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This is an answer to a question of Karl in the comments to my first answer to this question.
[EDIT: The following is a minimally simplified version of Proposition 3.3 of Hasett-Kovács04.]
Theorem. Let $X$ be a noetherian scheme, $r\in\mathbb N$, $Z\subseteq X$ a subscheme such that ${\rm codim}_XZ\geq r$, and $\mathscr F$ a coherent $\mathscr O_X$-module such that ${\rm supp}\,\mathscr F=X$ and $\mathscr F_x$ is $S_r$ for every $x\in Z$. Then
$$
\mathscr H^i_Z(X,\mathscr F)=0\quad\text{for $i=0,\ldots,r-1$}.
$$
Proof. Let $x\in Z$ and notice that we have the following equality of functors:
$$
H^0_x = H^0_x\circ \mathscr H^0_Z
$$
which induces a Grothendieck spectral sequence
$$
E^{p,q}_2= H^p_x \circ \mathscr H^q_Z \Rightarrow H^{p+q}_x.
$$
Now prove the statement using induction on $i$.
Suppose $\exists\,\sigma\in\mathscr H^0_Z(X,\mathcal F)$, $\sigma\neq 0$. Let $x\in Z$ be the general point of an irreducible component of ${\rm supp}\,\sigma$. Then $H^0_x(X, \mathscr H^0_Z(X,\mathscr F))\neq 0$ and hence $H^0_x(X,\mathscr F)\neq 0$. But this contradicts the assumption that $\mathscr F_x$ is $S_r$.
Now suppose that we already know that
$$
\mathcal H^i_Z(X,\mathscr F)=0\quad\text{for $i=0,\ldots,k-1$}
$$
for some $k<r$ and assume that $\mathscr H^k_Z(X,\mathscr F)\neq 0$. By the same argument as above we find a point such that $E^{0,k}_2=H^0_x(X,\mathscr H^k_Z(X,\mathscr F))\neq 0$. Since it is an $E^{0,k}$ term there are no differentials (including later pages of the spectral sequence) mapping to this term and all subsequent differentials mapping from this term map to something of the form $E^{p,q}$ with $0\leq q\leq k-1$. However those latter kind are zero by the inductive hypothesis. Therefore this implies that then $H^k_x(X,\mathscr F)\neq 0$ which is again a contradiction to the assumption that $\mathscr F_x$ is $S_r$. Q.E.D.
See also this MO answer
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edited Oct 31 2011 at 17:07 |
This is an answer to a question of Karl in the comments to my first answer to this question.
[EDIT: The following is a minimally simplified version of Proposition 3.3 of Hasett-Kovács04.]
Theorem. Let $X$ be a noetherian scheme, $r\in\mathbb N$, $Z\subseteq X$ a subscheme such that ${\rm codim}_XZ\geq r$, and $\mathscr F$ a coherent $\mathscr O_X$-module such that ${\rm supp}\,\mathscr F=X$ and $\mathscr F_x$ is $S_r$ for every $x\in Z$. Then
$$
\mathscr H^i_Z(X,\mathscr F)=0\quad\text{for $i=0,\ldots,r-1$}.
$$
Proof. Let $x\in Z$ and notice that we have the following equality of functors:
$$
H^0_x = H^0_x\circ \mathscr H^0_Z
$$
which induces a Grothendieck spectral sequence
$$
E^{p,q}_2= H^p_x \circ \mathscr H^q_Z \Rightarrow H^{p+q}_x.
$$
Now prove the statement using induction on $i$.
Suppose $\exists\,\sigma\in\mathscr H^0_Z(X,\mathcal F)$, $\sigma\neq 0$. Let $x\in Z$ be the general point of an irreducible component of ${\rm supp}\,\sigma$. Then $H^0_x(X, \mathscr H^0_Z(X,\mathscr F))\neq 0$ and hence $H^0_x(X,\mathscr F)\neq 0$. But this contradicts the assumption that $\mathscr F_x$ is $S_r$.
Now suppose that we already know that
$$
\mathcal H^i_Z(X,\mathscr F)=0\quad\text{for $i=0,\ldots,k-1$}
$$
for some $k<r$ and assume that $\mathscr H^k_Z(X,\mathscr F)\neq 0$. By the same argument as above we find a point such that $E^{0,k}_2=H^0_x(X,\mathscr H^k_Z(X,\mathscr F))\neq 0$. Since it is an $E^{0,k}$ term there are no differentials (including later pages of the spectral sequence) mapping to this term and all subsequent differentials mapping from this term map to something of the form $E^{p,q}$ with $0\leq q\leq k-1$. However those latter kind are zero by the inductive hypothesis. Therefore this implies that then $H^k_x(X,\mathscr F)\neq 0$ which is again a contradiction to the assumption that $\mathscr F_x$ is $S_r$. Q.E.D.
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edited Jun 13 2011 at 13:02 |
This is an answer to a question of Karl in the comments to my first answer to this question.
Theorem. Let $X$ be a noetherian scheme, $r\in\mathbb N$, $Z\subseteq X$ a subscheme such that ${\rm codim}_XZ\geq r$, and $\mathscr F$ a coherent $\mathscr O_X$-module such that ${\rm supp}\,\mathscr F=X$ and $\mathscr F_x$ is $S_r$ for every $x\in Z$. Then
$$
\mathscr H^i_Z(X,\mathscr F)=0\quad\text{for $i=0,\ldots,r-1$}.
$$
Proof. Let $x\in Z$ and notice that we have the following equality of functors:
$$
H^0_x = H^0_x\circ \mathscr H^0_Z
$$
which induces a Grothendieck spectral sequence
$$
E^{p,q}_2= H^p_x \circ \mathscr H^q_Z \Rightarrow H^{p+q}_x.
$$
Now prove the statement using induction on $i$.
Suppose $\exists\,\sigma\in\mathscr H^0_Z(X,\mathcal F)$, $\sigma\neq 0$. Let $x\in Z$ be the general point of an irreducible component of ${\rm supp}\,\sigma$. Then $H^0_x(X, \mathscr H^0_Z(X,\mathscr F))\neq 0$ and hence $H^0_x(X,\mathscr F)\neq 0$. But this contradicts the assumption that $\mathscr F_x$ is $S_r$.
Now suppose that we already know that
$$
\mathcal H^i_Z(X,\mathscr F)=0\quad\text{for $i=0,\ldots,k-1$}
$$
for some $k<r$ and assume that $\mathscr H^k_Z(X,\mathscr F)\neq 0$. By the same argument as above we find a point such that $E^{0,k}_2=H^0_x(X,\mathscr H^k_Z(X,\mathscr F))\neq 0$. Since it is an $E^{0,k}$ term there are no differentials (including later pages of the spectral sequence) mapping to this term and all subsequent differentials mapping from this term map to something of the form $E^{p,q}$ with $0\leq q\leq k-1$. However those latter kind are zero by the inductive hypothesis. Therefore this implies that then $H^k_x(X,\mathscr F)=0$ F)\neq 0$ which is again a contradiction to the assumption that $\mathscr F_x$ is $S_r$. Q.E.D.
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edited Feb 12 2011 at 22:03 |
This is an answer to a question of Karl in the comments to my first answer to this question.
Theorem. Let $X$ be a noetherian scheme, $r\in\mathbb N$, $Z\subseteq X$ a subscheme such that ${\rm codim}_XZ\geq r$, and $\mathcal \mathscr F$ a coherent $\mathcal \mathscr O_X$-module such that ${\rm supp}\,\mathcal supp}\,\mathscr F=X$ and $\mathcal \mathscr F_x$ is $S_r$ for every $x\in Z$. Then
$$
\mathcal H^i_Z(X,\mathcal mathscr H^i_Z(X,\mathscr F)=0\quad\text{for $i=0,\ldots,r-1$}.
$$
Proof. Let $x\in Z$ and notice that we have the following equality of functors:
$$
H^0_x = H^0_x\circ \mathcal mathscr H^0_Z
$$
which induces a Grothendieck spectral sequence
$$
E^{p,q}_2= H^p_x \circ \mathcal mathscr H^q_Z \Rightarrow H^{p+q}_x.
$$
Now prove the statement using induction on $i$.
Suppose $\exists\,\sigma\in\mathcal \exists\,\sigma\in\mathscr H^0_Z(X,\mathcal F)$, $\sigma\neq 0$. Let $x\in Z$ be the general point of an irreducible component of ${\rm supp}\,\sigma$. Then $H^0_x(X, \mathcal H^0_Z(X,\mathcal mathscr H^0_Z(X,\mathscr F))\neq 0$ and hence $H^0_x(X,\mathcal H^0_x(X,\mathscr F)\neq 0$. But this contradicts the assumption that $\mathcal \mathscr F_x$ is $S_r$.
Now suppose that we already know that
$$
\mathcal H^i_Z(X,\mathcal H^i_Z(X,\mathscr F)=0\quad\text{for $i=0,\ldots,k-1$}
$$
for some $k<r$ and assume that $\mathcal H^k_Z(X,\mathcal \mathscr H^k_Z(X,\mathscr F)\neq 0$. By the same argument as above we find a point such that $E^{0,k}_2=H^0_x(X,\mathcal H^k_Z(X,\mathcal E^{0,k}_2=H^0_x(X,\mathscr H^k_Z(X,\mathscr F))\neq 0$. Since it is an $E^{0,k}$ term there are no differentials (including later pages of the spectral sequence) mapping to this term and all subsequent differentials mapping from this term map to something of the form $E^{p,q}$ with $0\leq q\leq k-1$. However those latter kind are zero by the inductive hypothesis. Therefore this implies that then $H^k_x(X,\mathcal H^k_x(X,\mathscr F)=0$ which is again a contradiction to the assumption that $\mathcal \mathscr F_x$ is $S_r$. Q.E.D.
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edited Nov 11 2010 at 7:28 |
This is an attempt answer to prove a question of Karl in the following:comments to my first answer to this question.
Theorem. Let $X$ be a noetherian scheme, $r\in\mathbb N$, $Z\subseteq X$ a subscheme such that ${\rm codim}_XZ\geq r$, and $\mathcal F$ a coherent $\mathcal O_X$-module such that ${\rm supp}\,\mathcal F=X$ and $\mathcal F_x$ is $S_r$ for every $x\in Z$. Then
$$
\mathcal H^i_Z(X,\mathcal F)=0\quad\text{for $i=0,\ldots,r-1$}.
$$
Proof. Let $x\in Z$ and notice that we have the following equality of functors:
$$
H^0_x = H^0_x\circ \mathcal H^0_Z
$$
which induces a Grothendieck spectral sequence
$$
E^{p,q}_2= H^p_x \circ \mathcal H^q_Z \Rightarrow H^{p+q}_x.
$$
Now prove the statement using induction on $i$.
Suppose $\exists\,\sigma\in\mathcal H^0_Z(X,\mathcal F)$, $\sigma\neq 0$. Let $x\in Z$ be the general point of an irreducible component of ${\rm supp}\,\sigma$. Then $H^0_x(X, \mathcal H^0_Z(X,\mathcal F))\neq 0$ and hence $H^0_x(X,\mathcal F)\neq 0$. But this contradicts the assumption that $\mathcal F_x$ is $S_r$.
Now suppose that we already know that
$$
\mathcal H^i_Z(X,\mathcal F)=0\quad\text{for $i=0,\ldots,k-1$}
$$
for some $k<r$ and assume that $\mathcal H^k_Z(X,\mathcal F)\neq 0$. By the same argument as above we find a point such that $E^{0,k}_2=H^0_x(X,\mathcal H^k_Z(X,\mathcal F))\neq 0$. Since it is an $E^{0,k}$ term there are no differentials (including later pages of the spectral sequence) mapping to this term and all subsequent differentials mapping from this term map to something of the form $E^{p,q}$ with $0\leq q\leq k-1$. However those latter kind are zero by the inductive hypothesis. Therefore this implies that then $H^k_x(X,\mathcal F)=0$ which is again a contradiction to the assumption that $\mathcal F_x$ is $S_r$. Q.E.D.
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answered Nov 10 2010 at 22:59 |
This is an attempt to prove the following:
Theorem. Let $X$ be a noetherian scheme, $r\in\mathbb N$, $Z\subseteq X$ a subscheme such that ${\rm codim}_XZ\geq r$, and $\mathcal F$ a coherent $\mathcal O_X$-module such that ${\rm supp}\,\mathcal F=X$ and $\mathcal F_x$ is $S_r$ for every $x\in Z$. Then
$$
\mathcal H^i_Z(X,\mathcal F)=0\quad\text{for $i=0,\ldots,r-1$}.
$$
Proof. Let $x\in Z$ and notice that we have the following equality of functors:
$$
H^0_x = H^0_x\circ \mathcal H^0_Z
$$
which induces a Grothendieck spectral sequence
$$
E^{p,q}_2= H^p_x \circ \mathcal H^q_Z \Rightarrow H^{p+q}_x.
$$
Now prove the statement using induction on $i$.
Suppose $\exists\,\sigma\in\mathcal H^0_Z(X,\mathcal F)$, $\sigma\neq 0$. Let $x\in Z$ be the general point of an irreducible component of ${\rm supp}\,\sigma$. Then $H^0_x(X, \mathcal H^0_Z(X,\mathcal F))\neq 0$ and hence $H^0_x(X,\mathcal F)\neq 0$. But this contradicts the assumption that $\mathcal F_x$ is $S_r$.
Now suppose that we already know that
$$
\mathcal H^i_Z(X,\mathcal F)=0\quad\text{for $i=0,\ldots,k-1$}
$$
for some $k<r$ and assume that $\mathcal H^k_Z(X,\mathcal F)\neq 0$. By the same argument as above we find a point such that $E^{0,k}_2=H^0_x(X,\mathcal H^k_Z(X,\mathcal F))\neq 0$. Since it is an $E^{0,k}$ term there are no differentials (including later pages of the spectral sequence) mapping to this term and all subsequent differentials mapping from this term map to something of the form $E^{p,q}$ with $0\leq q\leq k-1$. However those latter kind are zero by the inductive hypothesis. Therefore this implies that then $H^k_x(X,\mathcal F)=0$ which is again a contradiction to the assumption that $\mathcal F_x$ is $S_r$. Q.E.D.
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