
Lemma 10.136.17. Let $R \to S$ be a smooth ring map. Given a commutative solid diagram

$\xymatrix{ S \ar[r] \ar@{-->}[rd] & A/I \\ R \ar[r] \ar[u] & A \ar[u] }$

where $I \subset A$ is a locally nilpotent ideal, a dotted arrow exists which makes the diagram commute.

Proof. By Lemma 10.136.14 we can extend the diagram to a commutative diagram

$\xymatrix{ S_0 \ar[r] & S \ar[r] \ar@{-->}[rd] & A/I \\ R_0 \ar[r] \ar[u] & R \ar[r] \ar[u] & A \ar[u] }$

with $R_0 \to S_0$ smooth, $R_0$ of finite type over $\mathbf{Z}$, and $S = S_0 \otimes _{R_0} R$. Let $x_1, \ldots , x_ n \in S_0$ be generators of $S_0$ over $R_0$. Let $a_1, \ldots , a_ n$ be elements of $A$ which map to the same elements in $A/I$ as the elements $x_1, \ldots , x_ n$. Denote $A_0 \subset A$ the subring generated by the image of $R_0$ and the elements $a_1, \ldots , a_ n$. Set $I_0 = A_0 \cap I$. Then $A_0/I_0 \subset A/I$ and $S_0 \to A/I$ maps into $A_0/I_0$. Thus it suffices to find the dotted arrow in the diagram

$\xymatrix{ S_0 \ar[r] \ar@{-->}[rd] & A_0/I_0 \\ R_0 \ar[r] \ar[u] & A_0 \ar[u] }$

The ring $A_0$ is of finite type over $\mathbf{Z}$ by construction. Hence $A_0$ is Noetherian, whence $I_0$ is nilpotent, see Lemma 10.31.5. Say $I_0^ n = 0$. By Proposition 10.136.13 we can successively lift the $R_0$-algebra map $S_0 \to A_0/I_0$ to $S_0 \to A_0/I_0^2$, $S_0 \to A_0/I_0^3$, $\ldots$, and finally $S_0 \to A_0/I_0^ n = A_0$. $\square$

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