Lemma 31.35.3. Let $X$ be a scheme. Let $\mathcal{F}$ be a finitely presented $\mathcal{O}_ X$-module. Let $U \subset X$ be a scheme theoretically dense open such that $\mathcal{F}|_ U$ is finite locally free of constant rank $r$. Then

1. the blowup $b : X' \to X$ of $X$ in the $r$th Fitting ideal of $\mathcal{F}$ is $U$-admissible,

2. the strict transform $\mathcal{F}'$ of $\mathcal{F}$ with respect to $b$ is locally free of rank $r$,

3. the kernel $\mathcal{K}$ of the surjection $b^*\mathcal{F} \to \mathcal{F}'$ is finitely presented and $\mathcal{K}|_ U = 0$,

4. $b^*\mathcal{F}$ and $\mathcal{K}$ are perfect $\mathcal{O}_{X'}$-modules of tor dimension $\leq 1$.

Proof. The ideal $\text{Fit}_ r(\mathcal{F})$ is of finite type by Lemma 31.9.2 and its restriction to $U$ is equal to $\mathcal{O}_ U$ by Lemma 31.9.5. Hence $b : X' \to X$ is $U$-admissible, see Definition 31.34.1.

By Lemma 31.9.5 the restriction of $\text{Fit}_{r - 1}(\mathcal{F})$ to $U$ is zero, and since $U$ is scheme theoretically dense we conclude that $\text{Fit}_{r - 1}(\mathcal{F}) = 0$ on all of $X$. Thus it follows from Lemma 31.9.5 that $\mathcal{F}$ is locally free of rank $r$ on the complement of subscheme cut out by the $r$th Fitting ideal of $\mathcal{F}$ (this complement may be bigger than $U$ which is why we had to do this step in the argument). Hence by Lemma 31.35.2 the strict transform

$b^*\mathcal{F} \longrightarrow \mathcal{F}'$

is locally free of rank $r$. The kernel $\mathcal{K}$ of this map is supported on the exceptional divisor of the blowup $b$ and hence $\mathcal{K}|_ U = 0$. Finally, since $\mathcal{F}'$ is finite locally free and since the displayed arrow is surjective, we can locally on $X'$ write $b^*\mathcal{F}$ as the direct sum of $\mathcal{K}$ and $\mathcal{F}'$. Since $b^*\mathcal{F}'$ is finitely presented (Modules, Lemma 17.11.4) the same is true for $\mathcal{K}$.

The statement on tor dimension follows from More on Algebra, Lemma 15.8.10. $\square$

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).