Why $p^i/p^{i+1}$ is a vector space over $O_k/p$

Question

Let $O_K$ be a number ring of number field $K$ and $p$ be a prime ideal of $O_K$. Consider the chain $$p\subseteq p^2\subseteq ...\subseteq p^v$$ If we choose $a\in p^i\setminus p^{i+1}$, and $b=(a)+p^{i+1}$, so, $p^{i+1}\subset b\subseteq p^i$ so, $b=p^i$.

So, $p^i/p^{i+1}$ is generated by a single element, $a\;mod\;p^{i+1}$. But why it generate $p^i/p^{i+1}$ over $O_K/p$ is what I am not getting.

This is a specific proof form Newurich, page-35, prop-6.1.

Answer 1

For $R$ a commutative ring and $I$ an ideal then $I^n/I^{n+1}$ is a $R/I$-module: it is clearly a $R$-module and for $b\in I$ we have $b . I^n/I^{n+1}=0$.

Then for $K$ a number field and $P$ a maximal ideal of $O_K$ the point is that $P$ becomes principal $=(\pi)$ in $O_K/P^{n+1}$, so does $P^n=(\pi^n)$, whence $b\in O_K\to \pi^n b \in P^n/P^{n+1}$ is surjective and $b\to \pi^n b$ is a $O_K/P$-module ($O_K/P$-vector space) isomorphism $O_K/P\to P^n/P^{n+1}$.

Compare with $R=\Bbb{Q}[x,y]$, the maximal ideal $m=(x,y)$ is not invertible, it is not principal in $R/m^{2}$ and $m/m^2 = \Bbb{Q}x +\Bbb{Q}y$ is a $2$-dimensional $\Bbb{Q}=R/m$-vector space.

Answer 2

Obviously $P^n/P^{n+1}$ is an $O/P$-module. But since $O$ is a Dedekind ring, $P$ is maximal, so that 1) $O/P$ is a field ; 2) there is no ideal which is strictly "sandwiched" between $P^n$ and $P^{n+1}$. This means that $P^n/P^{n+1}$ is a vector space of dimension 1 over the residual field $O/P$.