Length and valuation in a DVR

The case $n=0$ is $V\mathrm{/}(1)=0$, whose length is $0$ by [F1]. For $n=1$, the ideals of $V\mathrm{/}(\pi )$ correspond to the ideals of $V$ containing $(\pi )$. By [L2], those are only $(\pi )$ and $V$, so $V\mathrm{/}(\pi )$ is simple and has length $1$.

For each $n\ge 1$ there is a short exact sequence $0\to ({\pi }^n)\mathrm{/}({\pi }^{n+1})\to V\mathrm{/}({\pi }^{n+1})\to V\mathrm{/}({\pi }^n)\to 0$. Multiplication by ${\pi }^n$ induces an isomorphism $V\mathrm{/}(\pi )\cong ({\pi }^n)\mathrm{/}({\pi }^{n+1})$, so step 1.1 gives ${\mathrm{\ell }}_V(({\pi }^n)\mathrm{/}({\pi }^{n+1}))=1$. Therefore [L3] yields ${\mathrm{\ell }}_V(V\mathrm{/}({\pi }^{n+1}))={\mathrm{\ell }}_V(V\mathrm{/}({\pi }^n))+1$.

By induction on $n$, step 1.1 and step 2.1 give ${\mathrm{\ell }}_V(V\mathrm{/}({\pi }^n))=n$ for every $n\ge 0$.

Let $x\in V$ be nonzero, and write $x=u{\pi }^n$ as in [L1]. Multiplication by the unit $u^{-1}$ identifies the ideals $(x)$ and $({\pi }^n)$, so $V\mathrm{/}(x)\cong V\mathrm{/}({\pi }^n)$ as $V$-modules. Hence ${\mathrm{\ell }}_V(V\mathrm{/}(x))={\mathrm{\ell }}_V(V\mathrm{/}({\pi }^n))=n$.