Lemma 15.9.14. Let $A$ be a ring, let $I \subset A$ be an ideal. Consider a commutative diagram

$\xymatrix{ B \ar[rd] \\ A \ar[u] \ar[r] & A/I }$

where $B$ is a smooth $A$-algebra. Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and an $A$-algebra map $B \to A'$ lifting the ring map $B \to A/I$.

Proof. Let $J \subset B$ be the kernel of $B \to A/I$ so that $B/J = A/I$. By Algebra, Lemma 10.139.3 the sequence

$0 \to I/I^2 \to J/J^2 \to \Omega _{B/A} \otimes _ B B/J \to 0$

is split exact. Thus $\overline{P} = J/(J^2 + IB) = \Omega _{B/A} \otimes _ B B/J$ is a finite projective $A/I$-module. Choose an integer $n$ and a direct sum decomposition $A/I^{\oplus n} = \overline{P} \oplus \overline{K}$. By Lemma 15.9.11 we can find an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and a finite projective $A$-module $K$ which lifts $\overline{K}$. We may and do replace $A$ by $A'$. Set $B' = B \otimes _ A \text{Sym}_ A^*(K)$. Since $A \to \text{Sym}_ A^*(K)$ is smooth by Lemma 15.9.13 we see that $B \to B'$ is smooth which in turn implies that $A \to B'$ is smooth (see Algebra, Lemmas 10.137.4 and 10.137.13). Moreover the section $\text{Sym}^*_ A(K) \to A$ determines a section $B' \to B$ and we let $B' \to A/I$ be the composition $B' \to B \to A/I$. Let $J' \subset B'$ be the kernel of $B' \to A/I$. We have $JB' \subset J'$ and $B \otimes _ A K \subset J'$. These maps combine to give an isomorphism

$(A/I)^{\oplus n} \cong J/J^2 \oplus \overline{K} \longrightarrow J'/((J')^2 + IB')$

Thus, after replacing $B$ by $B'$ we may assume that $J/(J^2 + IB) = \Omega _{B/A} \otimes _ B B/J$ is a free $A/I$-module of rank $n$.

In this case, choose $f_1, \ldots , f_ n \in J$ which map to a basis of $J/(J^2 + IB)$. Consider the finitely presented $A$-algebra $C = B/(f_1, \ldots , f_ n)$. Note that we have an exact sequence

$0 \to H_1(L_{C/A}) \to (f_1, \ldots , f_ n)/(f_1, \ldots , f_ n)^2 \to \Omega _{B/A} \otimes _ B C \to \Omega _{C/A} \to 0$

see Algebra, Lemma 10.134.4 (note that $H_1(L_{B/A}) = 0$ and that $\Omega _{B/A}$ is finite projective, in particular flat so the Tor group vanishes). For any prime $\mathfrak q \supset J$ of $B$ the module $\Omega _{B/A, \mathfrak q}$ is free of rank $n$ because $\Omega _{B/A}$ is finite projective and because $\Omega _{B/A} \otimes _ B B/J$ is free of rank $n$ (see Algebra, Lemma 10.78.2). By our choice of $f_1, \ldots , f_ n$ the map

$\left((f_1, \ldots , f_ n)/(f_1, \ldots , f_ n)^2\right)_{\mathfrak q} \to \Omega _{B/A, \mathfrak q}$

is surjective modulo $J$. Hence we see that this map of modules over the local ring $C_{\mathfrak q}$ has to be an isomorphism (this is because by Nakayama's Algebra, Lemma 10.20.1 the map is surjective and then for example by Algebra, Lemma 10.16.4 because $((f_1, \ldots , f_ n)/(f_1, \ldots , f_ n)^2)_{\mathfrak q}$ is generated by $n$ elements the map is injective). Thus $H_1(L_{C/A})_{\mathfrak q} = 0$ and $\Omega _{C/A, \mathfrak q} = 0$. By Algebra, Lemma 10.137.12 we see that $A \to C$ is smooth at the prime $\overline{\mathfrak q}$ of $C$ corresponding to $\mathfrak q$. Since $\Omega _{C/A, \mathfrak q} = 0$ it is actually étale at $\overline{\mathfrak q}$. Thus $A \to C$ is étale at all primes of $C$ containing $JC$. By Lemma 15.9.4 we can find an $f \in C$ mapping to an invertible element of $C/JC$ such that $A \to C_ f$ is étale. By our choice of $f$ it is still true that $C_ f/JC_ f = A/I$. The map $C_ f/IC_ f \to A/I$ is surjective and étale by Algebra, Lemma 10.143.8. Hence $A/I$ is isomorphic to the localization of $C_ f/IC_ f$ at some element $g \in C$, see Algebra, Lemma 10.143.9. Set $A' = C_{fg}$ to conclude the proof. $\square$

Comment #2170 by JuanPablo on

I had a couple of things I didn't think were immediately clear so I'm detailing the arguments I found here.

"the module $\Omega_{B/A, \mathfrak q}$ is free of rank $n$ because $\Omega_{B/A}$ is finite projective and because $\Omega_{B/A} \otimes_B B/J$ is free of rank $n$"

This is because of lemma 10.77.4 (tag 00NZ).

"$\left((f_1, \ldots, f_n)/(f_1, \ldots, f_n)^2\right)_{\mathfrak q} \to \Omega_{B/A, \mathfrak q}$ is surjective modulo $I$."

Maybe here it's meant "$\left((f_1, \ldots, f_n)/(f_1, \ldots, f_n)^2\right)_{\mathfrak q} \to \Omega_{B/A, \mathfrak q}\otimes_{B_{\mathfrak q}} C_{\mathfrak q}$ is surjective modulo $J$."

"Hence we see that this map of modules over the local ring $C_{\mathfrak q}$ has to be an isomorphism."

This is because by Nakayama's lemma 10.19.1 (tag 00DV) the map is surjective . And then by lemma 10.15.4 (tag 05G8) and because $\left((f_1, \ldots, f_n)/(f_1, \ldots, f_n)^2\right)_{\mathfrak q}$ is generated by $n$ elements over $C_{\mathfrak q}$, the map is inyective.

Comment #2199 by on

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