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% 09/03/2023 12AM

\begin{lemma}
    $R$ an PID. Any submodule of $R^m$ is finitely
    generated.
\end{lemma}

\begin{proof}
    Example Sheet 4.
\end{proof}

\begin{flashcard}[free-basis-for-submodule-of-ed]
\begin{theorem}
    Let $R$ be a Euclidean Domain and $N \le R^m$.
    There is a free basis \cloze{$x_1, \ldots, x_m$ for
    $R^m$ such that $N$ is generated by $d_1 x_1,
    \ldots, d_t x_t$ for some $t \le m$ and $d_1,
    \ldots, d_t \in R$ with $d_1 \mid d_2 \mid
    \cdots \mid d_t$.}
\end{theorem}

\begin{proof}
    \cloze{
    By Lemma 16.3 we have $N = Ry_1 + \cdots +
    R y_n$ for some $n \le m$. Each $y_i$ belongs
    to $R^m$, so we can form an $m \times n$
    matrix
    \[ A = (y_1 | y_2 | \cdots | y_n) \]
    By Theorem 16.1, $A$ is equivalent to
    \[ A' = \diag(d_1, \ldots, d_t, 0, \ldots, 0) \]
    $A'$ obtained from $A$ by elementary row and
    column operations. Each \fcemph{row operation changes
    our choice of free basis} for $R^m$ and each
    \fcemph{column operation changes our set of
    generators}
    for $N$. Thus, after changing free basis of
    $R^m$ to $x_1, \ldots, x_m$ (say), the
    submodule $N$ is generated by $d_1 x_1, d_2
    x_2, \ldots, d_t x_t$ as claimed.
}
\end{proof}
\end{flashcard}

\begin{flashcard}[structure-theorem-for-ed]
\begin{theorem*}[Structure Theorem]
    \cloze{Let $R$ be a \fcemph{Euclidean Domain} and $M$ a
    finitely generated $R$-module. Then
    \[ M \cong R / (d_1) \oplus R / (d_2) \oplus
    \cdots \oplus R / (d_t) \oplus \ub{R \oplus
    \cdots \oplus R}_{\text{$k$ copies}} \]
    for some $0 \neq d_1 \in R$ with $d_1 \mid d_1
    \mid \cdots \mid d_t$ and $k \ge 0$. The $d_i$
    are called \emph{invariant factors}.}
\end{theorem*}

\begin{proof}
    \cloze{
    Since $M$ is finitely generated, there exists
    a surjective $R$-module homomorphism $\phi :
    R^m \to M$ for some $m$ (Lemma 14.1). By first
    isomorphism theorem, $M \cong R^m /
    \Ker(\phi)$. By Theorem 16.4, there exists a
    free basis $x_1, \ldots, x_m$ for $R^m$ such
    that $\Ker(\phi)$ is generated by $d_1 x_1,
    \ldots, d_t x_t$ with $d_1 \mid d_2 \mid
    \cdots \mid d_t$. Then
    \begin{align*}
        M
        &\cong \frac{R \oplus R \oplus \cdots
        \oplus R \oplus R \oplus \cdots \oplus
        R}{d_1 R \oplus d_2 R \oplus \cdots \oplus
        d_t R \oplus 0 \oplus \cdots \oplus 0} \\
        &\cong R / (d_1) \oplus R / (d_2) \oplus
        \cdots \oplus R / (d_t) \oplus R \oplus
        \cdots \oplus R
        &&\text{(by Lemma 15.1)}
        \qedhere
    \end{align*}
}
\end{proof}
\end{flashcard}

\begin{remark*}
    After deleting these $d_i$ which are units,
    the module $M$ uniquely determines the $d_i$
    (up to associates). Proof omitted.
\end{remark*}

\begin{flashcard}[torsion-iff-free-module-over-ed]
\begin{corollary}
    Let $R$ be a \cloze{\fcemph{Euclidean Domain}}. Then any
    finitely generated torsion-free $R$-module
    \cloze{is free.}
\end{corollary}

\begin{proof}
    \cloze{
    \prompt{(Using structure theorem):}
    $M$ torsion-free $\implies$ no submodules of
    the form $R / (d)$ with $d \neq 0$. Thus $M
    \cong R^m$ for some $m$.
}
\end{proof}
\end{flashcard}

\begin{example*}
    $R = \ZZ$. Consider the abelian group $G$
    generated by $a$ and $b$ subject to the
    relations $2a  +b = 0$, $-a + 2b = 0$. Then $G
    \cong \ZZ^2 / N$, where $N$ is generated by
    $(2, 1)$, $(-1, 2)$.
    \[ A =
    \begin{pmatrix}
        2 & -1 \\
        1 & 2
    \end{pmatrix}
    \qquad \text{ has SNF }\qquad
    \begin{pmatrix}
        1 & 0 \\
        0 & 5
    \end{pmatrix}
    \]
    Thus can change basis for $\ZZ^2$ such that
    $N$ is generated by $(1,0)$ and $(0,5)$. Thus
    \[ G \cong \ZZ^2 / N \cong \frac{\ZZ \oplus
    \ZZ}{\ZZ \oplus 5\ZZ} \cong \ZZ / 5\ZZ \]
\end{example*}

\noindent
More generally:

\begin{theorem*}[Structure theorem for finitely
    generated abelian groups]
    Any finitely generate abelian group $G$ is
    isomorphic to
    \[ \ZZ / d_1 \ZZ \oplus \cdots \oplus \ZZ /
    d_t \ZZ \oplus \ZZ \oplus \ZZ \oplus \cdots
    \oplus \ZZ \]
    where $d_1 \mid d_2 \mid \cdots \mid d_t$ and
    $r \ge 0$.
\end{theorem*}

\begin{proof}
    Take $R = \ZZ$ in structure theorem.
\end{proof}

\begin{remark*}
    The special case $G$ is finite (so $r = 0$)
    was quoted as Theorem 6.4.
\end{remark*}

\noindent
In \hyperlink{section.6}{Section 6}, we saw that
any finite abelian group can be written as a
product of $C_{p^i}$'s where $p$ is prime. To
generalise this we need:

\begin{flashcard}[direct-sum-of-coprime-ring-quotients]
\begin{lemma}
    Let $R$ be a \cloze{\fcemph{PID}} and $a, b \in R$ with
    \cloze{$\gcd(a, b) = 1$. Then}
    \[ R / (ab) \cong R / (a) \oplus R / (b) \]
    as $R$-modules.\fcscrap{ (Case $R = \ZZ$ was Lemma
    6.2).}
\end{lemma}

\begin{proof}
    \cloze{
    $R$ a PID $\implies (a, b) = (d)$ for some $d
    \in R$. But $\gcd(a, b) = 1$ hence $d$ a unit.
    So there exists $r, s \in R$ such that $ra +
    sb = 1$. Define an $R$-module homomorphism
    \[ \psi : R \to R / (a) \oplus R / (b) \qquad
    x \mapsto (x + (a), x + (b)) \]
    Then $\psi(sb) = (1 + (a), 0 + (b))$,
    $\psi(ra) = (0 + (a), 1 + (b))$. Thus
    \[ \psi(sbx + ray) = (x + (a), y + (b)) \]
    for any $x, y \in R$, so $\psi$ is
    surjective. Clearly $(ab) \le \Ker(\psi)$.
    Conversely, if $x \in \Ker(\psi)$, then $x \in
    (a) \cap (b)$ and
    \begin{align*}
        x
        &= x(ra + sb) \\
        &= \ub{r(ax)}_{\in (ab)} + \ub{s(xb)}_{\in
        (ab)} \\
        &\in (ab)
    \end{align*}
    Thus $\Ker(\psi) = (ab)$. Then by the First
    Isomorphism Theorem for rings, $R / (ab) \cong
    R / (a) \oplus R / (b)$.
}
\end{proof}
\end{flashcard}
