Equivalence classes of extensions of ${\bf Z}_m$ by ${\bf Z}$ Problem : What is the set of equivalence classes of extensions of ${\bf Z}_m$ by
${\bf Z}$ ?
Try : Note that $$ {\rm Ext}_{\bf Z}^1 (A:={\bf Z}_m,N)=N/mN $$
 where $N$ is an abelian. From definition of Ext by using cochain and long exact
 sequence theorem for cohomology, this can be proved. But by another
 definition, it is an equivalence classes of extensions $E$ : $$
  N \rightarrowtail E \twoheadrightarrow A $$
Note that if $N={\bf Z}$, there exist two choices : As far as I
know,
$$ E_1= {\bf Z},\ E_2={\bf Z}\times {\bf Z}_m$$
So $ {\rm Ext}_{\bf Z}^1 (A:={\bf Z}_m,N)={\bf Z}_2$ What I miss
something ?
 A: You want to compute $\def\Ext{\operatorname{Ext}}\def\Z{\mathbb{Z}}\Ext(\Z/m\Z,\Z)$ which is usually referred to as the group of extensions by $\Z/m\Z$ of $\Z$; some author exchanges by and of, but losing in clarity. These corresponds to classes of equivalences of exact sequences $0\to\Z\to G\to\Z/m\Z\to0$
Note that $\Ext(\Z,\Z/m\Z)$ is the trivial group, because $\Z$ is projective.
Consider the canonical exact sequence $0\to\Z\xrightarrow{\cdot m}\Z\to\Z/m\Z\to0$ (where $\cdot m$ means multiplication by $m$) and apply to it the functor $\def\Hom{\operatorname{Hom}}\Hom(-,\Z)$ that gives the long exact sequence
$$
0\to\Hom(\Z/m\Z,\Z)\to\Hom(\Z,\Z)\xrightarrow{\cdot m}\Hom(\Z,\Z)\to
\Ext(\Z/m\Z,\Z)\to\Ext(\Z,\Z)
$$
Now $\Hom(\Z/m\Z,\Z)=0$, $\Hom(\Z,\Z)\cong\Z$ and $\Ext(\Z,\Z)=0$, which proves that $\Ext(\Z/m\Z,\Z)\cong\Z/m\Z$.
Now, where are you doing a mistake?
Let's do it more generally; how can we build an exact sequence $0\to B\to G\to A\to0$ from an element of $\Ext(A,B)$ when this is defined as a derived functor? Consider a projective resolution $0\to F_1\to F_2\to A\to 0$, to get
$$
0\to \Hom(A,B)\to\Hom(F_2,B)\to\Hom(F_1,B)\to\Ext(A,B)\to0
$$
because $\Ext(F_2,B)=0$, being $F_2$ projective. An element $\tau\in\Ext(A,B)$ comes from some $f\colon F_1\to B$ and you can construct the pushout diagram
$$\require{AMScd}
\begin{CD}
0 @>>> F_1 @>>> F_2 @>>> A @>>> 0 \\
@. @VfVV @VVV @| @. \\
0 @>>> B @>>> G @>>> A @>>> 0
\end{CD}
$$
which is exactly what you need.
In the case of $A=\Z/m\Z$, you can take the above sequence and it should be clear how to proceed.
