I am confused about the calculation of $\text{Ext}(M,N)$. If $N$ is a fixed module and if we consider a projective resolution
$$\cdots \to C_1 \to C_0 \to M \to 0,$$ then $\text{Ext}^n(M,N)$ is the $n^{\text{th}}$ homology of
$$0 \to \text{Hom}(C_0,N) \to \text{Hom}(C_1,N)\to \cdots$$
Since the $\text{Hom}$ functor is left exact, we have
$$0 \to \text{Hom}(M,N) \to \text{Hom}(C_0,N) \to \text{Hom}(C_1,N)\to\cdots \to \text{Hom}(C_{n+1},N)$$ is exact (is this true?) so clearly $$\text{Ext}^0(M,N) = \text{Hom}(M,N),$$
but since the above sequence is exact, does this imply that $\text{Ext}^1(M,N)=0\,$? Any help would be greatly appreciated.
Best Answer
The $Hom$ functor is left exact but NOT exact. So it is not true that the above complex is exact, it's only a complex. your idea about the equality of "ext" and "hom" at point $0$ is correct because of "left exactness".
In fact it is exact till $0 \to \text{Hom}(M,N) \to \text{Hom}(C_0,N) \to \text{Hom}(C_1,N)$.