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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