| Version current |
Version 5 |
| \begin{thm} Let $R$ be a discrete valuation ring. Then all nonzero ideals of $R$ are powers of its maximal ideal $\smm$. |
\begin{thm} Let $R$ be a discrete valuation ring. Then all nonzero ideals of $R$ are powers of its maximal ideal $\smm$. |
| \end{thm} |
\end{thm} |
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| \textbf{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 $\pi\neq 0$. |
\textbf{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 |
Let $I=(\alpha)\subset R$ be any ideal; claim $(\alpha)=\smm^k$ for some $k$. By the Krull intersection theorem, we have |
| \[\bigcap_{n\geq 0}\smm^n=(0)\] |
\[\bigcap_{n\geq 0}\smm^n=(0)\] |
| 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 $(\alpha)=(\pi)^k$. |
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 $(\alpha)=(\pi)^k$. |
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| \begin{cor} Let $R$ be a Noetherian local ring with a principal maximal ideal. Then all nonzero ideals are powers of the maximal ideal $\smm$. |
\begin{cor} Let $R$ be a Noetherian local ring with a principal maximal ideal. Then all nonzero ideals are powers of the maximal ideal $\smm$. |
| \end{cor} |
\end{cor} |
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| \textbf{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 $k=\min(k_1,\ldots,k_n)$. |
\textbf{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 $k=\min(k_1,\ldots,k_n)$. |
