The rank-one root-power quotient is a finite-dimensional Hopf algebra

Work first over $R=\mathbb{C}[v,v^{-1},(v-v^{-1}{)}^{-1}]$ with $H_v=(K-K^{-1})\mathrm{/}(v-v^{-1})$. Combine adjacent powers of $K$ as elements $K^b$ of the group algebra $R[K,K^{-1}]$, and rewrite $K^{b}F\to v^{-2b}F{K}^b$, $E{K}^b\to v^{-2b}{K}^{b}E$, and $EF\to FE+H_v$. Count inversions in the order $F<K<E$, with a second measure equal to the number of $E$ and $F$ symbols. The first two rules decrease the inversion count; the $FE$ term of the third decreases it and its $H_v$ term lowers the second measure. Thus every word reduces after finitely many steps to an $R$-linear combination of the ordered monomials $F^a{K}^b{E}^c$. The same rules remain valid after $v=\zeta$ by [F2].

Define a unital algebra-map candidate $\mathrm{\Delta }$ and a counit candidate $\epsilon$ by the displayed formulas, with $\mathrm{\Delta }(1)=1\otimes 1$ and $\epsilon (1)=1$; define an anti-algebra-map candidate $S$ by the displayed formulas, with $S(1)=1$. For toral conjugation, direct multiplication gives $\mathrm{\Delta }(K)\mathrm{\Delta }(E)\mathrm{\Delta }(K^{-1})={\zeta }^2\mathrm{\Delta }(E)$ and $\mathrm{\Delta }(K)\mathrm{\Delta }(F)\mathrm{\Delta }(K^{-1})={\zeta }^{-2}\mathrm{\Delta }(F)$. For the mixed relation, expand the coproduct commutator: the two cross terms cancel since $EK={\zeta }^{-2}KE$ and $K^{-1}F={\zeta }^{2}F{K}^{-1}$, leaving $[E,F]\otimes K^{-1}+K\otimes [E,F]=H\otimes K^{-1}+K\otimes H=\mathrm{\Delta }(H)$. The counit assignments preserve each relation. For the antipode assignment, applying the anti-map to the two toral relations gives $-KE+{\zeta }^{2}EK=0$ and $-F{K}^{-1}+{\zeta }^{-2}{K}^{-1}F=0$. Applying it to $EF-FE-H$ gives $K^{-1}FEK-EF+H=FE-EF+H=0$. Thus $\mathrm{\Delta },\epsilon ,S$ descend to $D_{\zeta }$.

For $n\ge 1$, telescoping gives $E^{n}F-F{E}^n={\sum }_{j=0}^{n-1}{E}^j[E,F]E^{n-1-j}$; moving $K^{\pm 1}$ through $E^j$ by $E^{j}K={\zeta }^{-2j}K{E}^j$ and $E^j{K}^{-1}={\zeta }^{2j}{K}^{-1}{E}^j$ gives $E^{n}F-F{E}^n=\frac{{\sum }_{j=0}^{n-1}{\zeta }^{-2j}K-{\sum }_{j=0}^{n-1}{\zeta }^{2j}{K}^{-1}}{\zeta -{\zeta }^{-1}}{E}^{n-1}$. At $n=\mathrm{\ell }$ both geometric sums vanish because for $t={\zeta }^{\pm 2}$ one has $t\mathrm{\ne }1$, $t^{\mathrm{\ell }}=1$, and ${\sum }_{j=0}^{\mathrm{\ell }-1}{t}^j=(1-t^{\mathrm{\ell }})\mathrm{/}(1-t)=0$ by [F2]; hence $E^{\mathrm{\ell }}$ commutes with $F$ and it commutes with $E,K$ as well. Similarly, $F^{n}E-E{F}^n={\sum }_{j=0}^{n-1}{F}^j[F,E]F^{n-1-j}$; using $F^{j}K={\zeta }^{2j}K{F}^j$ and $F^j{K}^{-1}={\zeta }^{-2j}{K}^{-1}{F}^j$ gives $F^{n}E-E{F}^n=-\frac{{\sum }_{j=0}^{n-1}{\zeta }^{2j}K-{\sum }_{j=0}^{n-1}{\zeta }^{-2j}{K}^{-1}}{\zeta -{\zeta }^{-1}}{F}^{n-1}$, whose sums also vanish at $n=\mathrm{\ell }$, so $F^{\mathrm{\ell }}$ commutes with $E$ and already commutes with $F,K$. Finally $K^{\mathrm{\ell }}$ commutes with $E,F$ since ${\zeta }^{2\mathrm{\ell }}=1$. Thus $E^{\mathrm{\ell }},F^{\mathrm{\ell }},K^{\mathrm{\ell }}-1$ are central and $J_{\mathrm{\ell }}$ is a two-sided ideal.

The map from the rank-one algebra over $R$ presented by the three relations in [F1] with $v$ in place of $\zeta$ onto the $R$-subalgebra generated by $E,F,K^{\pm 1}$ is surjective by construction, and step 1.1 shows that its source is spanned by $F^a{K}^b{E}^c$. Their images are linearly independent in the generic algebra by [F4]; hence the map is injective and the $R$-source is free with that basis. Tensoring this free module along $v\mapsto \zeta$ preserves the basis. Thus $D_{\zeta }$ has the stated rank-one presentation and PBW basis.

On $K$ the two iterated coproducts are $K\otimes K\otimes K$. On $E$ they both equal $E\otimes K^{-1}\otimes K^{-1}+1\otimes E\otimes K^{-1}+1\otimes 1\otimes E$, and on $F$ they both equal $F\otimes 1\otimes 1+K\otimes F\otimes 1+K\otimes K\otimes F$; hence coassociativity holds on generators. The two counit composites give $K,E,F$ on each generator: for $E$ they are $0+E$ and $E+0$, and for $F$ they are $0+F$ and $F+0$, while $K$ is sent to $1\cdot K$ and $K\cdot 1$. Both counit composites and both iterated coproducts are algebra maps, so these identities hold on all of $D_{\zeta }$. For the antipode equations, on $K$ they are $K^{-1}K=1=K{K}^{-1}$; on $E$ they are $S(E)K^{-1}+E=0$ and $EK+S(E)=0$; on $F$ they are $S(F)+K^{-1}F=0$ and $F+KS(F)=0$. Since $S(1)=1$, both equations also hold on the unit. If both equations hold for $x$ and $y$, then anti-multiplicativity of $S$ and multiplicativity of $\mathrm{\Delta }$ give $\sum S(x_{(1)}{y}_{(1)})x_{(2)}{y}_{(2)}=\epsilon (x)\epsilon (y)1$ and $\sum x_{(1)}{y}_{(1)}S(x_{(2)}{y}_{(2)})=\epsilon (x)\epsilon (y)1$. Induction on word length proves the antipode equations on all of $D_{\zeta }$, so it is a Hopf algebra by [F3].

If $AB=tBA$, induction starts at $n=0$ with $(A+B{)}^0=1$ and $C_{0,0}=1$ and gives $(A+B{)}^n={\sum }_{r=0}^n{C}_{n,r}{B}^{n-r}{A}^r$, where $C_{n,0}=C_{n,n}=1$ and, for $0<r<n$, $C_{n,r}={\prod }_{j=1}^r(1-t^{n-r+j})\mathrm{/}(1-t^j)$. Multiplying the expansion on the right by $A+B$ shows the coefficient recurrence $C_{n+1,r}=C_{n,r-1}+t^r{C}_{n,r}$. For the product formula, putting those two terms over ${\prod }_{j=1}^r(1-t^j)$ leaves the common numerator ${\prod }_{k=n-r+2}^n(1-t^k)$ times $(1-t^r)+t^r(1-t^{n-r+1})=1-t^{n+1}$, exactly the product formula for $C_{n+1,r}$. Thus the formula follows inductively whenever its denominators are nonzero. For $t={\zeta }^2$ or $t={\zeta }^{-2}$ and $0<r<\mathrm{\ell }$, the denominator is nonzero and the numerator contains $1-t^{\mathrm{\ell }}=0$, so all intermediate coefficients vanish. In $\mathrm{\Delta }(E)$ the summands $A=E\otimes K^{-1}$ and $B=1\otimes E$ satisfy $AB={\zeta }^{-2}BA$; in $\mathrm{\Delta }(F)$ the summands $A=F\otimes 1$ and $B=K\otimes F$ satisfy $AB={\zeta }^{2}BA$. Applying the formula to these two pairs gives $\mathrm{\Delta }(E^{\mathrm{\ell }})=E^{\mathrm{\ell }}\otimes K^{-\mathrm{\ell }}+1\otimes E^{\mathrm{\ell }}$ and $\mathrm{\Delta }(F^{\mathrm{\ell }})=F^{\mathrm{\ell }}\otimes 1+K^{\mathrm{\ell }}\otimes F^{\mathrm{\ell }}$. Also $\mathrm{\Delta }(K^{\mathrm{\ell }}-1)=(K^{\mathrm{\ell }}-1)\otimes K^{\mathrm{\ell }}+1\otimes (K^{\mathrm{\ell }}-1)$. The counit kills all three generators. Since $\mathrm{\ell }$ is odd, $(EK{)}^{\mathrm{\ell }}={\zeta }^{-\mathrm{\ell }(\mathrm{\ell }-1)}{E}^{\mathrm{\ell }}{K}^{\mathrm{\ell }}$ and $(K^{-1}F{)}^{\mathrm{\ell }}={\zeta }^{\mathrm{\ell }(\mathrm{\ell }-1)}{F}^{\mathrm{\ell }}{K}^{-\mathrm{\ell }}$, with ${\zeta }^{\mathrm{\ell }(\mathrm{\ell }-1)}=1$, we have $S(E^{\mathrm{\ell }})=-E^{\mathrm{\ell }}{K}^{\mathrm{\ell }}$ and $S(F^{\mathrm{\ell }})=-K^{-\mathrm{\ell }}{F}^{\mathrm{\ell }}$, both in $J_{\mathrm{\ell }}$; also $S(K^{\mathrm{\ell }}-1)=K^{-\mathrm{\ell }}-1=-K^{-\mathrm{\ell }}(K^{\mathrm{\ell }}-1)\in J_{\mathrm{\ell }}$. The sum $J_{\mathrm{\ell }}\otimes D_{\zeta }+D_{\zeta }\otimes J_{\mathrm{\ell }}$ is a two-sided ideal of $D_{\zeta }\otimes D_{\zeta }$, so the coproduct inclusions for the three generators extend to all of $J_{\mathrm{\ell }}$ because $\mathrm{\Delta }$ is multiplicative. The counit kills the ideal because it kills its generators, and the anti-multiplicative antipode preserves the ideal because it sends each generator into $J_{\mathrm{\ell }}$. Therefore the quotient operations are well-defined and satisfy the Hopf identities inherited from $D_{\zeta }$.

By the PBW basis in step 2.1 and centrality in step 1.3, the ideal generated by $E^{\mathrm{\ell }},F^{\mathrm{\ell }}$ is spanned by monomials with $c\ge \mathrm{\ell }$ or $a\ge \mathrm{\ell }$, leaving a basis indexed by $0\le a,c<\mathrm{\ell }$ and $b\in \mathbb{Z}$. Modulo $K^{\mathrm{\ell }}-1$ one has $K^{-1}=K^{\mathrm{\ell }-1}$, so each toral coefficient is represented by a polynomial in $K$; division by the monic polynomial $K^{\mathrm{\ell }}-1$ gives a unique remainder of degree less than $\mathrm{\ell }$. Thus the quotient basis is exactly $F^a{K}^b{E}^c$ with $0\le a,b,c<\mathrm{\ell }$. There are ${\mathrm{\ell }}^3$ such elements, so the quotient has dimension ${\mathrm{\ell }}^3$.