I'm learning some properties of discrete valuation rings (DVR's further for geometrical use). By the way, a domain $R$ is said to be a DVR if there exists the so called uniformizing parameter $t$ such that each $a\in R\setminus\{0\}$ has the unique representation of the form $u\cdot t^k$ where $u$ is invertible and $k\in \mathbb{Z}_{\ge 0}$ (this is also equivalent for $A$ to be a noetherian local domain with principal maximal ideal that should be generated by the uniformizing parameter).

Suppose a DVR $R$ contains a field $\mathbb{k}$ as a subring: $\mathbb{k}\subset R$ and let $\mathfrak{m}$ be the maximal ideal in $R$. We also suppose the composition of the inclusion $\mathbb{k}\hookrightarrow R$ and of the projection $R\to R/ {\mathfrak{m}}$ to be an isomorphism. (For example, $R=\mathcal{O}_p(\mathbb{A}^1)$.)

The last means, in particular, that $\mathrm{dim}_{\mathbb{k}}(R/ {\mathfrak{m}})=1$.

I'm trying to prove that $\mathrm{dim}_{\mathbb{k}}(R/{\mathfrak{m}^n})=n$ for all natural $n$. Could you please give me a hint?

I also understand that this is the same as $\mathrm{dim}_{\mathbb{k}}(\mathfrak{m}^n/\mathfrak{m}^{n+1})=1$ because of exactness of the sequence $$0\to \mathfrak{m}^n/\mathfrak{m}^{n+1}\to R/\mathfrak{m}^{n+1}\to R/\mathfrak{m}^n\to 0.$$


$\mathfrak m^n/\mathfrak m^{n+1}=(t^n)/(t^{n+1})\simeq R/(t)=R/\mathfrak m$

  • $\begingroup$ Oh,really. Thanks a lot! $\endgroup$ – Mikolay Oct 8 '14 at 16:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.