Ordinary Specht modules are the signed conjugate Murphy cells

The integral map. Identify $C_{\mathrm{M}}(\mu {)}^{\theta }$ with $\theta (M(\mu ))\mathrm{/}\theta (K_{\mu })$, where $K_{\mu }=M(\mu )\cap I^{⊳\mu }$. By [F2], literal left multiplication by $q^{-N}{x}_{\lambda }{T}_{w_{\lambda }}$ kills $\theta (K_{\mu })$ and induces the surjection $F:C_{\mathrm{M}}(\mu {)}^{\theta }\to P(\lambda )$ sending the class of $x_{\mu }$ to $z_{\lambda }$. In particular $F(m)=q^{-N}{x}_{\lambda }{T}_{w_{\lambda }}\theta (\tilde{m})$ for any row-module lift, independently of that lift. Both modules are free of rank $f^{\mu }=f^{\lambda }$; finite basis preimages give a right inverse of the square matrix of $F$, so its determinant is a unit and $F$ is an isomorphism. Set ${\mathrm{\Phi }}_{\lambda }=F^{-1}$. Its $H$-linearity with the twisted target action gives ${\mathrm{\Phi }}_{\lambda }(z_{\lambda }h)=m_0\theta (h)$, where this expression uses the ordinary action on the underlying Murphy module. All these matrices and their inverses lie in the universal Laurent ring and specialize by [F8].

Generic uniqueness of the forms, after the integral isomorphism. Work over $A=\mathbb{Z}[v^{\pm 1}]$ with formal $q=v$ and $K=\mathbb{Q}(v)$. The matrix of the regular trace form $(a,b)\mapsto {\mathrm{Tr}}_K(L_{ab})$ has Laurent-polynomial entries. At $v=1$ its matrix is $n!$ times the permutation matrix $u\leftrightarrow u^{-1}$: in the regular group basis left multiplication by a nonidentity group element has trace zero, whereas the identity has trace $n!$. Its determinant is therefore nonzero, so the regular trace form is nonsingular over $K$. If $a\in J(H_K)$, every $ab$ is nilpotent by [F6], and $L_{ab}$ has trace zero; nondegeneracy forces $a=0$, so $H_K$ is semisimple. To justify the consequence used next, any quotient algebra of a finite-dimensional semisimple algebra has semisimple regular module: the images of its finitely many simple summands are simple or zero and span the quotient; selecting successively images not contained in their predecessors expresses it as a direct sum of simples. A square-zero two-sided ideal $J$ of such an algebra acts trivially on every simple module, since $JS$ is either $0$ or $S$ and the latter would force $J^{2}S=S$; it therefore annihilates the semisimple regular module and is zero. Each cellular layer is an ideal in a quotient of this semisimple algebra, with multiplication $(a\otimes b)(c\otimes d)=G_{\mu }(b,c)a\otimes d$. If $G_{\mu }$ had a nonzero radical, $(\mathrm{rad}{G}_{\mu })\otimes C_{\mathrm{M}}(\mu )$ would be a nonzero square-zero ideal of that quotient, stable under its surrounding algebra by the cell law and contravariance, a contradiction. Thus $G_{\mu }$ is nonsingular, and the cellular rank-one operators realize the full matrix algebra on $C_{\mathrm{M}}(\mu {)}_K$. Every module endomorphism is scalar because it commutes with all matrix units; this conclusion also holds on the $\theta$-twist, since $\theta (H_K)=H_K$. On that twist, any contravariant form $B^{\mathrm{\prime }}$ defines an operator $U$ by $G_{\mu }^{\theta }(Ux,y)=B^{\mathrm{\prime }}(x,y)$. Contravariance and nonsingularity imply that $U$ commutes with the module action, so $B^{\mathrm{\prime }}$ is a scalar multiple of $G_{\mu }^{\theta }$. No generic argument was used to establish the integral map or basis.

The exact generator computation. The expansion in [F2] has distinct distinguished row-module vectors and length $L+\mathrm{\ell }(u)$, so [F3] gives $\beta (z_{\lambda },z_{\lambda })={\sum }_{u\in S_{\mu }}{q}^{-2\mathrm{\ell }(u)}{q}^{L+\mathrm{\ell }(u)}=q^L{\sum }_{u\in S_{\mu }}{q}^{-\mathrm{\ell }(u)}=q^{L-N}{P}_{\mu }(q)\mathrm{.}$ The last equality pairs $u$ with $w_0(S_{\mu })u$. The row identity $x_{\mu }{T}_u=q^{\mathrm{\ell }(u)}{x}_{\mu }$ gives $x_{\mu }^2=P_{\mu }(q)x_{\mu }$, so the defining cell product in [F4] gives $G_{\mu }(m_0,m_0)=P_{\mu }(q)$. Therefore, with the specified normalization ${\mathrm{\Phi }}_{\lambda }(z_{\lambda })=m_0$, the only possible comparison unit is $u_{\lambda }(q)=q^{N-L}$. This computation uses the distinguished-basis definition of $\beta$, not any incorrectly unscaled trace identity.

Universal form identity. Transfer the DJ form to $C_{\mathrm{M}}(\mu {)}^{\theta }$ using $F={\mathrm{\Phi }}_{\lambda }^{-1}$: set $B^{\mathrm{\prime }}(c,d)=\beta (Fc,Fd)$. This is contravariant for the twisted action, as is $G_{\mu }^{\theta }$. Both are nonzero over $K$, since their respective generator values are $q^{L-N}{P}_{\mu }(q)$ and $P_{\mu }(q)$, nonzero in that field. Step 2.1 makes them proportional, and step 2.2 determines the scalar as $q^{N-L}$, proving $G_{\mu }^{\theta }({\mathrm{\Phi }}_{\lambda }x,{\mathrm{\Phi }}_{\lambda }y)=q^{N-L}\beta (x,y)$ over $K$. Both sides have matrices in $A$; the injection $A\to K$ proves their equality already in $A$, and then [F8] specializes it to every commutative $R$ with unit parameter. No cancellation of $P_{\mu }(q)$ takes place after specialization.

Radicals, divisibility, quotients and duality. The unit form identity and the isomorphism ${\mathrm{\Phi }}_{\lambda }$ identify the form radicals and yield $D_{\mathrm{D}\mathrm{J}}^{\lambda }\cong (D_{\mathrm{M}}^{{\lambda }^{\mathrm{\prime }}}{)}^{\theta }$, including zero quotients. Over a field, a vector outside a nonzero cell-form radical generates its whole cell module: choose a vector pairing to $1$ and use its finite basis expansion in the cell rank-one action $x{m}_{uv}=G_{\mu }(x,m_u)m_v$. Thus every proper submodule lies in that radical, making it the unique maximal submodule and the nonzero form quotient the simple head; transfer this conclusion through ${\mathrm{\Phi }}_{\lambda }$. If the form is identically zero only the zero form quotient is asserted, not a zero actual head. Over a PID with chosen prime $\pi$, the submodules $\{x:\beta (x,P(\lambda ))\subseteq {\pi }^{i}R\}$ for every $i\ge 0$ are identified with their Murphy counterparts because the forms differ by a unit; their reductions correspond as well. Applying step 1.1 to ${\lambda }^{\mathrm{\prime }}$ gives $P({\lambda }^{\mathrm{\prime }})\cong C_{\mathrm{M}}(\lambda {)}^{\theta }$. Compose with the same-shape perfect pairing [F7] and the commuting of star dual with $\theta$ in [F1] to obtain $P({\lambda }^{\mathrm{\prime }})\cong P(\lambda {)}^{\circ ,\theta }$. Empty shapes have $N=L=0$, $\mathrm{\Phi }=1$ and unit $1$. This proves all clauses, using generic semisimplicity only for uniqueness of forms on an already integral module. No choice principle is used.