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. $\square$

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