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 $n\geq 1$.

*You may be familiar only with the definition that an exact functor preserves short exact sequences. However, this implies that it preserves all exact sequences, since you can express exactness of any sequence in terms of certain other sequences being short exact. See my answer here for more details.

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