The Stacks project

Proof. Observe that $E$ locally on $X$ has the shape as stated in the lemma, see More on Algebra, Section 15.64, Cohomology, Section 20.47, and Derived Categories of Schemes, Section 36.10. Thus it suffices to prove that the ideal of minors is independent of the chosen representative. To do this, it suffices to check in local rings. Over a local ring $(R, \mathfrak m, \kappa )$ consider a bounded above complex

Denote $\text{Fit}_{k, p}(F^\bullet ) \subset R$ the ideal generated by the minors of size $k - n_{p + 1} + n_{p + 2} - \ldots + (-1)^{b - p} n_ b$ in the matrix of $d_ p$. Suppose some matrix coefficient of some differential of $F^\bullet $ is invertible. Then we pick a largest integer $i$ such that $d_ i$ has an invertible matrix coefficient. By Algebra, Lemma 10.102.2 the complex $F^\bullet $ is isomorphic to a direct sum of a trivial complex $\ldots \to 0 \to R \to R \to 0 \to \ldots $ with nonzero terms in degrees $i$ and $i + 1$ and a complex $(F')^\bullet $. We leave it to the reader to see that $\text{Fit}_{p, k}(F^\bullet ) = \text{Fit}_{p, k}((F')^\bullet )$; this is where the formula for the size of the minors is used. If $(F')^\bullet $ has another differential with an invertible matrix coefficient, we do it again, etc. Continuing in this manner, we eventually reach a complex $(F^\infty )^\bullet $ all of whose differentials have matrices with coefficients in $\mathfrak m$. Here you may have to do an infinite number of steps, but for any cutoff only a finite number of these steps affect the complex in degrees $\geq $ the cutoff. Thus the “limit” $(F^\infty )^\bullet $ is a well defined bounded above complex of finite free modules, comes equipped with a quasi-isomorphism $(F^\infty )^\bullet \to F^\bullet $ into the complex we started with, and $\text{Fit}_{p, k}(F^\bullet ) = \text{Fit}_{p, k}((F^\infty )^\bullet )$. Since the complex $(F^\infty )^\bullet $ is unique up to isomorphism by More on Algebra, Lemma 15.75.6 the proof is complete. $\square$