The Lusztig divided-power integral form $U_A$

The assignment $t\mapsto q$ defines a unital ring homomorphism ${\mathrm{\Lambda }}_1\to \mathbb{Q}(q)$ by the universal property of [F1], since $q$ is a unit of the field $\mathbb{Q}(q)$. It is injective: if ${\sum }_k{a}_k{q}^k=0$ with finitely many $a_k\mathrm{\ne }0$, choose $N\ge 0$ so that every $k+N\ge 0$; then the polynomial ${\sum }_k{a}_k{T}^{k+N}\in \mathbb{Q}[T]$ vanishes at the indeterminate $q$, so by [F2] it is the zero polynomial and all $a_k=0$. Hence its image $A$ is a subring of $\mathbb{Q}(q)$, and in particular $A$ is a commutative domain.

Each factor $N_{i,c,s}:=K_i{q}_i^{c-s+1}-K_i^{-1}{q}_i^{-c+s-1}$ is a $\mathbb{Q}(q)$-linear combination of the toral elements $K_{d_i{h}_i}$ and $K_{-d_i{h}_i}$, and the denominator $q_i^s-q_i^{-s}=q_i^{-s}(q_i^{2s}-1)$ is nonzero for $s\ge 1$: $q_i^{2s}=1$ would make the indeterminate $q$ a root of the nonzero polynomial $T^{2d_{i}s}-1$ by [F2]. Hence every factor, and therefore $T_{i,c,t}$, is a well-defined element of $U_q^0$; expanding the product and using $K_h{K}_{h^{\mathrm{\prime }}}=K_{h+h^{\mathrm{\prime }}}$ and $K_i{K}_i^{-1}=1$ of [F3], each $T_{i,c,t}$ is a finite $\mathbb{Q}(q)$-linear combination of the toral elements $K_{m{d}_i{h}_i}$ ($m\in \mathbb{Z}$), that is, a polynomial in $K_i^{\pm 1}$. Since $U_q^0$ is commutative by [F3], all $T_{i,c,t}$ commute with every $K_h$ and with one another.

The listed generators are homogeneous for the root grading of [F3]: $E_i^{(r)}=E_i^r\mathrm{/}[r{]}_i!$ has degree $r{\alpha }_i$ because $\mathrm{deg}{E}_i={\alpha }_i$ and the grading is multiplicative with $[r{]}_i!\mathrm{\ne }0$ by [F4]; likewise $F_i^{(r)}$ has degree $-r{\alpha }_i$; and $K_h^{\pm 1}$ and, by step 2.1, $T_{i,c,t}$ have degree $0$.

By definition $U_A$ consists of the finite $A$-linear combinations of finite products of the listed generators, the empty product $1$ included. Every finite product of homogeneous elements is homogeneous of the sum of their degrees (the grading of [F3] is multiplicative), so every element of $U_A$ is a finite sum of homogeneous elements of $U_A$. Since $U_q(\mathfrak{g})={⨁}_{\beta }{U}_q(\mathfrak{g})[\beta ]$ is a direct sum, this decomposition is unique, so $U_A={⨁}_{\beta }{U}_A[\beta ]$ with $U_A[\beta ]=U_A\cap U_q(\mathfrak{g})[\beta ]$; the same argument applies to the generated $A$-subalgebras $U_A^{\pm }$ (generators of degrees $\pm r{\alpha }_i$) and $U_A^0$ (degree-$0$ generators), which are therefore graded in the same way. Each $U_A[\beta ]$, being the intersection of the $A$-submodule $U_A$ with a $\mathbb{Q}(q)$-subspace, is an $A$-submodule.

Since $E_i=E_i^{(1)}$, $F_i=F_i^{(1)}$ by [F4] and the $K_h$ are among the listed generators, $U_A$ contains $E_i$, $F_i$ and $K_h$; consequently its $\mathbb{Q}(q)$-span contains all generators of $U_q(\mathfrak{g})$, so by [F5] the $\mathbb{Q}(q)$-span of $U_A$ is all of $U_q(\mathfrak{g})$.

The canonical multiplication map $\mathrm{\Phi }\text{\hspace{0.17em}⁣}:U_A{\otimes }_A\mathbb{Q}(q)\to U_q(\mathfrak{g})$, ${\sum }_i{u}_i\otimes s_i\mapsto {\sum }_i{s}_i{u}_i$ is surjective by step 5.1. It is injective by the argument of [F6]: given ${\sum }_i{u}_i\otimes s_i$ with $s_i=a_i{d}^{-1}$, $a_i\in A$, $0\mathrm{\ne }d\in A$, one has ${\sum }_i{u}_i\otimes s_i=\left({\sum }_i{a}_i{u}_i\right)\otimes d^{-1}$, and $\mathrm{\Phi }({\sum }_i{u}_i\otimes s_i)=d^{-1}{\sum }_i{a}_i{u}_i=0$ forces ${\sum }_i{a}_i{u}_i=0$ in the $\mathbb{Q}(q)$-vector space $U_q(\mathfrak{g})$ because $d\mathrm{\ne }0$; hence ${\sum }_i{u}_i\otimes s_i=0$. Thus $\mathrm{\Phi }$ is an isomorphism, which is claim (iii).

Finally, $A$ is not a principal ideal domain: the ideal $I=(2,q-1)\subseteq A$ is proper, since a relation $1=2a+(q-1)b$ would give $1=2\epsilon (a)$ under the ring homomorphism $\epsilon \text{\hspace{0.17em}⁣}:A\to \mathbb{Z}$ with $q\mapsto 1$ of [F1], in which $2a$ contributes an even integer. If $I=(d)$ were principal, then from $d\mid 2$ and the domain $A$ one gets $2=de$, and reducing modulo the ring homomorphism $A\to {\mathbb{F}}_2[q,q^{-1}]$, $q\mapsto q$ (again [F1]) gives $0=\bar{d}\bar{e}$; if $\bar{d}=0$ then $q-1=df$ reduces to $q+1=0$, a contradiction, so $\bar{d}\mathrm{\ne }0$ and hence $\bar{e}=0$, that is, $e=2e^{\mathrm{\prime }}$ for some $e^{\mathrm{\prime }}\in A$. Then $2=2d{e}^{\mathrm{\prime }}$ and therefore $d{e}^{\mathrm{\prime }}=1$ because $A$ is a domain with $2\mathrm{\ne }0$, so $d$ is a unit and $1\in I$, contradicting the properness of $I$. Hence no principal ideal equals $I$. This is why no freeness, independence or triangularity statement is attached to $U_A$ in this definition; those are separate theorems with their own hypotheses.