Derived functor of a derived functor

Given $F$ is a covariant additive functor from left R-module to a left S-module, show that

$\mathscr{L}_n(\mathscr{L_m}(F))=0$ if $m>0$ (where $\mathscr{L}$ refers to the derived functor).

I am trying to show this from induction, but I can't think of a projective resolution for $\mathscr{L}(F(B))$.

Given a projective resolution of $B$

$$P_n \to \ldots \to P_1 \to P_0 \to B \to 0$$ we get the n-th derived functor by taking the homology of

$$F(P_n) \to \ldots \to F(P_1) \to F(P_0) \to 0$$

I am now wondering - how do I form a projective resolution for $\mathscr{L}_m F(B)$?

The problem I can see is that $\mathscr{L}_0 F(B)$ is only right exact and $\mathscr{L}_n F$ is half exact.

(I think I should set this up as an induction of $n$).

Any hints? (On how to form the projective resolution)?

Solution 1:

The statement can be made more precise:

$$ L_n L_m F = \begin{cases} L_nF, & \text{if } m = 0, \\0, & \text{if }m \gt 0. \end{cases}$$

The point is that $L_mF$ is (co-)effaceable for $m \gt 0$, that is $L_mF(P) = 0$ for all projective $P$. See my answer here for some background on that.

First of all, it is easy to see that $L_nL_0F \cong L_nF$ by using universality. If $P_{\bullet} \twoheadrightarrow B$ is a projective resolution of $B$ then $L_{n}L_mF(B) = H_{n}(L_mF(P_{\bullet}))$. However, as $P_{n}$ is projective, we have $L_mF(P_n) = 0$ for $m \gt 0$ and hence $L_{n}L_mF(B) = 0$.