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Let $f:X\to B$ be a projective family of hypersurface of type $(d,n)$ in on some curve $B$, i.e., $X\subset B\times \mathbb P^N$ is closed subscheme, and $f:X\to B$ 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 $X'\subset B\times \mathbb P^{n+1}$ 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.

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    $\begingroup$ 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
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    $\begingroup$ 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 $X\subset B\times\mathbb{P}^{n+2}$. I do not know whether we can replace this with $n+1$. $\endgroup$ – Mohan Jan 6 at 20:16
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    $\begingroup$ On the other hand, $X$ can be embedded into a nontrivial $\mathbb{P}^{n+1}$-bundle. $\endgroup$ – Sasha Jan 7 at 8:25
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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.

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  • $\begingroup$ Sorry I cannot follow your last line: why $L\cong L^{-1}$? $\endgroup$ – 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
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    $\begingroup$ $O_X$ 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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