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

$$\bigcap _{n\ge 0}{𝔪}^n=(0)$$

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

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 $k=\min (k_1,\mathrm{\dots },k_n)$.