DVR cellular Jantzen filtration and exact Gram valuation

The rank-zero case. If $n=0$, then $M=0$ and $M^{\ast }=0$, the pairing is generically nondegenerate on the zero space, and $M(s)=M_F^s=0$ for every $s$. The empty Gram determinant is defined here by the empty Leibniz product to be $1$, whose valuation is $0$; the empty sum of elementary-divisor exponents is $0$. Thus the Gram valuation identity is $0=0$, the radical and stability assertions hold, and the conditional nonzero-cellular-form assertion has no rank-zero instance. For the remaining steps assume $n\ge 1$.

Smith reduction of the form. View $b$ as the $O$-linear map $\phi \text{\hspace{0.17em}⁣}:M\to M^{\ast }$, $\phi (x)=b(x,-)$, with matrix $G^{\mathsf{T}}$ in column coordinates by [F2]; nondegeneracy over $\mathrm{Frac}(O)$ gives $\det G\mathrm{\ne }0$, so $\phi$ has full rank. Choose a nonzero matrix entry of $\phi$ of least valuation ${\nu }_{\pi }$ and move it to position $(1,1)$ by permuting the bases, then multiply its row by a unit of $O$ to make it ${\pi }^{a_1}$; every entry of the matrix is divisible by ${\pi }^{a_1}$ by minimality (including zero entries). Subtracting suitable $O$-multiples of the first row from the other rows and of the first column from the other columns (all coefficients in $O$, all operations unimodular by [F1]) kills the first row and column outside position $(1,1)$, leaving a smaller matrix over $O$. Iterating over the remaining rows and columns produces bases $u_1,\dots ,u_n$ of $M$ and $v_1,\dots ,v_n$ of $M^{\ast }$ with $\phi (u_j)={\pi }^{a_j}{v}_j$ and $0\le a_1\le \cdots \le a_n$.

Reduction of the radical. For $\bar{x}\in M_F$ one has $b_F(\bar{x},M_F)=0$ if and only if $b(x,M)\subseteq \pi O$, i.e. $x\in M(1)$; hence $\mathrm{rad}(b_F)=(M(1)+\pi M)\mathrm{/}\pi M=M_F^1$ by [F3], the containment $\pi M\subseteq M(1)$ being immediate from $b(\pi x,M)\subseteq \pi O$.

Submodule stability. Let $a\in A$ and $x\in M(s)$; by the hypothesis $b(ax,y)=b(x,a^{\ast }y)\in {\pi }^{s}O$ for every $y\in M$, since $a^{\ast }y\in M$. Hence $ax\in M(s)$ and $M(s)$ is an $A$-submodule; consequently $M_F^s=(M(s)+\pi M)\mathrm{/}\pi M$ is an $A$-submodule of $M_F$, the $\pi M$-cosets of $M(s)$ being stable because $\pi M$ is stable.

The cellular standard module. In a cell layer, applying both axioms [F4] to $c_{us}{c}_{tv}$ shows that the product is $b_F(s,t)c_{uv}$ modulo higher shapes: right multiplication retains $u$, left multiplication retains $v$, and comparison makes the scalar independent of $u,v$. This is the cellular form; the cell action therefore sends a vector $w$ to $b_F(v,w)u$ in the left convention, with indices transposed in the right convention. Assume $M_F$ is a cellular standard module and $b_F$ is its nonzero cellular form, and let $N\subseteq M_F$ be a proper submodule. If some $w\in N$ lies outside $\mathrm{rad}(b_F)$, choose a label $v$ with $b_F(v,w)\mathrm{\ne }0$; by the rank-one action [F4] every basis vector $u$ equals $b_F(v,w{)}^{-1}$ times the image of $w$ under the cell element of the pair $(u,v)$, so $u\in N$ for every basis label $u$ and $N=M_F$, a contradiction. Thus every proper submodule is contained in $\mathrm{rad}(b_F)$, which is proper because the form is nonzero, and the same rank-one argument shows that the quotient $M_F\mathrm{/}\mathrm{rad}(b_F)$ is simple (it has no nonzero proper submodule). Hence $\mathrm{rad}(b_F)$ is the unique maximal submodule of $M_F$, i.e. its module radical by [F3].

The filtration in the Smith bases. In the bases of step 1.2 one computes for $s\ge 0$ that $b(u_j,M)={\pi }^{a_j}O$, so $M(s)={⨁}_{j=1}^n{\pi }^{\max (0,s-a_j)}O{u}_j,{\dim }_F{M}_F^s=\mathrm{\#}\{j:a_j\ge s\}\mathrm{.}$ In particular $M(0)=M$, so $M_F^0=M\mathrm{/}\pi M$; the reduced filtration terminates because every summand of $M(s)$ is contained in $\pi M$ for $s>{\max }_j{a}_j$, so $M_F^s=0$; and each $M_F^s$ is a subspace of the finite-dimensional $F$-space $M_F$.

The Gram valuation identity. By the matrix form of step 1.2, the determinant of $G$ equals that of $G^{\mathsf{T}}$ and hence $\det ({\pi }^{a_j}{\delta }_{jk})$ times a unit, since both change-of-basis matrices are invertible over $O$; hence, using additivity of ${\nu }_{\pi }$ by [F1], ${\nu }_{\pi }(\det G)={\sum }_{j=1}^n{a}_j={\sum }_{s>0}\mathrm{\#}\{j:a_j\ge s\}={\sum }_{s>0}{\dim }_F{M}_F^s,$ the middle equality counting each $a_j$ as the number of positive integers $s\le a_j$.

Conclusion. The rank-zero argument of step 1.1 and steps 1.3–1.5 and 2.1–3.1 establish all four assertions of the Statement: the identification of $M_F^0$ and $M_F^1=\mathrm{rad}(b_F)$, the termination together with the exact identity ${\sum }_{s>0}{\dim }_F{M}_F^s={\nu }_{\pi }(\det G)$, the submodule stability under an anti-involution action, and the equality of $\mathrm{rad}(b_F)$ with the module radical in the cellular standard case. The Smith reduction of step 1.2 is valid over any discrete valuation ring, including residue characteristic $2$, because it uses independent domain and codomain bases and no symmetric diagonalization.