Lemma 88.12.5. Let $\varphi : A \to B$ be an arrow of $\textit{WAdm}^{Noeth}$ satisfying the condition defined in Lemma 88.12.1. Then $A \to B$ is topologically of finite type.
Proof. Let $\mathfrak b \subset B$ be the ideal of topologically nilpotent elements. Choose $b_1, \ldots , b_ r \in B$ which map to generators of $B/\mathfrak b$ over $A$. Choose generators $b_{r + 1}, \ldots , b_ s$ of the ideal $\mathfrak b$. We claim that the image of
\[ \varphi : A[x_1, \ldots , x_ s] \longrightarrow B, \quad x_ i \longmapsto b_ i \]
has dense image. Namely, if $b \in \mathfrak b^ n$ for some $n \geq 0$, then we can write $b = \sum b_ E b_{r + 1}^{e_{r + 1}} \ldots b_ s^{e_ s}$ for multiindices $E = (e_{r + 1}, \ldots , e_ s)$ with $|E| = \sum e_ i = n$ and $b_ E \in B$. Next, we can write $b_ E = f_ E(b_1, \ldots , b_ r) + b'_ E$ with $b'_ E \in \mathfrak b$ and $f_ E \in A[x_1, \ldots , x_ r]$. Combined we obtain $b \in \mathop{\mathrm{Im}}(\varphi ) + \mathfrak b^{n + 1}$. By induction we see that $B = \mathop{\mathrm{Im}}(\varphi ) + \mathfrak b^ n$ for all $n \geq 0$ which mplies what we want as $\mathfrak b$ is an ideal of definition of $B$. $\square$
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Your email address will not be published.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)