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Let X be an integral quasi projective variety. Let Y be the normalisation. Given a line bundle on Y, does some power of it descends to X?

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  • $\begingroup$ Let $Y = E\times E$ for an elliptic curve $E$, and let $X$ be $Y$ with two fibers $E\times \{x_i\}$, $i=1, 2$ identified, where $x_1-x_2\in E$ is not a torsion point. The projection $\pi:Y\to X$ is the normalization map. No power of the Poincare bundle $L$ on $E$ descends to $X$. I'm not sure that such a quotient $X$ exists as a quasi-projective variety though... $\endgroup$ Jul 6, 2017 at 8:39
  • $\begingroup$ Of course it is easier to give a non-integral example: take $Y$ the disjoint union of two copies of $\mathbf{P}^2$, $L$ a line bundle on $Y$ whose restrictions to the two copies have different degrees. Let $X$ be obtained from $Y$ by identifying two lines, one in each $\mathbf{P}^2$. Then $X$ exists as a quasiprojective variety, $Y$ is its normalization, and no power of $L$ descends to $X$. $\endgroup$ Jul 6, 2017 at 8:46
  • $\begingroup$ @PiotrAchinger It is the product of a nodal curve and a smooth curve and thus is projective. $\endgroup$
    – Will Sawin
    Jul 6, 2017 at 8:54
  • $\begingroup$ @WillSawin ah yes of course! Thank you! $\endgroup$ Jul 6, 2017 at 8:57

1 Answer 1

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Posting the comment above as an answer:

The answer is yes for curves and no in higher dimension.

For curves, we have a short exact sequence $$0\to \mathcal{O}_X^*\to \pi_*\mathcal{O}_Y^*\to Q\to 0 $$ where $Q$ is supported in finitely many points. In particular, $H^1(X, Q)=0$. Moreover, $R^1\pi_* \mathcal{O}_Y^* = 0$, and thus $$ {\rm Pic}\, Y = H^1(Y, \mathcal{O}_Y^*) = H^1(X, \pi_* \mathcal{O}_Y^*). $$ We deduce that $\pi^*:{\rm Pic}\, X\to {\rm Pic}\, Y$ is surjective.

In higher dimension, consider the following example. Let $E$ be an elliptic curve, $x_1, x_2\in E$ two closed points such that the difference $x_1-x_2$ is not a torsion point. Let $C$ be the nodal curve $E/x_1\sim x_2$, and set $Y=E\times E$, $X=E\times C$, $\pi:Y\to X$ induced by the projection $E\to C$. Then $\pi$ is the normalization map of $X$. Let $L$ be the Poincare bundle on $Y=E\times E$; by definition, the restriction of $L$ to $E\times \{x\}$ is $\mathcal{O}_E(x-0)$. Suppose that $L^n =\pi^* M$ for some $n>0$ and some line bundle $M$ on $X$. Thus $L^n|_{E\times \{x_1\}}\cong L^n|_{E\times \{x_2\}}$, i.e., $\mathcal{O}_E(nx_1-0)\cong \mathcal{O}_E(nx_2-0)$ (here $nx_i$ means multiplication in the group law on $E$). This contradicts the fact that $x_1-x_2$ is non-torsion.

Of course it is easier to give a non-integral example: take $Y=$ the disjoint union of two copies of $\mathbf{P}^2$, $L$ a line bundle on $Y$ whose restrictions to the two copies have different degrees. Let $X$ be obtained from $Y$ by identifying two lines, one in each $\mathbf{P}^2$. Then $X$ exists as a quasiprojective variety, $Y$ is its normalization, and no positive power of $L$ descends to $X$

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