Saturated Weyl filtrations of finite dimensional integral based modules

If $x\in M_A$ is bar fixed and belongs to $b+vL$ for $b\in B$, then $x=b$; if it belongs to $vL$, then $x=0$. Indeed write $x={\sum }_{c\in B}{a}_{c}c$ with $a_c\in A$. Being in $L$ means each coefficient is regular at zero, so it lies in $\mathbb{Z}[v]$; being bar fixed means $a_c(v)=a_c(v^{-1})$, so each such polynomial is constant. Its residue therefore determines the entire coefficient. This proves both assertions without a positivity or norm assumption.

Choose a maximal label $\lambda$ in the finite weight support of $M$. Every vector of its basis subset $B_{\lambda }$ is highest. Generic rank-one integrability makes $\lambda$ dominant, and [F2] gives for each $b\in B_{\lambda }$ an injective homomorphism ${\varphi }_b:V(\lambda )\to M$ carrying its highest vector to $b$. Its integral lattice maps into $M_A$ because it is generated by integral divided powers; its bar commutes with the target bar since both commute with the $U$ action and fix the highest vector. The map also commutes with all crystal operators by the unique string decomposition, and therefore maps the source crystal lattice into $L$. Thus every source global vector maps to an integral bar-fixed vector congruent either to a member of $B$ or to zero modulo $vL$. Step 1.1 and field injectivity show that every image is a member of $B$, and never zero.

The basis images for distinct $b,c\in B_{\lambda }$ are disjoint. Otherwise a common image basis vector is congruent to a nonzero crystal lowering path from each of $b,c$. Apply the reversed raising path from $c$ to that common vector; it recovers $c$ in the residue crystal. Because ${\varphi }_b$ commutes with crystal operators, this produces a source crystal vector in $V(\lambda )$ at weight $\lambda$ whose image has residue $c$. The source highest-weight space has basis the highest vector alone, hence its image residue is $b$, forcing $b=c$. Consequently the direct sum of the ${\varphi }_b$ identifies the direct sum of actual integral highest lattices with the basis-subset lattice spanned by their disjoint images. This lattice is saturated, $U_A$-stable and has a free complementary basis quotient. No equality of arbitrary tensor-product Hom ranks before and after specialization has been used.

Remove that basis-subset submodule and apply [F1] to its based quotient. Its dimension is strictly smaller, and induction on the finite dimension constructs a finite filtration with sections equal to direct sums of integral highest lattices; refine each such section by its finitely many copies. The initial zero module has the empty filtration. In every short exact sequence of this filtration the quotient is free as an $A$-module, so choosing its displayed basis gives an $A$-linear splitting and tensoring with any specialized field preserves injectivity as well as surjectivity. The specialized sections are exactly the root Weyl lattices by definition. This proves the complete assertion, with Choice confined to [F4].