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Flatness of direct image sheaf over local artinian ring
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Let $pi:X to mboxSpec(mathbbC[t]/(t^2))$ be a smooth, projective morphism and $L$ be an invertible sheaf on $X$. Denote by $L_0$ the restriction of $L$ to the closed fiber, say $X_0$ of $pi$. Suppose that the natural morphism $H^0(L) to H^0(L_0)$ is surjective. Can we conclude that $pi_*L$ is flat over $mboxSpec(mathbbC[t]/(t^2))$?
ag.algebraic-geometry deformation-theory invertible-sheaves
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Let $pi:X to mboxSpec(mathbbC[t]/(t^2))$ be a smooth, projective morphism and $L$ be an invertible sheaf on $X$. Denote by $L_0$ the restriction of $L$ to the closed fiber, say $X_0$ of $pi$. Suppose that the natural morphism $H^0(L) to H^0(L_0)$ is surjective. Can we conclude that $pi_*L$ is flat over $mboxSpec(mathbbC[t]/(t^2))$?
ag.algebraic-geometry deformation-theory invertible-sheaves
asked 5 hours ago
Chen
54639
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $pi:X to mboxSpec(mathbbC[t]/(t^2))$ be a smooth, projective morphism and $L$ be an invertible sheaf on $X$. Denote by $L_0$ the restriction of $L$ to the closed fiber, say $X_0$ of $pi$. Suppose that the natural morphism $H^0(L) to H^0(L_0)$ is surjective. Can we conclude that $pi_*L$ is flat over $mboxSpec(mathbbC[t]/(t^2))$?
ag.algebraic-geometry deformation-theory invertible-sheaves
asked 5 hours ago
Chen
54639
Let $pi:X to mboxSpec(mathbbC[t]/(t^2))$ be a smooth, projective morphism and $L$ be an invertible sheaf on $X$. Denote by $L_0$ the restriction of $L$ to the closed fiber, say $X_0$ of $pi$. Suppose that the natural morphism $H^0(L) to H^0(L_0)$ is surjective. Can we conclude that $pi_*L$ is flat over $mboxSpec(mathbbC[t]/(t^2))$?
ag.algebraic-geometry deformation-theory invertible-sheaves
ag.algebraic-geometry deformation-theory invertible-sheaves
asked 5 hours ago
Chen
54639
asked 5 hours ago
Chen
54639
asked 5 hours ago
Chen
54639
asked 5 hours ago
Chen
54639
asked 5 hours ago
Chen
54639
54639
add a comment |Â
add a comment |Â
2 Answers
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This follows easily from the theory of modules over $R = mathbb C[t]/(t^2)$. Indeed, we have a short exact sequence
$$0 to H^0(mathscr L_0) to H^0(mathscr L) to H^0(mathscr L_0) to 0,$$
induced by the observation that $(t) cong R/(t)$ and your assumption that the map $H^0(mathscr L) to H^0(mathscr L_0)$ is surjective. Now we have the following lemma.
Lemma. Let $0 to N stackrel fto M stackrel gto N to 0$ be a short exact sequence of finitely generated $R$-modules, and assume that $fg colon M to M$ is multiplication by $t$. Then $M$ is free and $N = tM cong M/tM$.
Proof. Because $N stackrel fto M stackrel gto N$ is exact, $M stackrel gto N$ is surjective, and $N stackrel fto M$ is injective, the sequence
$$M stackrelfg longrightarrow M stackrelfglongrightarrow Mtag1label1$$
is exact. But we also have $fg = t$. From the structure theory of $R$-modules, we may write
$$M cong R^m oplus (R/(t))^n$$
for some $m,n in mathbb Z_geq 0$. If $n > 0$, then any nonzero element in $(R/(t))^n$ is killed by $t$ but not in the image of $t$, contradicting exactness of (ref1). We conclude that $n = 0$, so $M$ is free. The final statement follows since $N = operatornameim(fg) = tM$. $square$
answered 1 hour ago
R. van Dobben de Bruyn
9,63622959
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$newcommandCmathbbCnewcommandDC[t]/t^2newcommandimmathrmim,$A module $M$ over $D$ is flat if and only if the inclusion $(t)subset D$ remains injective after tensoring with $M$ over $D$. In other words, the map $M/tMxrightarrowt M$ is injective which is equivalent to saying that the inclusion $im (Mxrightarrowt M)subset ker, (Mxrightarrowt M)$ is equality. If $M$ is finite-dimensional as a $C$-vector space, then this condition is equivalent to $dimim t=dimker t=dim M-dim im t$ which can be rephrased as $dimim t=frac12dim M$(for a non-flat module the LHS is strictly smaller).
(All of the above is more or less tautological since any finitely generated module over $D$ is a direct sum of some number of copies of $C$ and a few copies of $D$ and a module is flat iff it is free)
Denote by $i:X_0to X$ the immersion of the closed fiber. we have an exact sequence of sheaves on $X$ $$0to Lotimes_D Cxrightarrowt Lto Lotimes_DCto 0$$ which can be rewritten as $$0to i_*L_0to Lto i_*L_0to 0$$ Applying $pi_*$ we get a sequence of $D$-modules which is in general not necessarily exact on the right, but you're assuming it is
$$0topi_*i_*L_0topi_*Ltopi_*i_*L_0to 0$$ Since $pi$ is projective, $pi_*L$ is a finitely generated $D$-module and from the exact sequence we see that $dim_C pi_*L=2dim H^0(L_0)$. A priori, this exact sequence doesn't imply that $pi_*L$ is flat, but let's recall the meaning of the arrow $pi_*i_*L_0topi_*L$. Consider multiplication by $t$ as an endomorphism of $L$. Since $L$ is flat over $D$, it factors as $$Ltwoheadrightarrow i_*L_0hookrightarrow L$$
Taking global sections and using your assumption again, we get $H^0(L_0)subsetim (pi_*Lxrightarrowtpi_*L)$ an this implies the desired equality $dimim t=frac12dim pi_*L$
answered 2 hours ago
SashaP
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
This follows easily from the theory of modules over $R = mathbb C[t]/(t^2)$. Indeed, we have a short exact sequence
$$0 to H^0(mathscr L_0) to H^0(mathscr L) to H^0(mathscr L_0) to 0,$$
induced by the observation that $(t) cong R/(t)$ and your assumption that the map $H^0(mathscr L) to H^0(mathscr L_0)$ is surjective. Now we have the following lemma.
Lemma. Let $0 to N stackrel fto M stackrel gto N to 0$ be a short exact sequence of finitely generated $R$-modules, and assume that $fg colon M to M$ is multiplication by $t$. Then $M$ is free and $N = tM cong M/tM$.
Proof. Because $N stackrel fto M stackrel gto N$ is exact, $M stackrel gto N$ is surjective, and $N stackrel fto M$ is injective, the sequence
$$M stackrelfg longrightarrow M stackrelfglongrightarrow Mtag1label1$$
is exact. But we also have $fg = t$. From the structure theory of $R$-modules, we may write
$$M cong R^m oplus (R/(t))^n$$
for some $m,n in mathbb Z_geq 0$. If $n > 0$, then any nonzero element in $(R/(t))^n$ is killed by $t$ but not in the image of $t$, contradicting exactness of (ref1). We conclude that $n = 0$, so $M$ is free. The final statement follows since $N = operatornameim(fg) = tM$. $square$
answered 1 hour ago
R. van Dobben de Bruyn
9,63622959
add a comment |Â
up vote
2
down vote
This follows easily from the theory of modules over $R = mathbb C[t]/(t^2)$. Indeed, we have a short exact sequence
$$0 to H^0(mathscr L_0) to H^0(mathscr L) to H^0(mathscr L_0) to 0,$$
induced by the observation that $(t) cong R/(t)$ and your assumption that the map $H^0(mathscr L) to H^0(mathscr L_0)$ is surjective. Now we have the following lemma.
Lemma. Let $0 to N stackrel fto M stackrel gto N to 0$ be a short exact sequence of finitely generated $R$-modules, and assume that $fg colon M to M$ is multiplication by $t$. Then $M$ is free and $N = tM cong M/tM$.
Proof. Because $N stackrel fto M stackrel gto N$ is exact, $M stackrel gto N$ is surjective, and $N stackrel fto M$ is injective, the sequence
$$M stackrelfg longrightarrow M stackrelfglongrightarrow Mtag1label1$$
is exact. But we also have $fg = t$. From the structure theory of $R$-modules, we may write
$$M cong R^m oplus (R/(t))^n$$
for some $m,n in mathbb Z_geq 0$. If $n > 0$, then any nonzero element in $(R/(t))^n$ is killed by $t$ but not in the image of $t$, contradicting exactness of (ref1). We conclude that $n = 0$, so $M$ is free. The final statement follows since $N = operatornameim(fg) = tM$. $square$
answered 1 hour ago
R. van Dobben de Bruyn
9,63622959
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This follows easily from the theory of modules over $R = mathbb C[t]/(t^2)$. Indeed, we have a short exact sequence
$$0 to H^0(mathscr L_0) to H^0(mathscr L) to H^0(mathscr L_0) to 0,$$
induced by the observation that $(t) cong R/(t)$ and your assumption that the map $H^0(mathscr L) to H^0(mathscr L_0)$ is surjective. Now we have the following lemma.
Lemma. Let $0 to N stackrel fto M stackrel gto N to 0$ be a short exact sequence of finitely generated $R$-modules, and assume that $fg colon M to M$ is multiplication by $t$. Then $M$ is free and $N = tM cong M/tM$.
Proof. Because $N stackrel fto M stackrel gto N$ is exact, $M stackrel gto N$ is surjective, and $N stackrel fto M$ is injective, the sequence
$$M stackrelfg longrightarrow M stackrelfglongrightarrow Mtag1label1$$
is exact. But we also have $fg = t$. From the structure theory of $R$-modules, we may write
$$M cong R^m oplus (R/(t))^n$$
for some $m,n in mathbb Z_geq 0$. If $n > 0$, then any nonzero element in $(R/(t))^n$ is killed by $t$ but not in the image of $t$, contradicting exactness of (ref1). We conclude that $n = 0$, so $M$ is free. The final statement follows since $N = operatornameim(fg) = tM$. $square$
answered 1 hour ago
R. van Dobben de Bruyn
9,63622959
This follows easily from the theory of modules over $R = mathbb C[t]/(t^2)$. Indeed, we have a short exact sequence
$$0 to H^0(mathscr L_0) to H^0(mathscr L) to H^0(mathscr L_0) to 0,$$
induced by the observation that $(t) cong R/(t)$ and your assumption that the map $H^0(mathscr L) to H^0(mathscr L_0)$ is surjective. Now we have the following lemma.
Lemma. Let $0 to N stackrel fto M stackrel gto N to 0$ be a short exact sequence of finitely generated $R$-modules, and assume that $fg colon M to M$ is multiplication by $t$. Then $M$ is free and $N = tM cong M/tM$.
Proof. Because $N stackrel fto M stackrel gto N$ is exact, $M stackrel gto N$ is surjective, and $N stackrel fto M$ is injective, the sequence
$$M stackrelfg longrightarrow M stackrelfglongrightarrow Mtag1label1$$
is exact. But we also have $fg = t$. From the structure theory of $R$-modules, we may write
$$M cong R^m oplus (R/(t))^n$$
for some $m,n in mathbb Z_geq 0$. If $n > 0$, then any nonzero element in $(R/(t))^n$ is killed by $t$ but not in the image of $t$, contradicting exactness of (ref1). We conclude that $n = 0$, so $M$ is free. The final statement follows since $N = operatornameim(fg) = tM$. $square$
answered 1 hour ago
R. van Dobben de Bruyn
9,63622959
answered 1 hour ago
R. van Dobben de Bruyn
9,63622959
answered 1 hour ago
R. van Dobben de Bruyn
9,63622959
answered 1 hour ago
R. van Dobben de Bruyn
9,63622959
9,63622959
add a comment |Â
add a comment |Â
up vote
1
down vote
$newcommandCmathbbCnewcommandDC[t]/t^2newcommandimmathrmim,$A module $M$ over $D$ is flat if and only if the inclusion $(t)subset D$ remains injective after tensoring with $M$ over $D$. In other words, the map $M/tMxrightarrowt M$ is injective which is equivalent to saying that the inclusion $im (Mxrightarrowt M)subset ker, (Mxrightarrowt M)$ is equality. If $M$ is finite-dimensional as a $C$-vector space, then this condition is equivalent to $dimim t=dimker t=dim M-dim im t$ which can be rephrased as $dimim t=frac12dim M$(for a non-flat module the LHS is strictly smaller).
(All of the above is more or less tautological since any finitely generated module over $D$ is a direct sum of some number of copies of $C$ and a few copies of $D$ and a module is flat iff it is free)
Denote by $i:X_0to X$ the immersion of the closed fiber. we have an exact sequence of sheaves on $X$ $$0to Lotimes_D Cxrightarrowt Lto Lotimes_DCto 0$$ which can be rewritten as $$0to i_*L_0to Lto i_*L_0to 0$$ Applying $pi_*$ we get a sequence of $D$-modules which is in general not necessarily exact on the right, but you're assuming it is
$$0topi_*i_*L_0topi_*Ltopi_*i_*L_0to 0$$ Since $pi$ is projective, $pi_*L$ is a finitely generated $D$-module and from the exact sequence we see that $dim_C pi_*L=2dim H^0(L_0)$. A priori, this exact sequence doesn't imply that $pi_*L$ is flat, but let's recall the meaning of the arrow $pi_*i_*L_0topi_*L$. Consider multiplication by $t$ as an endomorphism of $L$. Since $L$ is flat over $D$, it factors as $$Ltwoheadrightarrow i_*L_0hookrightarrow L$$
Taking global sections and using your assumption again, we get $H^0(L_0)subsetim (pi_*Lxrightarrowtpi_*L)$ an this implies the desired equality $dimim t=frac12dim pi_*L$
answered 2 hours ago
SashaP
2,69511631
add a comment |Â
up vote
1
down vote
$newcommandCmathbbCnewcommandDC[t]/t^2newcommandimmathrmim,$A module $M$ over $D$ is flat if and only if the inclusion $(t)subset D$ remains injective after tensoring with $M$ over $D$. In other words, the map $M/tMxrightarrowt M$ is injective which is equivalent to saying that the inclusion $im (Mxrightarrowt M)subset ker, (Mxrightarrowt M)$ is equality. If $M$ is finite-dimensional as a $C$-vector space, then this condition is equivalent to $dimim t=dimker t=dim M-dim im t$ which can be rephrased as $dimim t=frac12dim M$(for a non-flat module the LHS is strictly smaller).
(All of the above is more or less tautological since any finitely generated module over $D$ is a direct sum of some number of copies of $C$ and a few copies of $D$ and a module is flat iff it is free)
Denote by $i:X_0to X$ the immersion of the closed fiber. we have an exact sequence of sheaves on $X$ $$0to Lotimes_D Cxrightarrowt Lto Lotimes_DCto 0$$ which can be rewritten as $$0to i_*L_0to Lto i_*L_0to 0$$ Applying $pi_*$ we get a sequence of $D$-modules which is in general not necessarily exact on the right, but you're assuming it is
$$0topi_*i_*L_0topi_*Ltopi_*i_*L_0to 0$$ Since $pi$ is projective, $pi_*L$ is a finitely generated $D$-module and from the exact sequence we see that $dim_C pi_*L=2dim H^0(L_0)$. A priori, this exact sequence doesn't imply that $pi_*L$ is flat, but let's recall the meaning of the arrow $pi_*i_*L_0topi_*L$. Consider multiplication by $t$ as an endomorphism of $L$. Since $L$ is flat over $D$, it factors as $$Ltwoheadrightarrow i_*L_0hookrightarrow L$$
Taking global sections and using your assumption again, we get $H^0(L_0)subsetim (pi_*Lxrightarrowtpi_*L)$ an this implies the desired equality $dimim t=frac12dim pi_*L$
answered 2 hours ago
SashaP
2,69511631
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$newcommandCmathbbCnewcommandDC[t]/t^2newcommandimmathrmim,$A module $M$ over $D$ is flat if and only if the inclusion $(t)subset D$ remains injective after tensoring with $M$ over $D$. In other words, the map $M/tMxrightarrowt M$ is injective which is equivalent to saying that the inclusion $im (Mxrightarrowt M)subset ker, (Mxrightarrowt M)$ is equality. If $M$ is finite-dimensional as a $C$-vector space, then this condition is equivalent to $dimim t=dimker t=dim M-dim im t$ which can be rephrased as $dimim t=frac12dim M$(for a non-flat module the LHS is strictly smaller).
(All of the above is more or less tautological since any finitely generated module over $D$ is a direct sum of some number of copies of $C$ and a few copies of $D$ and a module is flat iff it is free)
Denote by $i:X_0to X$ the immersion of the closed fiber. we have an exact sequence of sheaves on $X$ $$0to Lotimes_D Cxrightarrowt Lto Lotimes_DCto 0$$ which can be rewritten as $$0to i_*L_0to Lto i_*L_0to 0$$ Applying $pi_*$ we get a sequence of $D$-modules which is in general not necessarily exact on the right, but you're assuming it is
$$0topi_*i_*L_0topi_*Ltopi_*i_*L_0to 0$$ Since $pi$ is projective, $pi_*L$ is a finitely generated $D$-module and from the exact sequence we see that $dim_C pi_*L=2dim H^0(L_0)$. A priori, this exact sequence doesn't imply that $pi_*L$ is flat, but let's recall the meaning of the arrow $pi_*i_*L_0topi_*L$. Consider multiplication by $t$ as an endomorphism of $L$. Since $L$ is flat over $D$, it factors as $$Ltwoheadrightarrow i_*L_0hookrightarrow L$$
Taking global sections and using your assumption again, we get $H^0(L_0)subsetim (pi_*Lxrightarrowtpi_*L)$ an this implies the desired equality $dimim t=frac12dim pi_*L$
answered 2 hours ago
SashaP
2,69511631
$newcommandCmathbbCnewcommandDC[t]/t^2newcommandimmathrmim,$A module $M$ over $D$ is flat if and only if the inclusion $(t)subset D$ remains injective after tensoring with $M$ over $D$. In other words, the map $M/tMxrightarrowt M$ is injective which is equivalent to saying that the inclusion $im (Mxrightarrowt M)subset ker, (Mxrightarrowt M)$ is equality. If $M$ is finite-dimensional as a $C$-vector space, then this condition is equivalent to $dimim t=dimker t=dim M-dim im t$ which can be rephrased as $dimim t=frac12dim M$(for a non-flat module the LHS is strictly smaller).
(All of the above is more or less tautological since any finitely generated module over $D$ is a direct sum of some number of copies of $C$ and a few copies of $D$ and a module is flat iff it is free)
Denote by $i:X_0to X$ the immersion of the closed fiber. we have an exact sequence of sheaves on $X$ $$0to Lotimes_D Cxrightarrowt Lto Lotimes_DCto 0$$ which can be rewritten as $$0to i_*L_0to Lto i_*L_0to 0$$ Applying $pi_*$ we get a sequence of $D$-modules which is in general not necessarily exact on the right, but you're assuming it is
$$0topi_*i_*L_0topi_*Ltopi_*i_*L_0to 0$$ Since $pi$ is projective, $pi_*L$ is a finitely generated $D$-module and from the exact sequence we see that $dim_C pi_*L=2dim H^0(L_0)$. A priori, this exact sequence doesn't imply that $pi_*L$ is flat, but let's recall the meaning of the arrow $pi_*i_*L_0topi_*L$. Consider multiplication by $t$ as an endomorphism of $L$. Since $L$ is flat over $D$, it factors as $$Ltwoheadrightarrow i_*L_0hookrightarrow L$$
Taking global sections and using your assumption again, we get $H^0(L_0)subsetim (pi_*Lxrightarrowtpi_*L)$ an this implies the desired equality $dimim t=frac12dim pi_*L$
answered 2 hours ago
SashaP
2,69511631
answered 2 hours ago
SashaP
2,69511631
answered 2 hours ago
SashaP
2,69511631
answered 2 hours ago
SashaP
2,69511631
2,69511631
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