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Tag 0CZN

Chapter 15: More on Algebra > Section 15.24: Blowing up and flatness

Lemma 15.24.4. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $k \geq 0$ and $I = \text{Fit}_k(M)$. Asssume 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 $$ M' = (M \otimes_R R')/a\text{-power torsion} $$ is locally free of rank $k$.

Proof. By Lemma 15.24.3 we have $\text{Fit}_k(M') = R'$. By Lemma 15.8.7 it suffices to show that $\text{Fit}_{k - 1}(M') = 0$. Recall that $R' \subset R'_a = R_a$, see Algebra, Lemma 10.69.2. 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 10.77.2) and the already cited Lemma 15.8.7. $\square$

    The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 5105–5116 (see updates for more information).

    \begin{lemma}
    \label{lemma-blowup-fitting-ideal-locally-free}
    Let $R$ be a ring. Let $M$ be a finite $R$-module.
    Let $k \geq 0$ and $I = \text{Fit}_k(M)$. Asssume 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
    $$
    M' = (M \otimes_R R')/a\text{-power torsion}
    $$
    is locally free of rank $k$.
    \end{lemma}
    
    \begin{proof}
    By Lemma \ref{lemma-blowup-fitting-ideal} we have
    $\text{Fit}_k(M') = R'$. By Lemma \ref{lemma-fitting-ideal-finite-locally-free}
    it suffices to show that $\text{Fit}_{k - 1}(M') = 0$.
    Recall that $R' \subset R'_a = R_a$, see
    Algebra, Lemma \ref{algebra-lemma-affine-blowup}.
    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 \ref{lemma-fitting-ideal-basics} part (3).
    This is true by our assumption that
    $M_a$ is finite locally free of rank $k$
    (see Algebra, Lemma \ref{algebra-lemma-finite-projective})
    and the already cited Lemma \ref{lemma-fitting-ideal-finite-locally-free}.
    \end{proof}

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