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.


All right, here's a case where thinking about it for a few days isn't enough to prod inspiration, but embarassing oneself in public is. I'm glad and actually, surprised I haven't done this on MO before.

Since $S$ is contained in $V$ and $W$ hence in $M$, we can mod out by $S$ to reduce to the case $S=0$. Since $W$ is contained in $\pi^{-1}(Q)$ where $\pi: M \to M/V$, we can shrink $M$ to $\pi^{-1}(Q)$ to reduce to the case $Q = M/V$.

Put together, the question is whether there's a submodule $W \leq M$ complementary to $V$, or whether the map $M \to M/V$ has an $A$-linear section.

Pulled back apart, the question is whether the map $\pi^{-1}(Q)/S \to \pi^{-1}(Q)/V = Q$ has a section. I'm going to take this as an answer.

Here's an example, to help expiate my sin. Let $A = k[x]$, $M = A/\langle x^2 \rangle$, and $V = xM$. Then the only $1$-d $A$-submodule of $M$ is $V$. There are two graded $1$-d submodules of $gr\ M$, namely $0 \oplus M/V$ and $V \oplus 0$ (the liftable one). Applying the criterion to the first of these two, we're looking for a section of $M \to M/V$, and there isn't one. To the second, we're looking for a section of $0 \to 0$, and there is.


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