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 $K^m$.


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.

  • $\begingroup$ OK, sorry, I don't have that much time at my disposal right now, so reading this will take a while. At the moment I have two questions: (1) why is $A=C\oplus B$ in Step 2? (2) Why is a flat module a direct limit of free modules? (This seems to be well-known, but I have no idea how to prove it.) $\endgroup$ – darij grinberg May 23 '11 at 10:01
  • 1
    $\begingroup$ (1) I just meant that if $B$ is a submodule of $A$ and $A/B$ is free then there exists a submodule $C\subset A$ such that $A$ is the direct sum of $B$ and $C$. I.e., the exact sequence $0\to B\to A\to A/B\to 0$ splits. (2) I don't know a reference, but the key point is that the category of all f.g. free modules over a given flat module is filtered. $\endgroup$ – Tom Goodwillie May 23 '11 at 11:25
  • $\begingroup$ (1) Ah, thanks, that's the standard argument which still holds if $A/B$ is projective. I should have realized it. (2) Thanks, I found it in literature. $\endgroup$ – darij grinberg May 23 '11 at 13:05
  • $\begingroup$ (2) was proven by Lazard, but a good reference is Lam's book about module theory. $\endgroup$ – Martin Brandenburg May 23 '11 at 14:58
  • $\begingroup$ It's an enjoyable exercise: Prove first that every module is the colimit of "all" finitely generated free modules over it, then (this is the more interesting part) prove that if the module is flat then the indexing category for that colimit is filtered. $\endgroup$ – Tom Goodwillie May 23 '11 at 23:08

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