Lemma 15.26.5. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $f \in R$ be an element such that $M_ f$ is finite locally free of rank $r$. Then there exists a finitely generated ideal $I \subset R$ with $V(f) = V(I)$ such that for all $a \in I$ with $R' = R[\frac{I}{a}]$ the strict transform

$M' = (M \otimes _ R R')/a\text{-power torsion}$

is locally free of rank $r$.

Proof. Choose a surjection $R^{\oplus n} \to M$. Choose a finite submodule $K \subset \mathop{\mathrm{Ker}}(R^{\oplus n} \to M)$ such that $R^{\oplus n}/K \to M$ becomes an isomorphism after inverting $f$. This is possible because $M_ f$ is of finite presentation for example by Algebra, Lemma 10.78.2. Set $M_1 = R^{\oplus n}/K$ and suppose we can prove the lemma for $M_1$. Say $I \subset R$ is the corresponding ideal. Then for $a \in I$ the map

$M_1' = (M_1 \otimes _ R R')/a\text{-power torsion} \longrightarrow M' = (M \otimes _ R R')/a\text{-power torsion}$

is surjective. It is also an isomorphism after inverting $a$ in $R'$ as $R'_ a = R_ f$, see Algebra, Lemma 10.70.7. But $a$ is a nonzerodivisor on $M'_1$, whence the displayed map is an isomorphism. Thus it suffices to prove the lemma in case $M$ is a finitely presented $R$-module.

Assume $M$ is a finitely presented $R$-module. Then $J = \text{Fit}_ r(M) \subset S$ is a finitely generated ideal. We claim that $I = fJ$ works.

We first check that $V(f) = V(I)$. The inclusion $V(f) \subset V(I)$ is clear. Conversely, if $f \not\in \mathfrak p$, then $\mathfrak p$ is not an element of $V(J)$ by Lemma 15.8.6. Thus $\mathfrak p \not\in V(fJ) = V(I)$.

Let $a \in I$ and set $R' = R[\frac{I}{a}]$. We may write $a = fb$ for some $b \in J$. By Algebra, Lemmas 10.70.2 and 10.70.8 we see that $J R' = b R'$ and $b$ is a nonzerodivisor in $R'$. Let $\mathfrak p' \subset R' = R[\frac{I}{a}]$ be a prime ideal. Then $JR'_{\mathfrak p'}$ is generated by $b$. It follows from Lemma 15.8.8 that $M'_{\mathfrak p'}$ can be generated by $r$ elements. Since $M'$ is finite, there exist $m_1, \ldots , m_ r \in M'$ and $g \in R'$, $g \not\in \mathfrak p'$ such that the corresponding map $(R')^{\oplus r} \to M'$ becomes surjective after inverting $g$.

Finally, consider the ideal $J' = \text{Fit}_{k - 1}(M')$. Note that $J' R'_ g$ is generated by the coefficients of relations between $m_1, \ldots , m_ r$ (compatibility of Fitting ideal with base change). Thus it suffices to show that $J' = 0$, see Lemma 15.8.7. Since $R'_ a = R_ f$ (Algebra, Lemma 10.70.7) and $M'_ a = M_ f$ is free of rank $r$ we see that $J'_ a = 0$. Since $a$ is a nonzerodivisor in $R'$ we conclude that $J' = 0$ and we win. $\square$

Comment #2646 by Axel on

3rd Paragraph of the proof: The reference to Tag 07ZA (Lemma 15.8.4) should point to Tag 07ZC (Lemma 15.8.6) instead.

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