Lemma 38.30.5. Let $A \to B$ be an étale ring map. Let $a \in A$ be a nonzerodivisor. Let $J \subset B$ be a finite type ideal with $V(J) \subset V(aB)$. For every $\mathfrak q \subset B$ there exists a finite type ideal $I \subset A$ with $V(I) \subset V(a)$ and $g \in B$, $g \not\in \mathfrak q$ such that $IB_ g = JJ'$ for some finite type ideal $J' \subset B_ g$.

**Proof.**
We may replace $B$ by a principal localization at an element $g \in B$, $g \not\in \mathfrak q$. Thus we may assume that $B$ is standard étale, see Algebra, Proposition 10.144.4. Thus we may assume $B$ is a localization of $C = A[x]/(f)$ for some monic $f \in A[x]$ of some degree $d$. Say $B = C_ h$ for some $h \in C$. Choose elements $h_1, \ldots , h_ n \in C$ which generate $J$ over $B$. The condition $V(J) \subset V(aB)$ signifies that $a^ m = \sum b_ i h_ i$ in $B$ for some large $m$. Set $h_{n + 1} = a^ m$. As in Lemma 38.30.4 we take $I = \text{Fit}_0(C/(h_1, \ldots , h_{r + 1}))$. Since the module $C/(h_1, \ldots , h_{r + 1})$ is annihilated by $a^ m$ we see that $a^{dm} \in I$ which implies that $V(I) \subset V(a)$.
$\square$

## 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Your email address will not be published.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)

There are also: