Proof. Let $\smm = (\pi)$ (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 $(\alpha)=\smm^k$ for some $k$ . By the Krull intersection theorem, we have $$ \bigcap_{n\geq 0}\smm^n=(0 $$ so that we may choose $k\geq 0$ with $\alpha\in \smm^k-\smm^{k+1}$ . Since $\alpha\in\smm^k$ , we have $\alpha = u\pi^k$ for $u\in R$ . $u\notin \smm$ , since otherwise $\alpha\in\smm^{k+1}$ , so that $\alpha$ is a unit (in a DVR, the maximal ideal consists precisely of the nonunits). Thus $(\alpha)=(\pi)^k$ .

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 $k=\min(k_1,\ldots,k_n)$ .