Does taking restriction and left derived functors commute?

Question

Let $R$ be a commutative Noetherian ring. Let Mod$(R)$ denote the category of $R$-modules, and mod$(R)$ denote the category of finitely generated $R$-modules; notice that both of these categories are abelian and has enough projective objects.

For a functor $F: \text{Mod}(R) \to \text{Mod}(R)$, if $F(M)\in$mod$(R)$ for every $M\in \text{mod}(R)$, then let $F|_\text{mod}$ denote the restricted functor $F|_\text{mod}: \text{mod}(R) \to \text{mod}(R)$ where for every $M,N \in \text{mod}(R)$, $F|_\text{mod}(M):=F(M)$ and $F|_\text{mod}(f):= F(f) $ for every $f\in \text{Hom}(M,N)$.

Now let $F: \text{Mod}(R) \to \text{Mod}(R)$ be an $R$-linear right exact functor satisfying $F(\text{mod}(R))\subseteq \text{mod}(R) $. So $F|_\text{mod}: \text{mod}(R) \to \text{mod}(R)$ is also right exact and $R$-linear.

We can consider the left derived functors (https://en.m.wikipedia.org/wiki/Derived_functor) $L_i F: \text{Mod}(R) \to \text{Mod}(R)$ and $L_i (F|_\text{mod}): \text{mod}(R) \to \text{mod}(R)$ . Since

$F(\text{mod}(R))\subseteq \text{mod}(R) $ holds, so

$L_iF(\text{mod}(R))\subseteq \text{mod}(R) $ holds for every $i\ge 0$, so for every $i\ge 0$, I can consider the restriction $(L_iF)|_\text{mod}: \text{mod}(R) \to \text{mod}(R)$.

My question is: Do we have an isomorphism of functors $L_i (F|_\text{mod}) \cong (L_iF)|_\text{mod}$ for all $i\ge 0$ ?

Since different choices of projective resolutions ultimately give Isomorphic derived functors, so I can clearly see that $L_i (F|_\text{mod}) (M) \cong (L_iF)|_\text{mod}(M)$ for all $M \in \text{mod}(R)$ and for all $i\ge 0$. However, I am not sure how to show if the corresponding functors are Isomorphic.

Please help.

Answer 1

The key property for defining left derived functors is that the homology modules don't depend on the particular projective resolution.

If you have a module $M$ and two projective resolutions thereof, say \begin{align} \dots\to P_n\to P_{n-1}\to\dots\to P_1&\to P_0\to M\to 0 \\[1ex] \dots\to P_n'\to P_{n-1}'\to\dots\to P_1'&\to P_0'\to M\to 0 \end{align} then upon applying $F$, the complexes you obtain have the same homology.

Since a finitely generated module has a projective resolution consisting of finitely projective modules, due to $R$ being Noetherian, the left derived functors of a functor that restricts to $\mathrm{mod}(R)$ are the same when computed over $\mathrm{Mod}(R)$ or when computed for the restriction.

Answer 2

I think the easiest way to understand what is going on is to think about the following more general situation.

Proposition. Let $\mathcal{C}, \mathcal{C}', \mathcal{D}, \mathcal{D}'$ be abelian categories, let $F : \mathcal{C} \to \mathcal{D}$ and $F' : \mathcal{C}' \to \mathcal{D}'$ be right exact functors, and let $I : \mathcal{C}' \to \mathcal{C}$ and $J : \mathcal{D}' \to \mathcal{D}$ be exact functors. Assuming $\mathcal{C}'$ and $\mathcal{C}$ have enough projectives, if $F I \cong J F'$ (as functors) and $I : \mathcal{C}' \to \mathcal{C}$ preserves projectives, then $(\mathrm{L}_* F) I \cong J (\mathrm{L}_* F')$.

The proof is easy. The composite functors $F I$ and $J F'$ are right exact and isomorphic by hypothesis, so their left derived functors must also be isomorphic, i.e. $\mathrm{L}_* (F I) \cong \mathrm{L}_* (J F')$. Since $J$ is exact, we have $\mathrm{L}_* (J F') \cong J (\mathrm{L}_* F')$. (Consider how each one is computed!) On the other hand, since $I$ is exact and preserves projectives, it preserves projective resolutions in particular, so $\mathrm{L}_* (F I) \cong (\mathrm{L}_* F) I$. Composing the three isomorphisms, we obtain $(\mathrm{L}_* F) I \cong J (\mathrm{L}_* F')$, as claimed.

The hypotheses of the proposition are not optimal, but the situation you are considering fits.