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Tag 0BBJ

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

Lemma 15.24.5. 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 $K \subset \mathop{\rm Ker}(R^{\oplus n} \to M)$ 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 10.77.2. 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 10.69.4. 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 15.8.4. 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 10.69.2 and 10.69.5 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 10.69.4) 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. $\square$

    The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 5136–5147 (see updates for more information).

    \begin{lemma}
    \label{lemma-blowup-module}
    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 $r$.
    \end{lemma}
    
    \begin{proof}
    Choose a surjection $R^{\oplus n} \to M$. Choose a finite submodule
    $K \subset \Ker(R^{\oplus n} \to M)$ 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 \ref{algebra-lemma-finite-projective}.
    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 \ref{algebra-lemma-blowup-in-principal}.
    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.
    
    \medskip\noindent
    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.
    
    \medskip\noindent
    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 \ref{lemma-fitting-ideal-basics}.
    Thus $\mathfrak p \not \in V(fJ) = V(I)$.
    
    \medskip\noindent
    Let $a \in I$ and set $R' = R[\frac{I}{a}]$. We may write $a = fb$
    for some $b \in J$. By Algebra, Lemmas \ref{algebra-lemma-affine-blowup} and
    \ref{algebra-lemma-blowup-add-principal} 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 \ref{lemma-principal-fitting-ideal}
    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 $g$.
    
    \medskip\noindent
    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 \ref{lemma-fitting-ideal-finite-locally-free}.
    Since $R'_a = R_f$ (Algebra, Lemma \ref{algebra-lemma-blowup-in-principal})
    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.
    \end{proof}

    Comments (1)

    Comment #2646 by Axel on July 14, 2017 a 6:55 pm UTC

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

    There are also 2 comments on Section 15.24: More on Algebra.

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