% vim: tw=50
% 02/03/2023 12AM

\begin{flashcard}[submodule-defn]
\begin{definition*}
    $M$ an $R$-module. $N \subset M$ is an
    $R$-submodule (written $N \le M$) if \cloze{it is a
    subgroup of $(M, +)$ and $r \cdot n \in N$ for
    all $r \in R$, $n \in N$.}
\end{definition*}
\end{flashcard}

\begin{hiddenflashcard}[how-to-check-submodule]
How to check if $N$ is a submodule of $M$? \\
\cloze{
\begin{enumerate}[(i)]
    \item $0 \in N$
    \item $n_1, n_2 \in N \implies n_1 + n_2 \in N$
    \item $n \in N, r \in R \implies r \cdot n \in N$
\end{enumerate}
}
\end{hiddenflashcard}

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item A subset of $R$ is an $R$-submodule
        \emph{precisely} when it is an ideal.
    \item When $R = F$ is a field, module $\equiv$
        vector space, submodule $\equiv$ vector
        subspace.
\end{enumerate}

\begin{flashcard}[module-quotient-defn]
\begin{definition*}
    If $N \le M$ an $R$-submodule, the quotient $M
    / N$ is \cloze{the quotient of groups under $+$ with
    \[ r \cdot (m + N) = rm + N \]
    This is well-defined, and makes $M / N$ an
    $R$-module.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[module-homomorphism-defn]
\begin{definition*}
    Let $M, N$ be $R$-modules. A function $f : M
    \to N$ is \cloze{an \emph{$R$-module
    homomorphism}} if
    it is \cloze{a homomorphism of abelian groups and
    \[ f(r \cdot m) = r \cdot f(m) \qquad \forall
    r \in R, m \in M \]}
\end{definition*}
\end{flashcard}

\begin{theorem*}[First isomorphism theorem]
    Let $f : M \to N$ be an $R$-module
    homomorphism. Then
    \begin{itemize}
        \item $\Ker(f) \defeq \{m \in M \mid f(m)
            = 0\} \le M$
        \item $\Im(f) \defeq \{f(m) \in N \mid m
            \in M\} \le N$
    \end{itemize}
    and $M / \Ker(f) \cong \Im(f)$.
\end{theorem*}

\begin{proof}
    Similar to before.
\end{proof}

\begin{theorem*}[Second isomorphism theorem]
    Let $A, B \le M$ be submodules. Then
    \[ A + B = \{a + b \mid a \in A, b \in B\} \le
    M \]
    \[ A \cap B \le M \]
    and
    \[ A / (A \cap B) \cong (A + B) / B \]
\end{theorem*}

\begin{proof}
    Apply first isomorphism theorem to the
    composite $A \injto M \injto M / B$.
\end{proof}

\myskip
For third isomorphism theorem, note that there
exists bijection
\[ \{\text{submodules of $M / N$}\}
\leftrightarrow \{\text{submodules of $M$
containing $N$}\} \]

\begin{theorem*}[Third isomorphism theorem]
    If $N \le L \le M$ are $R$-submodules of $M$,
    then
    \[ \frac{M / N}{L / N} \cong M / L \]
\end{theorem*}

\noindent
In particular, these apply to vector spaces
(compare with results from Linear Algebra).

\myskip
Let $M$ be an $R$-module. If $m \in M$, write $R_m
= \{rm \in M \mid r \in R\}$ -- submodule
generated by $m$. If $A, B \le M$, write
\[ A + B = \{a + b \mid a \in A, b \in B\} \le M \]

\begin{definition*}
    \begin{itemize}
        \item $M$ is cyclic if there exists $m \in
            M$ such that $M = R_m$.
        \item $M$ is finitely generated if there
            exists $m_1, \ldots, m_n \in M$ such
            that
            \[ M=  R_{m_1} + R_{m_2} + \cdots +
            R_{m_n} \]
    \end{itemize}
\end{definition*}

\begin{hiddenflashcard}[cyclic-module-defn]
\begin{definition*}[Cyclic Module]
    \cloze{
    $M$ is cyclic if there exists $m \in M$ such
    that $M = R_m$.
    }
\end{definition*}
\end{hiddenflashcard}

\begin{hiddenflashcard}[finitely-generated-module-defn]
\begin{definition*}[Finitely Generated Module]
    \cloze{
    $M$ is finitely generated if there exists $m_1,
    \ldots, m_n \in M$ such that
    \[ M = R_{m_1} + R_{m_2} + \cdots + R_{m_n} \]
    }
\end{definition*}
\end{hiddenflashcard}

\begin{flashcard}[cyclic-module-iff]
\begin{lemma}
    $M$ is cyclic if and only if \cloze{$M \cong R / I$
    for some $I \normalsub R$.}
\end{lemma}

\begin{proof}
    \begin{itemize}
        \item[$\Rightarrow$] \cloze{Suppose $M = R_m$.
            Then there is a surjective $R$-module
            homomorphism $R \to M$, $r \mapsto
            rm$. Its kernel is an $R$-submodule of
            $R$, i.e. an ideal. Then first
            isomorphism theorem gives $R / I \cong
            M$.}
        \item[$\Leftarrow$] \cloze{$R / I$ is generated as an
            $R$-module by $1_R + I$. \qedhere}
    \end{itemize}
\end{proof}
\end{flashcard}

\begin{flashcard}[finitely-generated-module-iff]
\begin{lemma}
    $M$ finitely generated if and only if \cloze{there
    exists a surjective $R$-module homomorphism $f
    : R^n \to M$ for some $n$.}
\end{lemma}

\begin{proof}
    \begin{itemize}
        \item[$\Rightarrow$] \cloze{If $M = R_{m_1} +
            R_{m_2} + \cdots + R_{m_n}$ define $f
            : R^n \to M$, $(r_1, \ldots, r_n)
            \mapsto \sum_{i = 1}^n r_i m_i$ a
            surjective $R$-module homomorphism.}
        \item[$\Leftarrow$] \cloze{Let $e_i = (0, \ldots, 1, \ldots,
            0) \in R^n$. (1 is in the $i$-th
            place). Given $f$, let $m_i \defeq
            f(e_i) \in M$. Then any $m \in M$ is
            of the form
            \begin{align*}
                f(r_1, \ldots, r_n)
                &= f \left( \sum_{i = 1}^n r_i e_i
                \right) \\
                &= \sum_{i = 1}^n r_i f(e_i) \\
                &= \sum_{i = 1}^n r_i m_i
            \end{align*}
            Thus $M = Rm_1 + \cdots + R m_n$.
            \qedhere}
    \end{itemize}
\end{proof}
\end{flashcard}

\begin{flashcard}[submodule-of-finitely-generated-is-finitely-generated]
\begin{corollary}
    Let $N \le M$ be an $R$-submodule. If $M$ is
    finitely generated, then $M / N$ is finitely
    generated.
\end{corollary}

\begin{proof}
    \cloze{
    Let $f : R^n \to M$ be a surjective $R$-module
    homomorphism. Then $R^n \to M \to M / N$ is a
    surjective $R$-module homomorphism.
}
\end{proof}
\end{flashcard}

\begin{hiddenflashcard}[submodule-of-finitely-generated-is-finitely-generated-counterexample]
\begin{lemma}
    Let $N \le M$ be an $R$-submodule. If $M$ is
    finitely generated, then $N$ is finitely
    generated.
\end{lemma}

\begin{proof}
    \cloze{
    This statement is actually \fcemph{false}. As
    a counter example, consider $R$ a non-Noetherian
    ring and $I \normalsub R$ a non-finitely generated
    ideal. Then $M = R$ is a finitely generated $R$-module
    and $N = I$ is a submodule which is not finitely
    generated.
    }
\end{proof}
\end{hiddenflashcard}

\begin{example*}[Counter-example]
    A submodule of a finitely generated module
    need not be finitely generated. Let $R$ be a
    non-Noetherian ring and $I \normalsub R$ a
    non-finitely generated ideal. Then $R$ is a
    finitely generated $R$-module and $I$ is a
    submodule which is not finitely generated.
\end{example*}

\begin{remark*}
    A submodule of a finitely generated module
    over a Noetherian ring is finitely generated
    (Examples Sheet 4).
\end{remark*}

\begin{lemma}
    Let $R$ be an integral domain. Then
    \[ \text{every submodule of a cyclic
    $R$-submodule is cyclic} \iff \text{$R$ is a
    PID} \]
\end{lemma}

\begin{proof}
    \begin{itemize}
        \item[$\Rightarrow$] $R$ is a cyclic
            $R$-module. Saying its submodules are
            cyclic precisely means that every
            ideal is principal.
        \item[$\Leftarrow$] If $M$ is a cyclic
            $R$-module, then $M \cong R / I$, $I
            \normalsub R$ by Lemma 14.1. Any
            submodule of $R / I$ is of the form $J
            / I$ for some ideal $J \normalsub R$
            and $I \le J$. $R$ a PID implies $J$
            principal hence $J / I$ is cyclic.
    \end{itemize}
\end{proof}

\begin{definition*}
    Let $M$ be an $R$-module.
    \begin{enumerate}[(i)]
        \item
            \begin{flashcard}[torsion-element-defn]
            An element $m \in M$ is torsion if
            \cloze{there exists $0 \neq r \in R$ with $rm
            = 0$.}
            \end{flashcard}
        \item
            \begin{flashcard}[torsion-module-defn]
            $M$ is a torsion module if \cloze{every $m
            \in M$ is torsion.}
            \end{flashcard}
        \item
            \begin{flashcard}[torsion-free-module-defn]
            $M$ is torsion free if \cloze{every $0 \neq
            m \in M$ is not torsion.}
            \end{flashcard}
    \end{enumerate}
\end{definition*}

\begin{example*}
    \begin{itemize}
        \item The torsion elements in a
            $\ZZ$-module ($=$ abelian group) are
            the elements of finite order.
        \item Any $F$-module ($=$ vector space) is
            torsion free.
    \end{itemize}
\end{example*}