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Let $R[x]$ be the polynomial ring over some principal ideal domain $R$. If $R[x]/I$ is free as a $R$-module for some ideal $I$, is $I$ a principal ideal which is generated by some monic polynomial in $R[x]$? Moreover, suppose that $f(x)\in R[x]$ is irreducible and $A=R[x]/(f(x)^i)$ is free as a $R$-module for some $i\geq 1$. Is any indecomposable $A$-module which is free as a $R$-module isomorphic to $R[x]/((f(x)^j)$ for some $j$ as an $A$-module?

Remark: The above questions are clearly true when $R$ is a field. But I am wondering whether they hold for arbitrary principal ideal domain $R$.

Improve this question
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    $\begingroup$ The answer to the first question is yes. For, let $I=(f_1,...,f_n)$. Let $c=\text{cont}(f_i)$ and write $f_i=cf'_i$. If $f'_i \not\in I$, then the class of $f'_i$ in $R[X]/I$ is $R$-torsion, in contradiction to the $R$-freeness of $R[X]/I$. Hence $f'_i\in I$ and wlog one can assume $\text{cont}(f_i)=1$. Let $q\in R[X]$ with $\text{cont}(q)=1$ such that $(f_1,...,f_n)=(q)$ in $\text{Quot}(R)[X]$. Write $f_i=gq$ for some $g\in \text{Quot}(R)[X]$. There is $c\in R$ such that $cg\in R[X]$ has content $1$. Thus $cf_i=(cg)q$ and by Gauss' lemma $c$ is a unit in $R$. Hence $I\subseteq (q)$ in $R[X]$. $\endgroup$
    – tj_
    Commented Mar 21, 2022 at 11:13
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    $\begingroup$ (cont.) By writing $q=\sum_ig_if_i$ with $g_i \in\text{Quot}(R)[X]$ we see that $cq\in I$ for some $c\in R, c\neq 0$. If $q\notin I$, then the class of $q$ in $R[X]/I$ is $R$-torsion in contradiction to the $R$-freeness of $R[X]/I$. Hence $q\in I$ and $I=(q)$. $\endgroup$
    – tj_
    Commented Mar 21, 2022 at 11:14
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    $\begingroup$ Just saw that question 1 also asks to show that $q$ above is monic. This can be seen as follows. Tensoring $R[X]/(q)$ with $\text{Quot}(R)$ shows that the $R$-rank of $R[X]/(q)$ equals the degree of $q$. Hence $R[X]/(q)$ is a finitely generated $R$-module and, in particular, $R[X]/(q)$ is an integral extension of $R$. Hence there is a monic $f\in R[X]$ s.t. $f(\bar{x})=\bar{0}$ in $R[X]/(q)$, i.e. there is $g\in R[X]$ s.t. $f(x)=gq$. Comparing leading coefficients shows that the leading coefficient of $q$ is a unit in $R$. $\endgroup$
    – tj_
    Commented Mar 21, 2022 at 15:23
  • $\begingroup$ @– tj_ Thanks for your affirmative answer for question 1. $\endgroup$
    – Master Gang
    Commented Mar 22, 2022 at 2:46

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