Let $G$ be your favorite complex semi-simple algebraic group, and let $B\supset T$ be your favorite Borus. For any $w\in W$, we have the Bruhat cell $BwB$, and its closure $\overline{BwB}$.

Now, it's very easy to write down some functions that cut out this variety. Let $V$ be any finite-dimensional representation of highest weight $\lambda$, and let $v$ be highest weight vector, and $\delta$ a non-zero functional killing all but the highest weight space. Then the generalized minors ~~$\omega_{w'\delta,v}(g)=w'\delta(gv)$~~ $\Delta_{w'\delta,v}(g)=w'\delta(gv)$ for all $w'$ with $w'\lambda > w\lambda$ all vanish on $BwB$ (since $BwBv$ in contained in the sum of weight spaces $\leq w\lambda$), and on no Bruhat cells $Bw'B$ with $w'>w$.

That is, the radical of the ideal generated by these functions is all functions vanishing on $\overline{BwB}$. In fact, it's enough to just consider $\lambda$ fundamental to get an ideal with the correct radical. So, my question is:

Is the ideal generated by these generalized minors already radical?