The Stacks project

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. $\square$

Comments (2)

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 is free of rank because is finite projective and because is free of rank "

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

" is surjective modulo ."

Maybe here it's meant " is surjective modulo ."

"Hence we see that this map of modules over the local ring 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 is generated by elements over , the map is inyective.

Comment #2199 by on

OK, thanks. I added your clarifications to the text here.

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