Central Hecke Jucys–Murphy scalars separate q-Schur residue contents

Work first over $A=\mathbb{Z}[u,u^{-1}]$ with fraction field $K=\mathbb{Q}(u)$. The regular trace pairing $(a,b)\mapsto \mathrm{Tr}({\mathrm{\ell }}_{ab})$ has entries in $A$ in the standard Hecke basis. At $u=1$ its matrix is $n!{\delta }_{v,w^{-1}}$, so its determinant is nonzero and it is nondegenerate over $K$. If $a$ belongs to the Jacobson radical, $ab$ is nilpotent for every $b$ by [F4]; its left multiplication is nilpotent and has trace zero. Nondegeneracy forces the radical to vanish. Thus $H_K$ is semisimple. Its shape-$\lambda$ cell layer has the multiplication $X{G}_{\lambda }Y$ of [F1]. If $G_{\lambda }$ were singular, the nonzero subspace $\{X:G_{\lambda }X=0\}$ would be a two-sided ideal with square zero, contradicting [F4]. Hence $G_{\lambda }$ is invertible and $X\mapsto G_{\lambda }X$ identifies this layer with a full matrix algebra over $K$.

Construct an injective right $S$-module map $j:W^{\lambda }\to {\mathrm{Hom}}_H(M,C_{\mathrm{M}}(\lambda ))$. Project the output of ${\phi }_{T^{\lambda }T}:M\to M^{\lambda }$ through $\pi :M^{\lambda }\to C_{\mathrm{M}}(\lambda )=M^{\lambda }\mathrm{/}(M^{\lambda }\cap I^{⊳\lambda })$. For $T$ of type $\nu$, the resulting map is zero on the other summands and sends $m_{\nu }$ to $m_T={\sum }_{t:\nu (t)=T}{m}_t$. Higher Schur maps with output type $\lambda$ have Hecke values of strictly dominant shape, so their output is killed by $\pi$. Thus this assignment respects the cell quotient and multiplication by any $s\in S$ becomes composition on the right: $j(ws)=j(w)\circ s$. For each fixed type the sets of standard $t$ with $\nu (t)=T$ are disjoint. Each is nonempty: standardize equal entries of a semistandard tableau in increasing column order; equal entries cannot share a column, row order is retained, and different entries occupy disjoint consecutive label blocks. Hence their sums $m_T$ are independent in the free Murphy cell basis, over every $R$. This proves injectivity, including zero type entries. No surjectivity onto all specialized Hom spaces is asserted.

For an integral symmetric polynomial $f$, $z_f$ is central by [F2]. Its action on the generic cell layer is a bimodule endomorphism of that full matrix algebra, hence scalar: its value at the identity commutes with every matrix and the rest follows by multiplication. Accordingly its matrix on $C_{\mathrm{M}}(\lambda {)}_K$ is scalar. The triangular matrices in [F2] give its diagonal as $f([c_t(1){]}_u,\dots ,[c_t(n){]}_u)$, equal to $a_{\lambda }(f)$ for every $t$ because $f$ is symmetric and the node multiset is fixed. Thus $z_f$ acts exactly by this scalar. Both sides have matrices over $A$, and injection into $K$ proves the identity before specialization. Every symmetric polynomial over an arbitrary $R$ is an $R$-linear combination of integral monomial symmetric polynomials: coefficients are constant on each monomial orbit. Apply the integral identities to these monomial polynomials and specialize $u\mapsto q$. Therefore every $z_f$ acts by $a_{\lambda }(f)$ on the actual Murphy cell over $R$, including singular parameters.

For central $z_f\in H$, right multiplication preserves every $M^{\alpha }$ and is $H$-linear. Every $H$-linear endomorphism $s$ of $M$ satisfies $s(m{z}_f)=s(m)z_f$, so ${\rho }_{z_f}$ is central in $S$. For $w\in W^{\lambda }$ and $m\in M$, step 1.2 and $H$-linearity give $j(w{\rho }_{z_f})(m)=j(w)(m{z}_f)=j(w)(m)z_f=a_{\lambda }(f)j(w)(m),$ the last equality by step 2.1. Injectivity gives $w{\rho }_{z_f}=a_{\lambda }(f)w$, proving the full ring-level assertion. If $M=0$ all cell assertions are empty. At rank zero the empty cell is $R$ with the empty polynomial evaluation, so the same conclusion holds.

Suppose now that $R=F$ and $L^{\mu }$ is a constituent of $W^{\lambda }$. Step 3.1 gives scalar $a_{\lambda }(f)$ on this constituent. Since $L^{\mu }$ is the nonzero quotient of $W^{\mu }$ by [F3], its scalar is also $a_{\mu }(f)$; hence they agree for every symmetric $f$. Taking all elementary symmetric polynomials gives ${\prod }_{A\in [\lambda ]}(X-[c(A){]}_q)={\prod }_{B\in [\mu ]}(X-[c(B){]}_q)\mathrm{.}$ By [F5], if $q=1$ the distinct $e$ residues are the distinct prime-field values $[c{]}_1$; if $q\mathrm{\ne }1$, $[a{]}_q=[b{]}_q$ is equivalent to $q^a=q^b$, hence to $a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}e)$. Unique factorization of these monic products into distinct field-linear factors therefore gives equality of every residue multiplicity, also when the field characteristic divides a multiplicity. There is no restriction on either label's Hecke-corner image, and no splitting extension, Gram division or choice principle enters the argument.