The Stacks project

Lemma 51.16.5. Let $(A, \mathfrak m)$ be a complete Noetherian local domain. Let $\mathfrak p \subset A$ be a prime ideal of dimension $1$. For every $n \geq 1$ there is an $m \geq n$ such that $\mathfrak p^{(m)} \subset \mathfrak p^ n$.

Proof. Recall that the symbolic power $\mathfrak p^{(m)}$ is defined as the kernel of $A \to A_\mathfrak p/\mathfrak p^ mA_\mathfrak p$. Since localization is exact we conclude that in the short exact sequence

\[ 0 \to \mathfrak a_ n \to A/\mathfrak p^ n \to A/\mathfrak p^{(n)} \to 0 \]

the support of $\mathfrak a_ n$ is contained in $\{ \mathfrak m\} $. In particular, the inverse system $(\mathfrak a_ n)$ is Mittag-Leffler as each $\mathfrak a_ n$ is an Artinian $A$-module. We conclude that the lemma is equivalent to the requirement that $\mathop{\mathrm{lim}}\nolimits \mathfrak a_ n = 0$. Let $f \in \mathop{\mathrm{lim}}\nolimits \mathfrak a_ n$. Then $f$ is an element of $A = \mathop{\mathrm{lim}}\nolimits A/\mathfrak p^ n$ (here we use that $A$ is complete) which maps to zero in the completion $A_\mathfrak p^\wedge $ of $A_\mathfrak p$. Since $A_\mathfrak p \to A_\mathfrak p^\wedge $ is faithfully flat, we see that $f$ maps to zero in $A_\mathfrak p$. Since $A$ is a domain we see that $f$ is zero as desired. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0EB5. Beware of the difference between the letter 'O' and the digit '0'.