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The exercise is stated as follows:

6.9. Let $X$ be an irreducible nonsingular curve in $\mathbf{P}^{3}$. Then for each $m\gg0$, there is a nonsingular surface $F$ of degree $m$ containing $X$. [Hint: Let $\pi: \tilde{\mathbf{P}} \rightarrow \mathbf{P}^{3}$ be the blowing-up of $X$ and let $Y=\pi^{-1}(X)$. Apply Bertini's theorem to the projective embedding of $\tilde{\mathbf{P}}$ corresponding to $\mathscr{I}_{Y} \otimes \pi^{*} \mathcal O_{\mathbf P^3}(m)$.]

By II Proposition 7,10 we know that there is a $m>0$ such that $\mathscr{I}_{Y} \otimes \pi^{*} \mathcal O_{\mathbf P^3}(m)$ is very ample, hence we can get a projective embedding $\theta:\tilde {\mathbf P}\to \mathbf P^N$ for some $N$. Apply Bertini's theorem I can get a hyperplane $H$ in $\mathbf P^N$ such that $H\cap \tilde{\mathbf P}$ is irreducible and nonsingular. But how can I show that $Y \subset H\cap \tilde{\mathbf P}$?

I know that $\mathscr I_Y$ can be identified with $\mathcal O_{\tilde{\mathbf P}}(1)$, and the ideal sheaf of $H\cap \tilde{\mathbf P}$ in $\tilde{\mathbf P}$ can be identified with $\theta^{-1}(\mathcal O_{\mathbf P^N}(-1))$. The only relationship of these invertible sheaves is $$\theta^*(\mathcal O_{\mathbf P^N}(1))=\mathscr{I}_{Y} \otimes \pi^{*} \mathcal O_{\mathbf P^3}(m).$$ What I need to show is $\theta^{-1}(\mathcal O_{\mathbf P^N}(-1))\subset \mathscr I_Y$, but how?

Any help is welcome, thanks!

Richard
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    Personally I would suggest doing this another way - the hint isn't so good in my opinion. Once you get $I_X(d)$ globally generated, run the proof of Bertini on that: consider the incidence correspondence of bad hyperplanes in $\Bbb P(I_X(d))\times \Bbb P^3$, compute the dimension of the fibers of the second projection (split in to cases based on $p\in X$ or not), and then conclude that the first projection can't be surjective. – KReiser Apr 24 '22 at 07:11
  • @KReiser Thanks for answering but I cannot understand your notation $\Bbb P(I_X(d))$. Besides, do you mean we should consider the $d$-uple embedding of $\Bbb P^3$ to find a hyperplane section that contains $X$? But why should we make $I_X(d)$ globally generated? Could you write a more concrete hint as an answer? I can try to finish the details. – Richard Apr 30 '22 at 12:24
  • For a vector space $V$, $\Bbb P(V)$ is the projectivization. If I wanted to consider the $d$-uple embedding, I would have said so. For why you need $I_X(d)$ globally generated, follow the proof outline. – KReiser Apr 30 '22 at 16:57
  • @KReiser Thank you for your hint! So we need to consider $\Sigma:={(H,x)\in\mathbb{P}H^0(I_X(m))\times\mathbb{P}^3:x\in H_{sing}}$ with two projections $\pi_1,\pi_2$. But how to compute the dimension of fiber of $\pi_2$? I know the case when we consider $\mathbb{P}H^0(\mathscr{O}_X(m))\cong\mathbb{P}^N$ instead of $\mathbb{P}H^0(I_X(m))$ which have fibers of dimension $N-4$. – WakeUp-X.Liu May 30 '23 at 03:25

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