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\begin{document}
\section*{A note on lifting isomorphisms of modules over PIDs}

\subsection*{\textit{Darij Grinberg}}

\subsubsection*{version 0.4, 8 Oct 2018}

This note has been written to supplement Keith Conrad's \cite{kc-ab}, though
it is largely independent of the latter. I am going to show some (rather
elementary) properties of free modules over PIDs and apply them to drop the
\textquotedblleft full submodule\textquotedblright\ resp. \textquotedblleft
nonzero determinant\textquotedblright\ restraints which qualify many
statements made in \cite{kc-ab}. Thanks are due to Keith Conrad for a
correction and helpful remarks.

{\small The LaTeX sourcecode of this note contains additional details of
proofs inside ``verlong'' environments (i. e., between ``\texttt{%
%TCIMACRO{\TEXTsymbol{\backslash}}%
%BeginExpansion
$\backslash$%
%EndExpansion
begin\{verlong\}}'' and ``\texttt{%
%TCIMACRO{\TEXTsymbol{\backslash}}%
%BeginExpansion
$\backslash$%
%EndExpansion
end\{verlong\}}''). I doubt they are of any use.}

\begin{center}
\textbf{\S 1. Some properties of modules}
\end{center}

Before we make any statements, let us clarify some notation:

\begin{quote}
\textbf{Convention.} In the following, $\mathbf{N}$ will denote the set of all
nonnegative integers. This set includes $0$.

\textbf{Definition.} An \textit{integral domain} is defined as a commutative
ring $R$ with unity such that $1\neq0$ in $R$ (that is, the unity of $R$ does
not equal the zero of $R$) and such that any two elements $a\in R$ and $b\in
R$ satisfying $ab=0$ must satisfy $\left(  a=0\text{ or }b=0\right)  $. A
\textit{principal ideal domain} is defined as an integral domain $R$ in which
every ideal is principal (i.e., every ideal $I$ of $R$ can be written in the
form $I=\left(  a\right)  $ for some $a\in R$). The word \textquotedblleft
PID\textquotedblright\ is an abbreviation for \textquotedblleft principal
ideal domain\textquotedblright.

\textbf{Definition.} Let $R$ be a commutative ring with unity. An $R$-module
is said to be \textit{finite free} if it has a finite basis.
\end{quote}

We are going to prove all but one (Lemma A.2) of the following seven
statements. Of course, very few of them are new.

\begin{quote}
\textbf{Lemma A.1.} Let $R$ be a commutative ring with unity (not necessarily
a PID). Let $S$ be a finite free $R$-module. Let $M$ be any $R$-module. Let
$f:M\rightarrow S$ be a surjective $R$-module homomorphism. Then, there exists
an $R$-submodule $N$ of $M$ such that $M=N\oplus\left(  \operatorname*{Ker}%
f\right)  $ (an internal direct sum) and such that $f\mid_{N}:N\rightarrow S$
is an $R$-module isomorphism.

\textbf{Lemma A.2.} Let $R$ be a PID. Every submodule of a finite free
$R$-module is finite free.

\textbf{Lemma A.3.} Let $R$ be a commutative ring with unity (not necessarily
a PID). Let $M$ be a finitely-generated $R$-module. Let $f$ be a surjective
$R$-module homomorphism $M\rightarrow M$. Then, $f$ is an $R$-module isomorphism.

\textbf{Lemma A.4.} Let $R$ be a commutative ring with unity (not necessarily
a PID). Let $n\in\mathbf{N}$ and $m\in\mathbf{N}$ be such that $n\neq m$.
Assume that $R^{n}\cong R^{m}$ as $R$-modules. Then, $R$ is the trivial ring.

\textbf{Lemma A.5.} Let $R$ be a commutative ring with unity (not necessarily
a PID). Let $S_{1}$, $S_{2}$ and $T$ be three finite free $R$-modules such
that $S_{1}\oplus T\cong S_{2}\oplus T$ as $R$-modules. Then, $S_{1}\cong
S_{2}$ as $R$-modules.

\textbf{Lemma A.6.} Let $R$ be a PID. Let $M$, $N_{1}$ and $N_{2}$ be three
finite free $R$-modules. Let $p_{1}:M\rightarrow N_{1}$ and $p_{2}%
:M\rightarrow N_{2}$ be two surjective $R$-module homomorphisms. Let
$j:N_{1}\rightarrow N_{2}$ be an $R$-module isomorphism. Then, there exists an
$R$-module isomorphism $k:M\rightarrow M$ such that $p_{2}\circ k=j\circ
p_{1}$.\ \ \ \ \footnote{In other words, there exists an $R$-module
isomorphism $k:M\rightarrow M$ which makes the diagram%
\[%
%TCIMACRO{\TeXButton{x}{\xymatrixcolsep{5pc}
%\xymatrix{
%M \ar@{.>}[r]^{k} \ar@{->>}[d]_{p_1} & M \ar@{->>}[d]^{p_2} \\
%N_1 \ar[r]_{j}^{\cong} & N_2
%}}}%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrix{
M \ar@{.>}[r]^{k} \ar@{->>}[d]_{p_1} & M \ar@{->>}[d]^{p_2} \\
N_1 \ar[r]_{j}^{\cong} & N_2
}%
%EndExpansion
\]
commute.}

\textbf{Proposition A.7.} Let $R$ be a PID. Let $M$ be a finite free
$R$-module. Let $A_{1}$ and $A_{2}$ be two $R$-module endomorphisms of $M$.
Then, $A_{1}\left(  M\right)  =A_{2}\left(  M\right)  $ if and only if there
exists a $U\in\operatorname*{GL}\left(  M\right)  $ satisfying $A_{1}=A_{2}U$.
(Note that $\operatorname*{GL}\left(  M\right)  $ denotes the group of all
$R$-module automorphisms of $M$. This group is isomorphic to the group of
invertible $n\times n$ matrices over $R$, where $n$ is the rank of the free
$R$-module $M$.)
\end{quote}

Proposition A.7 is the main result of this note, as it will help us generalize
\cite{kc-ab}.

Lemma A.1 is one incarnation of the well-known fact that free modules are
projective (and that projections on projective modules split); we will
nevertheless give a proof of it for the sake of completeness. (It should be
noticed that it doesn't really require commutativity of $R$.)

Lemma A.2 is a well-known fact and we will merely give references to its
proof, not least because its proofs are not too easy.

Lemma A.3 is a known fact and will be proven using Nakayama's lemma (other
proofs can be found in literature, to which we will give haphazard
references). It is a generalization of the known linear-algebraic fact that a
surjective endomorphism of a finite-dimensional vector space is always an
isomorphism. Notice that the analogous property of \textbf{injective}
endomorphisms does not generalize to modules over arbitrary rings.

Lemma A.4 is a quick and well-known corollary of Lemma A.3; it is the
so-called ``IBN property'' (``IBN'' stands for ``invariant basis number'') of
commutative rings (and holds not only for commutative rings, but also for
certain classes of noncommutative ones).

Lemma A.5 is a very easy consequence of Lemma A.4. It would fail if the word
``finite'' would be dropped from ``finite free'', and it would also fail if
``finite free'' would be replaced by ``finitely-generated''.

\begin{center}
\textbf{\S 2. Proofs}
\end{center}

\begin{proof}[Proof of Lemma A.1.]

\begin{verlong}
We have $S=f\left(  M\right)  $ (since $f$ is
surjective). Since $S$ is a finite free $R$-module, the $R$-module $S$ has a
finite basis. Let $\left(  e_{1},e_{2},...,e_{n}\right)  $ be such a basis.
Then, for every $i\in\left\{  1,2,...,n\right\}  $, we have $e_{i}\in
S=f\left(  M\right)  $. In other words, for every $i\in\left\{
1,2,...,n\right\}  $, there exists some element $x$ of $M$ such that
$e_{i}=f\left(  x\right)  $. Denote this $x$ by $x_{i}$. Then, for every
$i\in\left\{  1,2,...,n\right\}  $, the element $x_{i}$ of $M$ satisfies
$e_{i}=f\left(  x_{i}\right)  $.
\end{verlong}

\begin{vershort}
We have $S=f\left(  M\right)  $ (since $f$ is
surjective). Let $\left(  e_{1},e_{2},...,e_{n}\right)  $ be a finite basis of
the $R$-module $S$ (such a basis exists since $S$ is finite free). For every
$i\in\left\{  1,2,...,n\right\}  $, pick an element $x_{i}$ of $M$ satisfying
$e_{i}=f\left(  x_{i}\right)  $ (such an $x_{i}$ exists since $e_{i}\in
S=f\left(  M\right)  $).
\end{vershort}

\begin{verlong}
Now, recall that we can define an $R$-linear map from a free $R$-module by
specifying its values on the elements of a basis. Thus, we can define an
$R$-linear map from the free $R$-module $S$ by specifying its values on the
elements of the basis $\left(  e_{1},e_{2},...,e_{n}\right)  $ of $S$. Hence,
we can define an $R$-linear map $g:S\rightarrow M$ by setting%
\[
\left(  g\left(  e_{i}\right)  =x_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }%
i\in\left\{  1,2,...,n\right\}  \right)  .
\]
Consider the $R$-linear map $g:S\rightarrow M$ defined this way. Clearly,
$g\left(  S\right)  $ is an $R$-submodule of $M$.
\end{verlong}

\begin{vershort}
Now, recall that we can define an $R$-linear map from a free $R$-module by
specifying its values on the elements of a basis. This allows us to define an
$R$-linear map $g:S\rightarrow M$ by setting%
\[
\left(  g\left(  e_{i}\right)  =x_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }%
i\in\left\{  1,2,...,n\right\}  \right)  .
\]
Consider the $R$-linear map $g:S\rightarrow M$ defined this way. Clearly,
$g\left(  S\right)  $ is an $R$-submodule of $M$.
\end{vershort}

Every $i\in\left\{  1,2,...,n\right\}  $ satisfies%
\[
\left(  f\circ g\right)  \left(  e_{i}\right)  =f\left(  \underbrace{g\left(
e_{i}\right)  }_{=x_{i}}\right)  =f\left(  x_{i}\right)  =e_{i}%
=\operatorname*{id}\nolimits_{S}\left(  e_{i}\right)  .
\]
In other words, the two maps $f\circ g:S\rightarrow S$ and $\operatorname*{id}%
\nolimits_{S}:S\rightarrow S$ are equal to each other on each element of the
basis $\left(  e_{1},e_{2},...,e_{n}\right)  $ of $S$. Since these two maps
$f\circ g$ and $\operatorname*{id}\nolimits_{S}$ are both $R$-linear, this
yields that the two maps $f\circ g$ and $\operatorname*{id}\nolimits_{S}$ must
be identical (because any two linear maps which are equal to each other on
each element of a basis of their domain must be identical). In other words,
$f\circ g=\operatorname*{id}\nolimits_{S}$.

Now, let $x\in g\left(  S\right)  \cap\left(  \operatorname*{Ker}f\right)  $.
Then, $x\in g\left(  S\right)  \cap\left(  \operatorname*{Ker}f\right)
\subseteq g\left(  S\right)  $, so that there exists some $y\in S$ such that
$x=g\left(  y\right)  $. Consider this $y$. Then, $f\left(  \underbrace{x}%
_{=g\left(  y\right)  }\right)  =f\left(  g\left(  y\right)  \right)
=\underbrace{\left(  f\circ g\right)  }_{=\operatorname*{id}\nolimits_{S}%
}\left(  y\right)  =\operatorname*{id}\nolimits_{S}\left(  y\right)  =y$.
Compared with $f\left(  x\right)  =0$ (since $x\in g\left(  S\right)
\cap\left(  \operatorname*{Ker}f\right)  \subseteq\operatorname*{Ker}f$), this
yields $y=0$. Hence, $x=g\left(  \underbrace{y}_{=0}\right)  =g\left(
0\right)  =0$ (since $g$ is $R$-linear).

Now forget that we fixed $x$. We have thus shown that every $x\in g\left(
S\right)  \cap\left(  \operatorname*{Ker}f\right)  $ satisfies $x=0$. In other
words, $g\left(  S\right)  \cap\left(  \operatorname*{Ker}f\right)  =0$. The
internal direct sum $g\left(  S\right)  \oplus\left(  \operatorname*{Ker}%
f\right)  $ is thus well-defined.

Now, let $z\in M$. Then,
\begin{align*}
f\left(  z-g\left(  f\left(  z\right)  \right)  \right)   &  =f\left(
z\right)  -\underbrace{f\left(  g\left(  f\left(  z\right)  \right)  \right)
}_{\substack{=\left(  f\circ g\right)  \left(  f\left(  z\right)  \right)
\\=\operatorname*{id}\nolimits_{S}\left(  f\left(  z\right)  \right)
\\\text{(since }f\circ g=\operatorname*{id}\nolimits_{S}\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is }R\text{-linear}\right)
\\
&  =f\left(  z\right)  -\underbrace{\operatorname*{id}\nolimits_{S}\left(
f\left(  z\right)  \right)  }_{=f\left(  z\right)  }=f\left(  z\right)
-f\left(  z\right)  =0,
\end{align*}
so that $z-g\left(  f\left(  z\right)  \right)  \in\operatorname*{Ker}f$.
Thus,%
\[
z=g\left(  \underbrace{f\left(  z\right)  }_{\in S}\right)
+\underbrace{\left(  z-g\left(  f\left(  z\right)  \right)  \right)  }%
_{\in\operatorname*{Ker}f}\in g\left(  S\right)  +\left(  \operatorname*{Ker}%
f\right)  =g\left(  S\right)  \oplus\left(  \operatorname*{Ker}f\right)
\]
(since we know that the internal direct sum $g\left(  S\right)  \oplus\left(
\operatorname*{Ker}f\right)  $ is well-defined). Now forget that we fixed $z$.
We hence have proven that every $z\in M$ satisfies $z\in g\left(  S\right)
\oplus\left(  \operatorname*{Ker}f\right)  $. In other words, $M\subseteq
g\left(  S\right)  \oplus\left(  \operatorname*{Ker}f\right)  $. Combined with
$g\left(  S\right)  \oplus\left(  \operatorname*{Ker}f\right)  \subseteq M$
(which is obvious), this yields $M=g\left(  S\right)  \oplus\left(
\operatorname*{Ker}f\right)  $.

\begin{verlong}
We have%
\begin{align*}
\operatorname*{Ker}\left(  f\mid_{g\left(  S\right)  }\right)   &  =\left\{
x\in g\left(  S\right)  \ \mid\ \underbrace{\left(  f\mid_{g\left(  S\right)
}\right)  \left(  x\right)  }_{=f\left(  x\right)  }=0\right\} \\
&  =\left\{  x\in g\left(  S\right)  \ \mid\ f\left(  x\right)  =0\right\}
=g\left(  S\right)  \cap\underbrace{\left\{  x\in M\ \mid\ f\left(  x\right)
=0\right\}  }_{=\operatorname*{Ker}f}\\
&  =g\left(  S\right)  \cap\left(  \operatorname*{Ker}f\right)  =0.
\end{align*}
Thus, the $R$-linear map $f\mid_{g\left(  S\right)  }$ is injective.
\end{verlong}

\begin{vershort}
A moment of thought reveals that $\operatorname*{Ker}\left(  f\mid_{g\left(
S\right)  }\right)  =g\left(  S\right)  \cap\left(  \operatorname*{Ker}%
f\right)  $. In light of $g\left(  S\right)  \cap\left(  \operatorname*{Ker}%
f\right)  =0$, this rewrites as $\operatorname*{Ker}\left(  f\mid_{g\left(
S\right)  }\right)  =0$. Thus, the $R$-linear map $f\mid_{g\left(  S\right)
}$ is injective.
\end{vershort}

On the other hand,%
\[
\left(  f\mid_{g\left(  S\right)  }\right)  \left(  g\left(  S\right)
\right)  =f\left(  g\left(  S\right)  \right)  =\underbrace{\left(  f\circ
g\right)  }_{=\operatorname*{id}\nolimits_{S}}\left(  S\right)
=\operatorname*{id}\nolimits_{S}\left(  S\right)  =S.
\]
Thus, the $R$-linear map $f\mid_{g\left(  S\right)  }$ is surjective. Combined
with the fact that the $R$-linear map $f\mid_{g\left(  S\right)  }$ is
injective, this yields that the $R$-linear map $f\mid_{g\left(  S\right)  }$
is bijective. Thus, $f\mid_{g\left(  S\right)  }$ is an $R$-module isomorphism.

Altogether, we thus know that $g\left(  S\right)  $ is an $R$-submodule of $M$
such that $M=g\left(  S\right)  \oplus\left(  \operatorname*{Ker}f\right)  $
(an internal direct sum) and such that $f\mid_{g\left(  S\right)  }:g\left(
S\right)  \rightarrow S$ is an $R$-module isomorphism. Thus, there exists an
$R$-submodule $N$ of $M$ such that $M=N\oplus\left(  \operatorname*{Ker}%
f\right)  $ (an internal direct sum) and such that $f\mid_{N}:N\rightarrow S$
is an $R$-module isomorphism (namely, $N=g\left(  S\right)  $). This proves
Lemma A.1.
\end{proof}

\bigskip

\begin{proof}[Proof of Lemma A.2.] Lemma A.2 is probably the most well-known among
the many consequences of the Smith normal form. As such, it appears (in one or
another form) in almost every good book written about abstract algebra, and in
a multitude of lecture notes\footnote{Actually, googling for the statement of
Lemma A.2 is a good way to find lecture notes in abstract algebra.}. We will
not give a proof of Lemma A.2, but instead just refer to some of many places
where it is proven:

\begin{itemize}
\item Lemma A.2 follows from \cite[Theorem 8.25]{knapp}.

\item Lemma A.2 follows from \cite[Theorem 2.1]{kc-modPID}.

\item Lemma A.2 is part of \cite[Theorem 3.7.1]{brundan}.

\item Lemma A.2 is a consequence of \cite[Chapter XI, Theorem 12]%
{birkhoff-maclane}.

\item Lemma A.2 follows from the (somewhat stronger) Theorem 6.0.1 in Chapter
11 (\textquotedblleft Finitely-generated modules\textquotedblright) of Paul
Garrett's lecture notes \cite{garrett-aa}.

\item Lemma A.2 also follows from The Freedom Theorem in Lecture 5 of
McNulty's \cite{mcnulty}.

\item Lemma A.2 follows from \cite[Chapter 4, Corollary 4.6.2]{bash}.
\end{itemize}
\end{proof}

\bigskip

\begin{proof}[Proof of Lemma A.3.] There is no real need to give a proof of Lemma
A.3 here, since it is known: For example, it appears in \cite{kc-univid1} as
Theorem 5.3, with my $f$ renamed as $\varphi$. But let me give a slightly
different proof of Lemma A.3 (though a rather well-known one; for example, it
is identical with the one given in \cite[Theorem 0.3.2]{bash-ca},
\cite[Theorem 2.4]{mats} and \cite[Theorem 1]{szymiczek}). We will use the
following fact:

\begin{quote}
\textbf{Nakayama lemma:} Let $S$ be a commutative ring with unity, and let $N$
be a finitely-generated $S$-module. Let $I$ be an ideal of $S$ such that
$I\cdot N=N$. Then, there exists an $s\in S$ such that $s\equiv
1\operatorname{mod}I$ and $sN=0$.
\end{quote}

This is merely one of the many forms of the Nakayama lemma\footnote{The
statements referred to as the ``Nakayama lemma'' in literature are numerous,
and some of them are harder to derive from each other than prove from scratch.
Nevertheless, they are widely regarded as equivalent (not merely like any two
correct assertions). The form of the Nakayama lemma that we are using is
probably the one most familiar to commutative algebraists.
\par
The Nakayama lemma is also known as the ``NAK lemma'', with ``NAK''
abbreviating ``Nakayama-Azumaya-Krull'' rather than ``NAKayama''.}, but it is
the one most suitable for our needs. Proofs of the Nakayama lemma, in (more or
less) the form given above, can be found in \cite[Corollary 2.5]{am},
\cite[Theorem 0.3.1]{bash-ca}, \cite[Lemma 2]{szymiczek}, \cite[Theorem
2.2]{mats} and various other places.

Consider the situation of Lemma A.3. Let $T$ be an indeterminate, and consider
the polynomial ring $R\left[  T\right]  $. By the universal property of the
polynomial ring, there exists one and only one $R$-algebra homomorphism
$R\left[  T\right]  \rightarrow\operatorname*{End}M$ which sends $T$ to $f$.
Consider the $R$-module $M$ as an $R\left[  T\right]  $-module via this
homomorphism. Then, clearly, $T$ acts as $f$ on this $R\left[  T\right]
$-module $M$. In other words, $Tm=f\left(  m\right)  $ for every $m\in M$.

Let $I$ be the ideal $\left(  T\right)  $ of $R\left[  T\right]  $. Then,
$T\in I$, so that $T\cdot M\subseteq I\cdot M$. But since%
\[
T\cdot M=\left\{  \underbrace{Tm}_{=f\left(  m\right)  }\ \mid\ m\in
M\right\}  =\left\{  f\left(  m\right)  \ \mid\ m\in M\right\}  =f\left(
M\right)  =M
\]
(since $M$ is surjective), this rewrites as $M\subseteq I\cdot M$. Combined
with the obvious inclusion $I\cdot M\subseteq M$, this yields $I\cdot M=M$.
Hence, the Nakayama lemma (applied to $S=R\left[  T\right]  $ and $N=M$)
yields that there exists an $s\in R\left[  T\right]  $ such that
$s\equiv1\operatorname{mod}I$ and $sM=0$. Consider this $s$.

Since $s\equiv1\operatorname{mod}I$, there exists a $q\in I$ such that
$s=1+q$. Consider this $q$. Since $q\in I=\left(  T\right)  $, the polynomial
$q$ is divisible by $T$. Thus, there exists some $r\in R\left[  T\right]  $
such that $q=rT$. Consider this $r$. Then, $s=1+\underbrace{q}_{=rT}=1+rT$, so
that every $m\in M$ satisfies
\[
m-\left(  -rT\right)  m=m+rTm=\underbrace{\left(  1+rT\right)  }_{=s}m=sm=0
\]
(since $s\underbrace{m}_{\in M}\in sM=0$). In other words, every $m\in M$
satisfies
\[
m=\left(  -rT\right)  m=-r\underbrace{Tm}_{=f\left(  m\right)  }=-r\cdot
f\left(  m\right)  .
\]
Hence, every $m\in M$ such that $f\left(  m\right)  =0$ satisfies
$m=-r\cdot\underbrace{f\left(  m\right)  }_{=0}=-r\cdot0=0$. In other words,
$f$ is injective. Since we know that $f$ is surjective, this yields that $f$
is bijective. Combined with the fact that $f$ is an $R$-module homomorphism,
this yields that $f$ is an $R$-module isomorphism. Lemma A.3 is proven.
\end{proof}

\bigskip

\begin{proof}[Proof of Lemma A.4.]

\begin{verlong}
Since $n\neq m$, we have either $n<m$ or $m<n$.
Thus, we can WLOG assume that $n<m$ (since otherwise, we could just switch $n$ with
$m$). Assume this. Thus, $n<m$, so that $m>n$.
\end{verlong}

\begin{vershort}
Since $n\neq m$, we have either $n<m$ or $m<n$.
Since the scenery is symmetric in $n$ and $m$, we can WLOG assume that $n<m$.
Assume this.
\end{vershort}

Since $R^{n}\cong R^{m}$ as $R$-modules, there exists an isomorphism
$I:R^{n}\rightarrow R^{m}$. Consider this $I$. Since $I$ is an isomorphism, we
have $I\left(  R^{n}\right)  =R^{m}$.

Let $\left(  e_{1},e_{2},...,e_{n}\right)  $ be the standard basis of the free
$R$-module $R^{n}$. Let $\left(  f_{1},f_{2},...,f_{m}\right)  $ be the
standard basis of the free $R$-module $R^{m}$.

\begin{verlong}
Now, recall that we can define an $R$-linear map from a free $R$-module by
specifying its values on the elements of a basis. Thus, we can define an
$R$-linear map from the free $R$-module $R^{m}$ by specifying its values on
the elements of the basis $\left(  f_{1},f_{2},...,f_{m}\right)  $ of $R^{m}$.
Hence, we can define an $R$-linear map $g:R^{m}\rightarrow R^{n}$ by setting%
\begin{equation}
\left(  g\left(  f_{i}\right)  =\left\{
\begin{array}
[c]{l}%
e_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\leq n;\\
0,\ \ \ \ \ \ \ \ \ \ \text{if }i>n
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,...,m\right\}
\right)  . \label{lem.A.4.pf.1}%
\end{equation}
Consider the $R$-linear map $g:R^{m}\rightarrow R^{n}$ defined this way. For
every $i\in\left\{  1,2,...,n\right\}  $, we have $i\in\left\{
1,2,...,m\right\}  $ (since $1\leq i$ and $i\leq n<m$). Thus, for every
$i\in\left\{  1,2,...,n\right\}  $, the term $f_{i}$ is well-defined and
satisfies%
\[
g\left(  f_{i}\right)  =\left\{
\begin{array}
[c]{l}%
e_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\leq n;\\
0,\ \ \ \ \ \ \ \ \ \ \text{if }i>n
\end{array}
\right.  =e_{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i\leq n\right)  .
\]
It is now easy to see that $g\left(  R^{m}\right)  =R^{n}$%
\ \ \ \ \footnote{\textit{Proof.} Let $a\in R^{n}$ be arbitrary. Since the
$R$-module $R^{n}$ has basis $\left(  e_{1},e_{2},...,e_{n}\right)  $, we can
write $a$ in the form $a=\sum\limits_{i=1}^{n}a_{i}e_{i}$ for some $\left(
a_{1},a_{2},...,a_{n}\right)  \in R^{n}$. Consider this $\left(  a_{1}%
,a_{2},...,a_{n}\right)  \in R^{n}$. Since $g$ is $R$-linear, we have%
\[
g\left(  \sum\limits_{i=1}^{n}a_{i}f_{i}\right)  =\sum\limits_{i=1}^{n}%
a_{i}\underbrace{g\left(  f_{i}\right)  }_{\substack{=e_{i}\\\text{(since
}i\in\left\{  1,2,...,n\right\}  \text{)}}}=\sum\limits_{i=1}^{n}a_{i}%
e_{i}=a,
\]
so that $a=g\left(  \underbrace{\sum\limits_{i=1}^{n}a_{i}f_{i}}_{\in R^{m}%
}\right)  \in g\left(  R^{m}\right)  $. Now forget that we fixed $a$. We thus
have proven that every $a\in R^{n}$ satisfies $a\in g\left(  R^{m}\right)  $.
Hence, $R^{n}\subseteq g\left(  R^{m}\right)  $. Combined with the (obvious)
inclusion $g\left(  R^{m}\right)  \subseteq R^{n}$, this yields that $g\left(
R^{m}\right)  =R^{n}$, qed.}. Thus, $\left(  g\circ I\right)  \left(
R^{n}\right)  =g\left(  \underbrace{I\left(  R^{n}\right)  }_{=R^{m}}\right)
=g\left(  R^{m}\right)  =R^{n}$, so that $g\circ I$ is surjective. Since
$g\circ I$ is clearly an $R$-module homomorphism (because both $g$ and $I$ are
$R$-module homomorphisms), Lemma A.3 (applied to $M=R^{n}$ and $f=g\circ I$)
yields that $g\circ I$ is an $R$-module isomorphism. Thus, $g\circ I$ is
injective, so that $g$ is injective as well. But (\ref{lem.A.4.pf.1}) (applied
to $i=m$) yields%
\[
g\left(  f_{m}\right)  =\left\{
\begin{array}
[c]{l}%
e_{m},\ \ \ \ \ \ \ \ \ \ \text{if }m\leq n;\\
0,\ \ \ \ \ \ \ \ \ \ \text{if }m>n
\end{array}
\right.  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }m>n\right)  .
\]
Since $g$ is injective, this yields that $f_{m}=0$. Since $f_{m}$ is an
element of a basis of the $R$-module $R^{m}$ (namely, of the basis $\left(
f_{1},f_{2},...,f_{m}\right)  $), this yields that $1=0$ in $R$. In other
words, $R$ is the trivial ring. Lemma A.4 is proven.
\end{verlong}

\begin{vershort}
Now, recall that we can define an $R$-linear map from a free $R$-module by
specifying its values on the elements of a basis. Hence, we can define an
$R$-linear map $g:R^{m}\rightarrow R^{n}$ by setting%
\[
\left(  g\left(  f_{i}\right)  =\left\{
\begin{array}
[c]{l}%
e_{i},\ \ \ \ \ \ \ \ \ \ \text{if }i\leq n;\\
0,\ \ \ \ \ \ \ \ \ \ \text{if }i>n
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \text{for every }i\in\left\{  1,2,...,m\right\}
\right)  .
\]
Consider this map $g$. Since $n<m$, it is easily seen that the map $g$ is
surjective (since all elements of the basis $\left(  e_{1},e_{2}%
,...,e_{n}\right)  $ of $R^{n}$ occur as images of basis vectors $f_{i}$ under
$g$). Hence, $g\circ I:R^{m}\rightarrow R^{m}$ is also surjective (as $I$ is
an isomorphism). According to Lemma A.3 (applied to $M=R^{n}$ and $f=g\circ
I$), this entails that $g\circ I$ is an $R$-module isomorphism. Thus, $g\circ
I$ is injective, so that $g$ is injective as well. But the definition of $g$
yields $g\left(  f_{m}\right)  =0$ (since $m>n$), and thus (by the injectivity
of $g$) we have $f_{m}=0$. Since $f_{m}$ is an element of a basis of the
$R$-module $R^{m}$ (namely, of the basis $\left(  f_{1},f_{2},...,f_{m}%
\right)  $), this yields that $1=0$ in $R$. In other words, $R$ is the trivial
ring. Lemma A.4 is proven.
\end{vershort}

\end{proof}

An alternative proof of Lemma A.4 would be to apply \cite[Corollary
5.11]{kc-extmod}.

\bigskip

\begin{proof}[Proof of Lemma A.5.]

\begin{verlong}
Since $S_{1}$ is a finite free $R$-module, there
exists an $s_{1}\in\mathbf{N}$ such that $S_{1}\cong R^{s_{1}}$ as
$R$-modules. Consider this $s_{1}$.

Since $S_{2}$ is a finite free $R$-module, there exists an $s_{2}\in
\mathbf{N}$ such that $S_{2}\cong R^{s_{2}}$ as $R$-modules. Consider this
$s_{2}$.

If $s_{1}=s_{2}$, then Lemma A.5 holds (because if $s_{1}=s_{2}$, then we have
$R^{s_{1}}=R^{s_{2}}$ and therefore $S_{1}\cong R^{s_{1}}=R^{s_{2}}\cong
S_{2}$ as $R$-modules). Hence, let us now assume (for the rest of the proof of
Lemma A.5) that $s_{1}\neq s_{2}$.

Since $T$ is a finite free $R$-module, there exists a $t\in\mathbf{N}$ such
that $T\cong R^{t}$ as $R$-modules. Consider this $t$. Since $s_{1}\neq s_{2}%
$, we have $s_{1}+t\neq s_{2}+t$.

We have
\[
\underbrace{S_{1}}_{\cong R^{s_{1}}}\oplus\underbrace{T}_{\cong R^{t}}\cong
R^{s_{1}}\oplus R^{t}\cong R^{s_{1}+t}\ \ \ \ \ \ \ \ \ \ \text{as
}R\text{-modules,}%
\]
and%
\[
\underbrace{S_{2}}_{\cong R^{s_{2}}}\oplus\underbrace{T}_{\cong R^{t}}\cong
R^{s_{2}}\oplus R^{t}\cong R^{s_{2}+t}\ \ \ \ \ \ \ \ \ \ \text{as
}R\text{-modules.}%
\]
Thus, $R^{s_{1}+t}\cong S_{1}\oplus T\cong S_{2}\oplus T\cong R^{s_{2}+t}$ as
$R$-modules. Thus, Lemma A.4 (applied to $n=s_{1}+t$ and $m=s_{2}+t$) yields
that $R$ is the trivial ring. But since every module over the trivial ring is
$0$, this shows that $S_{1}=0$ and $S_{2}=0$, so that $S_{1}=0=S_{2}$ as
$R$-modules. This proves Lemma A.5.
\end{verlong}

\begin{vershort}
Since $S_{1}$, $S_{2}$ and $T$ are finite free
$R$-modules, there exist elements $s_{1}$, $s_{2}$ and $t$ of $\mathbf{N}$
such that $S_{1}\cong R^{s_{1}}$, $S_{2}\cong R^{s_{2}}$ and $T\cong R^{t}$ as
$R$-modules. Consider these $s_{1}$, $s_{2}$ and $t$.

If $s_{1}=s_{2}$, then Lemma A.5 is obvious (since $s_{1}=s_{2}$ leads to
$S_{1}\cong R^{s_{1}}=R^{s_{2}}\cong S_{2}$ as $R$-modules). Hence, let us now
assume (for the rest of the proof of Lemma A.5) that $s_{1}\neq s_{2}$. Thus,
$s_{1}+t\neq s_{2}+t$. But%
\[
\underbrace{S_{1}}_{\cong R^{s_{1}}}\oplus\underbrace{T}_{\cong R^{t}}\cong
R^{s_{1}}\oplus R^{t}\cong R^{s_{1}+t}\ \ \ \ \ \ \ \ \ \ \text{as
}R\text{-modules,}%
\]
and similarly $S_{2}\oplus T\cong R^{s_{2}+t}$ as $R$-modules. Thus,
$R^{s_{1}+t}\cong S_{1}\oplus T\cong S_{2}\oplus T\cong R^{s_{2}+t}$ as
$R$-modules. Thus, Lemma A.4 (applied to $n=s_{1}+t$ and $m=s_{2}+t$) yields
that $R$ is the trivial ring. But since every module over the trivial ring is
$0$, this shows that $S_{1}=0$ and $S_{2}=0$, so that $S_{1}=0=S_{2}$ as
$R$-modules. This proves Lemma A.5.
\end{vershort}

\end{proof}

\bigskip

\begin{proof}[Proof of Lemma A.6.]

\begin{verlong}
Applying Lemma A.1 to $S=N_{1}$ and $f=p_{1}$, we
obtain the following: There exists an $R$-submodule $N$ of $M$ such that
$M=N\oplus\left(  \operatorname*{Ker}\left(  p_{1}\right)  \right)  $ (an
internal direct sum) and such that $p_{1}\mid_{N}:N\rightarrow N_{1}$ is an
$R$-module isomorphism. Denote this $N$ by $J_{1}$. Then, $J_{1}$ is an
$R$-submodule of $M$ such that $M=J_{1}\oplus\left(  \operatorname*{Ker}%
\left(  p_{1}\right)  \right)  $ (an internal direct sum) and such that
$p_{1}\mid_{J_{1}}:J_{1}\rightarrow N_{1}$ is an $R$-module isomorphism.

Applying Lemma A.1 to $S=N_{2}$ and $f=p_{2}$, we obtain the following: There
exists an $R$-submodule $N$ of $M$ such that $M=N\oplus\left(
\operatorname*{Ker}\left(  p_{2}\right)  \right)  $ (an internal direct sum)
and such that $p_{2}\mid_{N}:N\rightarrow N_{2}$ is an $R$-module isomorphism.
Denote this $N$ by $J_{2}$. Then, $J_{2}$ is an $R$-submodule of $M$ such that
$M=J_{2}\oplus\left(  \operatorname*{Ker}\left(  p_{2}\right)  \right)  $ (an
internal direct sum) and such that $p_{2}\mid_{J_{2}}:J_{2}\rightarrow N_{2}$
is an $R$-module isomorphism.
\end{verlong}

\begin{vershort}
Applying Lemma A.1 to $S=N_{1}$ and $f=p_{1}$, we
obtain the following: There exists an $R$-submodule $J_{1}$ of $M$ such that
$M=J_{1}\oplus\left(  \operatorname*{Ker}\left(  p_{1}\right)  \right)  $ (an
internal direct sum) and such that $p_{1}\mid_{J_{1}}:J_{1}\rightarrow N_{1}$
is an $R$-module isomorphism. Fix such a $J_{1}$.

Similarly, fix an $R$-submodule $J_{2}$ of $M$ such that $M=J_{2}\oplus\left(
\operatorname*{Ker}\left(  p_{2}\right)  \right)  $ (an internal direct sum)
and such that $p_{2}\mid_{J_{2}}:J_{2}\rightarrow N_{2}$ is an $R$-module
isomorphism. The existence of such a $J_{2}$ is guaranteed by Lemma A.1 again.
\end{vershort}

\begin{verlong}
We know that every submodule of a finite free $R$-module is finite free (due
to Lemma A.2). Since $M$ is a finite free $R$-module, this yields that every
submodule of $M$ is finite free. Since $J_{1}$, $J_{2}$, $\operatorname*{Ker}%
\left(  p_{1}\right)  $ and $\operatorname*{Ker}\left(  p_{2}\right)  $ are
submodules of $M$, this yields that all $R$-modules $J_{1}$, $J_{2}$,
$\operatorname*{Ker}\left(  p_{1}\right)  $ and $\operatorname*{Ker}\left(
p_{2}\right)  $ are finite free.
\end{verlong}

\begin{vershort}
We know that every submodule of a finite free $R$-module is finite free (due
to Lemma A.2). Thus, the $R$-modules $J_{1}$, $J_{2}$, $\operatorname*{Ker}%
\left(  p_{1}\right)  $ and $\operatorname*{Ker}\left(  p_{2}\right)  $ are
finite free (being submodules of the finite free $R$-module $M$).
\end{vershort}

Since the map $p_{2}\mid_{J_{2}}:J_{2}\rightarrow N_{2}$ is an $R$-module
isomorphism, its inverse $\left(  p_{2}\mid_{J_{2}}\right)  ^{-1}%
:N_{2}\rightarrow J_{2}$ is well-defined and an $R$-module isomorphism as well.

Since all three maps $p_{1}\mid_{J_{1}}:J_{1}\rightarrow N_{1}$,
$j:N_{1}\rightarrow N_{2}$ and $\left(  p_{2}\mid_{J_{2}}\right)  ^{-1}%
:N_{2}\rightarrow J_{2}$ are $R$-module isomorphisms, their composition
$\left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\circ j\circ\left(  p_{1}\mid_{J_{1}%
}\right)  :J_{1}\rightarrow J_{2}$ must also be an $R$-module isomorphism.
Thus, $J_{2}\cong J_{1}$ as $R$-modules. Now,%
\begin{align*}
\left(  \operatorname*{Ker}\left(  p_{1}\right)  \right)  \oplus J_{1}  &
=J_{1}\oplus\left(  \operatorname*{Ker}\left(  p_{1}\right)  \right)
=M=\underbrace{J_{2}}_{\cong J_{1}}\oplus\left(  \operatorname*{Ker}\left(
p_{2}\right)  \right) \\
&  \cong J_{1}\oplus\left(  \operatorname*{Ker}\left(  p_{2}\right)  \right)
=\left(  \operatorname*{Ker}\left(  p_{2}\right)  \right)  \oplus J_{1}%
\end{align*}
as $R$-modules. Thus, Lemma A.5 (applied to $S_{1}=\operatorname*{Ker}\left(
p_{1}\right)  $, $S_{2}=\operatorname*{Ker}\left(  p_{2}\right)  $ and
$T=J_{1}$) yields that $\operatorname*{Ker}\left(  p_{1}\right)
\cong\operatorname*{Ker}\left(  p_{2}\right)  $ as $R$-modules. Thus, there
exists an $R$-module isomorphism $I:\operatorname*{Ker}\left(  p_{1}\right)
\rightarrow\operatorname*{Ker}\left(  p_{2}\right)  $. Consider this $I$.

\begin{verlong}
It is a well-known (and trivial) fact that the following holds: Whenever
$V_{1}$, $V_{2}$, $W_{1}$ and $W_{2}$ are four $R$-modules, and $\alpha
:V_{1}\rightarrow V_{2}$ and $\beta:W_{1}\rightarrow W_{2}$ are two $R$-module
isomorphisms, then $\alpha\oplus\beta:V_{1}\oplus W_{1}\rightarrow V_{2}\oplus
W_{2}$ is an $R$-module isomorphism. Applying this fact to $V_{1}=J_{1}$,
$V_{2}=\operatorname*{Ker}\left(  p_{1}\right)  $, $W_{1}=J_{2}$,
$W_{2}=\operatorname*{Ker}\left(  p_{2}\right)  $, $\alpha=\left(  p_{2}%
\mid_{J_{2}}\right)  ^{-1}\circ j\circ\left(  p_{1}\mid_{J_{1}}\right)  $ and
$\beta=I$, we obtain the following: The map
\[
\left(  \left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\circ j\circ\left(  p_{1}%
\mid_{J_{1}}\right)  \right)  \oplus I:J_{1}\oplus\left(  \operatorname*{Ker}%
\left(  p_{1}\right)  \right)  \rightarrow J_{2}\oplus\left(
\operatorname*{Ker}\left(  p_{2}\right)  \right)
\]
is an $R$-module isomorphism. Denote this map by $\Phi$. Then, $\Phi=\left(
\left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\circ j\circ\left(  p_{1}\mid_{J_{1}%
}\right)  \right)  \oplus I$, and thus every $\left(  u,v\right)  \in
J_{1}\times\left(  \operatorname*{Ker}\left(  p_{1}\right)  \right)  $
satisfies%
\begin{equation}
\Phi\left(  u,v\right)  =\left(  \left(  \left(  p_{2}\mid_{J_{2}}\right)
^{-1}\circ j\circ\left(  p_{1}\mid_{J_{1}}\right)  \right)  \left(  u\right)
,I\left(  v\right)  \right)  . \label{lem.A.6.pf.1}%
\end{equation}


Now, the map $\Phi$ is an $R$-module isomorphism from $J_{1}\oplus\left(
\operatorname*{Ker}\left(  p_{1}\right)  \right)  $ to $J_{2}\oplus\left(
\operatorname*{Ker}\left(  p_{2}\right)  \right)  $ (since $\Phi=\left(
\left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\circ j\circ\left(  p_{1}\mid_{J_{1}%
}\right)  \right)  \oplus I$, and since we know that the map $\left(  \left(
p_{2}\mid_{J_{2}}\right)  ^{-1}\circ j\circ\left(  p_{1}\mid_{J_{1}}\right)
\right)  \oplus I$ is an $R$-module isomorphism from $J_{1}\oplus\left(
\operatorname*{Ker}\left(  p_{1}\right)  \right)  $ to $J_{2}\oplus\left(
\operatorname*{Ker}\left(  p_{2}\right)  \right)  $). Since $J_{1}%
\oplus\left(  \operatorname*{Ker}\left(  p_{1}\right)  \right)  =M$ and
$J_{2}\oplus\left(  \operatorname*{Ker}\left(  p_{2}\right)  \right)  =M$,
this rewrites as follows: The map $\Phi$ is an $R$-module isomorphism from $M$
to $M$.
\end{verlong}

\begin{vershort}
Now, define a map%
\[
\Phi:J_{1}\oplus\left(  \operatorname*{Ker}\left(  p_{1}\right)  \right)
\rightarrow J_{2}\oplus\left(  \operatorname*{Ker}\left(  p_{2}\right)
\right)
\]
by%
\begin{align*}
\Phi\left(  u,v\right)   &  =\left(  \left(  \left(  p_{2}\mid_{J_{2}}\right)
^{-1}\circ j\circ\left(  p_{1}\mid_{J_{1}}\right)  \right)  \left(  u\right)
,I\left(  v\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(  u,v\right)  \in J_{1}%
\times\left(  \operatorname*{Ker}\left(  p_{1}\right)  \right)  .
\end{align*}
Clearly, $\Phi$ is an $R$-module isomorphism from $J_{1}\oplus\left(
\operatorname*{Ker}\left(  p_{1}\right)  \right)  $ to $J_{2}\oplus\left(
\operatorname*{Ker}\left(  p_{2}\right)  \right)  $ (in fact, $\Phi$ is the
direct sum of the isomorphisms $\left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\circ
j\circ\left(  p_{1}\mid_{J_{1}}\right)  :J_{1}\rightarrow J_{2}$ and
$I:\operatorname*{Ker}\left(  p_{1}\right)  \rightarrow\operatorname*{Ker}%
\left(  p_{2}\right)  $). Since both $J_{1}\oplus\left(  \operatorname*{Ker}%
\left(  p_{1}\right)  \right)  $ and $J_{2}\oplus\left(  \operatorname*{Ker}%
\left(  p_{2}\right)  \right)  $ are simply $M$, this means that $\Phi$ is an
$R$-module isomorphism from $M$ to $M$.
\end{vershort}

\begin{verlong}
Now, every $x\in M$ satisfies%
\[
\left(  p_{2}\circ\Phi\right)  \left(  x\right)  =\left(  j\circ p_{1}\right)
\left(  x\right)
\]
\footnote{\textit{Proof.} Let $x\in M$. Since $x\in M=J_{1}\oplus\left(
\operatorname*{Ker}\left(  p_{1}\right)  \right)  $, we can write $x$ in the
form $x=u+v$ for some $u\in J_{1}$ and $v\in\operatorname*{Ker}\left(
p_{1}\right)  $. Consider these $u$ and $v$. Then, $x=u+v$ corresponds to the
pair $\left(  u,v\right)  $ under the identification of the internal direct
sum $M=J_{1}\oplus\left(  \operatorname*{Ker}\left(  p_{1}\right)  \right)  $
with the external direct sum $J_{1}\oplus\left(  \operatorname*{Ker}\left(
p_{1}\right)  \right)  $. Hence, $x=\left(  u,v\right)  $. Thus,%
\begin{align*}
\left(  p_{2}\circ\Phi\right)  \left(  x\right)   &  =\left(  p_{2}\circ
\Phi\right)  \left(  u,v\right)  =p_{2}\left(  \Phi\left(  u,v\right)  \right)
\\
&  =p_{2}\left(  \left(  \underbrace{\left(  \left(  p_{2}\mid_{J_{2}}\right)
^{-1}\circ j\circ\left(  p_{1}\mid_{J_{1}}\right)  \right)  \left(  u\right)
}_{=\left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\left(  j\left(  \left(  p_{1}%
\mid_{J_{1}}\right)  \left(  u\right)  \right)  \right)  },I\left(  v\right)
\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{lem.A.6.pf.1}%
)}\right) \\
&  =p_{2}\left(  \left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\left(  j\left(
\underbrace{\left(  p_{1}\mid_{J_{1}}\right)  \left(  u\right)  }%
_{=p_{1}\left(  u\right)  }\right)  \right)  ,I\left(  v\right)  \right) \\
&  =p_{2}\underbrace{\left(  \left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\left(
j\left(  p_{1}\left(  u\right)  \right)  \right)  ,I\left(  v\right)  \right)
}_{\substack{=\left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\left(  j\left(
p_{1}\left(  u\right)  \right)  \right)  +I\left(  v\right)  \\\text{(because
the sum }\left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\left(  j\left(  p_{1}\left(
u\right)  \right)  \right)  +I\left(  v\right)  \\\text{corresponds to the
pair }\left(  \left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\left(  j\left(
p_{1}\left(  u\right)  \right)  \right)  ,I\left(  v\right)  \right)
\\\text{under the identification of the internal direct sum}\\M=J_{2}%
\oplus\left(  \operatorname*{Ker}\left(  p_{2}\right)  \right)  \text{ with
the external direct sum }J_{2}\oplus\left(  \operatorname*{Ker}\left(
p_{2}\right)  \right)  \text{)}}}\\
&  =p_{2}\left(  \left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\left(  j\left(
p_{1}\left(  u\right)  \right)  \right)  +I\left(  v\right)  \right) \\
&  =p_{2}\left(  \left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\left(  j\left(
p_{1}\left(  u\right)  \right)  \right)  \right)  +\underbrace{p_{2}\left(
I\left(  v\right)  \right)  }_{\substack{=0\\\text{(since }I\left(  v\right)
\in\operatorname*{Ker}\left(  p_{2}\right)  \text{ (since the target}%
\\\text{of }I\text{ is }\operatorname*{Ker}\left(  p_{2}\right)  \text{))}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }p_{2}\text{ is linear}\right) \\
&  =p_{2}\left(  \left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\left(  j\left(
p_{1}\left(  u\right)  \right)  \right)  \right)  =\left(  p_{2}\mid_{J_{2}%
}\right)  \left(  \left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\left(  j\left(
p_{1}\left(  u\right)  \right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  p_{2}\mid_{J_{2}}\right)
^{-1}\left(  j\left(  p_{1}\left(  u\right)  \right)  \right)  \in
J_{2}\right) \\
&  =j\left(  p_{1}\left(  u\right)  \right)  .
\end{align*}
On the other hand,%
\begin{align*}
p_{1}\left(  x\right)   &  =p_{1}\left(  u+v\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }x=u+v\right) \\
&  =p_{1}\left(  u\right)  +\underbrace{p_{1}\left(  v\right)  }%
_{\substack{=0\\\text{(since }v\in\operatorname*{Ker}\left(  p_{1}\right)
\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }p_{1}\text{ is
linear}\right) \\
&  =p_{1}\left(  u\right)  ,
\end{align*}
so that $p_{1}\left(  u\right)  =p_{1}\left(  x\right)  $. Thus,%
\[
\left(  p_{2}\circ\Phi\right)  \left(  x\right)  =j\left(  \underbrace{p_{1}%
\left(  u\right)  }_{=p_{1}\left(  x\right)  }\right)  =j\left(  p_{1}\left(
x\right)  \right)  =\left(  j\circ p_{1}\right)  \left(  x\right)  ,
\]
qed.}. In other words, $p_{2}\circ\Phi=j\circ p_{1}$.
\end{verlong}

\begin{vershort}
Now, every $x\in M$ satisfies%
\[
\left(  p_{2}\circ\Phi\right)  \left(  x\right)  =\left(  j\circ p_{1}\right)
\left(  x\right)
\]
(as revealed by a straightforward computation\footnote{\textit{Proof.} Let
$x\in M$. Since $x\in M=J_{1}\oplus\left(  \operatorname*{Ker}\left(
p_{1}\right)  \right)  $, we can write $x$ in the form $x=u+v$ for some $u\in
J_{1}$ and $v\in\operatorname*{Ker}\left(  p_{1}\right)  $. Consider these $u$
and $v$. Then, $x=u+v$ corresponds to the pair $\left(  u,v\right)  $ under
the identification of the internal direct sum $M=J_{1}\oplus\left(
\operatorname*{Ker}\left(  p_{1}\right)  \right)  $ with the external direct
sum $J_{1}\oplus\left(  \operatorname*{Ker}\left(  p_{1}\right)  \right)  $.
Hence, $x=\left(  u,v\right)  $. On the other hand, identifying the internal
direct sum $M=J_{2}\oplus\left(  \operatorname*{Ker}\left(  p_{2}\right)
\right)  $ with the external direct sum $J_{2}\oplus\left(
\operatorname*{Ker}\left(  p_{2}\right)  \right)  $, we have%
\[
\left(  \left(  \left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\circ j\circ\left(
p_{1}\mid_{J_{1}}\right)  \right)  \left(  u\right)  ,I\left(  v\right)
\right)  =\left(  \left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\circ j\circ\left(
p_{1}\mid_{J_{1}}\right)  \right)  \left(  u\right)  +I\left(  v\right)  ,
\]
so that%
\begin{align*}
\Phi\left(  \underbrace{x}_{=\left(  u,v\right)  }\right)   &  =\Phi\left(
u,v\right)  =\left(  \left(  \left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\circ
j\circ\left(  p_{1}\mid_{J_{1}}\right)  \right)  \left(  u\right)  ,I\left(
v\right)  \right) \\
&  =\left(  \left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\circ j\circ\left(
p_{1}\mid_{J_{1}}\right)  \right)  \left(  u\right)  +I\left(  v\right)  .
\end{align*}
Applying the map $p_{2}$ to both sides of this equality, we obtain%
\begin{align*}
\left(  p_{2}\circ\Phi\right)  \left(  x\right)   &  =p_{2}\left(  \left(
\left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\circ j\circ\left(  p_{1}\mid_{J_{1}%
}\right)  \right)  \left(  u\right)  +I\left(  v\right)  \right) \\
&  =p_{2}\left(  \left(  \left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\circ
j\circ\left(  p_{1}\mid_{J_{1}}\right)  \right)  \left(  u\right)  \right)
+\underbrace{p_{2}\left(  I\left(  v\right)  \right)  }%
_{\substack{=0\\\text{(since }I\left(  v\right)  \in\operatorname*{Ker}\left(
p_{2}\right)  \text{)}}}\\
&  =p_{2}\left(  \left(  \left(  p_{2}\mid_{J_{2}}\right)  ^{-1}\circ
j\circ\left(  p_{1}\mid_{J_{1}}\right)  \right)  \left(  u\right)  \right) \\
&  =\underbrace{\left(  p_{2}\circ\left(  p_{2}\mid_{J_{2}}\right)
^{-1}\right)  }_{=\left(  p_{2}\mid_{J_{2}}\right)  \circ\left(  p_{2}%
\mid_{J_{2}}\right)  ^{-1}=\operatorname*{id}}\left(  \left(  j\circ\left(
p_{1}\mid_{J_{1}}\right)  \right)  \left(  u\right)  \right) \\
&  =\left(  j\circ\left(  p_{1}\mid_{J_{1}}\right)  \right)  \left(  u\right)
=j\left(  p_{1}\left(  u\right)  \right)  .
\end{align*}
On the other hand, since $x=u+v$, we have%
\begin{align*}
p_{1}\left(  x\right)   &  =p_{1}\left(  u+v\right)  =p_{1}\left(  u\right)
+\underbrace{p_{1}\left(  v\right)  }_{\substack{=0\\\text{(since }%
v\in\operatorname*{Ker}\left(  p_{1}\right)  \text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }p_{1}\text{ is linear}\right) \\
&  =p_{1}\left(  u\right)  ,
\end{align*}
so that $p_{1}\left(  u\right)  =p_{1}\left(  x\right)  $. Thus, $\left(
p_{2}\circ\Phi\right)  \left(  x\right)  =j\left(  \underbrace{p_{1}\left(
u\right)  }_{=p_{1}\left(  x\right)  }\right)  =j\left(  p_{1}\left(
x\right)  \right)  =\left(  j\circ p_{1}\right)  \left(  x\right)  $, qed.}).
In other words, $p_{2}\circ\Phi=j\circ p_{1}$.
\end{vershort}

So we know that $\Phi$ is an $R$-module isomorphism from $M$ to $M$ such that
$p_{2}\circ\Phi=j\circ p_{1}$. Hence, there exists an $R$-module isomorphism
$k:M\rightarrow M$ such that $p_{2}\circ k=j\circ p_{1}$ (namely, $k=\Phi$).
This proves Lemma A.6.
\end{proof}

\bigskip

\begin{proof}[Proof of Proposition A.7.] First of all, if there exists a
$U\in\operatorname*{GL}\left(  M\right)  $ satisfying $A_{1}=A_{2}U$, then we
clearly have $A_{1}\left(  M\right)  =A_{2}\left(  M\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Assume that there exists a $U\in
\operatorname*{GL}\left(  M\right)  $ satisfying $A_{1}=A_{2}U$. Consider this
$U$. Since $U\in\operatorname*{GL}\left(  M\right)  $, we know that $U$ is an
$R$-module automorphism of $M$, so that $U\left(  M\right)  =M$. Now, from
$A_{1}=A_{2}U$, we conclude that $A_{1}\left(  M\right)  =\left(
A_{2}U\right)  \left(  M\right)  =A_{2}\left(  \underbrace{U\left(  M\right)
}_{=M}\right)  =A_{2}\left(  M\right)  $, qed.}.

\begin{verlong}
Now assume that $A_{1}\left(  M\right)  =A_{2}\left(  M\right)  $. We shall
prove that there exists a $U\in\operatorname*{GL}\left(  M\right)  $
satisfying $A_{1}=A_{2}U$.
\end{verlong}

\begin{vershort}
We now only need to prove the opposite direction . So assume that
$A_{1}\left(  M\right)  =A_{2}\left(  M\right)  $. We shall prove that there
exists a $U\in\operatorname*{GL}\left(  M\right)  $ satisfying $A_{1}=A_{2}U$.
\end{vershort}

Let $M^{\prime}$ be the $R$-submodule $A_{1}\left(  M\right)  $ of $M$.

\begin{verlong}
We know that every submodule of a finite free $R$-module is finite free (due
to Lemma A.2). Since $M$ is a finite free $R$-module, this yields that every
submodule of $M$ is finite free. Since $M^{\prime}$ is a submodule of $M$,
this yields that the $R$-module $M^{\prime}$ is finite free.
\end{verlong}

\begin{vershort}
By Lemma A.2, every submodule of a finite free $R$-module is finite free.
Thus, $M^{\prime}$ (being a submodule of the finite free $R$-module $M$) is
finite free.
\end{vershort}

Since $A_{1}\left(  M\right)  =M^{\prime}$, we can define a surjective
$R$-module homomorphism $p_{1}:M\rightarrow M^{\prime}$ by%
\begin{equation}
\left(  p_{1}\left(  x\right)  =A_{1}\left(  x\right)  \text{ for every }x\in
M\right)  . \label{lem.A.7.pf.1}%
\end{equation}
Consider this $p_{1}$.

Since $A_{2}\left(  M\right)  =A_{1}\left(  M\right)  =M^{\prime}$, we can
define a surjective $R$-module homomorphism $p_{2}:M\rightarrow M^{\prime}$
by
\begin{equation}
\left(  p_{2}\left(  x\right)  =A_{2}\left(  x\right)  \text{ for every }x\in
M\right)  . \label{lem.A.7.pf.2}%
\end{equation}
Consider this $p_{2}$.

Lemma A.6 (applied to $N_{1}=M^{\prime}$, $N_{2}=M^{\prime}$ and
$j=\operatorname*{id}\nolimits_{M^{\prime}}$) shows that there exists an
$R$-module isomorphism $k:M\rightarrow M$ such that $p_{2}\circ
k=\operatorname*{id}\nolimits_{M^{\prime}}\circ p_{1}$. Consider this $k$. We
know that $k$ is an $R$-module isomorphism $M\rightarrow M$, thus an
$R$-module automorphism of $M$. Thus, $k\in\operatorname*{GL}\left(  M\right)
$.

\begin{verlong}
Every $y\in M$ satisfies%
\begin{align*}
\left(  A_{2}\circ k\right)  \left(  y\right)   &  =A_{2}\left(  k\left(
y\right)  \right)  =p_{2}\left(  k\left(  y\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since (\ref{lem.A.7.pf.2}) (applied to
}x=k\left(  y\right)  \text{) yields }p_{2}\left(  k\left(  y\right)  \right)
=A_{2}\left(  k\left(  y\right)  \right)  \right) \\
&  =\underbrace{\left(  p_{2}\circ k\right)  }_{=\operatorname*{id}%
\nolimits_{M^{\prime}}\circ p_{1}=p_{1}}\left(  y\right)  =p_{1}\left(
y\right)  =A_{1}\left(  y\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{lem.A.7.pf.1}), applied to }x=y\right)  .
\end{align*}
In other words, $A_{2}\circ k=A_{1}$, so that $A_{1}=A_{2}\circ k$. That is,
we have $A_{1}=A_{2}k$ in the group $\operatorname*{GL}\left(  M\right)  $.
Hence, there exists a $U\in\operatorname*{GL}\left(  M\right)  $ satisfying
$A_{1}=A_{2}U$ (namely, $U=k$).
\end{verlong}

\begin{vershort}
But we have $A_{2}\circ k=A_{1}$. (Indeed, this follows from $p_{2}\circ
k=\operatorname*{id}\nolimits_{M^{\prime}}\circ p_{1}=p_{1}$ upon noticing
that the maps $p_{1}$ and $p_{2}$ are identical with the maps $A_{1}$ and
$A_{2}$, respectively, apart from having a different target set.)
\end{vershort}

\begin{verlong}
Now forget that we assumed that $A_{1}\left(  M\right)  =A_{2}\left(
M\right)  $. We have thus proven that if $A_{1}\left(  M\right)  =A_{2}\left(
M\right)  $, then there exists a $U\in\operatorname*{GL}\left(  M\right)  $
satisfying $A_{1}=A_{2}U$ (namely, $U=k$). This completes the proof of
Proposition A.7 (because we have also proven that if there exists a
$U\in\operatorname*{GL}\left(  M\right)  $ satisfying $A_{1}=A_{2}U$, then we
clearly have $A_{1}\left(  M\right)  =A_{2}\left(  M\right)  $).
\end{verlong}

\begin{vershort}
Thus, there exists a $U\in\operatorname*{GL}\left(  M\right)  $ satisfying
$A_{1}=A_{2}U$ (namely, $U=k$). This completes the proof of Proposition A.7.
\end{vershort}
\end{proof}

\bigskip

\begin{center}
\textbf{\S 3. Generalizing \cite{kc-ab}}
\end{center}

Proposition A.7 allows us to generalize the results of Conrad's \cite{kc-ab}.
First of all, our Proposition A.7 is obviously a generalization of \cite[Lemma
2]{kc-ab}. The notions of ``aligned bases'' and ``simultaneously aligned
bases'' defined in \cite{kc-ab} can be generalized:

\begin{quote}
\textbf{Definition.} Let $R$ be a PID, $n$ be a positive integer, and $M$ be a
finite free $R$-module of rank $n$.

\textbf{(a)} Let $M^{\prime}$ be a submodule of $M$ (not necessarily of rank
$n$). The structure theorem for modules over a PID (or the theory of the Smith
normal form) shows that there exists a basis $\left(  e_{1},e_{2}%
,...,e_{n}\right)  $ of $M$ and scalars $a_{1},a_{2},...,a_{n}$ such that
$M^{\prime}=\bigoplus\limits_{i=1}^{n}Ra_{i}e_{i}$. In such a situation, we
will say that $\left(  e_{1},e_{2},...,e_{n}\right)  $ and $\left(  a_{1}%
e_{1},a_{2}e_{2},...,a_{n}e_{n}\right)  $ are a pair of \textit{aligned} bases
for $M$ and $M^{\prime}$, whether or not $\left(  a_{1}e_{1},a_{2}%
e_{2},...,a_{n}e_{n}\right)  $ is a basis of $M^{\prime}$. (Of course,
$\left(  a_{k}e_{k}\right)  _{k\in\left\{  1,2,...,n\right\}  ;\ a_{k}\neq0}$
\textbf{is} a basis of $M^{\prime}$ in this case, so the only thing that can
keep the sequence $\left(  a_{1}e_{1},a_{2}e_{2},...,a_{n}e_{n}\right)  $ from
being a basis of $M^{\prime}$ are zero elements.)

\textbf{(b)} Let $M^{\prime}$ and $M^{\prime\prime}$ be two submodules of $M$.
If there exists a basis $\left(  e_{1},e_{2},...,e_{n}\right)  $ of $M$ and
scalars $a_{1}^{\prime},a_{2}^{\prime},...,a_{n}^{\prime}$ such that
$M^{\prime}=\bigoplus\limits_{i=1}^{n}Ra_{i}e_{i}$, as well as scalars
$a_{1}^{\prime\prime},a_{2}^{\prime\prime},...,a_{n}^{\prime\prime}$ such that
$M^{\prime\prime}=\bigoplus\limits_{i=1}^{n}Rb_{i}e_{i}$, then we say that
$M^{\prime}$ and $M^{\prime\prime}$ admit \textit{simultaneously aligned
bases}.
\end{quote}

With this definition, \cite[Theorem 3]{kc-ab} still holds without the
conditions $\det A\neq0$ and $\det B\neq0$, and \cite[Corollary 5]{kc-ab} is
still valid if the \textquotedblleft with nonzero
determinants\textquotedblright\ condition is removed. The proofs of these
results don't require any new arguments apart from replacing the (easy) proof
of \cite[Lemma 2]{kc-ab} by the (not so easy) proof of our Proposition A.7.

\begin{verlong}
Here is a comprehensive list of changes which must be made to \cite{kc-ab} in
order to obtain valid statements and proofs in the general case:

\begin{itemize}
\item Remove the words ``also having rank $n$ (to be called a \textit{full
submodule} of $M$)'' from the first paragraph of page 1.

\item Replace ``We call such a pair of bases of $M$ and $M^{\prime}$
\textit{aligned}'' by ``By abuse of notation, in such a situation we will say
that $\left(  e_{1},...,e_{n}\right)  $ and $\left(  a_{1}e_{1},...,a_{n}%
e_{n}\right)  $ form a pair of \textit{aligned} bases for $M$ and $M^{\prime}%
$, even though $\left(  a_{1}e_{1},...,a_{n}e_{n}\right)  $ needs not in
general be a basis of $M^{\prime}$ (but the nonzero vectors among $a_{1}%
e_{1},...,a_{n}e_{n}$ do form a basis of $M^{\prime}$)''.

\item Remove the word ``nonzero'' from ``nonzero scalars $a_{1},...,a_{n}$ in
$R$''.

\item Replace ``such that $a_{1}e_{1},...,a_{n}e_{n}$ is a basis of
$M^{\prime}$\ \ \ \ '' by ``such that $M^{\prime}=\bigoplus\limits_{i=1}%
^{n}Ra_{i}e_{i}$\ \ \ \ ''.

\item Remove the word ``full'' from ``Pick two full submodules of $M$, say
$M^{\prime}$ and $M^{\prime\prime}$\ \ \ \ ''.

\item Remove the word ``nonzero'' from ``two sets of nonzero''.

\item Remove the following two sentences:

``Of course if $R$ is a field then they do because the only full submodule of
$M$ is $M$, so the situation is trivial. To keep things interesting, we assume
from now on that $R$ is not a field, so $R$ contains prime elements.''

While it is true that the case of $R$ being a field is much easier than the
general case of $R$ being a PID, it cannot really be said that the former case
is trivial in the general situation (where our submodules $M^{\prime}$ and
$M^{\prime\prime}$ are not required to be full).

\item Remove all remaining appearances of the word ``full'' from the paper.

\item Remove the ``$\neq0$'' from ``and $\det A=a_{1}\cdots a_{n}\neq0$''.

\item Remove the words ``with nonzero determinant'' and ``with nonzero
determinants'' throughout the paper.

\item Replace the proof of Lemma 2 by the above proof of Proposition A.7.

\item Replace ``two sets of $n$ nonzero'' by ``two sets of $n$ elements''.

\item Remove ``with $\det A\neq0$ and $\det B\neq0$'' from Theorem 3.

\item Remove all appearances of the word ``nonzero'' from the proof of Theorem 3.

\item On page 3, replace ``a basis for the whole module that can be scaled to
a basis of each of the submodules'' by ``a basis $\left(  e_{1},e_{2}%
,...,e_{n}\right)  $ for the whole module such that, for each of our
submodules, there exist elements $a_{1},a_{2},...,a_{n}$ of $R$ such that the
submodule is $\bigoplus\limits_{i=1}^{n}Ra_{i}e_{i}$\ \ \ \ ''.

\item In the proof of Corollary 5, remove ``(that is, each $M_{j}$ has a basis
that can be scaled termwise to provide a basis of those submodules)''.
\end{itemize}
\end{verlong}

\begin{thebibliography}{99}                                                                                               %


\bibitem {kc-ab}Keith Conrad, \textit{Simultaneously aligned bases},
2013.\newline%
\url{http://www.math.uconn.edu/~kconrad/blurbs/}

\bibitem {knapp}Anthony W. Knapp, \textit{Basic Algebra}, digital second edition,
2016. \newline
\url{http://www.math.stonybrook.edu/~aknapp/download.html}

\bibitem {birkhoff-maclane}Saunders Mac Lane, Garrett Birkhoff,
\textit{Algebra}, 3rd edition, AMS 1999.

\bibitem {brundan}Jonathan Brundan, \textit{Math 647/8/9 notes, Fall
2004}.\newline%
\url{http://darkwing.uoregon.edu/~brundan/math647fall04/}

\bibitem {garrett-aa}Paul Garrett, \textit{Abstract Algebra}, course notes,
last updated 20 Oct 2012.\newline%
\url{http://www.math.umn.edu/~garrett/m/algebra/}

\bibitem {mcnulty}George F. McNulty, \textit{Rings and Modules}, Fall 2010
course notes, January 21, 2011.\newline%
\url{http://www.math.sc.edu/~mcnulty/algebra/grad/fall2010.pdf}

\bibitem {bash}Robert B. Ash, \textit{Abstract Algebra: The Basic Graduate
Year}, revised 11/02.\newline%
\url{http://www.math.uiuc.edu/~r-ash/Algebra.html}

\bibitem {bash-ca}Robert B. Ash, \textit{A Course In Commutative Algebra},
revised Jan 2006.\newline%
\url{http://www.math.uiuc.edu/~r-ash/ComAlg.html}

\bibitem {am}M. F. Atiyah, I. G. Macdonald, \textit{Introduction to
Commutative Algebra}, Addison-Wesley 1969.

\bibitem {mats}Hideyuki Matsumura, \textit{Commutative ring theory}, CUP 1989.

\bibitem {szymiczek}Kazimierz Szymiczek, \textit{NAK and injectivity of
surjections}, 2004.\newline%
\url{http://www.math.us.edu.pl/zatl/szymiczek/referaty/}

\bibitem {kc-modPID}Keith Conrad, \textit{Modules over a PID}, 2014.\newline%
\url{http://www.math.uconn.edu/~kconrad/blurbs/}

\bibitem {kc-univid1}Keith Conrad, \textit{Universal Identities I},
2013.\newline%
\url{http://www.math.uconn.edu/~kconrad/blurbs/}

\bibitem {kc-extmod}Keith Conrad, \textit{Exterior powers}, 2013.\newline%
\url{http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf}
\end{thebibliography}


\end{document}