Lemma 86.12.5. Let $\varphi : A \to B$ be an arrow of $\textit{WAdm}^{Noeth}$ satisfying the condition defined in Lemma 86.12.1. 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 $B$. $\square$

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