Lemma 15.11.2. Let $(A, I)$ be a pair with $I$ locally nilpotent. Then the functor $B \mapsto B/IB$ induces an equivalence between the category of étale algebras over $A$ and the category of étale algebras over $A/I$. Moreover, the pair is henselian.

Proof. Essential surjectivity holds by Algebra, Lemma 10.143.10. If $B$, $B'$ are étale over $A$ and $B/IB \to B'/IB'$ is a morphism of $A/I$-algebras, then we can lift this by Algebra, Lemma 10.138.17. Finally, suppose that $f, g : B \to B'$ are two $A$-algebra maps with $f \bmod I = g \bmod I$. Choose an idempotent $e \in B \otimes _ A B$ generating the kernel of the multiplication map $B \otimes _ A B \to B$, see Algebra, Lemmas 10.151.4 and 10.151.3 (to see that étale is unramified). Then $(f \otimes g)(e) \in IB'$. Since $IB'$ is locally nilpotent (Algebra, Lemma 10.32.3) this implies $(f \otimes g)(e) = 0$ by Algebra, Lemma 10.32.6. Thus $f = g$.

It is clear that $I$ is contained in the Jacobson radical of $A$. Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_0h_0$ be a factorization of $\overline{f} = f \bmod I$ with $g_0, h_0 \in A/I[T]$ monic generating the unit ideal in $A/I[T]$. By Lemma 15.9.5 there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ such that the factorization lifts to a factorization into monic polynomials over $A'$. By the above we have $A = A'$ and the factorization is over $A$. $\square$

Comment #8207 by Andrea Panontin on

I think it should be "Then $(f \otimes g)(e) \in IB'$. Since $IB'$ is locally nilpotent..." (i.e. change $IB$ to $IB'$ in both occurences).

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