Cyclotomic semistandard sums give the actual Schur endomorphism basis

Annihilator lemma. Let $m\in H_n$ be such that $m{H}_n$ is a direct summand of the codomain $H_n$, put $K=\mathrm{rAnn}(m)=\{a:ma=0\}$ and let $L_m(a)=ma$. The functional $f$ on $H_n$ vanishes on $K$ exactly when $f=g\circ L_m$ for some functional $g$ on $H_n$: one direction is immediate, and for the other $f$ descends to the split quotient $H_n\mathrm{/}K\cong m{H}_n$ and extends to $H_n$ by zero on a complementary summand. Hence, for $x$ with $\tau (xk)=0$ for all $k\in K$, the functional $a\mapsto \tau (xa)$ vanishes on $K$, so $\tau (xa)=g(ma)=\tau (yma)$ for the element $y$ representing $g$ under the perfect pairing [F4], and $\tau ((x-ym)a)=0$ for all $a$, so $x=ym\in H_{n}m$. Since $K$ is a right ideal, the converse also holds: if $x=ym$ then $\tau (xk)=\tau (ymk)=0$ for $k\in K$, and if $\tau (xk)=0$ for all $k\in K$ then $\tau ((xk)a)=0$ for all $a$ because $ka\in K$, so perfectness gives $xk=0$. Therefore $\mathrm{lAnn}(\mathrm{rAnn}(m))=H_{n}m$.

Evaluation is a bijection onto the intersection. The cyclic module $M_{\nu }=m_{\nu }{H}_n$ [F1] makes evaluation $f\mapsto f(m_{\nu })$ injective on ${\mathrm{Hom}}_{H_n}(M_{\nu },M_{\mu })$, since $f(m_{\nu }a)=f(m_{\nu })a$. Its image is exactly $\{x\in m_{\mu }{H}_n:xK=0\}$ with $K=\mathrm{rAnn}(m_{\nu })$: the inverse sends such an $x$ to the well-defined map $m_{\nu }a\mapsto xa$, which is a right $H_n$-map, and well-definedness is precisely $xK=0$. By step 1.1 with $m=m_{\nu }$ — whose image is a codomain direct summand by [F3] — this image is $m_{\mu }{H}_n\cap H_n{m}_{\nu }$, so ${\mathrm{Hom}}_{H_n}(M_{\nu },M_{\mu })\cong M_{\mu }\cap H_n{m}_{\nu }$.

The intersection basis. By [F3] the module $M_{\mu }$ has basis $m_{(S,t)}$ over semistandard $S$ of type $\mu$ and standard $t$, whose Murphy coordinates are constant on each admissible left fibre and zero off them ([F2]); applying the involution $\ast$ to the basis of $M_{\nu }$ gives a basis of $H_n{m}_{\nu }$ of the elements $m_{(s,T)}$ over semistandard $T$ of type $\nu$ and standard $s$, whose Murphy coordinates are constant on each admissible right fibre and zero off them. Hence an element of the intersection has Murphy coordinates that are zero unless the left fibre is admissible and the right fibre is admissible, and constant on each simultaneous pair of fibres. The rectangular sums $m_{(S,T)}$ have exactly these coordinate properties, with disjoint coefficient-$1$ pivots at the pairs $(s_0,T_0)$ of pivot refinements, and conversely every such coordinate family is the corresponding $R$-linear combination. Therefore the $m_{(S,T)}$ are an $R$-basis of $M_{\mu }\cap H_n{m}_{\nu }$.

The endomorphism basis. For $S,T$ of common shape $\lambda$ and types $\mu ,\nu$, formula ${\phi }_{ST}(m_{\alpha }h)={\delta }_{\alpha ,\nu }{m}_{(S,T)}h$ is well-defined: the right side depends on $m_{\alpha }h\in M_{\alpha }$ only through $h$ modulo $\mathrm{rAnn}(m_{\alpha })$, and for $\alpha \mathrm{\ne }\nu$ the map is defined to be zero, while for $\alpha =\nu$ well-definedness is exactly the coordinate property of step 3.1; multiplicativity in $h$ makes each ${\phi }_{ST}$ an $H_n$-map, and every endomorphism of $M$ is determined by its restrictions to the finitely many cyclic summands. Evaluation on each $M_{\nu }$ delivers by step 2.1 the whole intersection $M_{\mu }\cap H_n{m}_{\nu }$, so the ${\phi }_{ST}$ form an $R$-basis of $C_n={\mathrm{End}}_{H_n}(M)$, finite and free.

Cellular laws. Define an $R$-linear involution on the established endomorphism basis by ${\phi }_{ST}^{\ast }={\phi }_{TS}$. We verify anti-multiplicativity by evaluation; no generator-fixing star map on the right-module direct sum is assumed. If $m_{(S,T)}=m_{\mu }{h}_{ST}$ then $m_{(T,S)}=h_{ST}^{\ast }{m}_{\mu }$, and for a composable pair with $\mathrm{type}(T)=\mathrm{type}(U)$ the star of the evaluated product $m_{(S,T)}{h}_{UV}$ is $h_{UV}^{\ast }{m}_{(T,S)}=m_{(V,U)}{h}_{TS}$, while for non-composable types both products vanish; evaluation injectivity (step 4.1) gives $({\phi }_{ST}{\phi }_{UV}{)}^{\ast }={\phi }_{VU}{\phi }_{TS}$, hence ${\phi }_{ST}^{\ast }={\phi }_{TS}$. For the multiplication rule, evaluate ${\phi }_{ST}{\phi }_{UV}$ at the cyclic generator $m_{\mathrm{type}(V)}$ in its source summand: it is $m_{(S,T)}{h}_{UV}$ when $\mathrm{type}(U)=\mathrm{type}(T)$ and zero otherwise. The Murphy cell law [F2] expands this same-shape part as ${\sum }_{s\in \text{fibre }S}{\sum }_v{b}_{(T,v)}(h_{UV})m_{sv}$ with $b_{(T,v)}={\sum }_{t\in \text{fibre }T}{r}_{(t,v)}(h_{UV})$ independent of $s$ and $S$; because the product is an actual $H_n$-map and step 3.1 groups $v$ into its semistandard fibres, the coefficients are constant on each right fibre, so the coefficient of ${\phi }_{SW}$ depends on $T,W$ and the multiplier only, and all remaining terms have strictly higher shape. Thus the datum $\{{\phi }_{ST}\}$ with the dominance poset on shapes satisfies [F5]: it is a free dominance-cellular basis of $C_n$.

Specialization. Every step above is an identity of coefficients in the universal Laurent ring with denominators inverted only when they are units: the trace, the annihilator computation, the two one-sided bases, the intersection basis, and the structure constants all live over the universal ring and involve no field, dimension count, rationalization, or generic flatness. Hence applying any commutative base change in which the parameters are units — including the LM deformation DVR and arbitrary fields — preserves the well-definedness, the basis, the transpose rule, and the cellular multiplication constants.