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

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

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

## 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Your email address will not be published.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.

## Comments (0)