Projective family of hypersurfaces on curve admits some embedding?












2












$begingroup$


Let $f:Xto B$ be a projective family of hypersurface of type $(d,n)$ in on some curve $B$, i.e., $Xsubset Btimes mathbb P^N$ is closed subscheme, and $f:Xto B$ is flat, with every fiber isomorphic to $n$-dimensional hypersurface of degree $d$. I want to know the following:




Is it true that every fiber of $f$ admit an embedding to $mathbb P^{n+1}$? i.e. there exists some other flat family $f':X'to B$ with $X'subset Btimes mathbb P^{n+1}$ a divisor, such that the two family $f$ and $f'$ are isomorphic?




I tend to believe this is not true, but I want to know a counter-example. Any comments would be helpful.






algebraic-geometry







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What do you mean by "$n$-dimensional hypersurface in $mathbb{P}^N$"?
    $endgroup$
    – Sasha
    Jan 6 at 12:49










  • $begingroup$
    Note that even a smooth curve of genus 2 cannot be (isomorphically) embedded into $mathbb{P}^2$.
    $endgroup$
    – Sasha
    Jan 6 at 12:50










  • $begingroup$
    Dear @Sasha: I mean it is isomorphic to some hypersurface in $mathbb P^{n+1}subset mathbb P^N$. Here $mathbb P^{n+1}$ is a linear subspace of $mathbb P^N$. Sorry for didn't make this clear.
    $endgroup$
    – Akatsuki
    Jan 6 at 13:24






  • 2




    $begingroup$
    Clearly, this is true locally on $B$, that is there is an open cover of $B$ such that on each of these you have such an embedding. It is also true that you can assume $Xsubset Btimesmathbb{P}^{n+2}$. I do not know whether we can replace this with $n+1$.
    $endgroup$
    – Mohan
    Jan 6 at 20:16






  • 1




    $begingroup$
    On the other hand, $X$ can be embedded into a nontrivial $mathbb{P}^{n+1}$-bundle.
    $endgroup$
    – Sasha
    Jan 7 at 8:25
















2












$begingroup$


Let $f:Xto B$ be a projective family of hypersurface of type $(d,n)$ in on some curve $B$, i.e., $Xsubset Btimes mathbb P^N$ is closed subscheme, and $f:Xto B$ is flat, with every fiber isomorphic to $n$-dimensional hypersurface of degree $d$. I want to know the following:




Is it true that every fiber of $f$ admit an embedding to $mathbb P^{n+1}$? i.e. there exists some other flat family $f':X'to B$ with $X'subset Btimes mathbb P^{n+1}$ a divisor, such that the two family $f$ and $f'$ are isomorphic?




I tend to believe this is not true, but I want to know a counter-example. Any comments would be helpful.






algebraic-geometry







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What do you mean by "$n$-dimensional hypersurface in $mathbb{P}^N$"?
    $endgroup$
    – Sasha
    Jan 6 at 12:49










  • $begingroup$
    Note that even a smooth curve of genus 2 cannot be (isomorphically) embedded into $mathbb{P}^2$.
    $endgroup$
    – Sasha
    Jan 6 at 12:50










  • $begingroup$
    Dear @Sasha: I mean it is isomorphic to some hypersurface in $mathbb P^{n+1}subset mathbb P^N$. Here $mathbb P^{n+1}$ is a linear subspace of $mathbb P^N$. Sorry for didn't make this clear.
    $endgroup$
    – Akatsuki
    Jan 6 at 13:24






  • 2




    $begingroup$
    Clearly, this is true locally on $B$, that is there is an open cover of $B$ such that on each of these you have such an embedding. It is also true that you can assume $Xsubset Btimesmathbb{P}^{n+2}$. I do not know whether we can replace this with $n+1$.
    $endgroup$
    – Mohan
    Jan 6 at 20:16






  • 1




    $begingroup$
    On the other hand, $X$ can be embedded into a nontrivial $mathbb{P}^{n+1}$-bundle.
    $endgroup$
    – Sasha
    Jan 7 at 8:25














2












2








2





$begingroup$


Let $f:Xto B$ be a projective family of hypersurface of type $(d,n)$ in on some curve $B$, i.e., $Xsubset Btimes mathbb P^N$ is closed subscheme, and $f:Xto B$ is flat, with every fiber isomorphic to $n$-dimensional hypersurface of degree $d$. I want to know the following:




Is it true that every fiber of $f$ admit an embedding to $mathbb P^{n+1}$? i.e. there exists some other flat family $f':X'to B$ with $X'subset Btimes mathbb P^{n+1}$ a divisor, such that the two family $f$ and $f'$ are isomorphic?




I tend to believe this is not true, but I want to know a counter-example. Any comments would be helpful.






algebraic-geometry







share|cite|improve this question









$endgroup$




Let $f:Xto B$ be a projective family of hypersurface of type $(d,n)$ in on some curve $B$, i.e., $Xsubset Btimes mathbb P^N$ is closed subscheme, and $f:Xto B$ is flat, with every fiber isomorphic to $n$-dimensional hypersurface of degree $d$. I want to know the following:




Is it true that every fiber of $f$ admit an embedding to $mathbb P^{n+1}$? i.e. there exists some other flat family $f':X'to B$ with $X'subset Btimes mathbb P^{n+1}$ a divisor, such that the two family $f$ and $f'$ are isomorphic?




I tend to believe this is not true, but I want to know a counter-example. Any comments would be helpful.






algebraic-geometry




algebraic-geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 6 at 3:31









AkatsukiAkatsuki

1,1291726




1,1291726








  • 1




    $begingroup$
    What do you mean by "$n$-dimensional hypersurface in $mathbb{P}^N$"?
    $endgroup$
    – Sasha
    Jan 6 at 12:49










  • $begingroup$
    Note that even a smooth curve of genus 2 cannot be (isomorphically) embedded into $mathbb{P}^2$.
    $endgroup$
    – Sasha
    Jan 6 at 12:50










  • $begingroup$
    Dear @Sasha: I mean it is isomorphic to some hypersurface in $mathbb P^{n+1}subset mathbb P^N$. Here $mathbb P^{n+1}$ is a linear subspace of $mathbb P^N$. Sorry for didn't make this clear.
    $endgroup$
    – Akatsuki
    Jan 6 at 13:24






  • 2




    $begingroup$
    Clearly, this is true locally on $B$, that is there is an open cover of $B$ such that on each of these you have such an embedding. It is also true that you can assume $Xsubset Btimesmathbb{P}^{n+2}$. I do not know whether we can replace this with $n+1$.
    $endgroup$
    – Mohan
    Jan 6 at 20:16






  • 1




    $begingroup$
    On the other hand, $X$ can be embedded into a nontrivial $mathbb{P}^{n+1}$-bundle.
    $endgroup$
    – Sasha
    Jan 7 at 8:25














  • 1




    $begingroup$
    What do you mean by "$n$-dimensional hypersurface in $mathbb{P}^N$"?
    $endgroup$
    – Sasha
    Jan 6 at 12:49










  • $begingroup$
    Note that even a smooth curve of genus 2 cannot be (isomorphically) embedded into $mathbb{P}^2$.
    $endgroup$
    – Sasha
    Jan 6 at 12:50










  • $begingroup$
    Dear @Sasha: I mean it is isomorphic to some hypersurface in $mathbb P^{n+1}subset mathbb P^N$. Here $mathbb P^{n+1}$ is a linear subspace of $mathbb P^N$. Sorry for didn't make this clear.
    $endgroup$
    – Akatsuki
    Jan 6 at 13:24






  • 2




    $begingroup$
    Clearly, this is true locally on $B$, that is there is an open cover of $B$ such that on each of these you have such an embedding. It is also true that you can assume $Xsubset Btimesmathbb{P}^{n+2}$. I do not know whether we can replace this with $n+1$.
    $endgroup$
    – Mohan
    Jan 6 at 20:16






  • 1




    $begingroup$
    On the other hand, $X$ can be embedded into a nontrivial $mathbb{P}^{n+1}$-bundle.
    $endgroup$
    – Sasha
    Jan 7 at 8:25








1




1




$begingroup$
What do you mean by "$n$-dimensional hypersurface in $mathbb{P}^N$"?
$endgroup$
– Sasha
Jan 6 at 12:49




$begingroup$
What do you mean by "$n$-dimensional hypersurface in $mathbb{P}^N$"?
$endgroup$
– Sasha
Jan 6 at 12:49












$begingroup$
Note that even a smooth curve of genus 2 cannot be (isomorphically) embedded into $mathbb{P}^2$.
$endgroup$
– Sasha
Jan 6 at 12:50




$begingroup$
Note that even a smooth curve of genus 2 cannot be (isomorphically) embedded into $mathbb{P}^2$.
$endgroup$
– Sasha
Jan 6 at 12:50












$begingroup$
Dear @Sasha: I mean it is isomorphic to some hypersurface in $mathbb P^{n+1}subset mathbb P^N$. Here $mathbb P^{n+1}$ is a linear subspace of $mathbb P^N$. Sorry for didn't make this clear.
$endgroup$
– Akatsuki
Jan 6 at 13:24




$begingroup$
Dear @Sasha: I mean it is isomorphic to some hypersurface in $mathbb P^{n+1}subset mathbb P^N$. Here $mathbb P^{n+1}$ is a linear subspace of $mathbb P^N$. Sorry for didn't make this clear.
$endgroup$
– Akatsuki
Jan 6 at 13:24




2




2




$begingroup$
Clearly, this is true locally on $B$, that is there is an open cover of $B$ such that on each of these you have such an embedding. It is also true that you can assume $Xsubset Btimesmathbb{P}^{n+2}$. I do not know whether we can replace this with $n+1$.
$endgroup$
– Mohan
Jan 6 at 20:16




$begingroup$
Clearly, this is true locally on $B$, that is there is an open cover of $B$ such that on each of these you have such an embedding. It is also true that you can assume $Xsubset Btimesmathbb{P}^{n+2}$. I do not know whether we can replace this with $n+1$.
$endgroup$
– Mohan
Jan 6 at 20:16




1




1




$begingroup$
On the other hand, $X$ can be embedded into a nontrivial $mathbb{P}^{n+1}$-bundle.
$endgroup$
– Sasha
Jan 7 at 8:25




$begingroup$
On the other hand, $X$ can be embedded into a nontrivial $mathbb{P}^{n+1}$-bundle.
$endgroup$
– Sasha
Jan 7 at 8:25










1 Answer
1






active

oldest

votes


















2












$begingroup$

No, this is not possible. For example, let $f colon X to B$ be a nontrivial etale double cover. Every fiber is a union of two points, so it can be embedded into $mathbb{P}^1$ as a hypersurface. But there is no embedding $X subset B times mathbb{P}^1$.



Indeed, assume $X subset B times mathbb{P}^1$. Then $X$ is a divisor of relative degree 2 over $B$. Since $Pic(B times mathbb{P}^1) cong Pic(B) oplus Pic(mathbb{P}^1)$, the corresponding line bundle can be written as $L boxtimes O(2)$ for some $L in Pic(B)$, so that $X$ defines a nonzero element in
$$
H^0(B times mathbb{P}^1, L boxtimes O(2)) = H^0(B,L) otimes H^0(mathbb{P}^1,O(2)).
$$
In particular, $H^0(B,L) ne 0$.



On the other hand, we have a resolution
$$
0 to L^{-1} boxtimes O(-2) to O_{B times mathbb{P}^1} to O_X to 0,
$$
which implies that
$$
f_*O_X cong O_B oplus L^{-1}.
$$
Since $f$ is etale and nontrivial, the line bundle $L^{-1}$ on $B$ is a nontrivial element of order 2 in $Pic(B)$.
In particular, $L cong L^{-1}$ has no global sections.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry I cannot follow your last line: why $Lcong L^{-1}$?
    $endgroup$
    – Akatsuki
    Jan 13 at 10:21












  • $begingroup$
    Because $L$ is a line bundle of order 2.
    $endgroup$
    – Sasha
    Jan 13 at 14:47












  • $begingroup$
    So the question should be, why it is of order $2$?
    $endgroup$
    – Akatsuki
    Jan 13 at 14:58










  • $begingroup$
    Since we started with an etale double covering, the multiplication map $(f_*O_X) otimes (f_*O_X) to (f_*O_X)$ restricts to an isomorphism $L^{-1} otimes L^{-1} to O_B$.
    $endgroup$
    – Sasha
    Jan 13 at 15:06






  • 1




    $begingroup$
    $O_X$ is a coherent sheaf on $B times mathbb{P}^1$, the sequence is a locally free resolution for it.
    $endgroup$
    – Sasha
    Jan 13 at 16:43











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

No, this is not possible. For example, let $f colon X to B$ be a nontrivial etale double cover. Every fiber is a union of two points, so it can be embedded into $mathbb{P}^1$ as a hypersurface. But there is no embedding $X subset B times mathbb{P}^1$.



Indeed, assume $X subset B times mathbb{P}^1$. Then $X$ is a divisor of relative degree 2 over $B$. Since $Pic(B times mathbb{P}^1) cong Pic(B) oplus Pic(mathbb{P}^1)$, the corresponding line bundle can be written as $L boxtimes O(2)$ for some $L in Pic(B)$, so that $X$ defines a nonzero element in
$$
H^0(B times mathbb{P}^1, L boxtimes O(2)) = H^0(B,L) otimes H^0(mathbb{P}^1,O(2)).
$$
In particular, $H^0(B,L) ne 0$.



On the other hand, we have a resolution
$$
0 to L^{-1} boxtimes O(-2) to O_{B times mathbb{P}^1} to O_X to 0,
$$
which implies that
$$
f_*O_X cong O_B oplus L^{-1}.
$$
Since $f$ is etale and nontrivial, the line bundle $L^{-1}$ on $B$ is a nontrivial element of order 2 in $Pic(B)$.
In particular, $L cong L^{-1}$ has no global sections.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry I cannot follow your last line: why $Lcong L^{-1}$?
    $endgroup$
    – Akatsuki
    Jan 13 at 10:21












  • $begingroup$
    Because $L$ is a line bundle of order 2.
    $endgroup$
    – Sasha
    Jan 13 at 14:47












  • $begingroup$
    So the question should be, why it is of order $2$?
    $endgroup$
    – Akatsuki
    Jan 13 at 14:58










  • $begingroup$
    Since we started with an etale double covering, the multiplication map $(f_*O_X) otimes (f_*O_X) to (f_*O_X)$ restricts to an isomorphism $L^{-1} otimes L^{-1} to O_B$.
    $endgroup$
    – Sasha
    Jan 13 at 15:06






  • 1




    $begingroup$
    $O_X$ is a coherent sheaf on $B times mathbb{P}^1$, the sequence is a locally free resolution for it.
    $endgroup$
    – Sasha
    Jan 13 at 16:43
















2












$begingroup$

No, this is not possible. For example, let $f colon X to B$ be a nontrivial etale double cover. Every fiber is a union of two points, so it can be embedded into $mathbb{P}^1$ as a hypersurface. But there is no embedding $X subset B times mathbb{P}^1$.



Indeed, assume $X subset B times mathbb{P}^1$. Then $X$ is a divisor of relative degree 2 over $B$. Since $Pic(B times mathbb{P}^1) cong Pic(B) oplus Pic(mathbb{P}^1)$, the corresponding line bundle can be written as $L boxtimes O(2)$ for some $L in Pic(B)$, so that $X$ defines a nonzero element in
$$
H^0(B times mathbb{P}^1, L boxtimes O(2)) = H^0(B,L) otimes H^0(mathbb{P}^1,O(2)).
$$
In particular, $H^0(B,L) ne 0$.



On the other hand, we have a resolution
$$
0 to L^{-1} boxtimes O(-2) to O_{B times mathbb{P}^1} to O_X to 0,
$$
which implies that
$$
f_*O_X cong O_B oplus L^{-1}.
$$
Since $f$ is etale and nontrivial, the line bundle $L^{-1}$ on $B$ is a nontrivial element of order 2 in $Pic(B)$.
In particular, $L cong L^{-1}$ has no global sections.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry I cannot follow your last line: why $Lcong L^{-1}$?
    $endgroup$
    – Akatsuki
    Jan 13 at 10:21












  • $begingroup$
    Because $L$ is a line bundle of order 2.
    $endgroup$
    – Sasha
    Jan 13 at 14:47












  • $begingroup$
    So the question should be, why it is of order $2$?
    $endgroup$
    – Akatsuki
    Jan 13 at 14:58










  • $begingroup$
    Since we started with an etale double covering, the multiplication map $(f_*O_X) otimes (f_*O_X) to (f_*O_X)$ restricts to an isomorphism $L^{-1} otimes L^{-1} to O_B$.
    $endgroup$
    – Sasha
    Jan 13 at 15:06






  • 1




    $begingroup$
    $O_X$ is a coherent sheaf on $B times mathbb{P}^1$, the sequence is a locally free resolution for it.
    $endgroup$
    – Sasha
    Jan 13 at 16:43














2












2








2





$begingroup$

No, this is not possible. For example, let $f colon X to B$ be a nontrivial etale double cover. Every fiber is a union of two points, so it can be embedded into $mathbb{P}^1$ as a hypersurface. But there is no embedding $X subset B times mathbb{P}^1$.



Indeed, assume $X subset B times mathbb{P}^1$. Then $X$ is a divisor of relative degree 2 over $B$. Since $Pic(B times mathbb{P}^1) cong Pic(B) oplus Pic(mathbb{P}^1)$, the corresponding line bundle can be written as $L boxtimes O(2)$ for some $L in Pic(B)$, so that $X$ defines a nonzero element in
$$
H^0(B times mathbb{P}^1, L boxtimes O(2)) = H^0(B,L) otimes H^0(mathbb{P}^1,O(2)).
$$
In particular, $H^0(B,L) ne 0$.



On the other hand, we have a resolution
$$
0 to L^{-1} boxtimes O(-2) to O_{B times mathbb{P}^1} to O_X to 0,
$$
which implies that
$$
f_*O_X cong O_B oplus L^{-1}.
$$
Since $f$ is etale and nontrivial, the line bundle $L^{-1}$ on $B$ is a nontrivial element of order 2 in $Pic(B)$.
In particular, $L cong L^{-1}$ has no global sections.






share|cite|improve this answer









$endgroup$



No, this is not possible. For example, let $f colon X to B$ be a nontrivial etale double cover. Every fiber is a union of two points, so it can be embedded into $mathbb{P}^1$ as a hypersurface. But there is no embedding $X subset B times mathbb{P}^1$.



Indeed, assume $X subset B times mathbb{P}^1$. Then $X$ is a divisor of relative degree 2 over $B$. Since $Pic(B times mathbb{P}^1) cong Pic(B) oplus Pic(mathbb{P}^1)$, the corresponding line bundle can be written as $L boxtimes O(2)$ for some $L in Pic(B)$, so that $X$ defines a nonzero element in
$$
H^0(B times mathbb{P}^1, L boxtimes O(2)) = H^0(B,L) otimes H^0(mathbb{P}^1,O(2)).
$$
In particular, $H^0(B,L) ne 0$.



On the other hand, we have a resolution
$$
0 to L^{-1} boxtimes O(-2) to O_{B times mathbb{P}^1} to O_X to 0,
$$
which implies that
$$
f_*O_X cong O_B oplus L^{-1}.
$$
Since $f$ is etale and nontrivial, the line bundle $L^{-1}$ on $B$ is a nontrivial element of order 2 in $Pic(B)$.
In particular, $L cong L^{-1}$ has no global sections.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 9 at 6:43









SashaSasha

5,168139




5,168139












  • $begingroup$
    Sorry I cannot follow your last line: why $Lcong L^{-1}$?
    $endgroup$
    – Akatsuki
    Jan 13 at 10:21












  • $begingroup$
    Because $L$ is a line bundle of order 2.
    $endgroup$
    – Sasha
    Jan 13 at 14:47












  • $begingroup$
    So the question should be, why it is of order $2$?
    $endgroup$
    – Akatsuki
    Jan 13 at 14:58










  • $begingroup$
    Since we started with an etale double covering, the multiplication map $(f_*O_X) otimes (f_*O_X) to (f_*O_X)$ restricts to an isomorphism $L^{-1} otimes L^{-1} to O_B$.
    $endgroup$
    – Sasha
    Jan 13 at 15:06






  • 1




    $begingroup$
    $O_X$ is a coherent sheaf on $B times mathbb{P}^1$, the sequence is a locally free resolution for it.
    $endgroup$
    – Sasha
    Jan 13 at 16:43


















  • $begingroup$
    Sorry I cannot follow your last line: why $Lcong L^{-1}$?
    $endgroup$
    – Akatsuki
    Jan 13 at 10:21












  • $begingroup$
    Because $L$ is a line bundle of order 2.
    $endgroup$
    – Sasha
    Jan 13 at 14:47












  • $begingroup$
    So the question should be, why it is of order $2$?
    $endgroup$
    – Akatsuki
    Jan 13 at 14:58










  • $begingroup$
    Since we started with an etale double covering, the multiplication map $(f_*O_X) otimes (f_*O_X) to (f_*O_X)$ restricts to an isomorphism $L^{-1} otimes L^{-1} to O_B$.
    $endgroup$
    – Sasha
    Jan 13 at 15:06






  • 1




    $begingroup$
    $O_X$ is a coherent sheaf on $B times mathbb{P}^1$, the sequence is a locally free resolution for it.
    $endgroup$
    – Sasha
    Jan 13 at 16:43
















$begingroup$
Sorry I cannot follow your last line: why $Lcong L^{-1}$?
$endgroup$
– Akatsuki
Jan 13 at 10:21






$begingroup$
Sorry I cannot follow your last line: why $Lcong L^{-1}$?
$endgroup$
– Akatsuki
Jan 13 at 10:21














$begingroup$
Because $L$ is a line bundle of order 2.
$endgroup$
– Sasha
Jan 13 at 14:47






$begingroup$
Because $L$ is a line bundle of order 2.
$endgroup$
– Sasha
Jan 13 at 14:47














$begingroup$
So the question should be, why it is of order $2$?
$endgroup$
– Akatsuki
Jan 13 at 14:58




$begingroup$
So the question should be, why it is of order $2$?
$endgroup$
– Akatsuki
Jan 13 at 14:58












$begingroup$
Since we started with an etale double covering, the multiplication map $(f_*O_X) otimes (f_*O_X) to (f_*O_X)$ restricts to an isomorphism $L^{-1} otimes L^{-1} to O_B$.
$endgroup$
– Sasha
Jan 13 at 15:06




$begingroup$
Since we started with an etale double covering, the multiplication map $(f_*O_X) otimes (f_*O_X) to (f_*O_X)$ restricts to an isomorphism $L^{-1} otimes L^{-1} to O_B$.
$endgroup$
– Sasha
Jan 13 at 15:06




1




1




$begingroup$
$O_X$ is a coherent sheaf on $B times mathbb{P}^1$, the sequence is a locally free resolution for it.
$endgroup$
– Sasha
Jan 13 at 16:43




$begingroup$
$O_X$ is a coherent sheaf on $B times mathbb{P}^1$, the sequence is a locally free resolution for it.
$endgroup$
– Sasha
Jan 13 at 16:43


















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