Lemma 88.6.2. Let $I$ be an ideal of a Noetherian ring $A$. Let $C^ h$ be the henselization of a finite type $A$-algebra $C$ with respect to the ideal $IC$. Let $J \subset C^ h$ be an ideal. Then there exists a finite type $A$-algebra $B$ such that $B^\wedge \cong (C^ h/J)^\wedge $.
Proof. By More on Algebra, Lemma 15.12.4 the ring $C^ h$ is Noetherian. Say $J = (g_1, \ldots , g_ m)$. The ring $C^ h$ is a filtered colimit of étale $C$ algebras $C'$ such that $C/IC \to C'/IC'$ is an isomorphism (see proof of More on Algebra, Lemma 15.12.1). Pick an $C'$ such that $g_1, \ldots , g_ m$ are the images of $g'_1, \ldots , g'_ m \in C'$. Setting $B = C'/(g'_1, \ldots , g'_ m)$ we get a finite type $A$-algebra. Of course $(C, IC)$ and $C', IC')$ have the same henselizations and the same completions. It follows easily from this that $B^\wedge = (C^ h/J)^\wedge $. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)