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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 : https://math.stackexchange.com/posts/421742/edit

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  • $\begingroup$ For artin local $(R,m)$ and $k:=R/m$, a finite flat $R$-algebra $A$ has disc$_R(A)\in R^{\times}$ off disc$_k(A/mA)\ne 0$ iff $A/mA=\prod k_i$ for fields $k_i/k$ separable. Via the unique decomposition of $A$ as a product of $R$-flat local rings, a finite flat $R$-algebra $A$ with $K:=A/mA$ a field separable over $k$ is functorially determined 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 $\endgroup$
    – user30180
    Commented Jun 17, 2013 at 2:00
  • $\begingroup$ Hi, thank you very much for you answer. I'm not really sure how to fully write down your argument, especially the part where you say that you can decompose $A$ as a product of $R-flat$ local rings (maybe can you tell me how to do this in my case more specifically) and the part where you take the lift of $f$ and then lifting the root bu successive approximation. I guess this is some kind of Hensel lemma right ? Would you maybe consider writing an answer with a little bit more details ? $\endgroup$
    – vdd
    Commented Jun 17, 2013 at 7:45

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