The question is based on what I tried to solve two exercises in James E. Humphreys "Introduction to Lie Algebras and Representation Theory": chapter 26 exerise 1 and chapter 9 exercise 2.
I am looking at the Lie Algebra $L=\mathfrak{sl}(2, \mathbb{F})$ with standard basis $(x, \ y, \ h )$ and the root system $A_{1}=\{-\alpha, \alpha\}$ via the rootspace decomposition. $char\mathbb{F}=0$.
$( \ , \ )$ is a scalar product.
Let $\mathfrak{U}(L)_{\mathbb{Z}}$ (as of Kostant´s theorem) be the lattice in the universal enveloping algebra of $L$ $\mathfrak{U}(L)$ with $\mathbb{Z}$-basis consisting of all $f_{a}h_{b}e_{c}$, for $f_{a}=\dfrac{y^{a}}{a!}$, $h_{b}=\begin{pmatrix} h \\ b \end{pmatrix}$ and $e_{c}=\dfrac{x^{c}}{c!}$, $a, \ b, \ c \in\mathbb{Z}^{+}$.
I have already proved, that the $\mathbb{Z}$-span of the usual basis $(v_{0}, \dots, v_{m})$ ($v_{i}=\frac{1}{i!}y^{i}.v_{0}$) of the irreducible $L$-module $V(m)$ is invariant under $\mathfrak{U}(L)_{\mathbb{Z}}$. Now I am trying to show that this is not the case for the basis of $V(m)$ consisting of $w_{i}=i!\left[\dfrac{(\alpha, \alpha)^{i}}{2^{i}}\right]v_{i}$ (I am pretty sure, that $[ \ , \ ]$ denotes the floor function, but it is not named explicitly in the book, Humpheys "Introduction to Lie Algebras and Reprentation theory"), via a counterexample:
I used: $f_{a}.v_{i}=\begin{pmatrix} i+a \\ a \end{pmatrix}v_{i+a}$ (or $0$ if $a+i$ exceeds $m$).
Look at $V(3)$ with basis $w_{0}, \ w_{1}, \ w_{2}, \ w_{3}$. It is for $a=1$:
$f_{1}.w_{2}=y.w_{2}=2\left[\dfrac{(\alpha,\alpha)^{2}}{2^{2}}\right](y.v_{i})= 2\left[\dfrac{(\alpha,\alpha)^{2}}{2^{2}}\right]3v_{3}=\dfrac{\left[\dfrac{(\alpha,\alpha)^{2}}{2^{2}}\right]}{\left[\dfrac{(\alpha,\alpha)^{3}}{2^{3}}\right]}w_{3} $
Now I set $(\alpha,\alpha)=4$, then:
$=\dfrac{4}{8}w_{3}=\dfrac{1}{2}w_{3}$
what is not in the $\mathbb{Z}$-span of the $w_{i}$.
My question is, if it is simply possible to chose $(\alpha,\alpha)$ just like that, or if it is set in any way.
It is the same problem, when I try to draw the dual root system $A_{1}^{v}=\{\alpha^{v}, -\alpha^{v}\}$ of $A_{1}$. It is $\alpha^{v}=\dfrac{2\alpha}{(\alpha,\alpha)}$, so in this case I chose $(\alpha,\alpha)=2$ and got $A_{1}=A_{1}^{v}$.
Since in $A_{1}$ there are only the roots $\alpha, -\alpha$, the value of $(\alpha, \alpha)$ at least doesn´t matter for $A_{1}$ to be a root system, but the way I used it above seems quite blunt to me and I don´t know if it works so easy in this two cases. Thank you for helping me.