Lemma 15.26.4. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $k \geq 0$ and $I = \text{Fit}_ k(M)$. Assume that $M_\mathfrak p$ is free of rank $k$ for every $\mathfrak p \not\in V(I)$. Then for every $a \in I$ with $R' = R[\frac{I}{a}]$ the strict transform
is locally free of rank $k$.
Proof. By Lemma 15.26.3 we have $\text{Fit}_ k(M') = R'$. By Lemma 15.8.8 it suffices to show that $\text{Fit}_{k - 1}(M') = 0$. Recall that $R' \subset R'_ a = R_ a$, see Algebra, Lemma 10.70.2. Hence it suffices to prove that $\text{Fit}_{k - 1}(M')$ maps to zero in $R'_ a = R_ a$. Since clearly $(M')_ a = M_ a$ this reduces us to showing that $\text{Fit}_{k - 1}(M_ a) = 0$ because formation of Fitting ideals commutes with base change according to Lemma 15.8.4 part (3). This is true by our assumption that $M_ a$ is finite locally free of rank $k$ (see Algebra, Lemma 10.78.2) and the already cited Lemma 15.8.8. $\square$
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