$\operatorname{Ext}^2(A, \mathbb{Z})$ for $A$ abelian

Question

Why for all abelian groups $A$ the second ext-group $\operatorname{Ext}^2(A, \mathbb{Z})$ vanishes? Saw this 'fact' several times without any proof attached to.

Answer 1

Every abelian group $A$ has a free resolution of the form $0 \to \mathbb{Z}^{(n)} \to \mathbb{Z}^{(m)} \to A \to 0$ (i.e. a presentation). Here, $\mathbb{Z}^{(n)}$ means "the" free abelian group on a set of $n$ generators ($n$ may be an infinite cardinal). This comes from the fact that any subgroup of a free abelian group is free abelian: here is a proof from Wikipedia.

Then we compute $\operatorname{Ext}^i(A,\mathbb{Z})$ by applying $\operatorname{Hom}({-},\mathbb{Z})$ to the deleted resolution $0 \to \mathbb{Z}^{(n)} \to \mathbb{Z}^{(m)} \to 0$ and taking the $i$th homology group. There are only two nonzero terms in this deleted resolution, so $\operatorname{Ext}^i(A,\mathbb{Z})$ is automatically $0$ for $i \geq 2$!