 = \cdots =
    r_n = 0} \]
\end{definition*}
\end{flashcard}

\begin{flashcard}[module-freely-generated-defn]
\begin{definition*}
    A subset $S \subset M$ generates $M$ freely if
    \cloze{
    \begin{enumerate}[(i)]
        \item $S$ generates $M$, i.e. $\forall m
            \in M$, $m = \sum_{i = 1}^n r_i s_i$
            for some $r_i \in R$, $s_i \in S$.
        \item Any function $\psi : S \to N$ where
            $N$ is an $R$-module, extends to an
            $R$-module homomorphism $\theta : M
            \to N$. (Such an extension is unique
            by (i)).
    \end{enumerate}
    An $R$-module which is freely generated by
    some subset $S \subset M$ is called
    \emph{free} and $S$ is called a \emph{free
    basis}.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[subset-freely-generating-tfae]
\begin{proposition}
    For a subset $S = \{m_1, \ldots, m_n\} \subset
    M$, the following are equivalent:
    \begin{enumerate}[(i)]
        \item \cloze{$S$ generates $M$ freely.}
        \item \cloze{$S$ generates $M$ and $S$ is
            independent.}
        \item \cloze{Every element of $M$ can be written
            uniquely as
            \[ r_1 m_1 + \cdots r_n m_n \]
            for some $r_1, \ldots, r_n \in R$.}
        \item \cloze{The $R$-module homomorphism
            \begin{align*}
                R^n
                &\to M \\
                (r_1, \ldots, r_n)
                &\mapsto \sum_{i = 1}^n r_i m_i
            \end{align*}
            is an isomorphism.}
    \end{enumerate}
\end{proposition}
\end{flashcard}

\begin{proof}
    \begin{enumerate}
        \item[(i) $\implies$ (ii)] Let $S$
            generate $M$ freely. If $S$ is not
            independent, then $\exists r_1,
            \ldots, r_n \in R$ with $\sum r_i m_i
            = 0$ and some $r_j \neq 0$. Define
            $\psi : S \to R$
            \[ m_i \mapsto \begin{cases}
                1 & i = j \\
                0 & i \neq j
            \end{cases} \]
            extends to $R$-module homomorphism $M
            \to R$. Then
            \begin{align*}
                0
                &= \theta(0) \\
                &= \theta \left( \sum r_i m_i
                \right) \\
                &= \sum r_i \theta(m_i) \\
                &= r_i
            \end{align*}
            Thus $S$ is independent. The rest are
            exercises.
    \end{enumerate}
\end{proof}

\begin{example*}
    $A$ is non-trivial finite abelian group. Then
    $A$ is not a free $\ZZ$-module.
\end{example*}

\begin{example*}
    The set $\{2, 3\}$ generates $\ZZ$ as a
    $\ZZ$-module, but they are not independent
    since
    \[ (3) \cdot 2 + (-2) \cdot 3 = 0 \]
    Furthermore, no subset of $\{2, 3\}$ is a free
    basis, since $\{2\}$ and $\{3\}$ do not
    generate.
\end{example*}

\begin{flashcard}[invariance-of-dimension]
\begin{proposition}[Invariance of dimension]
    \cloze{
    Let $R$ be a non-zero ring. If $R^m \cong R^n$
    as $R$-modules then $m = n$.}
\end{proposition}

\begin{proof}
    \cloze{
    First, we introduce a general construction.
    Let $I \normalsub R$ and $M$ an $R$-module.
    Define
    \[ IM = \left\{ \sum a_i m_i : a_i \in I, m_i
    \in M\right\} \le M \]
    The quotient $M / IM$ is an $R/I$-module via
    \[ (r + I)(m + IM) = rm + IM \]
    Well-defined: if $b \in I$ then
    \[ b \cdot (m + IM) = bm + IM = 0 + IM \]
    Suppose $R^m \cong R^n$. Choose $I \normalsub
    R$ maximal ideal (user Zorn's Lemma and
    Example Sheet 2 Question 4). By the above, we
    get an isomorphism of $R / i$ module
    \[ (R / I)^m \cong R^m / IR^m \cong R^n / IR^n
    \cong (R / I)^n \]
    But $I \normalsub R$ is maximal hence $R / I$
    is a field. So $m = n$ by invariance of
    dimension for vector spaces.
}
\end{proof}
\end{flashcard}

\newpage
\section{The Structure Theorem and Applications}

Until further notice: $R$ is always a Euclidean
domain, $\phi : R \setminus \{0\} \to \ZZ_{\ge 0}$
Euclidean function. Let $A$ be an $m \times n$
matrix with entries in $R$.

\begin{definition*}
    The elementary row operations are:
    \begin{enumerate}[(ER1)]
        \item Add $\lambda$ times $i$-th row to
            $j$-th row ($\lambda \in R$, $i \neq
            j$).
        \item Swapping $i$-th and $j$-th rows.
        \item Multiply $i$-th row by $u \in
            R^\times$.
    \end{enumerate}
    Each of these can be realised by left
    multiplication by an $m \times m$ invertible
    matrix:
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/d2ce241eba8b11ed.png}
    \end{center}
\end{definition*}

\noindent
In particular, these operations are reversible.
Similarly, we can define elementary column operations
(EC1-3) -- realised b right multiplication by an
invertible $n \times n$ matrix.

\begin{flashcard}[equivalent-matrices]
\begin{definition*}[Equivalent matrices]
    \cloze{
    Two $m \times n$ matrices $A$ and $B$ are
    \emph{equivalent} if
    there exists a sequence of elementary row and
    column operations taking $A$ to $B$. If they
    are equivalent, then there exists (invertible) $P$, $Q$
    such that $B = QAP$.
}
\end{definition*}
\end{flashcard}