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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 $$0\to \mathfrak{m}^n/\mathfrak{m}^{n+1}\to R/\mathfrak{m}^{n+1}\to R/\mathfrak{m}^n\to 0.$$

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

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

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