Lemma 38.30.4. Let $A \to C$ be a finite locally free ring map of rank $d$. Let $h \in C$ be an element such that $C_ h$ is étale over $A$. Let $J \subset C$ be an ideal. Set $I = \text{Fit}_0(C/J)$ where we think of $C/J$ as a finite $A$-module. Then $IC_ h = JJ'$ for some ideal $J' \subset C_ h$. If $J$ is finitely generated so are $I$ and $J'$.

**Proof.**
We will use basic properties of Fitting ideals, see More on Algebra, Lemma 15.8.4. Then $IC$ is the Fitting ideal of $C/J \otimes _ A C$. Note that $C \to C \otimes _ A C$, $c \mapsto 1 \otimes c$ has a section (the multiplication map). By assumption $C \to C \otimes _ A C$ is étale at every prime in the image of $\mathop{\mathrm{Spec}}(C_ h)$ under this section. Hence the multiplication map $C \otimes _ A C_ h \to C_ h$ is étale in particular flat, see Algebra, Lemma 10.143.8. Hence there exists a $C_ h$-algebra such that $C \otimes _ A C_ h \cong C_ h \oplus C'$ as $C_ h$-algebras, see Algebra, Lemma 10.143.9. Thus $(C/J) \otimes _ A C_ h \cong (C_ h/J_ h) \oplus C'/I'$ as $C_ h$-modules for some ideal $I' \subset C'$. Hence $IC_ h = JJ'$ with $J' = \text{Fit}_0(C'/I')$ where we view $C'/J'$ as a $C_ h$-module.
$\square$

## 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Your email address will not be published.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: