The Stacks project

Proof. Since $I^ n = 0$ for some $n$, it follows by induction on $n$ that it suffices to consider the case where $I^2 = 0$. Let $\varphi : A \to \Lambda $ be an $R$-algebra map with $A$ of finite presentation over $R$. We have to find a factorization $A \to B \to \Lambda $ with $B$ smooth over $R$, see Algebra, Lemma 10.127.4. By Lemma 16.5.1 we may assume that $A = B/J$ with $B$ smooth over $R$ and $J \subset IB$ a finitely generated ideal. By Lemma 16.5.2 we can find a (possibly noncommutative) diagram

of $R$-algebras which commutes modulo $I$ and such that $\alpha (J) = 0$. The map

is a derivation over $R$ hence we can write it as $D = \xi \circ \text{d}_{B/R}$ for some $B$-linear map $\xi : \Omega _{B/R} \to I\Lambda $. Since $\Omega _{B/R}$ is a finite projective $B$-module we can write $\xi = \sum _{i = 1, \ldots , n} \epsilon _ i \Xi _ i$ for some $\epsilon _ i \in I$ and $B$-linear maps $\Xi _ i : \Omega _{B/R} \to \Lambda $. (Details omitted. Hint: write $\Omega _{B/R}$ as a direct sum of a finite free module to reduce to the finite free case.) We define

and we define $\beta ' : B'' \to \Lambda $ by $\beta $ on $B'$ and by

and $\alpha ' : B \to B''$ by

At this point the diagram

does commute. Moreover, it is direct from the definitions that $\alpha '(J) = 0$ as $I^2 = 0$. Hence the desired factorization. $\square$