Let $k$ be a commutative ring with $1$, and let $B$ be a submodule of a $k$-module $A$.


For every $n\in\mathbb N$ and every $k$-module $V$, let $K^n\left(V\right)$ denote the kernel of the canonical projection $V^{\otimes n}\to \mathrm{Sym}^n V$.


Let $m\in\mathbb N$ be such that $m\geq 2$.


Let $\rho$ be the canonical map $K^{m-1}\left(A\right)\otimes B\to A^{\otimes \left(m-1\right)}\otimes B$.


Let $\phi$ be the canonical map $A^{\otimes \left(m-1\right)}\otimes B \to A^{\otimes \left(m-1\right)}\otimes A = A^{\otimes m}$.


Let $\psi$ be the canonical map $A^{\otimes \left(m-2\right)} \otimes K^2\left(B\right) \to A^{\otimes \left(m-2\right)} \otimes B^{\otimes 2} \to A^{\otimes \left(m-2\right)} \otimes A^{\otimes 2} = A^{\otimes m}$.


Under what conditions do we have $K^m\left(A\right) \cap \phi\left(A^{\otimes \left(m-1\right)}\otimes B\right) = \phi\left(\rho\left(K^{m-1}\left(A\right)\otimes B\right)\right) + \psi\left(A^{\otimes \left(m-2\right)} \otimes K^2\left(B\right) \right)$ ?


(Note that when $k$ is a field, or under appropriately strong flatness conditions, we can think about the maps $\rho$, $\phi$ and $\psi$ as just being the obvious embeddings.)


I would like to see flatness or projectivity conditions, but at the moment I am not even sure that it holds when $k$ is a field. If you can find a condition better than "both $B$ and $A/B$ are projective", you have improved a result from my diploma thesis about relative Poincaré-Birkhoff-Witt theorems. But I think the question is interesting independently of Lie algebras, as it helps understanding $K^m$.



Commutator tensors and submodules
I believe that it holds whenever $A/B$ is flat.


Note (for use in Step 2) that the conclusion can be restated as follows: every element of the kernel of the canonical map $Sym^{n-1}A\otimes B\to Sym^nA$ is in the image of $Sym^{n-2}A\otimes K^2(B)$.


Step 1: True when $B=A$. 


This simply says that when you kill the image of $K^{n-1}(A)\otimes A$ in $A^{\otimes n}$ and also the image of $A^{\otimes n-2}\otimes K^2(A)$ then you have made $Sym^nA$. It's true because killing the first image makes one subgroup of $\Sigma\_n$ act trivially, killing the other makes another subgroup act trivially, and the two subgroups together generate the full group.


Step 2: True when $A/B$ is free.


Write $A=C\oplus B$. Split up $A^n$ into pieces according to how many $C$ and how many $B$ factors. Thus $A^{\otimes n}$ splits into pieces $Sym^pC\otimes Sym^qB$, one for each choice of $p+q=n$. The modules $Sym^{n-1}A\otimes B$ and $Sym^{n-2}A\otimes K^2(B)$ have similar splittings, and the maps to $Sym^nA$ respect the splittings, so what we have to show is:


Every element of the kernel of the canonical map $Sym^pC\otimes Sym^{q-1}B\otimes B\to Sym^pC\otimes Sym^qB$ is in the image of $Sym^pC\otimes Sym^{q-2}B\otimes K^2(B)$.


This reduces to Step 1, since $Sym^pC$ is free if $C$ is free.


Step 3: True when $A/B$ is flat.


Reduce to the free case: Express $A/B$ as a direct limit of free modules $F\_i$. Use this to write the pair $(A,B)$ as direct limit of the pairs $(A\times\_{A/B} F\_i,B)$, for each of which the conclusion holds. Now use that direct limits preserve tensor products and exact sequences.


