\documentclass{article}
\newcommand{\im}{\mbox{im}}
\newcommand{\Hom}{\mbox{Hom}}
\newtheorem{theorem}{Theorem}

\title{Proof: The Fundamental Theorem of Finitely Generated Modules
  over a Principal Ideal Domain (Dummit/Foote Version)}
\author{Po-Shen Loh}
\date{6 March 2001} 
\begin{document}
\maketitle

\begin{theorem}
  Let $R$ be a principal ideal domain, and let $M$ be a finitely
  generated free module with basis $\{x_k\}_n$.  Let $N$ be a
  submodule of $M$.  Then $N$ is free and there exists a basis
  $\{y_k\}_n$ of $M$ such that for some $\{a_k\} \in R$,
  $\{a_ky_k\}_m$ is a basis for $N$, where $m \leq n$.
\end{theorem}

\noindent\textbf{Proof of Theorem 1:}

Since $R$ is a PID, it is Noetherian; hence every set of ideals of $R$
has a maximal element.  Consider the set of ideals $\{\im \phi\}$
where $\phi \in \Hom_R (N, R)$.  Let the maximal element be $\im v$,
and let $\im v = (a)$.  We can do this since $R$ is a PID.  Let us
define $y$ s.t.\ $v(y) = a$.

Observe that for any other $\phi \in \Hom_R (N, R)$, $\phi(y)$ must be
divisible by $A$, because otherwise the PID-ness of $R$ would imply
that there existed some $\alpha, \beta \in R$ such that $\psi = \alpha
\phi + \beta v$ would evaluate to an element less than $A$; hence
$\psi$ would be the maximal element, contradiction.

In particular, the natural projection homomorphisms $\pi_i =
\mbox{proj onto basis elt} x_i$ satisfy this property, so $a |
\pi_i(y)$.  Decomposing $y$ into that sum, $y = \sum ab_ix_i$ for some
$b_i \in R$.  Notice that $a | y$, and define $y_1 = y/a$.  Now
$v(y_1) = 1$.

We next prove the decomposition $M = \langle y_1\rangle  \oplus \ker(v)$ and $N =
\langle y\rangle  \oplus (\ker(v) \cap N)$.  Since we have a PID, $ay_1 = 0
\Leftrightarrow a = 0$, so $v(y_1) = 1 \Rightarrow \langle y_1\rangle  \cap \ker(v)
= 0$, as desired.  The result for $N$ follows similarly.  And we can
write any element $x \in M$ as $x = [v(x)y_1] + [x - v(x)y_1]$; the
first summand is in $\langle y_1\rangle $ and the second summand maps to zero under
$v$.  Finally, consider $N$'s direct sum: $\forall n \in N$, write $n
= [v(n)y_1] + [n - v(n)y_1]$.  Imeediately we see that the second
summand satisfies our condition, and since $\im v = (a)$, $a | v(n)
\Rightarrow v(n)y_1 = ky$ for some $k \in R$, which completes our
proof.

\begin{theorem}
  Let $M$ be a finitely generated $R-$module.  Then there exists an
  invariant factor decomposition:

  \begin{displaymath}
    M = R/(a_1) \oplus R/(a_2) \oplus \cdots \oplus R/(a_t)
  \end{displaymath}
\end{theorem}

\noindent \textbf{Proof of Theorem 2}

Let $\{x_k\}_n$ generate $M$.  Consider the natural [surjective]
projection homomorphism $\pi : R^n \rightarrow M$.  By the First
Isomorphism Theorem, $M \cong R^n/\ker(\pi)$.  We can now use Theorem
1, as $\ker(\pi) \subset R^n$, so we can find some set of generators
$\{y_i\}_n$ for $R^n$ such that

\begin{eqnarray*}
  M &\cong& \frac{ \langle y_1\rangle  \oplus \langle y_2\rangle
    \oplus \cdots \oplus \langle y_n\rangle  }{ 
    \langle a_1y_1\rangle  \oplus \langle a_2y_2\rangle  \oplus \cdots
    \oplus \langle a_my_m\rangle  } \\
  &\cong&   \langle y_1\rangle /\langle a_1y_1\rangle  \oplus \cdots \oplus 
    \langle y_m\rangle /\langle a_my_m\rangle  \oplus \langle
    a_{m+1}\rangle  \oplus \cdots 
    \langle a_n\rangle \\
    &\cong& R/(a_1) \oplus R/(a_2) \oplus \cdots \oplus R/(a_m)
    \oplus R \oplus R \oplus \cdots \oplus R.
\end{eqnarray*}

And we are done.

\end{document}
