The Drinfeld pairing is nondegenerate in each root degree and matches the shuffle pairing

Transport makes $D$ homogeneous and perfect. By [F1] the form $B_K$ is nondegenerate on each finite-dimensional graded piece $U_q^{-}[-\nu ]$, and $\iota$ restricts to a $\mathbb{Q}(q)$-linear isomorphism $U_q^{+}[\nu ]\to U_q^{-}[-\nu ]$; multiplying by the nonzero scalar $d_{\nu }^{-1}$ does not change nondegeneracy. Hence $D$ is a homogeneous bilinear pairing perfect on every finite root degree, and its generator values are those displayed in [F1] and the Statement.

Descent to the Serre quotients. The half form $B_K$ is defined on the actual Serre quotient by The negative-half bilinear form is symmetric, nondegenerate and integral and is nondegenerate there; since $\iota$ identifies the positive Serre quotient with the negative one by the presentation comparison of The Lusztig–Kashiwara form, parameter and coproduct dictionary, the transport $D$ is defined on the quotient of each half and introduces no free-word radical.

The generic braided pairing equals $D$. Extracting the first simple-root component from the braided coproduct in [F2] gives the coefficient map $e_i^{\mathrm{\prime }}$: on a word $F_{j}y$ its recursion is ${\delta }_{ij}y+q_i^{-a_{ij}}{F}_j{e}_i^{\mathrm{\prime }}y$, by the twist in [F2]. Thus its pairing satisfies $D_{\mathrm{g}\mathrm{e}\mathrm{n}}(E_{i}x,y)=(q_i-q_i^{-1}{)}^{-1}{D}_{\mathrm{g}\mathrm{e}\mathrm{n}}(x,e_i^{\mathrm{\prime }}y)$. The definition $D=d_{\nu }^{-1}{B}_K(\iota x,y)$ satisfies the same recursion because $d_{\nu }\mathrm{/}{d}_{\nu -{\alpha }_i}=q_i-q_i^{-1}$. Homogeneity, the unit normalization and induction on word height give $D_{\mathrm{g}\mathrm{e}\mathrm{n}}=D$. In particular $D$ satisfies both braided adjunctions on the halves.

The simple-root Borel extension. For $\nu \in Q$ write $K_{\nu }:={\prod }_i{K}_i^{{\nu }_i}$. On normal-ordered homogeneous elements define $D_B(x{K}_{\nu },y{K}_{\mu }):=q^{(\nu ,\mu )}D(x,y)$, extending bilinearly; normal ordering in the Borel extensions makes this well defined (Triangular decomposition of a quantized enveloping algebra). If $r(x)=\sum x_1\otimes x_2$ and $r(y)=\sum y_1\otimes y_2$ are the braided coproducts, the generator formulas [F3] and their common twist give ${\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}x=\sum x_1{K}_{-\mathrm{\mid }{x}_2\mathrm{\mid }}\otimes x_2$ and $\mathrm{\Delta }y=\sum y_1{K}_{\mathrm{\mid }{y}_2\mathrm{\mid }}\otimes y_2$, where $\mathrm{\mid }{x}_2\mathrm{\mid },\mathrm{\mid }{y}_2\mathrm{\mid }\in Q_{+}$ are their positive root degrees. For the first Borel adjunction, the product $(x{K}_{\nu })(x^{\mathrm{\prime }}{K}_{{\nu }^{\mathrm{\prime }}})$ contributes $q^{(\nu ,\mathrm{\mid }{x}^{\mathrm{\prime }}\mathrm{\mid })}$, while pairing against $\mathrm{\Delta }(y{K}_{\mu })$ contributes $q^{(\nu ,\mathrm{\mid }{y}_2\mathrm{\mid }+\mu )+({\nu }^{\mathrm{\prime }},\mu )}$; nonzero terms have $\mathrm{\mid }{y}_2\mathrm{\mid }=\mathrm{\mid }{x}^{\mathrm{\prime }}\mathrm{\mid }$, so these factors agree. For the second adjunction, $(y{K}_{\mu })(y^{\mathrm{\prime }}{K}_{{\mu }^{\mathrm{\prime }}})$ contributes $q^{-(\mu ,\mathrm{\mid }{y}^{\mathrm{\prime }}\mathrm{\mid })}$, while ${\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}(x{K}_{\nu })$ contributes $q^{(\nu -\mathrm{\mid }{x}_2\mathrm{\mid },\mu )+(\nu ,{\mu }^{\mathrm{\prime }})}$; nonzero terms have $\mathrm{\mid }{x}_2\mathrm{\mid }=\mathrm{\mid }{y}^{\mathrm{\prime }}\mathrm{\mid }$, again giving agreement. The remaining factors are exactly the two braided adjunctions of step 1.3. This proves the displayed Borel identities, toral normalization $D_B(K_{\nu },K_{\mu })=q^{(\nu ,\mu )}$, mixed toral/root values zero and counit normalization. Conversely those identities and normal ordering force this extension, so it is the generic Borel pairing. All statements are over the field; no integral perfectness is inferred.

Conclusion. Steps 1.1–1.2 give (i), steps 1.3–2.1 give (ii), and step 2.1 records the exact field-level scope; in particular the rank-one values carry the factor $(q_i-q_i^{-1}{)}^{-n}$ and no lower-lattice integrality is asserted.