# ideals of a discrete valuation ring are powers of its maximal ideal

###### Theorem 1.

Let $R$ be a discrete valuation ring. Then all nonzero ideals of $R$ are powers of its maximal ideal $\mathfrak{m}$.

Proof. Let $\mathfrak{m}=(\pi)$ (that is, $\pi$ is a uniformizer for $R$). Assume that $R$ is not a field (in which case the result is trivial), so that $\pi\neq 0$. Let $I=(\alpha)\subset R$ be any ideal; claim $(\alpha)=\mathfrak{m}^{k}$ for some $k$. By the Krull intersection theorem, we have

 $\bigcap_{n\geq 0}\mathfrak{m}^{n}=(0)$

so that we may choose $k\geq 0$ with $\alpha\in\mathfrak{m}^{k}-\mathfrak{m}^{k+1}$. Since $\alpha\in\mathfrak{m}^{k}$, we have $\alpha=u\pi^{k}$ for $u\in R$. $u\notin\mathfrak{m}$, since otherwise $\alpha\in\mathfrak{m}^{k+1}$, so that $\alpha$ is a unit (in a DVR, the maximal ideal consists precisely of the nonunits). Thus $(\alpha)=(\pi)^{k}$.

###### Corollary 1.

Let $R$ be a Noetherian local ring with a principal maximal ideal. Then all nonzero ideals are powers of the maximal ideal $\mathfrak{m}$.

Proof. Let $I=(\alpha_{1},\ldots,\alpha_{n})$ be an ideal of $R$. Then by the above argument, for each $i$, $\alpha_{i}=u_{i}\pi^{k_{i}}$ for $u_{i}$ a unit, and thus $I=(\pi^{k_{1}},\ldots,\pi^{k_{n}})=(\pi^{k})$ for $k=\min(k_{1},\ldots,k_{n})$.

Title ideals of a discrete valuation ring are powers of its maximal ideal IdealsOfADiscreteValuationRingArePowersOfItsMaximalIdeal 2013-03-22 18:00:47 2013-03-22 18:00:47 rm50 (10146) rm50 (10146) 9 rm50 (10146) Theorem msc 13H10 msc 13F30 PAdicCanonicalForm IdealDecompositionInDedekindDomain