Let $p$ be a prime number and $n$ be an integer. Let $A$ be an $\mathbb{Z}/p^n\mathbb{Z}$-algebra of rank $2$ whose discriminant is non invertible. In Serre's book lecture on the mordell Weil theorem (in the appendix at the end, page $194$) he says that $A$ is completely determined by whether $A/pA \cong \mathbb{F}_p^2$ or $A/pA$ isom $Fp^2$ (I can't seem to make the latex work here hope you understand what I mean). The reason there are only those two possibilities for $A/pA$ is that $A/pA$ is a two dimensional reduced algebra over $\mathbb{F}_p$ and as such is a product of finite extensions of $\mathbb{F}_p$.

So my question is how do you prove serre's remark.

I asked this question on math.SE first and someone suggested that I post it here. I think it's probably a little bit easy for this site but anyway since I din't get any response over there so I'm trying here. Here is the link to the original question : http://math.stackexchange.com/posts/421742/edit

functoriallydetermined by $k$. Proof: Write $K=k(c)$ for a root $c$ of a monic irreducible separable $f$. Choose $F\in R[x]$ a monic lift of $f$. By successive approximation, $c$ lifts uniquely to a root of $F$ in $A$. Show $R[x]/(F)\rightarrow A$ is an isomorphism. Etc. QED – user30180 Jun 17 '13 at 2:00