Let $\Bbbk$ be an algebraically closed field of characteristic $p>0$. Let $G$ be a connected affine algebraic group over $\Bbbk$, and write $\Bbbk[G]$ for the coordinate ring of $G$.
Of course $G$ need not be smooth, so let $I\subseteq\Bbbk[G]$ denote the nilradical of $\Bbbk[G]$. My question is: if $f\in\Bbbk[G]\setminus I$ ($\textbf{this is set-minus!}$), then is it correct that $f$ is not a zero divisor in $\Bbbk[G]$?
My intuition is that something stronger is true: namely that there is a splitting of the natural surjection $\Bbbk[G]\to\Bbbk[G]/I$, and that under this splitting $\Bbbk[G]$ is a projective module over $\Bbbk[G]/I$. But perhaps my intuition is totally wrong :)
I think I answered my question; please confirm if the argument is correct!
Suppose we have nonzero $g\in I$ such that $fg=0$. Then there exists $k$ such that $g\in I^k$ but $g\notin I^{k+1}$. Thus $g$ defines an element of $I^{k}/I^{k+1}$, which is a $G_{red}=Spec(\Bbbk[G]/I)$ equivariant coherent sheaf on $G_{red}$, under left translation say. But this implies it is a vector bundle on $G_{red}$, and thus is torsion free. However, because $fg=0$, we would have that $g$ defines a nonzero torsion element of $I^k/I^{k+1}$, an impossibility.