Lemma 91.2.9. If $A \to B$ is a local complete intersection ring map, then there exists a solution to (91.2.0.1).

First proof. Write $B = A[x_1, \ldots , x_ n]/J$. By More on Algebra, Definition 15.33.2 the ideal $J$ is Koszul-regular. This implies $J$ is $H_1$-regular and quasi-regular, see More on Algebra, Section 15.32. Let $J' \subset A'[x_1, \ldots , x_ n]$ be the inverse image of $J$. Denote $I[x_1, \ldots , x_ n]$ the kernel of $A'[x_1, \ldots , x_ n] \to A[x_1, \ldots , x_ n]$. By More on Algebra, Lemma 15.32.5 we have $I[x_1, \ldots , x_ n] \cap (J')^2 = J'I[x_1, \ldots , x_ n] = JI[x_1, \ldots , x_ n]$. Hence we obtain a short exact sequence

$0 \to I \otimes _ A B \to J'/(J')^2 \to J/J^2 \to 0$

Since $J/J^2$ is projective (More on Algebra, Lemma 15.32.3) we can choose a splitting of this sequence

$J'/(J')^2 = I \otimes _ A B \oplus J/J^2$

Let $(J')^2 \subset J'' \subset J'$ be the elements which map to the second summand in the decomposition above. Then

$0 \to I \otimes _ A B \to A'[x_1, \ldots , x_ n]/J'' \to B \to 0$

is a solution to (91.2.0.1) with $N = I \otimes _ A B$. The general case is obtained by doing a pushout along the given map $I \otimes _ A B \to N$. $\square$

Second proof. Please read Remark 91.2.8 before reading this proof. By More on Algebra, Lemma 15.33.6 the maps $\mathop{N\! L}\nolimits _{A'/A} \otimes _ A B \to \mathop{N\! L}\nolimits _{B/A'} \to \mathop{N\! L}\nolimits _{B/A}$ do form a distinguished triangle in $D(B)$. Hence it suffices to show that $\mathop{\mathrm{Ext}}\nolimits ^2_{B/A}(\mathop{N\! L}\nolimits _{B/A}, N)$ vanishes. By More on Algebra, Lemma 15.85.4 the complex $\mathop{N\! L}\nolimits _{B/A}$ is perfect of tor-amplitude in $[-1, 0]$. This implies our $\mathop{\mathrm{Ext}}\nolimits ^2$ vanishes for example by More on Algebra, Lemma 15.76.1 part (1). $\square$

There are also:

• 2 comment(s) on Section 91.2: Deformations of rings and the naive cotangent complex

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).