Computing $\text{Ext}(\mathbb Z_p,\mathbb Z)$ It is well known that $\text{Ext}(\mathbb Z_p,\mathbb Z)$ is the trivial group, because $\mathbb Z_p$ is projective; this seems to be in contradiction with the Exercise 1.1 in Hilton - Stammbach, pag. 88: the two sequences
$$
0 \to \mathbb Z \xrightarrow{\mu} \mathbb Z \xrightarrow{\epsilon}\mathbb Z_3 \to 0
$$
and
$$
0 \to \mathbb Z \xrightarrow{\mu} \mathbb Z \xrightarrow{\epsilon'}\mathbb Z_3 \to 0
$$
where $\mu$ is the multiplication by 3 in $\mathbb Z$ and $\epsilon(n)=n\pmod{3}$, $\epsilon'(n)=n+1\pmod{3}$ are not equivalent, and then represent different elements in the set of extensions of $\mathbb Z_3$ and $\mathbb Z$... Where am I wrong?
 A: Here is one way to see why $\Bbb{Z}/p\Bbb{Z}$ isn't projective. Suppose you take $f$ to be the projection $f : \Bbb{Z} \to \Bbb{Z}/p\Bbb{Z}$. Now if $\Bbb{Z}/p\Bbb{Z}$ is projective then for every map $g : \Bbb{Z}/p\Bbb{Z} \to \Bbb{Z}$ we will get a map $h : \Bbb{Z}/p\Bbb{Z} \to \Bbb{Z}$ such that 
$$f \circ h = g.$$
However the problem now is that there are no homomorphisms from $\Bbb{Z}/p\Bbb{Z}$ to $\Bbb{Z}$ except the trivial homomorphism! 
IMO to compute the first Ext group in your problem, the easiest would be to look at the ses
$$0 \to \Bbb{Z} \stackrel{p\cdot}{\to} \Bbb{Z} \to \Bbb{Z}/p\Bbb{Z} \to 0$$
that gives rise to the following LES in Ext: 
$$0\to \textrm{Hom}(\Bbb{Z}/p\Bbb{Z},\Bbb{Z}) \to \textrm{Hom}(\Bbb{Z},\Bbb{Z}) \stackrel{p\cdot}{\to} \textrm{Hom}(\Bbb{Z},\Bbb{Z}) \to \textrm{Ext}^1(\Bbb{Z}/p\Bbb{Z},\Bbb{Z}) \to 0 \to 0\ldots $$
We have the zeros at the end because $\Bbb{Z}$ is a free abelian group and hence is projective. Since $\textrm{Hom}(\Bbb{Z},\Bbb{Z}) \cong \Bbb{Z}$ by the first isomorphism theorem we get 
$$\textrm{Ext}^1(\Bbb{Z}/p\Bbb{Z},\Bbb{Z}) \cong \Bbb{Z}/p\Bbb{Z}.$$
A: Well, $\mathbb Z_p=\mathbb Z/p\mathbb Z$ certainly isn't projective as a
$\mathbb Z$-module.
Added
Now $\mathrm{Ext}(\mathbb Z/p\mathbb Z,\mathbb Z)$ is cyclic of
order $p$. The zero element corresponds to the split extension of
$\mathbb Z$ by $\mathbb Z_p$. The other elements correspond to extensions
$$0\to \mathbb Z\to \mathbb Z\to \mathbb Z/p\mathbb Z\to0.\qquad\qquad\qquad(*)$$
In this exact sequence the surjection can be taken to map $1\in \mathbb Z$
to any given nonzero $a\in\mathbb Z/p\mathbb Z$. Thus there are
$p-1$ non-isomorphic extensions looking like $(*)$.
