# Projective family of hypersurfaces on curve admits some embedding?

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.

• What do you mean by "$n$-dimensional hypersurface in $\mathbb{P}^N$"? – Sasha Jan 6 '19 at 12:49
• Note that even a smooth curve of genus 2 cannot be (isomorphically) embedded into $\mathbb{P}^2$. – Sasha Jan 6 '19 at 12:50
• 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. – Akatsuki Jan 6 '19 at 13:24
• 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 $n+1$. – Mohan Jan 6 '19 at 20:16
• On the other hand, $X$ can be embedded into a nontrivial $\mathbb{P}^{n+1}$-bundle. – Sasha Jan 7 '19 at 8:25

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.
• Sorry I cannot follow your last line: why $L\cong L^{-1}$? – Akatsuki Jan 13 '19 at 10:21
• Because $L$ is a line bundle of order 2. – Sasha Jan 13 '19 at 14:47
• So the question should be, why it is of order $2$? – Akatsuki Jan 13 '19 at 14:58
• 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 $L^{-1} \otimes L^{-1} \to O_B$. – Sasha Jan 13 '19 at 15:06
• $O_X$ is a coherent sheaf on $B \times \mathbb{P}^1$, the sequence is a locally free resolution for it. – Sasha Jan 13 '19 at 16:43