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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)?

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On the first glance, it can't be true — or $L_m F$ would always be exact. What is the source of the question? –  Grigory M Jun 27 '11 at 8:53
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@Grigory: You don't need right exactness of $F$ to define its left derived functors. You only need that to ensure that $L_0F = F$. In general $L_0F$ is the best right exact approximation to $F$ (the right exact functors are a reflective subcategory of all additive functors and $L_0$ is the reflection). The precise formulation is a bit messier to state in detail than the customary one and that's probably the reason why most books don't deal with it. However, nothing deep is going on here (meaning in this comment). –  t.b. Jun 27 '11 at 11:16
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The question doesn't quite work as stated. If $F:R-\mathrm{mod}\to S-\mathrm{mod}$, then you won't even be able to apply $L_m(F)$ to $L_n(F)(M)$, as the former requires an $R$ module and you have an $S$ module. Therefore, we need $R=S$. –  Aaron Jun 27 '11 at 16:33
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@Grigory - this is "Homoloigcal Algebra" by Osborne - a book that has a high number of errors, so I guess this could be another. –  Juan S Jun 27 '11 at 23:40
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For reference: it's ex. 6.3a on page 160; and LF for not right-exact functors is, indeed, discussed shortly before that. BTW, 6.3b -- LnL0F=LnF -- is certainly (and obviously) true. –  Grigory M Jun 28 '11 at 6:58

1 Answer 1

up vote 8 down vote accepted

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 $L_{n}L_mF(B) = 0$.

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thank you for the answer. I am going to have a think about it and see if I can work it all out –  Juan S Jun 29 '11 at 8:29
    
@Qwirk: No problem, take your time. Just remember that $L_nG(B)$ is computed by taking a projective resolution of $B$, applying $G$ to it and taking homology. Then take $G = L_mF$. That $L_mF(P) = 0$ for $m \gt 0$ and $P$ projective follows immediately from that by taking the resolution $\cdots \to 0 \to P \twoheadrightarrow P$ of $P$. –  t.b. Jun 29 '11 at 8:40
    
@Qwirk: is there anything unclear in what I'm saying? I'd be happy to expand, but I don't really know what is missing. –  t.b. Jul 1 '11 at 8:17
    
no nothing, just haven't had much time to look at this recently! –  Juan S Jul 4 '11 at 12:03

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