Integral quantum highest modules are the normalized q-Schur Weyl cells

Tableau support. In any semistandard tableau, each column has at most $k$ entries in $\{1,\dots ,k\}$; summing over columns gives ${\sum }_{j\le k}{\mu }_j\le {\sum }_{j\le k}{\lambda }_j$. By [F4] the same inequality holds after permuting the content. Thus no weight $\lambda +a{\alpha }_i$, $a>0$, occurs. No weight $\lambda -a{\alpha }_i$ occurs when $a>m_i$: swapping its $i$th and $(i+1)$st coordinates makes its first $i$ coordinates sum to ${\sum }_{j\le i}{\lambda }_j+a-m_i$, violating that inequality. Negative coordinates also cannot occur. The type-$\lambda$ tableau is uniquely $T^{\lambda }$: the first content inequality is an equality forcing all $1$'s into row one, and successively equality for $k$ forces all $k$'s into row $k$. Hence the top weight line is $Ah$.

The highest map. The action in the Statement is a left action since $\tau$ reverses products. It has the stated weight labels by [F1]. Step 1.1 kills every $E_i^{(a)}h$ for $a>0$ and every $F_i^{(a)}h$ for $a>m_i$; the toral characters are those of weight $\lambda$. Thus [F3] gives a highest-module map $f:Q_A(\lambda )\to W_A^{\lambda }$. By the cell product and $B_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}(h,h)=1$, $h{\phi }_{T^{\lambda }T}=w_T$. Surjectivity in [F1] supplies an integral preimage under $\rho \tau$ of each of these endomorphisms, so $h$ generates the whole cell and $f$ is onto.

Cellular highest induction over any $R$. Put $I^{>\lambda }$ equal to the span of cellular shapes strictly dominating $\lambda$. This ideal is generated by the $1_{\eta }$ for $\eta >\lambda$: choose a finite order with larger shapes first, and use ${\phi }_{S,T^{\eta }}{1}_{\eta }{\phi }_{T^{\eta },T}\equiv {\phi }_{ST}$ modulo higher shapes, whose coefficient is $1$ by [F2]. Induction absorbs those higher terms in the already generated ideal; conversely each $1_{\eta }$ lies in $I^{>\lambda }$. The quotient $S_R{1}_{\lambda }\mathrm{/}{I}^{>\lambda }{1}_{\lambda }$ has exactly the basis ${\phi }_{T,T^{\lambda }}$ of shape $\lambda$: any semistandard type $\lambda$ in shape $\eta$ forces $\eta ⊵\lambda$ by the column count in step 1.1, and shape $\lambda$ has the single second tableau $T^{\lambda }$. Star identifies this left cell with the right cell acted on by $s^{\ast }$; the weight-unit map sending its weight-$\mu$ basis vector to $(c_{\mu }\mathrm{/}{c}_{\lambda })w_T$ instead identifies it with the action $s\cdot w=w{C}^{-1}{s}^{\ast }C$ of the Statement. Its highest vector maps to $h$.

Integral isomorphism. In type $A_{N-1}$ the finite Weyl alternating quotient of [F3], on composition labels of fixed total $r$, is $\det (x_i^{{\lambda }_j+N-j})\mathrm{/}\det (x_i^{N-j})$: the Weyl group permutes the $N$ coordinates, and multiplying numerator and denominator by the common monomial removes the half-sum shifts. This is the bialternant of [F4]. Consequently, for each weight $\mu$, [F3] and [F4] make the source and target finite free of the same rank $K_{\lambda \mu }$. The map in step 2.1 is graded, hence onto on each weight. Choose finite free bases and write its square matrix $D$; lifting the finitely many target basis vectors gives a matrix $P$ with $DP=I$. Therefore $\det (D)\det (P)=1$, and the adjugate formula gives an inverse matrix over $A$. Zero-rank weights have the unique empty inverse. Only finitely many weights occur, so these inverses assemble to $f^{-1}$. This proves the entire quotient presentation and the basis $b_T$. Concretely a preimage under the surjection of ${\phi }_{T^{\lambda }T}$, followed by integral triangular spanning and highest-character evaluation, gives a representative of $b_T$ in $U_A^{-}$; different preimages give the same quotient class. No generic-rank argument replaces the proved integral surjection.

Let $X=U_{R}x$ be the cyclic highest module with the composition support and full toral characters in the Statement. By [F6], every divided highest relation holds, and the highest-vector map $Q_R(\lambda )\to X$ is a surjective $U_R$-module map. The integral isomorphism of step 3.1 and its inverse remain inverses after arbitrary base change. Thus this is a quotient of the actual normalized Weyl cell. In particular $X$ factors through $S_R$: [F1] makes $\ker \rho$ stable under $\tau$, so it annihilates that cell and every quotient. No factorization of $X$ was assumed. Via the unit identification of step 2.2, the corresponding left-cell quotient map preserves the highest generator. Apply this argument also to $Y=S_R{\otimes }_{S_R^{+}}{R}_{\lambda }$: its canonical generator has the stated toral character, is killed by all positive divided powers, and is fixed by $1_{\lambda }$. The last assertion follows because the integral toral projection ${\prod }_j\left[\begin{array}{c}{K}_j;0\\ {\lambda }_j\end{array}\right]$ is $1_{\lambda }$ and has highest-character value $1$. Thus $Y$ is a quotient of the left cell. Conversely the highest vector of that cell defines a surjection $Y$ onto it. Both compositions fix the cyclic highest generators, hence are identities, proving the induction equivalence. This is a direct cellular proof of the universal property.

Construct the form on the actual cell first. Give weight $\mu$ the scalar $d_{\mu }=c_{\lambda }\mathrm{/}{c}_{\mu }$ times $B_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}$ and pair distinct weights by zero. Weight spaces are orthogonal for $B_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}$ because its self-adjoint projections $1_{\mu }$ are orthogonal; thus the new form is symmetric and integral and has value $1$ on $h$. For $s=\rho (u)$, [F1] gives $\rho (\tau (u){)}^{\ast }=Cs{C}^{-1}$. On a homogeneous component, right multiplication by this conjugate contributes $c_{\nu }\mathrm{/}{c}_{\mu }$ when $s$ takes right weight $\nu$ to right weight $\mu$. Hence $d_{\nu }(c_{\nu }\mathrm{/}{c}_{\mu })=d_{\mu }$, and cellular contravariance gives $B(u\cdot a,b)=B(a,\tau (u)\cdot b)$. Linearity handles sums of homogeneous components. Pull this form back by the integral isomorphism $f$.

Highest coefficient and uniqueness. For $x,y\in U_A^{-}$, contravariance gives $B(xz,yz)=B(z,\tau (x)yz)$. Orthogonality and the top line $Az$ make the last expression exactly $[z](\tau (x)yz)$. This proves the displayed coefficient construction descends through the complete quotient, including both arguments, because the already constructed form is defined on that quotient. Conversely the same equality is forced for any homogeneous normalized contravariant form, and negative words span by [F3], proving uniqueness. The construction has used only Laurent units and the integral action; it has not restricted a tensor form or divided by a tensor highest norm.

Base change and singular forms. Tensor the proved isomorphism and its inverse with any $R$; the quotient is its specialized presentation by right exactness of tensor product, the basis remains the specialized $b_T$, and the integral form matrices and highest projection specialize entrywise. The weight factors $d_{\mu }$ remain units. Thus a vector pairs to zero for the specialized form exactly when its image pairs to zero for the specialized cellular form, even if the Gram determinant is zero. Steps 2.2 and 4.1 used the same free cellular basis and unit coefficients over $R$, so their universal property also holds without flatness. At $r=0$ the unique empty highest cell and quotient are $R$ with form $1$; at $N=1$ the only degree-$r$ cell and quotient are likewise rank one. These prove all claims with only the inherited Choice of [C1].