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\begin{document}

\title{Powers of matrices with all principal minors equal to $1$}
\author{Darij Grinberg, Hamesh M. Hamesh\thanks{This paper was written
by GPT-5.5 in June 2026 with some amount of strategic prompting.
It was then edited by the (first) author to improve writing.\\
This work is in the public domain.}}
\date{\today}
\maketitle

\begin{abstract}
Consider a square matrix $A$ whose all principal minors are equal to $1$.
Over a field, this property is inherited by any power of $A$, but this
is not the case over an arbitrary commutative ring.
We show that it is the case over any regular ring, and also over the ring
$\mathbb{Z} / d$ for any integer $d$, and in some other settings
(quotients of Pr\"ufer domains and principal quotients of normal domains).
This generalizes Problem B5 of the 2021 Putnam contest.

Over arbitrary commutative rings, we identify a stronger property that is
always inherited by powers: We say that a matrix
$A = \left(a_{i,j}\right)_{i,j\in\left[n\right]}$
is strongly $1$-principled if all its diagonal entries are $1$ and
if all the cyclic products $a_{i_1, i_2} a_{i_2, i_3} \cdots a_{i_k, i_1}$
with $k>1$ vanish.
We show that the latter products are always integral over the ideal
generated by the principal minors of $A$ minus $1$.
\end{abstract}

\section{Definitions and the main result}

Throughout this note, rings are commutative, associative and unital. For
$n\in\mathbb{N}$, we set $\ive{n}:=\left\{  1,2,\ldots,n\right\}  $.

If $A=\tup{a_{i,j}}_{i,j\in\ive{n}}$ is an $n\times n$-matrix
over a ring $R$ and if $S\subseteq\ive{n}$, then $A_{S}$ denotes the
principal submatrix of $A$ with row and column set $S$. That is,
$A_{S}=\operatorname{sub}_{S}^{S}A$ in the notation of
\cite{GrinbergPrincipal}. For instance,
$\begin{pmatrix}
a & b & c\\
a^{\prime} & b^{\prime} & c^{\prime}\\
a^{\prime\prime} & b^{\prime\prime} & c^{\prime\prime}%
\end{pmatrix}_{\left\{  1,3\right\}  }
= \begin{pmatrix}
a & c\\
a^{\prime\prime} & c^{\prime\prime}
\end{pmatrix}$.
The \emph{principal minors} of an $n\times n$-matrix $A$ are its
minors $\det\left(  A_{S}\right)  $ for all $S\subseteq\ive{n}$. We
use the convention that the empty determinant is $1$.

Problem B5 of the 2021 Putnam contest \cite[Problem B5]{Putnam2021} asserts
that if $A\in\mathbb{Z}^{n\times n}$ is an integer matrix whose all principal
minors are odd, then all powers $A^{m}$ of $A$ have the same property. By
reducing the matrix modulo $2$, this can be restated as follows: If
$A\in\left(  \mathbb{Z}/2\right)  ^{n\times n}$ is a matrix over the
two-element field $\mathbb{Z}/2$ whose all principal minors equal $1$, then
all its powers $A^{m}$ have the same property. This suggests a generalization
to arbitrary commutative rings instead of $\mathbb{Z}/2$; however, this
generalization was disproved in \cite[\S 6]{GrinbergPrincipal} for $4\times
4$-matrices over a certain finite ring.\footnote{That said, a part of the
generalization is true over any $R$ (see \cite[Theorem 5.2]{GrinbergPrincipal}%
): If $A\in R^{n\times n}$ is a matrix whose all principal minors equal $1$,
then all \textbf{diagonal entries} of its powers $A^{m}$ equal $1$ as well
(even though some principal minors of $A^{m}$ may differ from $1$).}

In this note, we shall show that this generalization is nevertheless true if
we replace $\mathbb{Z}/2$ by any ring of the form $\mathbb{Z}/d$ with
$d$ a positive integer. More generally, it is true over every quotient $D/I$
whose kernel ideal $I$ is integrally closed in $D$. This includes quotients of
Pr\"ufer domains by arbitrary ideals and quotients of normal domains by
principal ideals.

We introduce some terminology for the type of matrices we will study.

\begin{definition}
Let $R$ be a commutative ring, and let $A=\tup{a_{i,j}}
_{i,j\in\ive{n}}$ be an $n\times n$-matrix over $R$. We say that $A$ is
\emph{$1$-principled} if
\[
\det A_{S}=1 \qquad \text{for every }S\subseteq\ive{n}.
\]

\end{definition}

Note that the diagonal entries of an $n\times n$-matrix are its $1\times1$
principal minors. Hence, the diagonal entries of a $1$-principled matrix are
$1$.

\begin{definition}
Let $S$ be a set. A \emph{cycle} on $S$ will mean a $k$-tuple%
\[
C=\left(  i_{1},i_{2},\ldots,i_{k}\right)
\]
where $k>0$ and where $i_{1},i_{2},\ldots,i_{k}$ are distinct elements of
$S$. To be more precise, the cycle will be not this $k$-tuple itself, but
rather its equivalence class under cyclic rotation (i.e., we will count
$\left(  i_{1},i_{2},\ldots,i_{k}\right)  $ and $\left(  i_{2},i_{3}%
,\ldots,i_{k},i_{1}\right)  $ as being the same cycle). This cycle is said
to have \emph{length} $k$, \emph{vertices} $i_1, i_2, \ldots, i_k$ and
\emph{arcs} $\tup{i_1, i_2},\ \tup{i_2, i_3},\ \ldots,\ \tup{i_k, i_1}$;
furthermore, we call it \emph{nontrivial} if $k>1$ (that is,
if the cycle has more than one arc).

\end{definition}

\begin{definition}
Let $A = \tup{a_{i,j}}_{i,j\in\ive{n}}$ be an $n\times n$-matrix
over a commutative ring $R$.

\begin{enumerate}
\item[\textbf{(a)}]
The \emph{$A$-weight} of a cycle $C=\left(  i_{1},i_{2},\ldots,i_{k}\right)$
on $\ive{n}$ is defined to be
\[
w_{A}(C):=a_{i_{1},i_{2}}a_{i_{2},i_{3}}\cdots a_{i_{k-1},i_{k}}a_{i_{k}%
,i_{1}}.
\]


\item[\textbf{(b)}] We say that $A$ is \emph{strongly $1$-principled} if
\[
a_{i,i}=1\qquad\text{for all }i\in\ive{n}
\]
(that is, all diagonal entries of $A$ are $1$)
and
\[
w_{A}(C)=0\qquad\text{for each nontrivial cycle }C\text{ on }\ive{n}
\]
(that is, each nontrivial cycle on $\ive{n}$ has $A$-weight $0$).

\end{enumerate}
\end{definition}

\begin{example}
\label{exa.unitri}
If a matrix $A = \tup{a_{i,j}}_{i,j\in\ive{n}}$
is unitriangular (i.e., triangular and satisfies $a_{i,i} = 1$
for all $i \in \ive{n}$), then $A$ is strongly $1$-principled.
Indeed, any nontrivial cycle on $\ive{n}$ has an arc $\tup{i,j}$
with $i<j$ and an arc $\tup{i,j}$ with $i>j$, and at least one
of these arcs will satisfy $a_{i,j} = 0$; thus, the $A$-weight
of the cycle is $0$.

However, there are strongly $1$-principled matrices that are
not unitriangular, such as
\[
A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}
\qquad \text{ over }  R = \ZZ / 4.
\]
\end{example}

Our main result is the following.

\begin{theorem}
\label{thm.main} Let $D$ be a commutative ring, let $I$ be an integrally
closed ideal of $D$, and set $R:=D/I$. Let $A$ be a $1$-principled
matrix over $R$. Then $A^{m}$ is $1$-principled for every $m\in\mathbb{N}$.
\end{theorem}

The notion of an integrally closed ideal will be recalled in
Section~\ref{sec.integral-closure}. The proof relies on three independent
results, all of which hold over any (commutative) ring:
First, every strongly $1$-principled matrix
is $1$-principled (Proposition~\ref{prop.strong-implies-principled}), although
the converse does not always hold.
Second, strongly $1$-principled
matrices are stable under powers (Proposition~\ref{prop.strong-powers}).
Third, every nontrivial cycle's
$A$-weight is integral over the ideal generated by the principal-minor
defects (i.e., the principal minors minus $1$)
(Theorem~\ref{thm.cycle-weight-integral}).
In the setting of Theorem~\ref{thm.main}, this forces these weights to
belong to $I$, since $I$ is integrally closed.
After we prove all the results we have described, we shall discuss some
examples of integrally closed ideals.
Our proof yields a new solution to Problem B5 of the 2021 Putnam contest
(which is the claim of Theorem~\ref{thm.main} for $D = \ZZ$ and $I = 2\ZZ$).

\section{\texorpdfstring{Strongly $1$-principled matrices}{Strongly
1-principled matrices}}

We begin with the theory of strongly $1$-principled matrices.

\begin{proposition}
\label{prop.strong-implies-principled} Every strongly $1$-principled matrix
over a commutative ring is $1$-principled.
\end{proposition}

\begin{proof}
Let $A=\tup{a_{i,j}}_{i,j\in\ive{n}}$ be a strongly
$1$-principled $n\times n$-matrix. Let $S\subseteq\ive{n}$. We expand
\begin{equation}
\det A_{S}
= \sum_{\pi\in\Symm_S}\sgn\tup{\pi}
\prod_{i\in S}a_{i,\pi(i)}
\label{eq.lem.strong-implies-principled.1}
\end{equation}
(where $\Symm_S$ is the group of all permutations of $S$). The
identity permutation $\id \in \Symm_S$ contributes
\[
\sgn\tup{\id} \prod_{i\in S}a_{i,\id(i)}
= \prod_{i\in S}a_{i,i}
= 1 \qquad \tup{\text{since $a_{i,i} = 1$ for all $i$}}
\]
to the right-hand side of (\ref{eq.lem.strong-implies-principled.1}). Every
non-identity permutation $\pi \in \Symm_S$
has at least one nontrivial cycle $\left(  i_1, i_2, \ldots, i_k\right)$
in its cycle decomposition. The corresponding addend on the right-hand side of
(\ref{eq.lem.strong-implies-principled.1}) therefore contains, as a factor,
the $A$-weight of this nontrivial cycle. This factor is $0$, since $A$ is strongly
$1$-principled. Hence all non-identity addends on the right-hand side of
(\ref{eq.lem.strong-implies-principled.1}) vanish, and we conclude that $\det
A_{S}=1$. That is, $A$ is $1$-principled.
\end{proof}

We shall use some basic graph theory (see, e.g., \cite[Chapter 4]{22s}). Let
$K_{n}^{\rightarrow}$ be the simple digraph (i.e., directed graph) with $n$
vertices $1,2,\ldots,n$ and $n^{2}$ arcs $\left(  i,j\right)  $ for all
$i,j\in\ive{n}$. What we called \textquotedblleft cycles on
$\ive{n}$\textquotedblright\ above are exactly the cycles of
$K_{n}^{\rightarrow}$ (considered up to cyclic
rotation). We make the following elementary observation about walks.

\begin{lemma}
\label{lem.closed-walk-has-cycle} Let
\[
i_{0}\rightarrow i_{1}\rightarrow\cdots\rightarrow i_{m}=i_{0}%
\]
be a closed walk of a simple digraph. If this closed walk is not the
stationary walk $i_{0}\rightarrow i_{0}\rightarrow\cdots\rightarrow i_{0}$,
then it contains a nontrivial cycle. (\textquotedblleft
Contains\textquotedblright\ means that each arc of the cycle is an arc of the walk.)
\end{lemma}

\begin{proof}
Delete all loops\footnote{Recall that a \emph{loop} means an arc of the form
$\tup{v,v}$ for some vertex $v$.} from the closed walk. Since the original walk is not
stationary, some arcs remain upon this deletion.
Starting at any remaining arc and following the
walk cyclically, eventually a vertex is repeated. The part of the walk between
the first occurrence of this vertex and its next occurrence is a closed walk
with no repeated internal vertices. This closed walk is therefore a cycle,
and moreover a nontrivial cycle (since all loops have been deleted).
\end{proof}

\begin{proposition}
\label{prop.strong-powers} Let $A$ be a strongly $1$-principled matrix over a
commutative ring $R$. Then $A^{m}$ is strongly $1$-principled for every
$m\in\mathbb{N}$.
\end{proposition}

\begin{proof}
Let $m\in\mathbb{N}$. The case $m=0$ is clear, since $A^{0}=I_{n}$. Assume
$m>0$.

Write the $n\times n$-matrix $A$ as $A=\tup{a_{i,j}}_{i,j\in\ive{n}}$.
We also use $B_{i,j}$ to refer to the $\left(  i,j\right)  $-th entry of
any matrix $B$; thus, $A_{i,j}=a_{i,j}$ for all $i,j\in\ive{n}$.
Since $A$ is strongly $1$-principled, all diagonal entries of $A$ are $1$: For
each $i\in\ive{n}$, we have
\begin{equation}
a_{i,i} %=\det\left(  A_{\left\{  i\right\}  }\right)
=1.\label{pf.lem.strong-powers.aii}%
\end{equation}


First, we show that every diagonal entry of $A^{m}$ is $1$. Fix $i\in
\ive{n}$. It is well-known (see, e.g., \cite[last sentence of \S 1]{Zeil};
also, the weighted version of \cite[Theorem 4.5.10]{22s}) that for any two
vertices $i$ and $j$ of $K_{n}^{\rightarrow}$, we have
\footnote{A reader unfamiliar with this formula \eqref{pf.lem.strong-powers.1}
can easily prove it by induction on $m$ using $A^m = A A^{m-1}$.}
\begin{equation}
\left(  A^{m}\right)_{i,j}
= \sum_{\substack{i=i_{0}\rightarrow i_{1}\rightarrow\cdots
\rightarrow i_{m}=j\\\text{is a walk of }K_{n}^{\rightarrow}}}
a_{i_{0},i_{1}}a_{i_{1},i_{2}}\cdots a_{i_{m-1},i_{m}}
.\label{pf.lem.strong-powers.1}
\end{equation}
Let us refer to the product $a_{i_{0},i_{1}}a_{i_{1},i_{2}}\cdots
a_{i_{m-1},i_{m}}$ in this sum as the \emph{$A$-weight} of the walk
$i=i_{0}\rightarrow i_{1}\rightarrow\cdots\rightarrow i_{m}=j$. Thus,
(\ref{pf.lem.strong-powers.1}) says that
\begin{equation}
\left(  A^{m}\right)_{i,j}
= \tup{\text{sum of the $A$-weights of all length-$m$ walks from $i$ to $j$}}.
\qquad\qquad
\label{pf.lem.strong-powers.2}
\end{equation}

In particular, $\left(  A^{m}\right)_{i,i}$ is the sum of the $A$-weights of
all length-$m$ walks from $i$ to $i$.
The stationary walk $i=i\rightarrow i\rightarrow\cdots\rightarrow i=i$
contributes $1$ to this sum, since its $A$-weight is
$a_{i,i}a_{i,i}\cdots a_{i,i}=a_{i,i}^{m}=1$ (by
(\ref{pf.lem.strong-powers.aii})).
Each of the other length-$m$ walks from $i$ to $i$ contains a nontrivial
cycle by Lemma~\ref{lem.closed-walk-has-cycle} (since it is a closed walk but
not stationary).
Since the latter cycle has $A$-weight $0$ (because $A$ is
strongly $1$-principled), we conclude that the walk that contains it must have
$A$-weight $0$ as well (indeed, since $R$ is commutative, the $A$-weight of the
cycle is a factor of the $A$-weight of the walk).
Thus, $\left(  A^{m}\right)_{i,i}$ is the sum of a single $1$
(corresponding to the stationary walk
$i=i\rightarrow i\rightarrow\cdots\rightarrow i=i$) and a lot of $0$'s (coming
from all the other walks). Therefore,
\begin{equation}
\left(  A^{m}\right)_{i,i}=1.
\label{pf.lem.strong-powers.3}
\end{equation}

Forget that we fixed $i$. So we have proved \eqref{pf.lem.strong-powers.3} for
each $i \in \ive{n}$.

It remains to prove that every nontrivial cycle on $\ive{n}$ has $A^{m}%
$-weight $0$. Let%
\[
C=\left(  i_{1},i_{2},\ldots,i_{k}\right)
\]
be a nontrivial cycle on $\ive{n}$, with length $k>1$. We must show that
$w_{A^m} \tup{C} = 0$.

Note that $i_1 \neq i_2$ (by the definition of a cycle, since $k>1$).

Set $i_{k+1}=i_{1}$ (that is, read the indices cyclically modulo $k$).
Then,
\begin{align*}
w_{A^m} \tup{C}
&=
\left(  A^{m}\right)_{i_{1},i_{2}}\left(  A^{m}\right)_{i_{2},i_{3}}%
\cdots\left(  A^{m}\right)_{i_{k},i_{1}}
= \prod_{j=1}^{k}\left(
A^{m}\right)_{i_{j},i_{j+1}} \\
&= \prod_{j=1}^{k}
\tup{\text{sum of the $A$-weights of all length-$m$ walks from $i_j$ to $i_{j+1}$}}
\end{align*}
(by \eqref{pf.lem.strong-powers.2}).
Expanding this product, we obtain a
sum over all $k$-tuples of length-$m$ walks from $i_{j}$ to $i_{j+1}$ for each
$j\in\ive{k}$. The addend corresponding to such a $k$-tuple is the
product of the $A$-weights of all these walks; but this is, of course, the
$A$-weight of the closed walk (of length $km$) obtained by concatenating these
$k$ walks. This closed walk is not stationary
(since $i_1 \neq i_2$ are two distinct vertices on it).
Hence it contains a nontrivial
cycle, again by Lemma~\ref{lem.closed-walk-has-cycle}. The $A$-weight of
this cycle is $0$ since $A$ is strongly $1$-principled; but it is a factor of
the $A$-weight of the walk. Thus, the whole closed walk has $A$-weight $0$ as
well.

Thus, we have shown that $w_{A^m} \tup{C}$ is a sum of $A$-weights of certain
closed walks, but each of these closed walks has $A$-weight $0$. Hence,
\[
w_{A^m} \tup{C} = 0.
\]
Combined with \eqref{pf.lem.strong-powers.3},
this completes the proof of the fact that $A^{m}$ is strongly $1$-principled.
The proposition is proved.
\end{proof}

\begin{corollary}
\label{cor.strong-powers-principled} Let $A$ be a strongly $1$-principled
matrix over a commutative ring $R$. Then $A^{m}$ is $1$-principled for every
$m\in\mathbb{N}$.
\end{corollary}

\begin{proof}
The matrix $A^m$ is strongly $1$-principled by
Proposition~\ref{prop.strong-powers}, and therefore is $1$-principled
by Proposition~\ref{prop.strong-implies-principled}.
\end{proof}

\begin{remark}

Proposition~\ref{prop.strong-powers} is genuinely a statement about powers,
not about products. Even commuting products of strongly $1$-principled matrices
need not be strongly $1$-principled.

For an explicit counterexample, let $R = \ZZ / 2 \ZZ = \mathbb F_2$. Let
\[
J=\begin{pmatrix}
1&1\\
1&1
\end{pmatrix},
\]
so that $J^2=0$. Now define the block matrices
\[
A=\begin{pmatrix}
I_2&0\\
J&I_2
\end{pmatrix}
=\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
1&1&1&0\\
1&1&0&1
\end{pmatrix},
\qquad
B=\begin{pmatrix}
I_2&J\\
0&I_2
\end{pmatrix}
=\begin{pmatrix}
1&0&1&1\\
0&1&1&1\\
0&0&1&0\\
0&0&0&1
\end{pmatrix}.
\]
Both $A$ and $B$ are unitriangular and thus strongly $1$-principled
(by Example~\ref{exa.unitri}).
% : the non-loop arcs of $A$ all go
% from the lower block to the upper block, while the non-loop arcs of $B$ all go
% from the upper block to the lower block. Thus $A$ has no nontrivial cycle of nonzero $A$-weight, and $B$ has no
% nontrivial cycle of nonzero $B$-weight.

However,
\[
AB=\begin{pmatrix}
I_2&J\\
J&I_2+J^2
\end{pmatrix}
=\begin{pmatrix}
I_2&J\\
J&I_2
\end{pmatrix}
=\begin{pmatrix}
I_2+J^2&J\\
J&I_2
\end{pmatrix}=BA
=\begin{pmatrix}
1&0&1&1\\
0&1&1&1\\
1&1&1&0\\
1&1&0&1
\end{pmatrix}.
\]
This common product is not strongly $1$-principled, since the nontrivial cycle
$\tup{1,3}$ has $AB$-weight $1$.
\end{remark}

\section{Integral closure and cycle weights}\label{sec.integral-closure}

We now turn towards the harder, ``upstream'' direction, from
$1$-principled to strongly $1$-principled. As we already mentioned,
in general, a $1$-principled matrix is not always strongly
$1$-principled; a counterexample is constructed in the footnote
in \cite[\S 6]{GrinbergPrincipal}.
However, $1$-principledness implies a weaker version of strong
$1$-principledness, which we will later leverage to obtain the
full property under certain conditions.

We will need the classical notion of \emph{integrality} over an ideal,
which we will now briefly recall;
see \cite{HunekeSwanson} for a much more extensive treatment.

\begin{definition}
Let $R$ be a commutative ring, let $I$ be an ideal of $R$, and let $x\in R$.
We say that $x$ is \emph{integral over $I$} if there exist a positive integer
$d$ and elements $c_j\in I^j$ for all $j\in\ive{d}$ such that
\[
x^d+c_1x^{d-1}+c_2x^{d-2}+\cdots+c_dx^0=0.
\]
The set of all elements of $R$ that are integral over $I$ is called the
\emph{integral closure} of $I$ and is denoted by $\overline I$. The ideal $I$
is said to be \emph{integrally closed} if $\overline I=I$.
\end{definition}

Thus, an element is integral over the zero ideal if and only if it is
nilpotent. We shall use the following standard properties of integral closure
of ideals.

\begin{lemma}
\label{lem.integral-closure-basics}
Let $R$ be a commutative ring, and let $I$ and $J$ be ideals of $R$. Then:
\begin{enumerate}
\item[\textbf{(a)}] The set $\overline I$ is an ideal of $R$ containing $I$
as a subset.

\item[\textbf{(b)}] If $I\subseteq J$, then $\overline I\subseteq\overline J$.

\item[\textbf{(c)}] The ideal $\overline I$ is integrally closed; that is,
\[
\overline{\overline I}=\overline I.
\]

\item[\textbf{(d)}] If $I\subseteq J\subseteq\overline I$, then
$\overline J=\overline I$.
\end{enumerate}
\end{lemma}

\begin{proof}
Parts \textup{(a)} and \textup{(c)} are
\cite[Corollary 1.3.1]{HunekeSwanson}.
Part \textup{(b)} is trivial.
Part \textup{(d)} follows from \textup{(b)} and \textup{(c)}: the
inclusions $I\subseteq J\subseteq\overline I$ yield
\[
\overline I\subseteq\overline J\subseteq
\overline{\overline I}=\overline I.
\qedhere
\]
\end{proof}

We next isolate the combinatorial ingredient. A \emph{Hamilton cycle on a
finite set $S$} means a cycle on $S$ whose vertices are all the elements
of $S$. In other words, it means a Hamilton
cycle of $K_S^\to$, where $K_S^\to$ is the digraph whose vertices are the
elements of $S$ and whose arcs are all the pairs $\tup{s,t}$ for
$s,t\in S$. We observe two obvious facts:
\begin{enumerate}
\item Each Hamilton cycle on a subset $S$ of $\ive{n}$
    is a cycle on $\ive{n}$.
    Conversely, each cycle on $\ive{n}$ is a Hamilton cycle on
    $S$, where $S$ is the set of vertices of this cycle.
\item A cycle on a finite set $S$ is Hamilton if and only
    if it has length $\abs{S}$.
\end{enumerate}

We identify each cycle with its set of arcs (since the latter set
uniquely determines the former cycle).
The \emph{multiset union} $H \uplus H'$ of two cycles $H$ and $H'$
is defined to be a multidigraph (i.e., a directed multigraph) whose multiset
of arcs is obtained by combining the sets of arcs of $H$ and of $H'$.
(If an arc appears in both $H$ and $H'$, then it will appear twice in this
multiset union.)

\begin{lemma}
\label{lem.two-Hamilton-cycles}
Let $S$ be a finite set, and let $H$ and $H'$ be two distinct Hamilton cycles
on $S$. Then the multiset union of (the sets of arcs of) $H$ and $H'$ can be
partitioned into at least three cycles, each having length strictly smaller
than $\left|S\right|$.
\end{lemma}

\begin{example}
Let $S=\left\{1,2,3,4,5\right\}$, and let
\[
H=\tup{1,2,3,4,5}
\qquad\text{and}\qquad
H'=\tup{1,3,5,2,4}
\]
be two Hamilton cycles on $S$. Their multiset union has arcs
\begin{align*}
\underbrace{(1,2),(2,3),(3,4),(4,5),(5,1)}_{\text{arcs of }H},
\underbrace{(1,3),(3,5),(5,2),(2,4),(4,1)}_{\text{arcs of }H'}.
\end{align*}
These ten arcs can be partitioned into the following three cycles:
\[
\tup{1,2,4},\qquad \tup{1,3,4,5},\qquad \tup{2,3,5}.
\]
The following two pictures show the same multidigraph $H\uplus H'$ twice.
In the left picture, the arcs of $H$ are red and the arcs of $H'$ are blue.
In the right picture, the three colors show the three cycles just listed.
\begin{center}
\begin{tikzpicture}[scale=1.15,
  vertex/.style={circle,draw,fill=white,inner sep=1.2pt,font=\footnotesize},
  arr/.style={-{Latex[length=2mm]},line width=0.8pt}]
\foreach \name/\theta in {1/90,2/18,3/-54,4/-126,5/162}
  \node[vertex] (\name) at (\theta:1.35) {$\name$};
\foreach \u/\v in {1/2,2/3,3/4,4/5,5/1}
  \path[arr,draw=red] (\u) edge (\v);
\foreach \u/\v in {1/3,3/5,5/2,2/4,4/1}
  \path[arr,draw=blue] (\u) edge (\v);
\end{tikzpicture}
\qquad\qquad
\begin{tikzpicture}[scale=1.15,
  vertex/.style={circle,draw,fill=white,inner sep=1.2pt,font=\footnotesize},
  arr/.style={-{Latex[length=2mm]},line width=0.8pt}]
\foreach \name/\theta in {1/90,2/18,3/-54,4/-126,5/162}
  \node[vertex] (\name) at (\theta:1.35) {$\name$};
\foreach \u/\v in {1/2,2/4,4/1}
  \path[arr,draw=ForestGreen] (\u) edge (\v);
\foreach \u/\v in {1/3,3/4,4/5,5/1}
  \path[arr,draw=Purple] (\u) edge (\v);
\foreach \u/\v in {2/3,3/5,5/2}
  \path[arr,draw=Brown] (\u) edge (\v);
\end{tikzpicture}
\end{center}
Indeed, these three cycles use the following arcs:
\begin{align*}
\tup{1,2,4} &: (1,2),(2,4),(4,1),\\
\tup{1,3,4,5} &: (1,3),(3,4),(4,5),(5,1),\\
\tup{2,3,5} &: (2,3),(3,5),(5,2).
\end{align*}
Thus each arc of $H\uplus H'$ is used exactly once. None of the three
cycles is Hamilton on $S$, since their lengths are $3,4,3$, respectively.
\end{example}

\begin{proof}[Proof of Lemma~\ref{lem.two-Hamilton-cycles}.]
Choose a vertex $v\in S$ whose outgoing arcs in $H$ and $H'$ are distinct.%
\footnote{Such a vertex must exist, since $H$ and $H'$ are distinct.}
Write these arcs as
\[
e=(v,a)\in H\qquad\text{and}\qquad e'=(v,b)\in H',
\]
where $a\neq b$. Let $P'$ be the directed path in $H'$ from $a$ back to $v$,
and let $P$ be the directed path in $H$ from $b$ back to $v$. Then the
walks
\begin{align*}
C &:= P'e \qquad \text{(that is, the path $P'$ followed by the arc $e$)}
\qquad \text{ and } \\
C' &:= Pe' \qquad \text{(that is, the path $P$ followed by the arc $e'$)}
\end{align*}
are cycles. Both have length at most $\left|S\right| - 1$. Indeed,
if the cycle $C$ had length $\geq \left|S\right|$, then it would be
Hamilton, and thus the path $P'$ would contain every vertex.
Hence the complementary path in $H'$ from $v$ to $a$ would consist
of a single arc. This arc would be the outgoing arc $(v,b)$ of $v$ in $H'$,
forcing $a=b$. The same argument applies to $C'$.

The cycles $C$ and $C'$ are arc-disjoint in the multidigraph
$H\uplus H'$.
Indeed:
\begin{enumerate}
\item The only $H$-arc used by $C$ is $e$, while the $H$-arcs
      used by $C'$ lie in $P$, which does not use $e$ because
      $P$ ends at $v$. Thus, $C$ and $C'$ have no $H$-arcs in
      common.
\item Similarly, $C$ and $C'$ have no $H'$-arcs in common.
\end{enumerate}
% only $H'$-arc used by $C'$ is $e'$, while the $H'$-arcs used by $C$ lie in
% $P'$, which does not use $e'$.

Remove the arcs of the two cycles $C$ and $C'$ from this multidigraph
$H\uplus H'$.
The remaining multidigraph is balanced\footnote{A
multidigraph is said to be \emph{balanced} if for each vertex $v$, the
indegree of $v$ equals the outdegree of $v$.}, because both the
original multidigraph $H\uplus H'$ and the removed union $C\cup C'$ are balanced.
Hence its arcs can be partitioned into cycles\footnote{We are using
a well-known result saying that the multiset of arcs of a balanced multidigraph
can be partitioned into cycles. This can be proved in many ways, e.g.:
Start walking at any non-isolated vertex; each time you enter a vertex,
the balancedness will ensure that you will be able to exit again; sooner
or later you will run into a cycle. Whenever this happens, remove the cycle
from the digraph, and repeat the same procedure. Removing a cycle leaves
the digraph balanced, so this algorithm will continue until the digraph
has no arcs left; at that point, the cycles obtained will form a partition
of the set of all arcs.}. None of these new cycles contains $v$, since
both outgoing arcs from $v$ in $H\uplus H'$
have been removed when we removed the arcs of $C$ and $C'$. Thus
every new cycle has length at most $\left|S\right| - 1$.

We have therefore partitioned the $2\left|S\right|$ arcs of
$H\uplus H'$ into cycles of length at most
$\left|S\right|-1$. Consequently, the number of these cycles
is at least
\[
\frac{2\left|S\right|}{\left|S\right|-1} > 2,
\]
that is, at least $3$ (since it is an integer).
This proves the lemma.
\end{proof}

Let $A=\tup{a_{i,j}}_{i,j\in\ive{n}}$ be a matrix over a commutative ring
$R$. For every integer $r\geq2$, let $K_{<r}(A)$ denote the ideal of $R$
generated by the $A$-weights of all nontrivial cycles on $\ive{n}$
whose length is strictly smaller than $r$.
Thus, $K_{<2}(A)=0$ (since a nontrivial cycle cannot have length
smaller than $2$).

\begin{lemma}
\label{lem.products-Hamilton-cycles}
Let $S\subseteq\ive{n}$ have size $r\geq2$, and let
$H_1,H_2,\ldots,H_t$ be $t$ distinct Hamilton cycles on $S$, where $t\geq2$.
Then
\[
w_A(H_1)w_A(H_2)\cdots w_A(H_t)\in K_{<r}(A)^t.
\]
\end{lemma}

\begin{proof}
Pick any $i \neq j$ in $\ive{t}$.
By Lemma~\ref{lem.two-Hamilton-cycles}, the multiset union of the two distinct
Hamilton cycles $H_i$ and $H_j$ on $S$ can be partitioned into at least three
cycles of length strictly smaller than $r$. These latter cycles are all nontrivial,
since they cannot contain any loops (indeed, all their arcs must be arcs of the
original two Hamilton cycles, but those did not contain any loops).
Hence, their $A$-weights belong to $K_{<r}(A)$.
Since the product of the $A$-weights is unchanged by repartitioning the same
multiset of arcs, this shows that
\begin{align}
w_A(H_i)w_A(H_j)\in K_{<r}(A)^3.
\label{pf.lem.products-Hamilton-cycles.4}
\end{align}
Thus, we have proved \eqref{pf.lem.products-Hamilton-cycles.4} for any $i \neq j$
in $\ive{t}$.

Now, pair off $2\left\lfloor t/2\right\rfloor$ of the $t$ Hamilton cycles
$H_1,H_2,\ldots,H_t$. Multiplying
the inclusions \eqref{pf.lem.products-Hamilton-cycles.4} for these pairs,
and multiplying by the $A$-weight of the remaining cycle if $t$ is odd, gives
\[
w_A(H_1)w_A(H_2)\cdots w_A(H_t)
\in K_{<r}(A)^{3\left\lfloor t/2\right\rfloor}.
\]
Since $3\left\lfloor t/2\right\rfloor\geq t$ for every $t\geq2$, the right
hand side is contained in $K_{<r}(A)^t$.
\end{proof}

We can now state the universal algebraic result. Define the
\emph{principal-minor-defect ideal} of $A$ to be the ideal
\[
J(A):=\left(\det A_S-1\ \middle|\ S\subseteq\ive{n}\right)\subseteq R
\]
(that is, the ideal of $R$ generated by all $2^n$ differences
$\det A_S - 1$, where $S$ ranges over the subsets of $\ive{n}$).

\begin{theorem}
\label{thm.cycle-weight-integral}
Let $A$ be an $n\times n$-matrix over a commutative ring $R$.
Then the $A$-weight of every
nontrivial cycle on $\ive{n}$ is integral over $J(A)$. Equivalently,
every nontrivial cycle $C$ on $\ive{n}$ satisfies
\[
w_A(C)\in\overline{J(A)}.
\]
\end{theorem}

\begin{proof}
We must show that each nontrivial cycle $H$ on $\ive{n}$
satisfies $w_A\tup{H} \in \overline{J(A)}$.
In other words, we must show that for each $r\geq 2$ and each
$r$-element subset $S$ of $\ive{n}$, each Hamilton cycle $H$ on $S$
satisfies $w_A\tup{H} \in \overline{J(A)}$
(since any nontrivial cycle is a Hamilton cycle on a subset of
size $\geq 2$).

We use strong induction on $r$.
Fix an $r$-element subset
$S\subseteq\ive{n}$, where $r\geq2$, and let $\mathcal H_S$ be the set of all
Hamilton cycles on $S$. For every $H\in\mathcal H_S$, set
\[
z_H:=(-1)^{r-1}w_A(H).
\]
The sign $(-1)^{r-1}$ is the sign of the cyclic permutation corresponding to
$H$ (that is, of the permutation of $S$ that sends each vertex of $H$
to the next vertex that follows it on $H$).

Set
\[
K:=K_{<r}(A)
\qquad\text{and}\qquad
L:=J(A)+K.
\]
We first show that each $z_H$ is integral over $L$. Let
$e_t$ be the $t$-th elementary symmetric polynomial in the elements
$z_H$ for $H\in\mathcal H_S$.
In particular, $e_0 = 1$ and $e_1 = \sum_{H\in\mathcal H_S}z_H$.

We shall now use the determinant expansion
\eqref{eq.lem.strong-implies-principled.1} to show that
\begin{equation}
e_1 \in L.
\label{eq.Hamilton-e1-in-L}
\end{equation}

\begin{proof}[Proof of \eqref{eq.Hamilton-e1-in-L}.]
For each $i \in S$, we have $a_{i,i} - 1 \in J(A)$
(since $a_{i,i}$ is the principal minor $\det A_{\set{i}}$ of $A$)
and thus $a_{i,i} - 1 \in J(A) \subseteq L$,
so that $a_{i,i} \equiv 1 \mod J(A)$.
Hence, $\prod_{i\in S}a_{i,i} \equiv \prod_{i\in S} 1 = 1 \mod L$.
Thus, the addend corresponding to $\pi = \id \in \Symm_S$
on the right-hand side of \eqref{eq.lem.strong-implies-principled.1}
is $\equiv 1 \mod L$.
Each of the remaining addends in that sum
\begin{enumerate}
\item either corresponds to a permutation $\pi$ that is an
      $r$-cycle, and thus is equal to
      $\sgn\tup{\pi} \prod_{i\in S} a_{i, \pi(i)}
      = \tup{-1}^{r-1} w_A\tup{H} = z_H$
      for a Hamilton cycle $H \in \mathcal{H}_S$;
\item or corresponds to a permutation $\pi$ that has at least
      two cycles (all of which must therefore have length
      smaller than $r$, and at least one of which must be
      nontrivial because $\pi \neq \id$),
      and therefore contains the $A$-weight
      of a nontrivial cycle of length strictly smaller than $r$
      as a factor; therefore this addend belongs to
      $K_{<r}(A) = K \subseteq L$.
\end{enumerate}
Hence, reduced modulo $L$, the equality
\eqref{eq.lem.strong-implies-principled.1} becomes
\[
\det A_{S}
\equiv 1 + \sum_{H \in \mathcal{H}_S} z_H \mod L.
\]
Therefore,
\[
\sum_{H \in \mathcal{H}_S} z_H \equiv \det A_S - 1 \equiv 0 \mod L
\]
(since the definition of $J(A)$ yields
$\det A_{S} - 1 \in J(A) \subseteq L$).
That is, $\sum_{H \in \mathcal{H}_S} z_H \in L$.
In other words, $e_1 \in L$
(since $e_1 = \sum_{H \in \mathcal{H}_S} z_H$).
This proves \eqref{eq.Hamilton-e1-in-L}.
\end{proof}

Now we shall generalize \eqref{eq.Hamilton-e1-in-L}
by showing that
\begin{align}
e_t \in L^t \qquad \text{ for each } t \geq 0.
\label{eq.Hamilton-et-in-Lt}
\end{align}

\begin{proof}[Proof of \eqref{eq.Hamilton-et-in-Lt}.]
If $t = 0$, then this is obvious (since $L^0 = R$).
If $t = 1$, then it follows from \eqref{eq.Hamilton-e1-in-L}.
Thus, assume that $t \geq 2$ henceforth.
Now, $e_t$ is defined as the sum of the $t$-wise products
of the $z_H$'s with $H \in \mathcal{H}_S$.
Each addend in this sum is, up to sign, a product of the
$A$-weights of $t$ distinct Hamilton cycles on $S$
(since the $z_H$'s are, up to sign, the $A$-weights of these
cycles).
But Lemma~\ref{lem.products-Hamilton-cycles} shows that
each such product belongs to $K_{<r}(A)^t$.
Therefore, their sum $e_t$ belongs to $K_{<r}(A)^t$ as well,
and thus also to $L^t$
(since $K_{<r}(A) = K \subseteq L$).
This proves \eqref{eq.Hamilton-et-in-Lt}.
\end{proof}

But Vi\`ete's formulas show that every $z_H$ is a root of
the monic polynomial
\[
\prod_{G\in\mathcal H_S}(X-z_G)
=X^{\left|\mathcal H_S\right|}-e_1X^{\left|\mathcal H_S\right|-1}
+e_2X^{\left|\mathcal H_S\right|-2}-\cdots
\in R\left[X\right],
\]
whose $X^{\left|\mathcal H_S\right|-t}$-coefficient belongs to $L^t$
(by \eqref{eq.Hamilton-et-in-Lt}).
Thus every $z_H$ is integral over $L$.
Therefore, every $w_A(H)$ is integral over $L$ as well (since
$z_H$ is $w_A(H)$ up to sign).
In other words,
\begin{align}
w_A(H) \in \overline L
\qquad \text{ for each }
H \in \mathcal{H}_S.
\label{eq.wAH-in-L1}
\end{align}

By the induction hypothesis, the $A$-weights of all nontrivial cycles of
length strictly smaller than $r$ belong to $\overline{J(A)}$. Since
$\overline{J(A)}$ is an ideal, this yields
\[
K\subseteq\overline{J(A)}
\]
(since $K = K_{<r}(A)$ is the ideal generated by these $A$-weights).
Hence
$L = J(A)+K \subseteq \overline{J(A)}$
(since $J(A) \subseteq \overline{J(A)}$ and $K\subseteq\overline{J(A)}$).
Thus,
\[
J(A) \subseteq L \subseteq \overline{J(A)}.
\]
Hence, Lemma~\ref{lem.integral-closure-basics} \textup{(d)} now
gives $\overline{L} = \overline{J(A)}$.
Thus, \eqref{eq.wAH-in-L1} rewrites as
\begin{align*}
w_A(H) \in \overline{J(A)}
\qquad \text{ for each }
H \in \mathcal{H}_S.
\end{align*}
In other words, each Hamilton cycle $H$ on $S$ satisfies
$w_A(H) \in \overline{J(A)}$.
This completes the induction.
\end{proof}

\begin{corollary}
\label{cor.principled-cycle-nilpotent}
Let $A$ be a $1$-principled $n\times n$-matrix over a
commutative ring $R$. Then the
$A$-weight of every nontrivial cycle on $\ive{n}$ is nilpotent.
\end{corollary}

\begin{proof}
In this case $J(A)=0$ (since $A$ is $1$-principled, so that all
generators of $J(A)$ are $0$).
Thus, Theorem~\ref{thm.cycle-weight-integral} shows that
every nontrivial cycle's $A$-weight is integral over the
zero ideal, and therefore nilpotent.
\end{proof}

\begin{corollary}
\label{cor.reduced-principled-strong}
Over a reduced commutative ring, every $1$-principled matrix is strongly
$1$-principled. Consequently, every power of a $1$-principled matrix over a
reduced ring is $1$-principled.
\end{corollary}

\begin{proof}
A reduced ring has no nonzero nilpotent elements, so the first claim follows
from Corollary~\ref{cor.principled-cycle-nilpotent}. The second then follows
from Corollary~\ref{cor.strong-powers-principled}.
\end{proof}



\section{Quotients by integrally closed ideals}

\subsection{The general case}

The universal theorem from the preceding section has the following immediate
consequence.

\begin{theorem}
\label{thm.integrally-closed-quotient-strong}
Let $D$ be a commutative ring, let $I$ be an integrally closed ideal of
$D$, and set $R:=D/I$. Let $A$ be a $1$-principled matrix over $R$. Then $A$
is strongly $1$-principled.
\end{theorem}

\begin{proof}
Choose a matrix $\widetilde A$ over $D$ whose image modulo $I$ is $A$. Since
$A$ is $1$-principled, we have
\[
\det \widetilde A_S-1\in I
\qquad\text{for every }S\subseteq\ive{n}.
\]
Thus $J(\widetilde A)\subseteq I$. By
Theorem~\ref{thm.cycle-weight-integral}, every nontrivial cycle
$C$ on $\ive{n}$ satisfies
$w_{\widetilde A}(C) \in \overline{J(\widetilde A)} \subseteq
\overline{I}$ (by Lemma~\ref{lem.integral-closure-basics} (b),
since $J(\widetilde A)\subseteq I$)
and therefore $w_{\widetilde A}(C) \in \overline{I} = I$
(since $I$ is integrally closed).
Reducing modulo $I$, we obtain $w_A\tup{C} = 0$
for every nontrivial cycle $C$. Also, the
diagonal entries of $A$ are $1$, since they are its $1\times1$ principal
minors. Hence $A$ is strongly $1$-principled.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm.main}]
By Theorem~\ref{thm.integrally-closed-quotient-strong}, the matrix $A$ is
strongly $1$-principled. Hence Corollary~\ref{cor.strong-powers-principled}
shows that $A^m$ is $1$-principled for every $m\in\mathbb N$.
\end{proof}

\subsection{\texorpdfstring{The rings $\mathbb Z/d$}{The rings Z/d}}

The case of the ring $\mathbb Z/d = \ZZ / d \ZZ$ admits a particularly
elementary application of Theorem~\ref{thm.cycle-weight-integral}.

\begin{lemma}
\label{lem.Z-ideals-integrally-closed}
Every ideal of $\mathbb Z$ is integrally closed.
\end{lemma}

\begin{proof}
Every ideal of $\mathbb Z$ has the form $d\mathbb Z$ for some
$d\in\mathbb N$. The claim is clear for $d=0$, since the only nilpotent
integer is $0$. Assume that $d>0$, and let $x\in\mathbb Z$ be integral over
$d\mathbb Z$. Thus, for some positive integer $r$, we have
\begin{align}
x^r+c_1x^{r-1}+\cdots+c_rx^0=0
\qquad \text{with }c_j\in \tup{d\ZZ}^j = d^j\mathbb Z.
\label{pf.lem.Z-ideals-integrally-closed.1}
\end{align}
Write $c_j=d^jb_j$ with $b_j\in\mathbb Z$, and put $y=x/d\in\mathbb Q$.
Dividing the equation \eqref{pf.lem.Z-ideals-integrally-closed.1}
by $d^r$ gives
\begin{align}
y^r+b_1y^{r-1}+\cdots+b_ry^0=0.
\label{pf.lem.Z-ideals-integrally-closed.2}
\end{align}
Thus $y$ is a rational root of a monic polynomial in $\mathbb Z[X]$. To see
directly that $y\in\mathbb Z$, write $y=a/b$ in lowest terms with $b>0$.
After multiplication by $b^r$, the equation
\eqref{pf.lem.Z-ideals-integrally-closed.2} shows that $b\mid a^r$.
Since $\gcd(a,b)=1$, this forces $b=1$. Hence $y\in\mathbb Z$, so
$x=dy\in d\mathbb Z$.
\end{proof}

\begin{corollary}
\label{cor.zmod-n}
Let $d$ be a positive integer, and let $A$ be a $1$-principled matrix over
$\mathbb Z/d\mathbb Z$. Then $A^m$ is $1$-principled for every
$m\in\mathbb N$.
\end{corollary}

\begin{proof}
The ideal $d \ZZ$ of $\ZZ$ is integrally closed by
Lemma~\ref{lem.Z-ideals-integrally-closed}.
Hence, Theorem~\ref{thm.main}
(applied to $D = \ZZ$ and $I = d\ZZ$)
shows that $A^m$ is $1$-principled for every $m\in\mathbb N$.
% Lift $A$ to an integer matrix $\widetilde A$. Since $A$ is $1$-principled,
% we have $J(\widetilde A)\subseteq d\mathbb Z$. By
% Theorem~\ref{thm.cycle-weight-integral}, every nontrivial cycle's $\widetilde A$-weight
% is integral over $J(\widetilde A)$, and hence, directly from
% the definition, integral over $d\mathbb Z$. Lemma~\ref{lem.Z-ideals-integrally-closed}
% therefore shows that every such nontrivial cycle's $\widetilde A$-weight belongs to $d\mathbb Z$.
% In other words, every nontrivial cycle's $A$-weight is $0$.
% That is, $A$ is strongly $1$-principled. Now apply
% Corollary~\ref{cor.strong-powers-principled}.
\end{proof}

\begin{remark}
Our above proof of Corollary~\ref{cor.zmod-n} is
self-contained.
Indeed, the only result we have used without proof is
Lemma~\ref{lem.integral-closure-basics}, which in this
case is being applied to $R = \ZZ$; but this
lemma is trivial when all ideals of $R$ are integrally
closed.
% Once a cycle's $\widetilde A$-weight is known to be
% integral over $J(\widetilde A)\subseteq d\mathbb Z$, the same equation already
% shows that it is integral over $d\mathbb Z$, and the elementary proof of
% Lemma~\ref{lem.Z-ideals-integrally-closed} finishes the argument.
\end{remark}

\subsection{Pr\"ufer domains}

We next give a broad class of domains for which every ideal is integrally
closed.

\begin{definition}
Let $D$ be an integral domain with fraction field $K$. A nonzero fractional
ideal $L$ of $D$ is called \emph{invertible} if there exists a fractional
ideal $M$ such that $LM=D$. The domain $D$ is called a \emph{Pr\"ufer domain}
if every nonzero finitely generated ideal of $D$ is invertible.
\end{definition}

For background on
Pr\"ufer domains and their equivalent characterizations, see
\cite[\S XII.3]{LomQui} and \cite{FontanaHuckabaPapick}.

\begin{lemma}
\label{lem.Prufer-ideals-integrally-closed}
Every ideal of a Pr\"ufer domain is integrally closed.
\end{lemma}

\begin{proof}
This is part of \cite[Theorem XII.3.2]{LomQui}, but we give
a proof for the sake of completeness.

Let $I$ be an ideal of a Pr\"ufer domain $D$, and let $x\in D$ be integral
over $I$. We must show that $x \in I$.

Choose an equation
\begin{align}
x^r+c_1x^{r-1}+\cdots+c_rx^0=0
\qquad \text{with } c_j\in I^j
\label{pf.Pruefer.eq1}
\end{align}
(since $x$ is integral over $I$).
Only finitely many elements of $I$ are needed to express all the $c_j$ as
sums of products of $j$ elements of $I$. Let $J\subseteq I$ be the finitely
generated ideal generated by these elements. Then $c_j\in J^j$ for every
$j$, so $x$ is integral over $J$.

Set $L:=J+xD$. If $J=0$, then \eqref{pf.Pruefer.eq1} gives $x^r=0$, whence
$x=0$ since $D$ is a domain; thus the claim is clear. Hence we may assume
that $J\neq0$. Then both $J$ and $L$ are nonzero finitely generated ideals,
and therefore invertible. Moreover, \eqref{pf.Pruefer.eq1} yields
\[
x^r\in JL^{r-1}.
\]
Any other product of $r$ generators of $L=J+xD$ already contains a factor
from $J$ and thus belongs to $JL^{r-1}$ as well.
Hence, $L^r \subseteq JL^{r-1}$. Since the opposite inclusion
is obvious, we thus have shown that
\[
L^r=JL^{r-1}.
\]
Since $L^{r-1}$ is invertible, we may cancel it and obtain $L=J$. In
particular, $x\in L=J\subseteq I$. Thus $I$ is integrally closed.
\end{proof}

\begin{corollary}
\label{cor.Prufer-quotient}
Let $D$ be a Pr\"ufer domain, let $I$ be an ideal of $D$, and let $A$
be a $1$-principled matrix over $D/I$. Then $A^m$ is $1$-principled for every
$m\in\mathbb N$.
\end{corollary}

\begin{proof}
Combine Lemma~\ref{lem.Prufer-ideals-integrally-closed} with
Theorem~\ref{thm.main}.
\end{proof}

\begin{remark}
Valuation domains are precisely the local Pr\"ufer domains
(see \cite[\S XII.3]{LomQui}). Thus quotients of
valuation domains are included in Corollary~\ref{cor.Prufer-quotient}.
\end{remark}

\subsection{Normal domains}

Recall that a normal domain is a domain that is integrally closed in its
fraction field. Every principal ideal of a normal domain is integrally closed;
see \cite[Proposition 1.5.2]{HunekeSwanson}. Therefore
Theorem~\ref{thm.main} yields the following.

\begin{corollary}
\label{cor.normal-principal-quotient}
Let $D$ be a normal domain, let $f\in D$ be a nonzero nonunit, and let $A$ be a
$1$-principled matrix over $D/fD$. Then $A^m$ is $1$-principled for every
$m\in\mathbb N$.
\end{corollary}

\begin{remark}
The hypothesis that the kernel ideal be integrally closed is the exact input
needed by Theorem~\ref{thm.integrally-closed-quotient-strong}. One does not
need every ideal of the ambient ring to be integrally closed. Thus the normal
domain corollary applies to principal quotients even though arbitrary ideals
of a normal domain need not be integrally closed.
\end{remark}

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\end{document}