\documentclass[12pt]{article}%
\usepackage[margin=1.2in]{geometry}
\usepackage{hyperref}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{comment}
\usepackage{amsfonts}
\usepackage{graphicx}%
\setcounter{MaxMatrixCols}{30}
%TCIDATA{OutputFilter=latex2.dll}
%TCIDATA{Version=5.50.0.2960}
%TCIDATA{LastRevised=Sunday, July 12, 2026 15:23:44}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%BeginMSIPreambleData
\providecommand{\U}[1]{\protect\rule{.1in}{.1in}}
%EndMSIPreambleData
\newtheoremstyle{plainsl}
{8pt plus 2pt minus 4pt}
{8pt plus 2pt minus 4pt}
{\slshape}
{0pt}
{\bfseries}
{.}
{5pt plus 1pt minus 1pt}
{}
\theoremstyle{plainsl}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\newenvironment{noncompile}{}{}
\excludecomment{noncompile}
\newcommand{\NN}{\mathbb N}
\newcommand{\cont}{\operatorname{cont}}
\newcommand{\coeff}{\operatorname{coeff}}
\let\sumnonlimits\sum
\let\prodnonlimits\prod
\let\cupnonlimits\bigcup
\let\capnonlimits\bigcap
\renewcommand{\sum}{\sumnonlimits\limits}
\renewcommand{\prod}{\prodnonlimits\limits}
\renewcommand{\bigcup}{\cupnonlimits\limits}
\renewcommand{\bigcap}{\capnonlimits\limits}
\iffalse
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\fi
\begin{document}

\title{A Universal Noncommutative Splitting Algebra for Polynomials}
\author{GPT-5.5, edited by Darij Grinberg}
\date{\today}
\maketitle

\begin{abstract}
\textbf{Abstract.} Let $R$ be a commutative ring. We show that every
homogeneous polynomial $f\in R[x_{1},\ldots,x_{m}]$ of degree $n$ splits into
a product of $n$ homogeneous linear forms over a suitable noncommutative ring
extension $S$ of $R$ (that is, over a noncommutative $R$-algebra $S$ whose
structure morphism $R\rightarrow S$ is injective). The algebra $S$ is
universal for such factorizations and is free as an $R$-module.

The proof uses Bergman's Diamond Lemma: we define $S$ by generators and
relations, and the relations form a terminating reduction system that has no
ambiguities and thus is confluent.

An analogous result is also shown for inhomogeneous polynomials (with
inhomogeneous factors). This easily follows from the homogeneous case by
homogenizing and then setting the homogenizing variable equal to $1$. \medskip

\textbf{Manifest.} This note was written by GPT-5.5 in response to my
prompting it with MathOverflow question \#27675. There is nothing deep or
difficult here; GPT's main contribution was abstracting away from the
unnecessary specifics of the question. -- DG\footnote{This work is in the
public domain.}

\end{abstract}

\section{Introduction}

If $f(t)\in R[t]$ is a univariate polynomial over a commutative ring $R$, then
one can construct a commutative $R$-algebra $S$ whose structure map
$R\rightarrow S$ is injective (i.e., the ring $R$ canonically embeds into $S$)
and such that $f(t)$ splits into linear factors in $S[t]$. This fact is
well-known when $f\left(  t\right)  $ is monic (in which case we can obtain
$S$ by successively adjoining roots of $f\left(  t\right)  $ one by one using
the classical $R\left[  t\right]  /\left(  f\left(  t\right)  \right)  $
construction), and has been proved in \cite[Theorem 2.1]{GrinbergFactorpoly}
in the general case. This can be viewed as an analogue of splitting fields for
arbitrary commutative rings (even though this construction does not always
produce a field $S$ when $R$ is a field).

This fact cannot be extended to arbitrary multivariate polynomials $f\left(
x_{1},x_{2},\ldots,x_{m}\right)  $ while keeping $S$ commutative. The reason
is that a multivariate form which is a product of linear forms satisfies
polynomial equations in its coefficients. Already for quadratic forms, a
product of two linear forms (over a ring in which $2$ is invertible) has a
vanishing catalecticant\footnote{Explicitly: If a quadratic form
$ax^{2}+by^{2}+cz^{2}+2dyz+2ezx+2fxy\in R\left[  x,y,z\right]  $ factors as a
product $\left(  \alpha x+\beta y+\gamma z\right)  \left(  \alpha^{\prime
}x+\beta^{\prime}y+\gamma^{\prime}z\right)  $ over a commutative ring, then
$\det\left(
\begin{array}
[c]{ccc}%
a & f & e\\
f & b & d\\
e & d & c
\end{array}
\right)  =0$.}, while a generic quadratic form does not. Thus a general
quadratic form cannot be made into a product of linear forms merely by
enlarging the commutative coefficient ring.

In the present note, we show that by allowing $S$ to be noncommutative, we can
remove all such obstructions and recover a multivariate version of the
splitting algebra $S$. The construction of this algebra $S$ is the usual
\textquotedblleft make a wish\textquotedblright\ universal construction: The
variables $x_{i}$ remain central, but the coefficients of the desired linear
factors need not commute with each other. When $f$ is homogeneous of degree
$n$, we introduce independent coefficients
\[
a_{i}^{(k)}\qquad\left(  1\leq i\leq m,\ 1\leq k\leq n\right)
\]
for the $k$-th linear factor and impose the coefficient identities forced by
\[
f=\prod_{k=1}^{n}\left(  \sum_{i=1}^{m}x_{i}a_{i}^{(k)}\right)  .
\]
This gives a universal noncommutative coefficient algebra. We show that this
algebra is free as an $R$-module; hence the map from $R$ into it is injective.

The freeness proof is the most nontrivial step, but even that is
straightforward from the right viewpoint. We regard the defining relations as
a reduction system in the sense of Bergman's Diamond Lemma
\cite{BergmanDiamond}. For each coefficient relation, we choose a monic
leading word. Every leading word has the same superscript pattern
$1,2,\ldots,n$. Consequently, a proper suffix of one leading word cannot be a
proper prefix of another, and one leading word cannot properly contain
another. Thus there are no overlap ambiguities and no inclusion ambiguities,
in Bergman's terminology. The irreducible words therefore form an $R$-basis.

After proving these results, we derive an inhomogeneous version, in which $f$
is no longer required to be homogeneous (and its factors now are allowed to
have constant terms). This follows easily from the homogeneous case:
homogenize $f$ by adding a new variable $x_{0}$, apply the homogeneous
theorem, and then set $x_{0}=1$. The final section records the generic
determinant as a special case, answering
\href{https://mathoverflow.net/questions/27675/splitting-the-determinant-polynomial-into-linear-factors-a-dedekind-problem}{MathOverflow
question \#27675}.

\section{The homogeneous construction}

Let $R$ be a commutative ring, and let $m,n\in\mathbb{N}$, where $\mathbb{N}$
is the set of all nonnegative integers. Assume that $n$ is positive. Let
\[
f\in R[x_{1},x_{2},\ldots,x_{m}]
\]
be a homogeneous polynomial of degree $n$. Write $f$ as
\[
f=\sum_{\substack{\alpha\in\mathbb{N}^{m};\\|\alpha|=n}}c_{\alpha}x^{\alpha},
\]
where for each $m$-tuple $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{m}\right)  \in\mathbb{N}^{m}$ we set
\[
|\alpha|=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{m},\qquad x^{\alpha}%
=x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}\cdots x_{m}^{\alpha_{m}},
\]
and where $c_{\alpha}=[x^{\alpha}]f\in R$ are the coefficients of $f$.

Let
\[
F=R\left\langle a_{i}^{(k)}\mid1\leq i\leq m,\ 1\leq k\leq n\right\rangle
\]
be the free associative unital $R$-algebra on the $mn$ generators
$a_{i}^{\left(  k\right)  }$ with $1\leq i\leq m$ and $1\leq k\leq n$. As an
$R$-module, $F$ is free, with a basis consisting of all \emph{words} (i.e.,
noncommutative monomials) in these $mn$ generators.

We shall construct a quotient $R$-algebra of $F$ over which the polynomial $f$
factors as $\prod_{k=1}^{n}\left(  \sum_{i=1}^{m}x_{i}a_{i}^{(k)}\right)  $.
(So the parenthesized superscripts $\left(  k\right)  $ in $a_{i}^{\left(
k\right)  }$ distinguish the factors of this factorization.)

For an $n$-tuple
\[
\mathbf{i}=(i_{1},i_{2},\ldots,i_{n})\in\{1,2,\ldots,m\}^{n},
\]
define its \emph{content} $\operatorname{cont}(\mathbf{i})\in\mathbb{N}^{m}$
to be the $m$-tuple whose $j$-th entry (for each $1\leq j\leq m$) is%
\[
\operatorname{cont}(\mathbf{i})_{j}=\#\{k\mid i_{k}=j\}
\]
(so that $x^{\operatorname*{cont}\left(  \mathbf{i}\right)  }=x_{i_{1}%
}x_{i_{2}}\cdots x_{i_{n}}$), and we define the \emph{layered word}
\[
w(\mathbf{i})=a_{i_{1}}^{(1)}a_{i_{2}}^{(2)}\cdots a_{i_{n}}^{(n)}%
\qquad\left(  \text{a word in }F\right)  .
\]
For instance, if $m=3$ and $n=4$ and $\mathbf{i}=\left(  1,3,2,3\right)  $,
then $\operatorname*{cont}\left(  \mathbf{i}\right)  =\left(  1,1,2\right)  $
and $w\left(  \mathbf{i}\right)  =a_{1}^{\left(  1\right)  }a_{3}^{\left(
2\right)  }a_{2}^{\left(  3\right)  }a_{3}^{\left(  4\right)  }$.

For each $\alpha\in\mathbb{N}^{m}$ with $|\alpha|=n$, set
\begin{equation}
r_{\alpha}=\sum_{\substack{\mathbf{i}\in\{1,2,\ldots,m\}^{n}%
;\\\operatorname{cont}(\mathbf{i})=\alpha}}w(\mathbf{i})-c_{\alpha}.
\label{eq.ral=}%
\end{equation}
For example, if $m=3$ and $n=3$, then%
\begin{align*}
r_{\left(  2,0,1\right)  }  &  =w\left(  \left(  1,1,3\right)  \right)
+w\left(  \left(  1,3,1\right)  \right)  +w\left(  \left(  3,1,1\right)
\right)  -c_{\left(  2,0,1\right)  }\\
&  =a_{1}^{\left(  1\right)  }a_{1}^{\left(  2\right)  }a_{3}^{\left(
3\right)  }+a_{1}^{\left(  1\right)  }a_{3}^{\left(  2\right)  }a_{1}^{\left(
3\right)  }+a_{3}^{\left(  1\right)  }a_{1}^{\left(  2\right)  }a_{1}^{\left(
3\right)  }-c_{\left(  2,0,1\right)  }.
\end{align*}


Finally define the quotient $R$-algebra%
\[
S_{f}=F/\underbrace{\left(  r_{\alpha}\mid\alpha\in\mathbb{N}^{m}%
,\ |\alpha|=n\right)  }_{\text{ideal of }F\text{ generated by all these
}r_{\alpha}}\ \ \ \ \ \ \ \ \ \ \text{of }F.
\]


Consider the polynomial ring $S_{f}[x_{1},x_{2},\ldots,x_{m}]$ in $m$
indeterminates $x_{1},x_{2},\ldots,x_{m}$ over $S_{f}$. (These indeterminates
are supposed to commute with everything, even though $S_{f}$ is not
commutative.) Then, by construction, in $S_{f}[x_{1},x_{2},\ldots,x_{m}]$ we
have\footnote{Noncommutative products of the form $\prod_{k=1}^{n}u_{k}$ are
always understood to mean $u_{1}u_{2}\cdots u_{n}$.}%
\begin{equation}
f=\prod_{k=1}^{n}\left(  \sum_{i=1}^{m}x_{i}a_{i}^{(k)}\right)  .
\label{eq.f=prod-in-Sf}%
\end{equation}
Indeed, the right hand side of this equality rewrites as follows:%
\begin{align*}
\prod_{k=1}^{n}\left(  \sum_{i=1}^{m}x_{i}a_{i}^{(k)}\right)   &
=\sum_{\mathbf{i}=(i_{1},i_{2},\ldots,i_{n})\in\{1,2,\ldots,m\}^{n}%
}\underbrace{x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}}_{=x^{\operatorname*{cont}%
\left(  \mathbf{i}\right)  }}\underbrace{a_{i_{1}}^{\left(  1\right)
}a_{i_{2}}^{\left(  2\right)  }\cdots a_{i_{n}}^{\left(  n\right)  }%
}_{=w\left(  \mathbf{i}\right)  }\\
&  =\sum_{\mathbf{i}\in\{1,2,\ldots,m\}^{n}}x^{\operatorname*{cont}\left(
\mathbf{i}\right)  }w\left(  \mathbf{i}\right)  =\sum_{\substack{\alpha
\in\mathbb{N}^{m};\\|\alpha|=n}}x^{\alpha}\underbrace{\sum
_{\substack{\mathbf{i}\in\{1,2,\ldots,m\}^{n};\\\operatorname{cont}%
(\mathbf{i})=\alpha}}w(\mathbf{i})}_{\substack{=r_{\alpha}+c_{\alpha
}\\\text{(by \eqref{eq.ral=})}}}\\
&  =\sum_{\substack{\alpha\in\mathbb{N}^{m};\\|\alpha|=n}}x^{\alpha
}\underbrace{\left(  r_{\alpha}+c_{\alpha}\right)  }_{\substack{=c_{\alpha
}\text{ in }S_{f}\\\text{(since we factored out }r_{\alpha}\text{ to get
}S_{f}\text{)}}}=\sum_{\substack{\alpha\in\mathbb{N}^{m};\\|\alpha
|=n}}x^{\alpha}c_{\alpha}=\sum_{\substack{\alpha\in\mathbb{N}^{m}%
;\\|\alpha|=n}}c_{\alpha}x^{\alpha}=f.
\end{align*}


The construction of $S_{f}$ is universal in the evident sense:

\begin{proposition}
[Universal property of $S_{f}$]\label{prop.Sf-univers}Let $T$ be any
associative unital $R$-algebra with elements $b_{i}^{(k)}\in T$ satisfying
\begin{equation}
f=\prod_{k=1}^{n}\left(  \sum_{i=1}^{m}x_{i}b_{i}^{(k)}\right)  \quad\text{in
}T[x_{1},x_{2},\ldots,x_{m}]. \label{eq.f=prod-in-T}%
\end{equation}
Then there is a unique $R$-algebra homomorphism $S_{f}\rightarrow T$ sending
$a_{i}^{(k)}$ to $b_{i}^{(k)}$.
\end{proposition}

\begin{proof}
By running the above proof of (\ref{eq.f=prod-in-Sf}) in reverse, we can see
that the equality (\ref{eq.f=prod-in-T}) is equivalent to having all the
coefficientwise equalities
\[
\sum_{\substack{\mathbf{i}\in\{1,2,\ldots,m\}^{n};\\\operatorname{cont}%
(\mathbf{i})=\alpha}}w_{T}(\mathbf{i})=c_{\alpha}\ \ \ \ \ \ \ \ \ \ \text{for
all }\alpha\in\mathbb{N}^{m}\text{ satisfying }\left\vert \alpha\right\vert
=n
\]
hold in $T$, where $w_{T}(\mathbf{i})=b_{i_{1}}^{(1)}b_{i_{2}}^{(2)}\cdots
b_{i_{n}}^{(n)}$ for any $n$-tuple $\mathbf{i}=(i_{1},i_{2},\ldots,i_{n}%
)\in\{1,2,\ldots,m\}^{n}$. But the latter coefficientwise equalities are
simply saying that the $R$-algebra morphism $F\rightarrow T$ that sends each
$a_{i}^{\left(  k\right)  }$ to $b_{i}^{\left(  k\right)  }$ annihilates all
the $r_{\alpha}\in F$. And this, in turn, means that the morphism
$F\rightarrow T$ factors through the quotient algebra $F/\left(  r_{\alpha
}\mid\alpha\in\mathbb{N}^{m},\ |\alpha|=n\right)  =S_{f}$.
\end{proof}

\section{The reduction system}

Our goal in this little note is to prove that the structure map of the
$R$-algebra $S_{f}$ (that is, the $R$-algebra morphism $R\rightarrow
S_{f},\ r\mapsto r\cdot1$) is injective. This will allow us to view $S_{f}$ as
a ring extension of $R$ (that is, as a ring that contains $R$ as a subring).

We shall use Bergman's terminology for reduction systems in free associative
algebras \cite[Section~1]{BergmanDiamond}. Thus a \emph{reduction system} is a
set of pairs $\left(  W,g\right)  $, where $W$ is a word and $g$ is a linear
combination of smaller words (with respect to some order that we shall specify
below). Each of these pairs $\left(  W,g\right)  $ is called a \emph{reduction
rule}\footnote{Bergman does not use this terminology; instead he writes $S$
for the set of all these pairs $\left(  W,g\right)  $. (Of course, this is not
our $S$.)}, and is written as $W\longrightarrow g$; we refer to $W$ as its
\emph{leading word} and to $g$ as its \emph{tail}. The possible conflicts
between two such rules are the \emph{overlap ambiguities} and \emph{inclusion
ambiguities} defined in \cite[Section~1]{BergmanDiamond}; they happen when the
leading words $W$ and $W^{\prime}$ of two distinct reduction rules $\left(
W,g\right)  $ and $\left(  W^{\prime},g^{\prime}\right)  $ have a
prefix-suffix collision (i.e., a nonempty proper suffix of $W$ equals a
nonempty proper prefix of $W^{\prime}$) or when one of them is a contiguous
subword of the other. Given a reduction system in a free algebra, the
\emph{difference ideal} of this system means the ideal generated by the $W-g$
for all reduction rules $\left(  W,g\right)  $, and the \emph{difference
quotient} of this system means the quotient of the free algebra by this ideal.
A word $w$ is said to be \emph{irreducible} with respect to the reduction
system if none of the leading words of the system appears as a contiguous
subword of $w$. The crux of \cite[\S 1]{BergmanDiamond} is a sufficient
criterion (\cite[Theorem 1.2 (a) $\Longrightarrow$ (c)]{BergmanDiamond}) for
when the irreducible words (or, more precisely, their canonical projections)
form a basis of the difference quotient. Namely, this happens whenever

\begin{enumerate}
\item there is a \emph{semigroup partial ordering} consistent with the
reduction system and satisfying the descending chain condition, and

\item all ambiguities of the reduction system are \emph{resolvable}.
\end{enumerate}

\noindent We refer to \cite[\S 1]{BergmanDiamond} for the meanings of these
concepts. Note that if no ambiguities exist, then all ambiguities are
resolvable (for vacuous reasons); such situations are particularly simple (and
we are lucky enough that we will end up in precisely such a situation).

Let us apply this in our free algebra $F$. Our alphabet $\mathfrak{A}$
consists of all the $mn$ symbols $a_{i}^{\left(  k\right)  }$ with $1\leq
i\leq m$ and $1\leq k\leq n$. We order words over this alphabet by
degree-lexicographic order (i.e., shorter words are always smaller than longer
ones, but words of equal length are compared lexicographically), where the
lexicographic order is such that%
\begin{align*}
&  \ a_{1}^{\left(  1\right)  }<a_{1}^{\left(  2\right)  }<\cdots
<a_{1}^{\left(  n\right)  }\\
<  &  \ a_{2}^{\left(  1\right)  }<a_{2}^{\left(  2\right)  }<\cdots
<a_{2}^{\left(  n\right)  }\\
<  &  \ \cdots\\
<  &  \ a_{m}^{\left(  1\right)  }<a_{m}^{\left(  2\right)  }<\cdots
<a_{m}^{\left(  n\right)  }.
\end{align*}
\footnote{Actually, many other choices of order on the indeterminates would
work equally well; all we need is that the order relation between two
indeterminates $a_{i}^{\left(  k\right)  }$ and $a_{j}^{\left(  \ell\right)
}$ with $i\neq j$ depends only on $i$ and $j$ and not on $k$ and $\ell$. We
chose the above ordering merely for its familiarity.} This total order on
words is a semigroup partial ordering (i.e., if $w\leq w^{\prime}$, then
$uwv\leq uwv^{\prime}$ for any words $w,w^{\prime},u,v$), and satisfies the
descending chain condition. Thus, Bergman's \cite[Theorem 1.2 (a)
$\Longrightarrow$ (c)]{BergmanDiamond} applies to any reduction system that is
consistent with it and has only resolvable ambiguities.

For each $m$-tuple $\alpha\in\mathbb{N}^{m}$ with $|\alpha|=n$, let%
\[
\mathbf{i}^{\alpha}=(i_{1}^{\alpha},i_{2}^{\alpha},\ldots,i_{n}^{\alpha})
\]
be the lexicographically largest of all $n$-tuples $\mathbf{i}\in\left\{
1,2,\ldots,m\right\}  ^{n}$ satisfying $\operatorname{cont}(\mathbf{i}%
)=\alpha$. Equivalently, the sequence $i_{1}^{\alpha},i_{2}^{\alpha}%
,\ldots,i_{n}^{\alpha}$ is weakly decreasing with respect to our chosen order
and satisfies $\operatorname{cont}(\mathbf{i}^{\alpha})=\alpha$. (This is
equivalent because the $n$-tuples $\mathbf{i}\in\left\{  1,2,\ldots,m\right\}
^{n}$ satisfying $\operatorname{cont}(\mathbf{i})=\alpha$ form an orbit under
the permutation action of the symmetric group $S_{n}$, and the
lexicographically largest among them is the one whose entries weakly
decrease.) Define the word%
\[
W_{\alpha}=w(\mathbf{i}^{\alpha})=a_{i_{1}^{\alpha}}^{(1)}a_{i_{2}^{\alpha}%
}^{(2)}\cdots a_{i_{n}^{\alpha}}^{(n)}.
\]
Then,%
\[
r_{\alpha}=\sum_{\substack{\mathbf{i}\in\{1,2,\ldots,m\}^{n}%
;\\\operatorname{cont}(\mathbf{i})=\alpha}}w(\mathbf{i})-c_{\alpha}=W_{\alpha
}+\sum_{\substack{\mathbf{i}\in\{1,2,\ldots,m\}^{n};\\\operatorname{cont}%
(\mathbf{i})=\alpha;\\\mathbf{i}\neq\mathbf{i}^{\alpha}}}w(\mathbf{i}%
)-c_{\alpha}%
\]
(here, we have split off the addend $w\left(  \mathbf{i}^{\alpha}\right)
=W_{\alpha}$ from the sum). Thus, the relation $r_{\alpha}=0$ gives the
reduction rule
\[
W_{\alpha}\longrightarrow c_{\alpha}-\sum_{\substack{\mathbf{i}\in
\{1,2,\ldots,m\}^{n};\\\operatorname{cont}(\mathbf{i})=\alpha;\\\mathbf{i}%
\neq\mathbf{i}^{\alpha}}}w(\mathbf{i}).
\]
This is a reduction rule over $R$. Its leading word is $W_{\alpha}$, whereas
its tail is an $R$-linear combination of words that are strictly smaller than
$W_{\alpha}$ in the chosen order (indeed, each $w\left(  \mathbf{i}\right)  $
is lexicographically smaller than $W_{\alpha}$ while having the same length as
$W_{\alpha}$, whereas the constant term $c_{\alpha}$ is smaller by
degree\footnote{NB: Here we are using $n>0$.}). Thus, if we consider the
reduction system that consists of all these reduction rules (for all
$\alpha\in\mathbb{N}^{m}$ satisfying $\left\vert \alpha\right\vert =n$), then
our chosen order on words is consistent with this reduction system.

The difference ideal of this reduction system is the ideal generated by all
the differences
\[
W_{\alpha}-\left(  c_{\alpha}-\sum_{\substack{\mathbf{i}\in\{1,2,\ldots
,m\}^{n};\\\operatorname{cont}(\mathbf{i})=\alpha;\\\mathbf{i}\neq
\mathbf{i}^{\alpha}}}w(\mathbf{i})\right)  =W_{\alpha}+\sum
_{\substack{\mathbf{i}\in\{1,2,\ldots,m\}^{n};\\\operatorname{cont}%
(\mathbf{i})=\alpha;\\\mathbf{i}\neq\mathbf{i}^{\alpha}}}w(\mathbf{i}%
)-c_{\alpha}=r_{\alpha}.
\]
Hence, the difference quotient is precisely $F/\left(  r_{\alpha}\mid\alpha
\in\mathbb{N}^{m},\ |\alpha|=n\right)  =S_{f}$. Crucially, our reduction
system has the following nice property, which will allow us to obtain an
$R$-module basis for $S_{f}$ from it using \cite[Theorem 1.2 (a)
$\Longrightarrow$ (c)]{BergmanDiamond}:

\begin{lemma}
\label{lem.no}Our reduction system has no ambiguities.
\end{lemma}

\begin{proof}
Each leading word $W_{\alpha}$ has superscript pattern $1,2,\ldots,n$ (meaning
that it has the form $a_{i_{1}}^{\left(  1\right)  }a_{i_{2}}^{\left(
2\right)  }\cdots a_{i_{n}}^{\left(  n\right)  }$ for some $i_{1},i_{2}%
,\ldots,i_{n}$).

An overlap ambiguity would require a nonempty proper suffix of a leading word
$W_{\alpha}$ to equal a nonempty proper prefix of a leading word $W_{\beta}$.
But such an equality is impossible, since the former suffix would begin with a
letter with superscript $r\geq2$ (where $r$ is the position of $W_{\alpha}$ at
which it starts), while the latter prefix would begin with a letter with
superscript $1$ (the first letter of $W_{\beta}$). Thus, there are no overlap ambiguities.

Inclusion ambiguities don't exist either. Indeed, an inclusion ambiguity would
require one of our leading words $W_{\alpha}$ to be a contiguous subword of
another. Since all our leading words have the same length, this could only
happen if the leading words were equal; but this cannot happen (because
$W_{\alpha}=W_{\beta}$ is easily seen to imply $\alpha=\beta$).

Thus, our reduction system has no ambiguities.
\end{proof}

Lemma \ref{lem.no} shows that all ambiguities of our reduction system are
resolvable (since there are none). Hence, Bergman's Diamond Lemma
\cite[Theorem~1.2 (c)]{BergmanDiamond} applies to it, and shows that the
irreducible words (i.e. the words containing no $W_{\alpha}$ as a contiguous
subword) form an $R$-basis of the quotient $R$-module $S_{f}$. We summarize:

\begin{theorem}
\label{thm:homogeneous-free} The algebra $S_{f}$ is a free $R$-module. More
precisely, an $R$-basis of $S_{f}$ is given by the images of all words in the
alphabet $a_{i}^{(k)}$ which do not contain any of the leading words
$W_{\alpha}$ as a (contiguous) subword.
\end{theorem}

As a consequence:

\begin{theorem}
\label{thm:homogeneous} The natural map $R\rightarrow S_{f}$ is injective.
Consequently, the natural map
\begin{equation}
R[x_{1},x_{2},\ldots,x_{m}]\longrightarrow S_{f}[x_{1},x_{2},\ldots,x_{m}]
\label{eq.thm:homogeneous.map}%
\end{equation}
is injective, where the variables $x_{i}$ are central over $S_{f}$.
\end{theorem}

\begin{proof}
The empty word is irreducible, and hence its image is one of the basis vectors
of the $R$-module basis of $S_{f}$ found in Theorem \ref{thm:homogeneous-free}%
. Thus no nonzero element of $R$ can vanish in $S_{f}$. Moreover, the basis of
$S_{f}$ found in Theorem \ref{thm:homogeneous-free} gives an $R$-linear
projection $S_{f}\rightarrow R$ sending the empty word to $1$ and all other
irreducible words to $0$. Extending this projection coefficientwise to
polynomials gives an $R[x_{1},x_{2},\ldots,x_{m}]$-linear projection
$S_{f}[x_{1},x_{2},\ldots,x_{m}]\rightarrow R[x_{1},x_{2},\ldots,x_{m}]$ that
splits the map (\ref{eq.thm:homogeneous.map}). Hence the latter map is injective.
\end{proof}

Thus every homogeneous degree-$n$ polynomial over $R$ splits into $n$
homogeneous linear factors after an injective extension of the coefficient
ring to a noncommutative $R$-algebra.

\section{The inhomogeneous case}

Now we turn to a slightly more general situation. Let
\[
f\in R[x_{1},x_{2},\ldots,x_{m}]
\]
be any polynomial of degree at most $n$ (not necessarily homogeneous). Write
it in the form%
\[
f=\sum_{\substack{\beta\in\mathbb{N}^{m};\\|\beta|\leq n}}c_{\beta}x^{\beta}.
\]
Define the degree-$n$ homogenization of $f$ to be the polynomial%
\[
F(x_{0},x_{1},\ldots,x_{m}):=\sum_{\substack{\beta\in\mathbb{N}^{m}%
;\\|\beta|\leq n}}c_{\beta}x_{0}^{n-|\beta|}x^{\beta}\in R[x_{0},x_{1}%
,\ldots,x_{m}].
\]
This is homogeneous of degree $n$. Applying the homogeneous construction to
$F$, we obtain a noncommutative $R$-algebra $S_{F}$, free as an $R$-module,
and a factorization
\begin{equation}
F(x_{0},x_{1},\ldots,x_{m})=\prod_{k=1}^{n}\left(  \sum_{i=0}^{m}x_{i}%
a_{i}^{(k)}\right)  \label{eq.inhg.homogenized}%
\end{equation}
in $S_{F}[x_{0},x_{1},\ldots,x_{m}]$, where the variables $x_{0},x_{1}%
,\ldots,x_{m}$ are central.

Now set $x_{0}=1$ (that is, apply the canonical left-$S_{F}$-linear
$R$-algebra morphism $S_{F}[x_{0},x_{1},\ldots,x_{m}]\rightarrow S_{F}%
[x_{1},x_{2},\ldots,x_{m}]$ that sends each monomial $x_{0}^{i_{0}}%
x_{1}^{i_{1}}\cdots x_{m}^{i_{m}}$ to $x_{1}^{i_{1}}x_{2}^{i_{2}}\cdots
x_{m}^{i_{m}}$) to (\ref{eq.inhg.homogenized}). Thus,
(\ref{eq.inhg.homogenized}) transforms into%
\begin{equation}
f(x_{1},x_{2},\ldots,x_{m})=\prod_{k=1}^{n}\left(  a_{0}^{(k)}+\sum_{i=1}%
^{m}x_{i}a_{i}^{(k)}\right)  \label{eq.inhg.dehomogenized}%
\end{equation}
in $S_{F}[x_{1},x_{2},\ldots,x_{m}]$. This shows that the polynomial $f$
splits into $n$ affine-linear factors over the ring $S_{F}$, which is an
injective noncommutative ring extension of $R$. So we have shown the following:

\begin{theorem}
\label{thm:inhomogeneous} Let $R$ be a commutative ring and let $f\in
R[x_{1},x_{2},\ldots,x_{m}]$ be a polynomial of degree at most $n$, where $n$
is a positive integer. Then there exists an associative unital $R$-algebra
$S$, free as an $R$-module, such that
\[
R[x_{1},x_{2},\ldots,x_{m}]\hookrightarrow S[x_{1},x_{2},\ldots,x_{m}]
\]
injectively and such that $f$ splits in $S[x_{1},x_{2},\ldots,x_{m}]$ as a
product of $n$ affine-linear factors.
\end{theorem}

\begin{proof}
Take $S=S_{F}$, where $F$ is the degree-$n$ homogenization of $f$. Then,
(\ref{eq.inhg.dehomogenized}) gives the factorization. Since $S_{F}$ is free
as an $R$-module (by Theorem \ref{thm:homogeneous-free}), the same
irreducible-word projection used in Theorem \ref{thm:homogeneous} shows that
$R[x_{1},x_{2},\ldots,x_{m}]\rightarrow S_{F}[x_{1},x_{2},\ldots,x_{m}]$ is injective.
\end{proof}

\begin{remark}
If $f\neq0$, one may take $n=\deg f$ in Theorem \ref{thm:inhomogeneous}. If
$f=0$, then any positive $n$ may be chosen.

Theorem \ref{thm:homogeneous} can be recovered from Theorem
\ref{thm:inhomogeneous} by comparing the degree-$n$ components of both sides.
(Per se, Theorem \ref{thm:inhomogeneous} only gives a factorization into
affine-linear factors; but projecting down to the $n$-th degree component will
make all the constant terms go away.)
\end{remark}

\section{The determinant as a special case}

Applying Theorem \ref{thm:homogeneous} to $R=\mathbb{Z}$, $m=n^{2}$, and
$f=\det\left(  (x_{ij})_{1\leq i,j\leq n}\right)  $, we obtain a universal
splitting algebra for the generic determinant polynomial $\det\left(
(x_{ij})_{1\leq i,j\leq n}\right)  $. The variables are indexed by the matrix
positions $(i,j)$, and the coefficients $c_{\alpha}$ are $0,1,-1$: they are
nonzero precisely when $\alpha$ is the characteristic function of the graph of
a permutation $\sigma\in S_{n}$, in which case $c_{\alpha}=\operatorname{sgn}%
(\sigma)$.

The proof above shows that the corresponding algebra $S_{f}=S_{\det}$ is a
free $\mathbb{Z}$-module, and hence that
\[
\mathbb{Z}[x_{ij}\mid1\leq i,j\leq n]\hookrightarrow S_{\det}[x_{ij}\mid1\leq
i,j\leq n]
\]
is injective. This answers
\href{https://mathoverflow.net/questions/27675/splitting-the-determinant-polynomial-into-linear-factors-a-dedekind-problem}{MathOverflow
question \#27675} in the positive.

\begin{thebibliography}{9}                                                                                                %


\bibitem {BergmanDiamond}%
\href{https://doi.org/10.1016/0001-8708(78)90010-5}{George M. Bergman,
\emph{The Diamond Lemma for Ring Theory}, Advances in Mathematics \textbf{29}
(1978), 178--218.}\newline See
\url{https://math.berkeley.edu/~gbergman/papers/updates/diamond.html} for errata.

\bibitem {GrinbergFactorpoly}GPT-5.5 with prompting by Darij Grinberg,
\emph{Splitting a univariate polynomial over a commutative extension of the
coefficient ring}, 4 July 2026, \url{https://www.cip.ifi.lmu.de/~grinberg/algebra/factorpoly-gpt.pdf}.
\end{thebibliography}


\end{document}