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!

  • $\begingroup$ Consider $H^1(\Bbb Z/p\Bbb Z,\Bbb Z/p\Bbb Z)$. $\endgroup$ Jan 5 '20 at 8:07
  • $\begingroup$ @LordSharktheUnknown In the first question, does the proof complete? $\endgroup$
    – Minh
    Jan 5 '20 at 8:53
  • $\begingroup$ 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 $\ge2$. $\endgroup$ Jan 5 '20 at 8:55
  • $\begingroup$ @LordSharktheUnknown I know the fact, but I'm trying to prove this. $\endgroup$
    – 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 $\mathrm{Ext}^1$.


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