# Show that $\text{Ext}^n (X,Y) = 0$ for all $n \ge 2$ and give an example in which $X$ and $Y$ be two Abelian groups but $\text{Ext} (X,Y) \ne 0$

Problems: Let $$X$$ and $$Y$$ be two Abelian groups. Show that $$\text{Ext}^n (X,Y) = 0$$ for all $$n \ge 2$$. Give an example in which $$X$$ and $$Y$$ be two Abelian groups but $$\text{Ext} (X,Y) \ne 0$$.

My attempt: Consider the exact sequence $$K \colon 0 \rightarrow A \rightarrow F \rightarrow X \rightarrow 0$$ $$F$$ be a free Abelian group, implies $$A$$ is a free group. Hence, $$K$$ be a projective resolution of $$X$$. We have $$\text{Hom}(K,Y) \colon 0 \rightarrow \text{Hom}(A,Y) \rightarrow \text{Hom}(F,Y) \rightarrow \text{Hom}(X,Y) \rightarrow 0$$ is also an exact sequence.

Could you give me some suggestion to continue the proof? Thank all!

• Consider $H^1(\Bbb Z/p\Bbb Z,\Bbb Z/p\Bbb Z)$. Jan 5 '20 at 8:07
• @LordSharktheUnknown In the first question, does the proof complete?
– Minh
Jan 5 '20 at 8:53
• You get Ext by computing the homology of $\text{Hom}(K^\bullet, Y)$ where $K^\bullet$ is a projective resolution of $X$. But your projective resolution of $X$ is zero in dimensions $\ge2$. Jan 5 '20 at 8:55
• @LordSharktheUnknown I know the fact, but I'm trying to prove this.
– Minh
Jan 5 '20 at 8:59

You are on the right line by finding a projective resolution of $$X$$, since $$\mathrm{Hom}(-,Y)$$ is a contravariant left exact functor, and so to compute the right derived functors of it we need to apply it to a projective resolution of $$X$$. Really, the projective resolution of $$X$$ is the chain complex $$0\leftarrow F \leftarrow A \leftarrow 0$$ where $$F$$ is in degree $$0$$. Then, when we apply $$\mathrm{Hom}(-,Y)$$ to this projective resolution, we get a cochain complex $$0 \rightarrow \mathrm{Hom}(F,Y) \rightarrow \mathrm{Hom}(A,Y) \rightarrow 0$$ where $$\mathrm{Hom}(F,Y)$$ is in degree $$0$$. Now the group $$\mathrm{Ext}^n(X,Y)$$ is the $$n$$-th cohomology group of this complex. But since the complex is $$0$$ in degrees $$\ge2$$, that must mean $$\mathrm{Ext}^n(X,Y)=0$$ for all $$n\ge 2$$.
The exact sequence $$0\rightarrow\mathrm{Hom}(X,Y) \rightarrow \mathrm{Hom}(F,Y) \rightarrow \mathrm{Hom}(A,Y)$$ shows, as always, that $$\mathrm{Ext}^0(X,Y)=\mathrm{Hom}(X,Y)$$. So the only interesting ext group to calculate in this case is $$\mathrm{Ext}^1$$.