Let $k$ be a field, $A$ a $k$-algebra (probably noncommutative), and $M$ an $A$-module that's finite-dimensional as a vector space over $k$.

Let $Gr(M;k)$ denote the set of all $k$-subspaces of $M$, i.e. the disjoint union of Grassmannians, and $Gr(M;A) \subseteq Gr(M;k)$ denote the subscheme consisting of $A$-submodules.

If $V < M$ is an $A$-submodule, let $gr\ M := V \oplus (M/V)$, the associated graded. There's a natural constructible (i.e. discontinuous) map $Gr(M; A) \to Gr(gr\ M; A)$ taking $W \mapsto ((W\cap V), (image: W\to M\twoheadrightarrow M/V))$.

What is the image of this map, or, Which graded submodules of $gr\ M$ lift to submodules of $M$?

I believe I have proved that the image of this map is closed. (Sketch: consider the Rees family over ${\mathbb A}^1$ degenerating $M$ to $gr\ M$. Replace each fiber with $Gr(\bullet;A)$ of the fiber, obtaining a family $F$. Let $F'$ be the closure of the subfamily over ${\mathbb G}_m$, so flat over ${\mathbb A}^1$. Then the Bia\l ynicki-Birula map from $F' \to F_0 = Gr(gr\ M)$ has image $F'_0$, even if we restrict the map to the fiber $Gr(M;A)$ over $1$. Hence the map $Gr(M;A) \to Gr(gr\ M;A)$ has image $F'_0$, which is closed in $F_0$.) So one can rephrase:

What are the equations on a pair of $A$-submodules $(S \leq V, Q \leq M/V)$ that ensure existence of a $W \leq M$ with $gr\ W = S\oplus Q$?

Because the proof sketch relies on excising spurious components over $0$ of a Rees family, it sounds like a Gr\"obner basis calculation, but I didn't see which one.