TITLE: In a principal ideal ring, is every nonzero prime ideal maximal?
QUESTION [19 upvotes]: Inspired by this question, I was wondering whether from just the hypothesis that $A[X]$ is a nontrivial (commutative) principal ideal ring (so without supposing it is a domain) one can deduce that $A$ is a field. One possibility to prove that would be to use the fact, if it is one, that in principal ideal rings nonzero prime ideals are maximal. Namely, if $\def\p{\mathfrak p}\p\subset A$ is a prime ideal, then $(A/\p)[X]$, which is a quotient of $A[X]$ that is an integral domain but not a field, would have to be $A[X]$ itself, so $\p=(0)$, and $A$ having no nonzero prime ideals would have to be a field.
I think I can prove this, but I don't really like the argument I found, and since commutative algebra with zero divisors is not my speciality, I might have tripped up. So my question is are the proposition and its proof below correct, and if so is there a more elementary proof?
Notably I tried, under the hypothesis that $R$ is a principal ideal ring and $\p\subset R$ a prime ideal, and supposing $R\supset I=xR\supset\p=pR$, to use the existence of $y$ with $xy=p$ to arrive at a contradiction; but from $\p$ prime I only get $yR=pR$, and since over rings with zero divisors principal ideals may concide without their generators being associated, I don't see how to conclude.
(Proposed but false)
Proposition. Any nonzero prime ideal $\p$ of a principal ideal ring $R$ is maximal in $R$.
Proof. According to the Zariski-Samuel theorem $R$ is isomorphic to a finite direct product of rings that are either a principal ideal domain or a special principal ring. Every ideal $I$ of a finite product $R_1\times\cdots\times R_n$ is equal to a product $I_1\times\cdots\times I_n$ of ideals of the respective rings, where $I_k$ is the projection of $I$ to $R_k$ (clearly $I\subseteq I_1\times\cdots\times I_n$, but also $(0,\ldots,0,1,0,\ldots,0)I=\{0\}\times\cdots\{0\}\times I_k\times\{0\}\times\cdots\times\{0\}\subseteq I$) and since a product of more than one nontrivial ring is never an integral domain, 
$\p$ has to be of the form $R_1\times\cdots\times R_{k-1}\times\p_0\times R_{k+1}\times\cdots\times R_n$ where $\p_0\subset R_k$ is a prime ideal. So we have reduced to proving the proposition in the cases that $R$ is a principal ideal domain or a special principal ring. The former case is well known, and in a nontrivial special principal ring every ideal is a power of the unique maximal ideal $\def\m{\mathfrak m}\m$; putting $\p=\m^k$ a generator of $\m$ gives a nilpotent element of the integral domain $R/\p$, which must be zero, so $\p=\m$ as desired. QED

REPLY [5 votes]: Below is a much more general characterization of PIR semigroup-ring $\rm\,R[S].\,$ Note that the polynomial ring $\rm\,R[x]\,$ is the special case when $\rm\:S \cong \Bbb N,\,$ and the Laurent polynomial ring $\rm\:R[x,x^{-1}]\:$ is the special case $\rm\:S\cong \Bbb Z.\:$ See this answer for much more.
Theorem $\ $ TFAE for a semigroup ring R[S], with unitary ring R, and nonzero torsion-free cancellative monoid S.   
$\rm(1)\ \ $  R[S] is a PIR (Principal Ideal Ring)
$\rm(2)\ \ $ R[S] is a general ZPI-ring  (i.e. a Dedekind ring, see below)
$(3)\ \ $ R[S] is a multiplication ring  (i.e.  $\rm\ I \supset\ J \Rightarrow\ I\ |\ J\ $  for ideals $\rm\:I,J\,)$
$(4)\ \ $ R is a finite direct sum of fields, and $\rm\,S\,$ is isomorphic to $\,\mathbb Z\,$ or $\,\mathbb N\,$ 
A  general ZPI-ring  is a ring-theoretic analog of a Dedekind domain 
i.e. a ring where every ideal is a finite product of prime ideals. 
A unitary ring R is a general ZPI-ring $\iff$ R is a finite direct sum 
of Dedekind domains and special primary rings (aka SPIR = special PIR) 
i.e. local PIRs with nilpotent max ideals. ZPI comes from the German 
phrase "Zerlegung in Primideale" = factorization in prime ideals. 
The classical results on Dedekind domains were extended to rings 
with zero divisors by S. Mori circa 1940, then later by K. Asano 
and, more recently, by R. Gilmer. See Gilmer's book "Commutative 
Semigroup Rings" section 18 (and section 13 for the domain case).