The formal quantum Serre half embeds in the shuffle algebra and is degreewise free

By [F2], the generator assignment of [F1] defines $p_{\mathrm{\hslash }}:U_{\mathrm{\hslash }}{\mathfrak{b}}^{+}\to \mathcal{V}$. Its restriction to the Cartan coefficient algebra $A_C$ is the identity, and its image contains every one-letter word $[v_i]$. Every element of $\mathcal{V}=⟨V⟩{\otimes }_R{A}_C$ is a finite sum of products of a letter-generated element and an element of $A_C$, so $p_{\mathrm{\hslash }}$ is surjective. In the symmetric formula, $[m{]}_i=q_i^{-(m-1)}(1+q_i^2+\cdots +q_i^{2m-2})$ specializes to $m$ at $q_i=1$; therefore $[m{]}_i!$ specializes to $m!$ and each Gaussian coefficient specializes to $\left(\genfrac{}{}{0px}{}{m}{r}\right)$. By [F3], the reduced source is $U{\mathfrak{b}}^{+}$ and reduction defines a surjective graded algebra map $p:U{\mathfrak{b}}^{+}\to {\mathcal{V}}_0:=\mathcal{V}\mathrm{/}\mathrm{\hslash }\mathcal{V}$.

The ring $R$ is a PID: if $I\mathrm{\ne }0$ is an ideal, the orders of its nonzero elements have a least value $m\in \mathbb{N}$; choose $f\in I$ of order $m$. It has the form $f={\mathrm{\hslash }}^{m}u$ with $u$ a unit by [F22], so ${\mathrm{\hslash }}^m\in I$, while every $g\in I$ has order at least $m$ and is divisible by ${\mathrm{\hslash }}^m$. Thus $I=({\mathrm{\hslash }}^m)$; the zero ideal is principal as well, and [F6] says $R$ is a domain, so [F13] applies.

Each generator $h_i,D_j,e_i$ of the classical Borel has primitive coproduct by [F17], and its image under $p$ is primitive in ${\mathcal{V}}_0$: this follows for Cartan symbols from [F1] and for $[v_i]$ by reducing its displayed coproduct modulo $\mathrm{\hslash }$. Hence $p$ commutes with coproduct and counit on generators, and therefore on the generated algebra. Thus $p$ is a surjective Hopf map; since $U{\mathfrak{b}}^{+}$ is cocommutative by [F17], so is ${\mathcal{V}}_0$. This argument establishes Hopf compatibility only after reduction and makes no Hopf claim about $p_{\mathrm{\hslash }}$.

Put $J=\ker p$ and $\mathfrak{j}=J\cap {\mathfrak{b}}^{+}$. As the kernel of a Hopf map, $J$ is a two-sided ideal and has zero counit. For $x\in J$, $(p\otimes p)\mathrm{\Delta }(x)=0$; right exactness of tensor products and surjectivity of $p$ give $\ker (p\otimes p)=J\otimes U{\mathfrak{b}}^{+}+U{\mathfrak{b}}^{+}\otimes J$, so $J$ is a coideal. Under the stated AC hypothesis, apply [F4] to obtain $J=U{\mathfrak{b}}^{+}\mathfrak{j}=\mathfrak{j}U{\mathfrak{b}}^{+}$ with $\mathfrak{j}$ a Lie ideal. The universal property [F15] identifies the quotient $U{\mathfrak{b}}^{+}\mathrm{/}J$ with $U\mathfrak{a}$, where $\mathfrak{a}={\mathfrak{b}}^{+}\mathrm{/}\mathfrak{j}$, and $p$ with the quotient map. AC is used here only through [F4]; all other choices below are finite-dimensional or explicitly specified.

The first-order skew part $\delta (\bar{x})=((\mathrm{\Delta }(x)-{\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}(x))\mathrm{/}\mathrm{\hslash })<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>\mathrm{\hslash }$ of the formal Borel coproduct is well-defined on ${\mathcal{V}}_0$: its numerator reduces to zero by step 2.1, and changing a lift by $\mathrm{\hslash }y$ changes the quotient by $\mathrm{\Delta }(y)-{\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}(y)$, which is zero modulo $\mathrm{\hslash }$ because ${\mathcal{V}}_0$ is cocommutative. The quotient is unique because $R$ is a domain and the coefficientwise word–Cartan module is $\mathrm{\hslash }$-torsion-free. To see this, each $⟨V⟩[\gamma ]$ is a submodule of a finite free word module and hence finite free by [F23] and step 1.2; tensoring each such component with the coefficientwise torsion-free $A_C$ in [F1] preserves $\mathrm{\hslash }$-torsion-freeness, as do its coefficientwise completed tensor powers. Put $D=\mathrm{\Delta }-{\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}$ and let $c$ cyclically permute three tensor factors. Coassociativity gives $(1+c+c^2)(D\otimes \mathrm{id})D=0$: expanding $D$ gives four permutations of $(\mathrm{\Delta }\otimes \mathrm{id})\mathrm{\Delta }=(\mathrm{id}\otimes \mathrm{\Delta })\mathrm{\Delta }$, whose cyclic sums cancel in pairs. Since $D$ is divisible by $\mathrm{\hslash }$, division by ${\mathrm{\hslash }}^2$ and reduction prove co-Jacobi for $\delta$, and flip gives antisymmetry. For primitive $a,b\in {\mathcal{V}}_0$, subtracting the two algebra-map commutator identities for $\mathrm{\Delta }$ and ${\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}$, dividing by $\mathrm{\hslash }$, and reducing gives $\delta ([a,b])=[a\otimes 1+1\otimes a,\delta (b)]-[b\otimes 1+1\otimes b,\delta (a)]$, the required $1$-cocycle rule. Directly from [F1], $\delta ({\bar{h}}_i)=\delta ({\bar{D}}_j)=0$ and $\delta (\overline{[v_i]})=d_i\overline{[v_i]}\wedge {\bar{h}}_i$. These values lie in ${\bigwedge }^2\mathfrak{a}$; since $\mathfrak{a}$ is generated by their images and the cobracket is a $1$-cocycle, $\delta$ restricts to a Lie-bialgebra cobracket on $\mathfrak{a}$. The transposed bracket in [F5] gives the same formulas on ${\mathfrak{b}}^{+}$: its Cartan cobracket is zero and its explicit rescaled-Manin normalization gives ${\delta }_{{\mathfrak{b}}^{+}}(e_i)=d_i{e}_i\wedge h_i$. Thus $p{\mathrm{\mid }}_{{\mathfrak{b}}^{+}}:{\mathfrak{b}}^{+}\to \mathfrak{a}$ is a Lie-bialgebra map, because both cobrackets obey the $1$-cocycle rule and agree on the Cartan and simple generators.

The graded dual $p^{\ast }:{\mathfrak{a}}^{\mathrm{g}\mathrm{r}\mathrm{\prime }}\hookrightarrow ({\mathfrak{b}}^{+}{)}^{\mathrm{g}\mathrm{r}\mathrm{\prime }}\cong {\mathfrak{b}}^{-}$ is injective and is a Lie-algebra map by step 3.2 and [F5]; every root component is finite-dimensional. Its image contains the full Cartan subalgebra because $p$ maps the classical Cartan basis to the independent polynomial Cartan variables, and it contains each $f_i$ because $p(e_i)=\overline{[v_i]}\mathrm{\ne }0$ in the one-dimensional simple-root component. Since ${\mathfrak{b}}^{-}$ is generated by its Cartan and the $f_i$ by the separate Serre presentation [F3], the image is all of ${\mathfrak{b}}^{-}$. Dualizing each finite-dimensional root component shows $p{\mathrm{\mid }}_{{\mathfrak{b}}^{+}}$ is an isomorphism, so $\mathfrak{j}=0$ and $p:U{\mathfrak{b}}^{+}\to {\mathcal{V}}_0$ is an isomorphism.

The map $p$ preserves the ${\mathbb{N}}^I$ grading and sends $e_i$ to $\overline{[v_i]}$. The module decomposition in [F1] identifies the image of $⟨V⟩$ in ${\mathcal{V}}_0$ with $⟨V⟩\mathrm{/}\mathrm{\hslash }⟨V⟩$. Restricting the isomorphism of step 4.1 to the positive subalgebra proves $U_{\mathrm{\hslash }}{\mathfrak{n}}^{+}\mathrm{/}\mathrm{\hslash }{U}_{\mathrm{\hslash }}{\mathfrak{n}}^{+}\cong U{\mathfrak{n}}^{+}$ and identifies that classical half with $⟨V⟩\mathrm{/}\mathrm{\hslash }⟨V⟩$.

Fix $\alpha \in Q_{+}$. Since $I$ is finite by [F18], the word space $\mathrm{S}\mathrm{h}(V)[\alpha ]$ has finitely many words and is finite free over $R$, so $N_{\alpha }:=⟨V⟩[\alpha ]$ is finite-rank free by [F23]. The presented component $M_{\alpha }:=U_{\mathrm{\hslash }}{\mathfrak{n}}^{+}[\alpha ]$ is finitely generated, since it is a quotient of the finite free span of words of degree $\alpha$. By [F7] and [F14] over the PID of step 1.2, write $M_{\alpha }\cong R^p\oplus T$, with $T\cong {⨁}_{j=1}^{t}R\mathrm{/}({\mathrm{\hslash }}^{m_j})$ for positive integers $m_j$: every nonzero nonunit of $R$ is a unit times a power of $\mathrm{\hslash }$. Since $M_{\alpha }\mathrm{/}\mathrm{\hslash }{M}_{\alpha }\cong U{\mathfrak{n}}^{+}[\alpha ]$ by step 5.1, if $d={\dim }_{\mathbb{C}}U{\mathfrak{n}}^{+}[\alpha ]$, then $d=p+t$.

The surjection $p_{\mathrm{\hslash }}$ restricts to $M_{\alpha }\twoheadrightarrow N_{\alpha }$, and $N_{\alpha }$ has rank $d$ because its reduction is the classical component in step 5.1. The target is torsion-free, so this map kills $T$; reducing the induced surjection $R^p\twoheadrightarrow N_{\alpha }$ modulo $\mathrm{\hslash }$ gives $p\ge d$. Together with $d=p+t$, this forces $t=0$ and $p=d$. At $\alpha =0$, both components are $R\cdot 1$ and the map sends unit to unit. If $d=0$, the decomposition gives $M_{\alpha }=N_{\alpha }=0$. For $d>0$, choose bases: the surjection is a square matrix whose determinant reduces to a nonzero determinant over $\mathbb{C}$, hence is a unit by [F22]; [F10] makes it invertible. Therefore $M_{\alpha }\to N_{\alpha }$ is an isomorphism for every $\alpha$, proving (i) and the first equality in (iii).

Put $K=\mathrm{Frac}(R)$. Evaluation $q^{\mathrm{\prime }}\mapsto e^{\mathrm{\hslash }}$ embeds $\mathbb{C}[q^{\mathrm{\prime }}]$ into $R$: for a nonzero polynomial, factor out its maximal power of $(q^{\mathrm{\prime }}-1)$; the remaining factor has nonzero value at $1$, while $e^{\mathrm{\hslash }}-1\mathrm{\ne }0$, so its evaluation is nonzero in the domain $R$. By [F19] and [F21], this substitution sends $q_i^{\mathrm{\prime }}=q^{\mathrm{\prime }{d}_i}$ to $e^{d_i\mathrm{\hslash }}$; [F12] extends the injection to $\mathbb{C}(q^{\mathrm{\prime }})\hookrightarrow K$. For fixed $\alpha$, the relations in that degree are the columns of a finite matrix over $\mathbb{C}[q^{\mathrm{\prime }\pm 1}]$; finiteness follows because there are finitely many words of degree $\alpha$, and [F24] makes every Gaussian entry Laurent polynomial. Substituting $q^{\mathrm{\prime }}=e^{\mathrm{\hslash }}$ gives the formal presentation matrix for $M_{\alpha }$. By [F16], its quotient after extension to $K$ is the quotient by the same specialized columns. A field embedding preserves which matrix minors vanish, so the generic and formal matrices have the same rank; [F11] and step 7.1 give ${\dim }_{\mathbb{C}(q^{\mathrm{\prime }})}{U}_{q^{\mathrm{\prime }}}{\mathfrak{n}}^{+}[\alpha ]={\dim }_K(M_{\alpha }{\otimes }_{R}K)=d$. This proves the remaining equality in (iii) and completes all assertions. The theorem is not an iff statement, so there are no reverse implications to prove.