The prime-power numerical Serre half is the integral specialization

For an empty index set interpret both empty-generator halves as their coefficient ring; the assertion is then the identity map. Otherwise choose a regular Cartan datum without changing the matrix: take $P=P^{\vee }={\mathbb{Z}}^{2I}$ with evaluation pairing, $h_i=(e_i,0)$, and ${\alpha }_j=((a_{ij}{)}_i,e_j)$. The simple roots and coroots are independent and their pairings are $a_{ij}$, as required by Symmetrizable Cartan data for quantum groups. The separate generic halves and their divided-power subalgebras depend only on these matrix relations, so this introduces no restriction. There is a canonical surjection $\pi :S_{\nu }\to H_{\nu }^{+}$. The actual half satisfies every numerical Serre relation by evaluation of their generic polynomial identities. It is generated by the images of the $E_i$, since the divided-power generators equal $E_i^n\mathrm{/}[n{]}_{\nu }!$ and the denominators are nonzero. Suppose for contradiction that $\ker \pi \mathrm{\ne }0$, and choose a nonzero homogeneous $z\in \ker \pi$ of least positive height, of degree $\beta ={\sum }_i{n}_i{\alpha }_i$. The degree-zero map is injective, so this height is positive. Renaming $E_i$ as $F_i$ identifies this kernel with the residual kernel of [F2]; hence $(\beta ,\beta )=2\mathrm{ht}\beta >0$.

Both skew derivatives of $z$ lie in $\ker \pi$. Here is a direct justification that requires no numerical presentation comparison. For homogeneous $x$ in the actual positive half, with $i$-weight $m$, define the two word deletions $D_i,D_i^{\mathrm{\prime }}$ with product rules from [F3]. The source mixed relations give $[x,F_i]=(D_i^{\mathrm{\prime }}(x)K_i-K_i^{-1}{D}_i(x))\mathrm{/}(\nu -{\nu }^{-1})$; it is true on a simple generator and its extension to a product follows by expanding the commutator and crossing the torus. The deletion formulas also define $D_i,D_i^{\mathrm{\prime }}$ on $S_{\nu }$ by [F3]. Apply the mixed identity to the word expression $z$ whose image in the actual half is zero. Independence of the toral factors $K_i,K_i^{-1}$ in [F4], after moving them to one side, forces $\pi (D_{i}z)=\pi (D_i^{\mathrm{\prime }}z)=0$. Each derivative lowers height, so minimality of $z$ gives $D_{i}z=D_i^{\mathrm{\prime }}z=0$ in $S_{\nu }$, for every $i$.

Extend scalars from $K$ to $\mathbb{C}$; the nonzero vector stays nonzero because this is a field extension. Apply [F3] for every $i$ to the common-kernel vector of step 2.1. It gives $\beta (h_i)={\sum }_j{a}_{ij}{n}_j\le 0$. Symmetry of the Cartan matrix and its normalization $({\alpha }_i,{\alpha }_j)=a_{ij}$ yield $(\beta ,\beta )={\sum }_i{n}_i\beta (h_i)\le 0$. This contradicts the strictly positive value in step 1.1. Therefore the assumed kernel vanishes, and the canonical surjection is an isomorphism.

Now set $t={\nu }^{-1}$ and repeat steps 1.1–3.1 with $t$ in place of $\nu$. The same proof applies: $t>0$, $t\mathrm{\ne }1$, the symmetric quantum integers and Serre coefficients are invariant under parameter inversion, [F1] supplies the nondegenerate pairing and PBW ranks at both specializations, [F2] gives the Casimir constraint for $U_t$, [F3] applies at $q=t$, and [F4] supplies the specialized toral and triangular factors. Thus the least-height contradiction proves injectivity of the canonical map at $t={\nu }^{-1}$ directly. The remaining form and Green-pairing assertions are exactly the inverse-parameter assertions in [F1]; under the generator renaming $F_i\leftrightarrow E_i$, they have the displayed values and coproduct twist. No semilinear map between coefficient-field fibers is used.