What is $R^0F(M)$, $R^nF$ being the right derived functor?

Question

Let $0 \rightarrow M \rightarrow I_0 \rightarrow I_1 \rightarrow \dots$ be an injective resolution of module $M$.

Let $F$ be a left exact functor. Let $I$ be the complex $0\rightarrow I_1 \rightarrow I_1 \rightarrow \dots$.

According to Lang's book on p. 791, $R^nF(M)=H^n(F(I))$, in other words, the $n$-th homology of the complex $0\rightarrow I_1 \rightarrow I_1 \rightarrow \dots$.

But he then said that "we see that there is an isomorphism $R^0F(M)=F(M)$". How did he get that? I thought $R^0F(M)$ is $Ker(I_0 \rightarrow I_1)$ only.

Thanks in advance.

Answer 1

From left exactness. As $$0\to M\to I_0\to I_1$$ is exact, so is $$0\to F(M)\to F(I_0)\to F(I_1).$$ So $F(M)$ is naturally isomorphic to the kernel of $F(I_0)\to F(I_1)$, which by definition is $R^0F(M)$.