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?



Descent of a line bundle from normalisation
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 $X$


