Let $G = GL_n(\mathbb{K})$, where $\mathbb{K}$ is a field of characteristic $p > 0$. Let $\mathfrak{g} = \mathfrak{gl}_n(\mathbb{C})$. Let $e_{ij}$ denote the elementary matrices which are a basis for $\mathfrak{g}$. The Kostant $\mathbb{Z}$ form is the $\mathbb{Z}$-subalgebra $U_\mathbb{Z}$ of $U(\mathfrak{g})$ generated by elements $e_{ij}^r/r!$ for $1 \leq i \neq j \leq n$, $r \geq 0$ and ${h_i \choose r}$ for $1 \leq i \leq n$, $r \geq 0$ and $h_i = e_{ii}$. This is a free $\mathbb{Z}$ module with $\mathbb{Z}$-basis given by monomials of the form:
$\prod e_{ij}^{a_{ij}}/a_{ij}! \prod {h_i \choose r_i}$
with $a_{ij}, r_i \in \mathbb{Z}_{\geq 0}$, product taken in any fixed order. Finally, let $U_\mathbb{K} = \mathbb{K} \otimes_\mathbb{Z} \mathbb{U}_\mathbb{Z}$. It can be shown that this is isomorphic to $\mathrm{Dist}(G)$.
My question is how does one interpret these denominators, especially since for $r \geq p$ it looks like we are dividing by zero? For instance, is the following computation valid? Let us identify $1 \otimes x$ with $x$, and let $ p = 3$.
$e_{45}e_{14}^2/2! - e_{14}^2/2!e_{45} = 1/2! ([e_{45},e_{14}]e_{14} + e_{14}[e_{45}, e_{14}]) = 1/2!(-2e_{15}) = -e_{15}$
