# 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^{} $\mathrm{m}$.

Proof. Let $\U0001d52a=(\pi )$ (that is, $\pi $ is a uniformizer for $R$). Assume that $R$ is not a field (in which case the result is trivial), so that $\pi \ne 0$. Let $I=(\alpha )\subset R$ be any ideal; claim $(\alpha )={\U0001d52a}^{k}$ for some $k$. By the Krull intersection theorem, we have

$$\bigcap _{n\ge 0}{\U0001d52a}^{n}=(0)$$ |

so that we may choose $k\ge 0$ with $\alpha \in {\U0001d52a}^{k}-{\U0001d52a}^{k+1}$. Since $\alpha \in {\U0001d52a}^{k}$, we have $\alpha =u{\pi}^{k}$ for $u\in R$. $u\notin \U0001d52a$, since otherwise $\alpha \in {\U0001d52a}^{k+1}$, 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 $\mathrm{m}$.

Proof. Let $I=({\alpha}_{1},\mathrm{\dots},{\alpha}_{n})$ 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 $I=({\pi}^{{k}_{1}},\mathrm{\dots},{\pi}^{{k}_{n}})=({\pi}^{k})$ for $k=\mathrm{min}({k}_{1},\mathrm{\dots},{k}_{n})$.

Title | ideals of a discrete valuation ring are powers of its maximal ideal |
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Canonical name | IdealsOfADiscreteValuationRingArePowersOfItsMaximalIdeal |

Date of creation | 2013-03-22 18:00:47 |

Last modified on | 2013-03-22 18:00:47 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 9 |

Author | rm50 (10146) |

Entry type | Theorem |

Classification | msc 13H10 |

Classification | msc 13F30 |

Related topic | PAdicCanonicalForm |

Related topic | IdealDecompositionInDedekindDomain |