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Suppose that $X \subset \mathbb{P}^3$ is a smooth complex projective surface of degree $d \geq 4$. Then the Noether-Lefschetz theorem states that if $X$ is very general, then $\mathrm{Pic}(\mathbb{P}^3) \rightarrow \mathrm{Pic}(X)$ is an isomorphism, so in particular,

$$\mathrm{Pic}(X) \cong \mathbb{Z} \cdot \mathcal{O}_X(1).$$

Of course, there are special surfaces for which this fails and the Picard number jumps. However, I was wondering: are there any special surfaces $X$ for which $\mathcal{O}_X(1)$ is not primitive in $\mathrm{Pic}(X)$? In other words, can there exist a line bundle $L$ on $X$ and a positive integer $r \geq 2$ such that $L^{\otimes r} \cong \mathcal{O}_X(1)$? Examples of this or references to a proof of the contrary would be appreciated.

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This cannot happen. Grothendieck proved that for any such surface, the natural map $\operatorname{Pic} \mathbb{P}^3\to \operatorname{Pic} \hat{X}$ is an isomorphism where $\hat{X}$ is the formal completion. So it suffices to prove that the cokernel of the natural map $\operatorname{Pic}\hat{X}\to \operatorname{Pic} X$ is torsion free. If $X _n$ denotes the $n^ {th}$ order thickening defined by the ideal $I^n$ where $I$ defines $X$, we have a natural exact sequence, $0\to I^n/I^{n+1}\to \mathcal{O}_{X_{n+1}}^*\to\mathcal{O}_{X_n}^*\to 0$. Taking cohomologies, we see that the desired cokernel is filtered by subspaces of $H^2(I^n/I^{n+1})$ and since we are over complex numbers, these are torsion free.

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  • $\begingroup$ That's a neat argument, thanks! $\endgroup$
    – Irwin
    Commented Sep 4, 2022 at 15:46

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