# Why are all of these Ext's not zero?

My confusion has stemmed from the fact that apparently for projective module P that $Ext^{n}_{R}(P,N)=0$ for any $R$-module $N$ and $n\geq 1$. Note that all modules being discussed are $R$ modules where $R$ is a commutative ring.

My understanding of the functor $Ext$ is that if we want to find $Ext^{n}_{R}(A,B)$ then we first need to construct an injective resolution of B. I.e, we need an exact sequence

$$0\rightarrow B \rightarrow I^{0} \rightarrow I^{1} \rightarrow \cdots$$

where each $I^{i}$ is injective. Then we need to apply $Hom(A,-)$ to this sequence (except A). Doing so leads to the following sequence

$$0\rightarrow Hom(A,I^{0}) \rightarrow Hom(A,I^{1}) \rightarrow Hom(A,I^{2}) \rightarrow \cdots$$

Since $Hom(A,-)$ is a left exact functor The above sequence should be exact t every point except at $Hom(A,I^{0})$. To get $Ext_{R}^{n}(A,B)$ we take the nth cohomology of this sequence. Wouldn't this be zero except for $n=0$ since the sequence is still exact? I feel this is wrong, but I can't see why...

In any case my two questions are

Where has my understanding gone wrong in the above?

Why is it for a projective module $P$ that $Ext_{R}^{n}(P,N)=0$ for any $R$-module $N$ and $N\geq 1$?. (Note I am okay with n=1 case).

If $F$ is a left exact functor, it means that if $$0\to A\to B\to C$$ is exact then $$0\to F(A)\to F(B)\to F(C)$$ is exact. It does not mean that if you have a longer exact sequence $$0\to A\to B\to C\to D\to E\to\dots$$ then the entire sequence $$0\to F(A)\to F(B)\to F(C)\to F(D)\to F(E)\to\dots$$ is exact.
So in your case, with $\operatorname{Hom}(A,-)$ being left-exact, you would get only that $$0\to\operatorname{Hom}(A,B)\to\operatorname{Hom}(A,I^0)\to\operatorname{Hom}(A,I^1)$$ is exact. When you drop the first term, you get no exactness at all, so you can't deduce that $\operatorname{Ext}^n(A,B)=0$ for any value of $n$.
On the other hand, if $A$ is projective, then the functor $\operatorname{Hom}(A,-)$ is not just left-exact but exact, and so it preserves all exact sequences.* So when $A$ is projective, your argument does work, and shows that $\operatorname{Ext}^n(A,B)=0$ for all $n\geq 1$.