I'm learning about ext groups right now, but the example I have to work with is confusing. I was wondering if someone could take a look at it, and help fill in the gaps.
Let $A=\mathbb{Z}/8\mathbb{Z}$, $M=\mathbb{Z}/2\mathbb{Z}$, and $N \in$ mod $A$. I want to calculate Ext$_A^1(M,N)$. The kernel of the surjection $\mathbb{Z}/8\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$ is isomorphic to $\mathbb{Z}/4\mathbb{Z}$, and so we get the exact sequence $$0 \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/8\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0 $$ Applying the Hom$_A(-,N)$ functor, we get $$0 \to \text{Hom}_A(\mathbb{Z}/2\mathbb{Z},N)\to \text{Hom}_A(\mathbb{Z}/8\mathbb{Z},N) \to \text{Hom}_A(\mathbb{Z}/4\mathbb{Z},N) \to \text{Ext}_A(\mathbb{Z}/2\mathbb{Z},N)\to 0. $$ I'm not really sure what it means to "apply" the Hom$_A(-,N)$ functor, but this is the lesser of my questions. My main question is as follows:
The example ends by saying: Hom$_A(\mathbb{Z}/8\mathbb{Z},N) \cong N$ and Hom$_A(\mathbb{Z}/4\mathbb{Z},N) \cong N[4]=\{x \in N:x4=0\}.$ And therefore, $$\text{Ext}_A(\mathbb{Z}/2\mathbb{Z},N) \cong \frac{N[4]}{2N}. $$
It's obvious why Hom$_A(\mathbb{Z}/8\mathbb{Z},N) \cong N$, however, I'm not sure why Hom$_A(\mathbb{Z}/4\mathbb{Z},N)\cong N[4]$. Moreover, where does the $2N$ come from? And why, then, is the ext group isomorphic to the quotient of $N[4]$ and $2N$?
Bear with me because it's been a while - perhaps others could fill in anything missing - I am conjecturing what is happening at a couple of places.
$Hom_{A}(-,N)$ is a contravariant functor, that is why when applied to your original SES, the direction gets reversed. Moreover, if it were exact (the resulting 5 term sequence being exact) there would be no need for $Ext$ here. The new term $Ext$ basically measures the failure of $Hom$ to be exact and that is why you now get a 6 term exact sequence.
Since the new sequence is exact, the map to $Ext$ is surjective and so $Ext$ is isomorphic to the prior module mod the image of the map coming into it. In fact that defines $Ext$.
So now by the first identification you noted you have (simplifying notation):
$N\rightarrow Hom(\mathbb{Z}_4,N)\rightarrow Ext\rightarrow 0$
and then
$N\rightarrow N[4]\rightarrow Ext\rightarrow 0$
by the isomorphism you questioned - this is likely straightforward - I've not tried it.
In the original SES, the first nonzero map is multiplication by $2$, which should translate to $2$ as the first arrow here (or rather its dual):
$N\xrightarrow{2} N[4]\rightarrow Ext\rightarrow 0$
So now we get $Ext \cong N[4]/Im (2)$ and since $Im(2)\cong2N$ you are done once you check the details that I omitted or only suggested were true.