I'm stuck in the exercise 1 of section 6.6, Derivative Functors of Jacobson's book "Basic Algebra II". It states:
If $M$ is a projective module, then $(L_0 F)M = FM$ and $(L_n F) M = 0, \, \forall n \geq 1.$, where $L_n F$ denotes the left derived functor of $F$.
My attempt: Given a projective resolution $(C, \epsilon)$ of $M$, i.e. $$ \cdots \to C_2 \overset{d_2}{\rightarrow} C_1\overset{d_1}{\rightarrow} C_0 \overset{\epsilon}{\rightarrow} M\to 0 \quad \quad (I) $$ is an exact sequence and each $C_n, \, \neq 0$ is a projective module. Applyng the functor $F$ to the complex $C$, we have $$ \cdots \to FC_2 \overset{Fd_2}{\rightarrow} FC_1\overset{Fd_1}{\rightarrow} FC_0 \overset{Fd_0}{\rightarrow} 0 \quad \quad (II)$$ and for the sequence $(I)$: $$ \cdots \to FC_2 \overset{Fd_2}{\rightarrow} FC_1\overset{Fd_1}{\rightarrow} FC_0 \overset{F\epsilon}{\rightarrow} FM \to 0. $$
We also have that $(L_0 F)M = H_0(FC) = \ker Fd_0 \,/\, \text{im}\, Fd_1 = FC_0 \,/\, \text{im}\, Fd_1$ and that $(L_n F)M = H_n(FC) = \ker Fd_n \,/\, \text{im}\, Fd_{n+1}$.
How can I proceed with this question?
The key idea is that you can use whichever projective resolution of $M$ you want to compute $(L_nF)(M)$. In particular, when $M$ itself is projective, there's a really easy projective resolution: $$\cdots\to 0\to 0\to 0\to M\stackrel{1}\to M\to 0,$$ with $C_0=M$ and $C_n=0$ for all $n\geq 1$. Applying the definition, we see that $(L_nF)(M)$ is the $n$th homology of the chain complex $$\cdots\to F(0)\to F(0)\to F(0)\to F(M)\to 0.$$ Since $F(0)=0$, the $n$th term of this complex is $0$ for all $n\geq 1$, and so the $n$th homology is also $0$ for all $n\geq 1$.