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    \begin{document} 

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    \title{Factorization of Monic Polynomials} 
    \author{William J. Heinzer} 
    \address{Mathematics Department, Purdue University, West Lafayette, 
    IN 47907-1395} 
    %\curraddr{Mathematics Department, Purdue University, West Lafayette, 
    %IN 47907-1395} 
    \email{heinzer@math.purdue.edu} 

    \author{David C. Lantz} 
    \address{Mathematics Department, Colgate University, Hamilton, 
    NY 13346-1398} 
    %\curraddr{Mathematics Department, Colgate University, Hamilton, 
    %NY 13346-1398} 
    \email{dlantz@mail.colgate.edu} 
    %    \thanks will become a 1st page footnote. 
    \thanks{The second author is grateful for the hospitality and 
    support of Purdue University while this work was done.} 

    %    General info 
    \subjclass{Primary 13B25, 13G05, 13J1. } 

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    \keywords{Hensel's Lemma, monic polynomial, comaximal ideals, 
    H-prime, integral upper} 



    \begin{abstract} We prove a uniqueness result about the 
    factorization of a monic polynomial over a general 
    commutative ring into comaximal factors. 
    We apply this result to address several questions 
    raised by Steve McAdam. These questions, inspired by 
    Hensel's Lemma, concern properties of prime 
    ideals and the factoring of monic polynomials modulo 
    prime ideals. 
    \end{abstract} 

    \maketitle 
    \section{Introduction} 

    \baselineskip 16 pt 

    There is an interesting relationship  between the factorization 
    of monic polynomials and the behavior of prime ideals in 
    integral extensions.  This is illustrated for example by the 
    well-known result of Nagata \cite[(43.12)]{N} that asserts that 
    a quasilocal integral domain $R$ satisfies Hensel's Lemma if and 
    only if every extension domain integral over $R$ is quasilocal. 
    Other references that deal with this relationship include 
    the papers \cite{HW} and \cite{M1}.  
    Recent work of Steve McAdam \cite{M1}, \cite{M2}, \cite{Mabs} on 
    this topic is the motivation for  
    our interest in the matters considered here. For a prime ideal 
    contained in the Jacobson radical of an integral domain, 
    McAdam \cite{M1} introduces the concepts  of H-prime, weak-H-prime 
    and quasi-H-prime. The H-primes are precisely those for which a 
    version of Hensel's Lemma holds. The other definitions reflect a 
    careful analysis of the comaximal factorization of monic polynomials. 

    In Theorem~\ref{CMF} we make use of a  famous theorem of 
    Quillen-Suslin, a key ingredient in their resolution of 
    the Serre  Conjecture,  to prove a uniqueness 
    result concerning comaximal factors of a monic polynomial over a 
    general commutative ring. We apply this result to 
    prove in Theorems~\ref{WHH} and \ref{QHH} that the concepts of 
    H-prime,  weak H-prime and   quasi-H-prime  are equivalent. 

    All rings considered here are commutative with unity. 
    Two  general references for our notation and terminology 
    are \cite{N} and \cite{K}. 


    \section{Comaximal factors of monic polynomials} 


    \begin{rem}\label{QS} 
    Let $I,J$ be ideals in a ring $S$ for which $IJ=fS$ where $f$ is 
    a nonzerodivisor in $S$.  Then $I,J$ are invertible ideals, i.e., 
    rank-1 projective $S$-modules. Moreover, using subscript $f$ to 
    denote passing to the ring of fractions with respect to 
    the multiplicatively closed system generated by $f$, we have 
    $I_f = S_f = J_f$.  Suppose in particular that $S=R[X]$, where 
    $R$ is a ring and $X$ is an indeterminate over $R$, and that 
    $f$ is monic in $R[X]$. Then by the Quillen--Suslin theorem 
    (\cite{S}; \cite{Q}; \cite[Chapter IV, Theorem~3.14]{K}), 
    $I,J$ are free $R[X]$-modules, i.e., principal ideals generated 
    by nonzerodivisors, and there are generators $p,q$ of $I,J$ 
    respectively for which $f=pq$. 
    \end{rem} 

    \begin{thm}\label{CMF} 
    Let $R$ be a ring and $X$ be an indeterminate over 
    $R$. Let $g,h$ be comaximal monic polynomials in $R[X]$, and suppose 
    that the monic polynomial $f$ in $R[X]$ is such that $gh\in fR[X]$. 
    Then $f$ has a factorization of the form 
    $f=pq$ where $g\in pR[X]$ 
    and $h\in qR[X]$.  In particular, if $f$ is irreducible in $R[X]$, 
    then either $g\in fR[X]$ or $h\in fR[X]$. 
    \end{thm} 

    \begin{proof} 
    Let $R[X]_g$ and $R[X]_h$ denote the localizations of $R[X]$ at the 
    multiplicatively closed systems generated by $g$ and $h$  respectively, 
    and let $I := fR[X]_g\cap R[X]$ and $J := fR[X]_h\cap R[X]$. 
    Suppose $g(X)h(X)=f(X)k(X)$, where $k(X) \in R[X]$.  Then because 
    \begin{displaymath} 
    h(X)=f(X)k(X)/g(X)\in I \ \ {\rm and}\ \ g(X) 
    =f(X)k(X)/h(X)\in J\ , 
    \end{displaymath} 
    the ideals $I$ and $J$ are comaximal in $R[X]$; so their intersection, 
    which is 
    \begin{displaymath} 
    fR[X]_g\cap fR[X]_h\cap R[X] = fR[X] \ , 
    \end{displaymath} 
    is their product.  By Remark~\ref{QS}, 
    there are generators $q(X),p(X)$ of $I,J$ respectively 
    for which $f=pq$, $g\in pR[X]$ and $h\in qR[X]$. 
    \end{proof} 

  \begin{rem} 
  With the notation of Theorem \ref{CMF}, if $R$ is 
  an integral domain, the polynomials $p(X), q(X) \in R[X]$ 
  such that $f = pq$ can clearly be chosen to be 
  monic. More generally, if $\Spec R$ is connected, 
  then $p(X)$ and $q(X)$ can be chosen to be monic.  To see 
  this, we use the fact that for any ring $S$ and nonconstant 
  polynomial $a(X) \in  S[X]$, if $S[X]/(a)$ is a free $S$-module of 
  rank~$n$, then there exists a  monic polynomial $b$ of 
  degree~$n$ in the ideal $(a)$ \cite[Prop. 2.2, page~44]{K}; 
  and because $S[X]/(b)$ is also a free $S$-module  of rank~$n$
  of which $S[X]/(a)$ is a homomorphic image, $b$ is a generator of
  $(a)$.  Now in our present
  context, we have $R[X]/(f) \cong R[X]/(p) \oplus R[X]/(q)$, 
  so $R[X]/(p)$ and $R[X]/(q)$ are locally free $R$-modules.  Thus, 
  $\Spec R$ is covered by neighborhoods $\Spec R_i$ on which 
  the extension of $(p)$ is  generated by monic polynomials. 
  Because $\Spec R$ is connected,  
  $R[X]/(p)$ has constant rank and these monic 
  polynomials all have the same degree $n$ and up to 
units of $R_i$ are extensions 
of polynomials in $(p)$ of degree $n$. An 
  appropriate  $R$-linear combination of these polynomials in $(p)$ 
  gives a monic polynomial of degree $n$ in 
  the  ideal $(p)$, and this monic  polynomial 
  generates $(p)$.  Similarly, the ideal $(q)$ is generated 
  by a monic polynomial if  $\Spec R$ is connected.  
  \end{rem} 



    \begin{exmp} 
    The hypotheses in Theorem~\ref{CMF} that we are working with monic 
    polynomials is necessary:  Indeed, let $D$ be a Dedekind domain 
    that is not a principal ideal domain. Let $P$ be a maximal ideal 
    of $D$ that is not principal. Then there exists an irreducible 
    element $f$ in $P - P^2$. Write $fD = PQ_1^{e_1}\cdots Q_n^{e_n}$, 
    where the $Q_i$'s are distinct maximal ideals; because $P$ is 
    not principal, $n > 0$. Let 
    $g \in PQ_1^{e_1} \cdots Q_{n-1}^{e_{n-1}} - Q_n$, and choose 
    $h \in Q_n^{e_n}$ but not in any of the maximal ideals containing 
    $g$. Then $gD + hD = D$ and $gh \in fD$, but $g, h \not\in fD$. 
    For a specific example, consider the Dedekind domain  $D =  
    {\mathbb Z}[\sqrt{-5}]$ (\cite[page 417]{A}). We have  $2\cdot3 
    = (1+\sqrt{-5})(1-\sqrt{-5})$ in this $D$, and 2 and 3 are 
    comaximal, but neither of the irreducible factors on the right side 
    of the equation divides either 2 or 3. 
    \end{exmp} 

    \begin{exmp} 
    The hypothesis that $g,h$ are comaximal is very necessary: 
    Let $R$ be any integral domain that is not integrally closed, let 
    $a$ be an element of the field of fractions of 
    $R$ that  is integral over $R$ 
    but not in $R$, and let $f\in R[X]$ be a monic polynomial of minimal 
    degree of which $a$ is a root. Then $f(X)=(X-a)g(X)$ for some 
    polynomial $g(X)$ over the integral closure of $R$.  Let $b,c$ be 
    nonzero elements of $R$ for which $ba\in R$ and $cg(X)\in R[X]$. 
    Then $f$ is irreducible in $R[X]$ and 
    \begin{eqnarray*} 
    & &f(X)(f(X)+b(X-a)+cg(X)+bc) \\ 
    & & \qquad\qquad = (f(X)+b(X-a))(f(X)+cg(X))\ , 
    \end{eqnarray*} 
    but $f(X)$ divides neither of the factors on the right side of the 
    equation. 

    Thus over any domain $R$ that is not integrally 
    closed, there exist monic polynomials $f, g, h \in R[X]$ such 
    that $f$ is irreducible and $gh \in fR[X]$, but 
    $g,h \not\in fR[X]$. For $R$ an 
    integrally closed domain this phenomenon is not possible; for in 
    this case a monic irreducible in $R[X]$ generates a prime ideal. 
    \end{exmp} 
      

    \section{Henselian-like conditions} 


    In \cite{M1}, McAdam uses the following definitions: 

    \begin{defn} 
    Let $P$ be a prime contained in the Jacobson radical of an 
    integral domain $R$. Then $P$ is 
    \begin{itemize} 
    \item[(a)] an \textit{H-prime} if, for every list of nonconstant 
    monic polynomials $f,g,h$ in $R[X]$ such that $gR[X]+hR[X]=R[X]$ and 
    $f-gh\in PR[X]$, there exist monic $p,q$ in $R[X]$ for which $f=pq$, 
    and $g-p,h-q\in PR[X]$; 
    \item[(b)] a \textit{weak-H-prime} if, for every list of nonconstant 
    monic polynomials $f,g,h$ in $R[X]$ such that $gR[X]+hR[X]=R[X]$ and 
    $f-gh\in PR[X]$, $f$ is reducible; and 
    \item[(c)] a \textit{quasi-H-prime} if, for every list of nonconstant 
    monic polynomials $f,g,h$ in $R[X]$ such that $gR[X]+hR[X]=R[X]$ and 
    $f-gh\in PR[X]$,  and for every prime ideal $K$ in $R[X]$ lying over 0 
    in $R$ and having  $f \in K$, either $K+gR[X]=R[X]$ or $K+hR[X]=R[X]$. 
    \end{itemize} 
    \end{defn} 

    \begin{thm}\label{WHH} 
    A weak-H-prime is an H-prime (and of course conversely). 
    \end{thm} 

    \begin{proof} 
    Let $P$ be a weak-H-prime in the domain $R$, and let $f,g,h$ be 
    nonconstant monic polynomials in $R[X]$ for which $f-gh\in PR[X]$ 
    and $g,h$ generate the unit ideal in $R[X]$. 
    Denote by overbars images mod $PR[X]$,  let 
    $\overline R := R/P$  and identify $R[X]/PR[X] \cong \overline R[X]$. 
    For each monic irreducible factor $p$ of $f$, we show that 
    either $\overline g$ or $\overline h$ is in $\overline p\overline R[X]$:

    We have $\overline g\overline h \in \overline p \overline R[X]$ 
    and $\overline g$ and $\overline h$ are comaximal in $\overline R[X]$. 
    Hence by Theorem \ref{CMF}, $\overline p$ factors into 
    a monic factor of $\overline g$ and a monic factor of $\overline h$. 
    The latter factors are comaximal because $\overline g$ and 
    $\overline h$ are comaximal.  If both  factors were nonconstant, 
    then because $P$ is a weak-H-prime, $p$ would be reducible; so one 
    of the  factors is a constant, i.e., 1, 
    and hence either $\overline g \in \overline p\overline R[X]$ 
    or $\overline h \in \overline p\overline R[X]$. 

    We proceed by induction on the number $n$ of irreducible 
    factors of $f$. 
    The case where $n = 1$ is clear. 
    Assume the theorem holds for monic polynomials $f$ 
    having $n$ irreducible factors. Suppose 
    $f' = pf$, where $p$ is irreducible and monic, and 
    that $\overline{f'}  =  \overline{g'}\overline{h'}$, 
    where $g', h'$ are nonconstant, monic and comaximal. 
    Then by the last paragraph we may assume $\overline{g'} 
    \in \overline p\overline R[X]$. 
    Let $g$ in $R[X]$ be such that 
    $\overline{g'} = \overline p\overline g$, and set 
    $h = h'$. Then $\overline f = \overline g\overline h$ 
    and 
    \begin{displaymath} 
    R[X] = g'R[X] + h'R[X] \subseteq gR[X] + hR[X] + PR[X] 
    \subseteq R[X]\ . 
    \end{displaymath} 
    Suppose  there is a maximal ideal $M$ of $R[X]$ that 
    contains both $g$ and $h$. Then because $M$ contains 
    monic polynomials, it meets $R$ in a maximal ideal and 
    so contains $P$, a contradiction. Therefore $g, h$ are 
    comaximal.  By the induction hypothesis, there exist 
    monic polynomials $g_1, h_1$ in $R[X]$ for which 
    $f = g_1h_1$, $\overline{g_1} = \overline g$, and 
    $\overline{h_1} = \overline h$; so 
    $f' = pf = pg_1h_1$, $\overline{pg_1} 
    = \overline p\overline g = \overline{g'}$ and 
    $\overline{h_1} = \overline h = \overline{h'}$. This 
    completes the induction and the proof. 
    \end{proof} 

    Much  of the work needed to prove Theorem \ref{QHH} is done 
    by McAdam in \cite{M1}; it merely remains for us to make a 
    few observations and apply the Quillen-Suslin Theorem. 

    \begin{thm}\label{QHH} 
    A quasi-H-prime is an H-prime, and conversely. 
    \end{thm} 

    \begin{proof} 
    McAdam proves in  \cite[Proposition~(2.1)]{M1} that an 
    H-prime is a quasi-H-prime; so it remains to prove 
    that a quasi-H-prime is an H-prime. 
    Let $P$ be a quasi-H-prime in the domain $R$, and let $f,g,h$ be 
    nonconstant monic polynomials in $R[X]$ such that $f-gh\in PR[X]$ 
    and $gR[X]+hR[X] =R[X]$.  Then by \cite[Proposition~(2.7)]{M1}, 
    there are ideals $I,J$ properly containing $fR[X]$ 
    for which 
    $$ 
    \frac{I}{fR[X]} \oplus \frac{J}{fR[X]} = \frac{R[X]}{fR[X]}. 
    $$ 
    It follows that $I$ and $J$ are comaximal in $R[X]$  and 
    intersect in $fR[X]$. Thus as in the proof of Theorem \ref{CMF}, 
    $IJ= fR[X]$, so by  Remark~\ref{QS}, 
    $f$ is the product of generators of the principal ideals $I,J$. 
    Neither generator can be a unit, because $I,J$ both properly 
    contain $fR[X]$ and their product is $fR[X]$. Thus, $f$ is reducible; 
    so $P$ is a weak-H-prime and hence by Theorem~\ref{WHH} an H-prime. 
    \end{proof} 

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    \end{thebibliography} 

    \end{document} 



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