# How to calculate $\text{Ext}(M,N)$?

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.

• You seem to have misunderstood the meaning of left exactness. The indicated complex of $\mathrm{Hom}$-modules is not exact. – Zhen Lin Mar 3 '14 at 16:59
• Hi @Lms, you may want to learn about tex for your next post. Also, any thoughts about why you think this might hold and work you have done so far would help people give better answers – Joe Tait Mar 3 '14 at 17:03
• How is the left exactness working with an infinite sequence for a contravariant functor? – Mai Mar 3 '14 at 17:09
• Here is an example which may help you see your error. Take the ring to be $A=k[x]/x^2$ and $M=N=k$ with $x$ acting as zero (except $C_0 \to k$). Take the resolution to be $C_i=A$ with all maps multiplication by $x$. Then the complex of homs is not exact, indeed the induced maps are zero and every ext group is one-dimensional. – Matthew Towers Mar 3 '14 at 17:40
• So, left exactness can only apply on short exact sequences, not long exact sequences – Mai Mar 3 '14 at 18:02

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 $0 \to \text{Hom}(M,N) \to \text{Hom}(C_0,N) \to \text{Hom}(C_1,N)$.