I would like to establish what the set of units are in the quantized enveloping algebra $U_q(\mathfrak{sl}_2)$. First, I recall the definition of the quantized enveloping algebra- throughout the ground field will be taken to be $\mathbb{C}$ and $q\in \mathbb{C}$ is not a root of unity.
Definition: $U_q(\mathfrak{sl}_2)$ is the associative unital algebra generated by the four variables $E,F,K,K^{-1}$ subject to the defining relations:
\begin{align*} KK^{-1}=&K^{-1}K=1\\ KEK^{-1}=q^2E, &\quad KFK^{-1}=q^{-2}F\\ [E,F]=&\frac{K-K^{-1}}{q-q^{-1}}. \end{align*} I will often use the notation $U_q:=U_q(\mathfrak{sl}_2)$ for simplicity. $U_q(\mathfrak{sl}_2)$ is a Hopf algebra, and can be viewed as a deformation of the enveloping algebra $U(\mathfrak{sl}_2)$- but both these things will not be too important for this particular problem, it will be the algebraic of structure of $U_q(\mathfrak{sl}_2)$ that plays a role here.
It is well known that $\{F^iK^{\ell}E^j:i,j\in \mathbb{N}_0, \ell \in \mathbb{Z}\}$ is a basis of $Uq(\mathfrak{sl}_2)$, it has no zero divisors, and that the units of $U_q(\mathfrak{sl}_2)$ are the elements of the form $\lambda K^{\ell}$ with $\lambda \in \mathbb{C}^{\times}$ and $\ell \in \mathbb{Z}$, one way this can be established is by establishing a formula for $E^i F^j$ in terms of the basis and one can impose an ordering on monomials in such a way that degree arguments can be used. See for example This paper Lemma 2.1 on page 4. This is probably the preferred method for someone learning from Jantzen's book "Lectures on Quantum Groups", since there the desired formulas and gradings are discussed.
I would like to attempt a different proof of this result more in the spirit of Kassel's book "Quantum groups".
Idea
There it is proved that $U_q$ is an iterated Ore extension. To summarize: we set $A_0=\mathbb{C}[K,K^{-1}]$ and form the Ore extension $A_1=A_0[F,\alpha_1,0]$ where $\alpha_1$ is the automorphism of $A_0$ determined by $\alpha_1(K)=q^2K$. It turns out that $A_1$ is the algebra generated by $F,K,K^{-1}$ subject to the relation $FK=q^2KF$.One can then form the Ore extension $A_2=A_1[E,\alpha_2,\delta]$ where $\alpha_2$ is determined by $\alpha_2(F^jK^{\ell})=q^{-2\ell}F^j K^{\ell}$ and $\delta$ is an $\alpha_2$ derivation of $A_1$ with $\delta(K)=\frac{K-K^{-1}}{q-q^{-1}}$. It is not too difficult to see that $A_2\cong U_q(\mathfrak{sl}_2).$
My idea is that it should be easier to determine what the units are if we view $U_q$ as the Ore extension $A_2$. Specifically, as a set $A_2=A_1[E,\alpha_2,\delta]$ is just $A_1[E]$ (polynomials in one indeterminate), with a different algebra structure, $A_2$ has no zero divisors and shares the nice property of behaving well with respect to degrees: If $P$ and $Q$ are in $A_2$, then $\deg(PQ)=\deg(P)+\deg(Q)$. Thus if $P$ is a unit in $A_2$ with inverse $Q$, then from this we can deduce that $P$ must belong to $A_1$. We can actually be more specific: If $P=\sum_{i=0}^n a_i E^i,\enspace Q=\sum_{i=0}^m b_iE^i$ then $PQ=\sum_{i=0}^{n+m}c_i E^i$ where: $$c_i=\sum_{p=0}^i a_p \sum_{k=0}^p S_{p,k}(b_{i-p+k}) $$ where $S_{n,k}$ is the linear endomorphism of $A_1$ defined as the sum of all $n \choose k$ compositions of $k$ copies of $\delta$ and of $n-k$ copies of $\alpha_2$ (see Kassel Corollary I.7.4).
From this we get that $a_0$ must be a unit in $A_1$. Thus we have reduced the argument to finding the units of $A_1=A_0[F,\alpha_1,0]$, but by an identical argument we can conclude that the units of $A_1$ are the units of $A_0$, i.e the units of $\mathbb{C}[K,K^{-1}]$, which are precisely the elements of the form $\lambda K^{\ell}$ with $\ell\in \mathbb{Z}$ and $\lambda$ a nonzero scalar (this is not difficult to prove but I omit it here).
Closing Thoughts This is my first exposure to Ore extensions, but I am fairly certain I have applied the theory correctly. Nonetheless, I would appreciate a second look, or any comments on the work.