Let $B$ be a connected, solvable algebraic group (of dimension 2) acting on a projective variety $Y$ (of dimension $3$): for instance, let $B$ be a Borel subgroup of a reductive algebraic group $G$. 


**Question1**: Is it always true that there is a $B$-invariant complete flag (i.e. codimension $1$, irreducible and $B$-invariant subvarieties), $Y \supset Y\_1\supset Y\_2 \supset Y\_3$? For my purposes I can assume that there is a point in $Y$ with finite stabliser group and therefore can choose the orbit-closure of that point to be $Y\_2$. How about $Y\_1$? Is there always a point with $1$-dimensional stabiliser group of the $B$-action on the just described $Y\_2$? Obviously there will always exist a $B$-fixed point $Y\_3$, by Borel's fixed point theorem, if $Y\_1$ is a projective variety on which $B$ acts. 


**Question2**: Does this generalize to arbitrary dimension $n$ of $Y$ and any reductive algebraic group $G$ with a Borel subgroup $B$? 



B-invariant subvarieties
Here is an argument which avoids the use of Hilbert schemes and works for any complete (irreducible) variety $Y$.


As in Jason's answer it suffices to construct a $B$-stable subset $Y\_1$ of codimension $1$, provided that $\dim Y\ge1$. For this we consider two cases:


1. $Y$ does not contain an open $B$-orbit. By Rosenlicht's theorem, there exists an open $B$-stable subset $Y\_0\subseteq Y$ such that the orbit space $\pi:Y\_0\to Z:=Y\_0/B$ exists. Our assumption implies $\dim Z>0$. Therefore, $Z$ contains a subvariety $Z\_1$ of codimension $1$. Now take for $Y\_1$ the closure of $\pi^{-1}(Z\_1)$ in $Y$.
2. $Y$ does contain an open $B$-orbit $Y\_0$. Then the above argument is not applicable since $Z$ is a point. Instead we use the well-known fact that homogeneous spaces for solvable groups are affine. Thus $Y\_0$ is affine. But it not complete since $\dim Y\_0\ge1$. Thus the complement $C$ of $Y\_0$ in $Y$ is non-empty. We conclude with another well-known fact, namely that the complement of an affine open set is always pure of codimension one. So, take for $Y\_1$ any component of $C$.


