3
$\begingroup$

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!

$\endgroup$
4
  • $\begingroup$ 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
1
$\begingroup$

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 $\mathrm{Ext}^1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.