The Jantzen filtration of a lifted cell module

Let $R$ be a principal ideal domain (Principal ideal domain) with a chosen prime element $\pi$, let $F:=R\mathrm{/}\pi R$ be the residue field, and let $M$ be a free $R$-module of finite rank (The free module on a set and its standard basis) equipped with a symmetric $R$-bilinear form $⟨,⟩:M\times M\to R$ whose extension to the fraction field $K=\mathrm{Frac}(R)$ is non-degenerate (equivalently $\det G\mathrm{\ne }0$ for a Gram matrix $G$ of the form). For $i\ge 1$ put $M(i):=\{x\in M:⟨x,y⟩\in {\pi }^{i}R\text{ for all }y\in M\},M_F(i):=(M(i)+\pi M)\mathrm{/}\pi M\subseteq M_F:=M{\otimes }_{R}F\mathrm{.}$ Then $M(0)=M$, the $M(i)$ are $R$-submodules with $M(i+1)\subseteq M(i)$, the descending chain $M_F=M_F(0)\supseteq M_F(1)\supseteq M_F(2)\supseteq \cdots$ stabilises at $0$ (that is, $M_F(z)=0$ for all sufficiently large $z$, because $\det G\mathrm{\ne }0$), and $M_F(1)=\mathrm{r}\mathrm{a}\mathrm{d}{M}_F$, the radical of the reduced form. This finite descending filtration is the Jantzen filtration of $M$ with respect to $(R,\pi ,⟨,⟩)$. The hypotheses are exactly: $R$ a PID, $\pi$ a prime element, the form non-degenerate over $\mathrm{Frac}(R)$; no valuation-ring or DVR hypothesis is substituted, and the prime element $\pi$, its ideal $\pi R$ and the fraction field $K$ are kept distinct. Each $M(i)$ is a submodule: for the cell forms on this page the associativity identity $⟨xa,y⟩=⟨x,y{a}^{\ast }⟩$ shows closedness under the action. For $0\mathrm{\ne }r\in R$, define ${\nu }_{\pi }(r)=\max \{k\ge 0:{\pi }^k\mid r\}$. This maximum is finite: otherwise the uniquely determined elements $r\mathrm{/}{\pi }^k$ give an ascending chain of ideals whose union is principal, say generated by an element in its $N$-th ideal. The chain then stabilizes at $N$, whereas $(r\mathrm{/}{\pi }^N)=(r\mathrm{/}{\pi }^{N+1})$ and cancellation would make $\pi$ a unit. For termination, put $a={\nu }_{\pi }(\det G)$. If $x\in M(i)$, its coordinate column $v$ satisfies $Gv\in {\pi }^i{R}^d$, so the adjugate identity gives $(\det G)v\in {\pi }^i{R}^d$. For $i>a$, primality of $\pi$ forces every coordinate of $v$ into $\pi R$. Thus $M(i)\subseteq \pi M$ and $M_F(i)=0$. In rank zero the filtration is already zero. Finally a reduced vector belongs to the reduced form radical exactly when any lift pairs with every vector of $M$ into $\pi R$, which is exactly membership in $M(1)$. No choice principle is used.