Find a basis for $B/\mathfrak{B}^e$

Question

In the context of algebraic integers, I would like to solve te following problem. Let $A \subset B$ be two rings, $\mathfrak{p}$ a prime ideal of $A$ and $\mathfrak{B}$ a prime ideal of $B$ lying above $\mathfrak{p}$ (those prime ideals are maximal). Let $\beta$ be an element of $B$ such that its residue class modulo $\mathfrak{B}$ generates $B/\mathfrak{B}$ over $A/\mathfrak{p}$. Let $\Pi$ be an element of $\mathfrak{B}$ but $\Pi \notin \mathfrak{B}^2$.

I would like to prove that the elements $\beta^j\Pi^i$ generate $B/\mathfrak{B}^e$ as a vector space over $A/\mathfrak{p}$.

I do already that we have the following decomposition.

$B \supset \mathfrak{B} \supset \mathfrak{B}^2 \supset \dots \supset \mathfrak{B}^e$

Where $\mathfrak{B}^{i+1}/\mathfrak{B}^i$ is isomorphic to $B/\mathfrak{B}$ with isomorphism 'multiplying be $\Pi^i$'.

What I would like to prove seems quite obvious but I keep failing to find a rigorous argument.

Answer 1

I have been stuck on a similar question myself. I'll write some of what I have so far, and hopefully you will find it useful.

As you said, the map $B/\mathfrak B \rightarrow \mathfrak B^k/\mathfrak B^{k+1}$ given by $a + \mathfrak B \mapsto a \Pi^k + \mathfrak B^k$ is a well defined isomorphism of $A/P$-modules.

The sequence $$ 0 \rightarrow \mathfrak B/ \mathfrak B^2 \rightarrow B/\mathfrak B^2 \rightarrow B/ \mathfrak B \rightarrow 0$$ is exact with $B/ \mathfrak B$ projective (any module over a field is projective, hence free), so we can conclude that $B/ \mathfrak B^2 \cong B/ \mathfrak B \oplus \mathfrak B/ \mathfrak B^2$. More generally for $1 \leq k \leq e$ the sequence $$ 0 \rightarrow \mathfrak B^{k-1}/ \mathfrak B^k \rightarrow B/ \mathfrak B^k \rightarrow B/ \mathfrak B^{k-1} \rightarrow 0 $$ is split exact, giving us $B/ \mathfrak B^k \cong B/ \mathfrak B^{k-1} \oplus \mathfrak B^{k-1}/ \mathfrak B^{k}$, which by induction is isomorphic to $B/ \mathfrak B \oplus \mathfrak B / \mathfrak B^2 \oplus \cdots \oplus \mathfrak B^{k-1}/\mathfrak B^k$.

The isomorphisms here aren't explicit, so perhaps by working out the details of the where the maps go you can see $B/ \mathfrak B^e$ as an internal direct sum of $A/P$-modules rather than an external direct sum as I've constructed.