Lemma 88.17.7. Let $A \to B \to C$ be arrows in $\textit{WAdm}^{Noeth}$ which are adic and topologically of finite type. If $B \to C$ is rig-smooth, then the kernel of the map
(see Lemma 88.3.5) is annihilated by an ideal of definition.
Proof. Let $\overline{\mathfrak q} \subset C$ be a prime ideal which does not contain an ideal of definition. Since the modules in question are finite it suffices to show that
is injective. As in the proof of Lemma 88.3.5 choose presentations $B = A\{ x_1, \ldots , x_ r\} /J$, $C = B\{ y_1, \ldots , y_ s\} /J'$, and $C = A\{ x_1, \ldots , x_ r, y_1, \ldots , y_ s\} /K$. Looking at the diagram in the proof of Lemma 88.3.5 we see that it suffices to show that $J/J^2 \otimes _ B C \to K/K^2$ is injective after localization at the prime ideal $\mathfrak q \subset A\{ x_1, \ldots , x_ r, y_1, \ldots , y_ s\} $ corresponding to $\overline{\mathfrak q}$. Please compare with More on Algebra, Lemma 15.33.6 and its proof. This is the same as asking $J/KJ \to K/K^2$ to be injective after localization at $\mathfrak q$. Equivalently, we have to show that $J_\mathfrak q \cap K^2_\mathfrak q = (KJ)_\mathfrak q$. By Lemma 88.17.6 we know that $(K/J)_\mathfrak q = J'_\mathfrak q$ is generated by a regular sequence. Hence the desired intersection property follows from More on Algebra, Lemma 15.32.5 (and the fact that an ideal generated by a regular sequence is $H_1$-regular, see More on Algebra, Section 15.32). $\square$
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