Lemma 10.64.3. Let $R \to S$ be flat ring map. Let $\mathfrak p \subset R$ be a prime such that $\mathfrak q = \mathfrak p S$ is a prime of $S$. Then $\mathfrak p^{(n)} S = \mathfrak q^{(n)}$.

Proof. Since $\mathfrak p^{(n)} = \mathop{\mathrm{Ker}}(R \to R_\mathfrak p/\mathfrak p^ nR_\mathfrak p)$ we see using flatness that $\mathfrak p^{(n)} S$ is the kernel of the map $S \to S_\mathfrak p/\mathfrak p^ nS_\mathfrak p$. On the other hand $\mathfrak q^{(n)}$ is the kernel of the map $S \to S_\mathfrak q/\mathfrak q^ nS_\mathfrak q = S_\mathfrak q/\mathfrak p^ nS_\mathfrak q$. Hence it suffices to show that

$S_\mathfrak p/\mathfrak p^ nS_\mathfrak p \longrightarrow S_\mathfrak q/\mathfrak p^ nS_\mathfrak q$

is injective. Observe that the right hand module is the localization of the left hand module by elements $f \in S$, $f \not\in \mathfrak q$. Thus it suffices to show these elements are nonzerodivisors on $S_\mathfrak p/\mathfrak p^ nS_\mathfrak p$. By flatness, the module $S_\mathfrak p/\mathfrak p^ nS_\mathfrak p$ has a finite filtration whose subquotients are

$\mathfrak p^ iS_\mathfrak p/\mathfrak p^{i + 1}S_\mathfrak p \cong \mathfrak p^ iR_\mathfrak p/\mathfrak p^{i + 1}R_\mathfrak p \otimes _{R_\mathfrak p} S_\mathfrak p \cong V \otimes _{\kappa (\mathfrak p)} (S/\mathfrak q)_\mathfrak p$

where $V$ is a $\kappa (\mathfrak p)$ vector space. Thus $f$ acts invertibly as desired. $\square$

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