Lemma 10.143.11. Consider a commutative diagram

with exact rows where $B' \to B$ and $A' \to A$ are surjective ring maps whose kernels are ideals of square zero. If $A \to B$ is étale, and $J = I \otimes _ A B$, then $A' \to B'$ is étale.

Lemma 10.143.11. Consider a commutative diagram

\[ \xymatrix{ 0 \ar[r] & J \ar[r] & B' \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & I \ar[r] \ar[u] & A' \ar[r] \ar[u] & A \ar[r] \ar[u] & 0 } \]

with exact rows where $B' \to B$ and $A' \to A$ are surjective ring maps whose kernels are ideals of square zero. If $A \to B$ is étale, and $J = I \otimes _ A B$, then $A' \to B'$ is étale.

**Proof.**
By Lemma 10.143.10 there exists an étale ring map $A' \to C$ such that $C/IC = B$. Then $A' \to C$ is formally smooth (by Proposition 10.138.13) hence we get an $A'$-algebra map $\varphi : C \to B'$. Since $A' \to C$ is flat we have $I \otimes _ A B = I \otimes _ A C/IC = IC$. Hence the assumption that $J = I \otimes _ A B$ implies that $\varphi $ induces an isomorphism $IC \to J$ and an isomorphism $C/IC \to B'/IB'$, whence $\varphi $ is an isomorphism.
$\square$

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