Lemma 15.26.3. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $k \geq 0$ and $I = \text{Fit}_ k(M)$. For every $a \in I$ with $R' = R[\frac{I}{a}]$ the strict transform
has $\text{Fit}_ k(M') = R'$.
Lemma 15.26.3. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $k \geq 0$ and $I = \text{Fit}_ k(M)$. For every $a \in I$ with $R' = R[\frac{I}{a}]$ the strict transform
has $\text{Fit}_ k(M') = R'$.
Proof. First observe that $\text{Fit}_ k(M \otimes _ R R') = IR' = aR'$. The first equality by Lemma 15.8.4 part (3) and the second equality by Algebra, Lemma 10.70.2. From Lemma 15.8.9 and exactness of localization we see that $M'_{\mathfrak p'}$ can be generated by $\leq k$ elements for every prime $\mathfrak p'$ of $R'$. Then $\text{Fit}_ k(M') = R'$ for example by Lemma 15.8.7. $\square$
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