The Stacks project

Proof. Let $T$ be the $m \times m$ matrix with entries $\partial f_ j/\partial x_ i$ for $1 \leq i, j \leq n$. Let $K \in D(B)$ be represented by the complex $T : B^{\oplus m} \to B^{\oplus m}$ with terms sitting in degrees $-1$ and $0$. By More on Algebra, Lemmas 15.84.12 we have $g : K \to K$ is zero in $D(B)$. Set $J = (f_1, \ldots , f_ m)$. Recall that $\mathop{N\! L}\nolimits _{B/A}$ is homotopy equivalent to $J/J^2 \to \bigoplus _{i = 1, \ldots , n} B\text{d}x_ i$, see Algebra, Section 10.134. Denote $L$ the complex $J/J^2 \to \bigoplus _{i = 1, \ldots , m} B\text{d}x_ i$ to that we have the quotient map $\mathop{N\! L}\nolimits _{B/A} \to L$. We also have a surjective map of complexes $K \to L$ by sending the $j$th basis element in the term $B^{\oplus m}$ in degree $-1$ to the class of $f_ j$ in $J/J^2$. Picture

From More on Algebra, Lemma 15.84.8 we conclude that multiplication by $g$ on $L$ is $0$ in $D(B)$. On the other hand, the distinguished triangle $B^{\oplus n - m}[0] \to \mathop{N\! L}\nolimits _{B/A} \to L$ shows that $\mathop{\mathrm{Ext}}\nolimits ^1_ B(L, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$ is surjective for every $B$-module $N$ and hence annihilated by $g$. This proves part (1). If $n = m$ then $\mathop{N\! L}\nolimits _{B/A} = L$ and we see that (2) holds. $\square$