Descent of a line bundle from normalisation

Question

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?

Answer 1

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 $X$