Suppose $X$ is a projective variety, and $D$ is a Cartier divisor on $X$. Is it possible (or this is always true) that there is a (ramified) double cover $\pi: \tilde{X} \to X$ such that $\frac{1}{2}\pi^*(D)$ is also a Cartier divisor?
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This is not always possible. Take $X=\mathbb{P}^2$, $D=$ a line. If $\pi ^*D$ is twice a Cartier divisor on $\tilde{X} $, this holds also on the normalization of $\tilde{X} $, so you can assume that $\tilde{X} $ is normal. Then $(\frac{1}{2}\pi ^*D)^2 = \frac{1}{2}D^2=\frac{1}{2}$; but on a normal surface the intersection of two Cartier divisors is an integer.
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1$\begingroup$ No, it is $2\times \frac{1}{4}=\frac{1}{2} $ ($2\times $ because of the double covering). $\endgroup$– abxCommented Nov 29, 2014 at 19:15
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$\begingroup$ @abx Thank you for your answer! Maybe this is silly, but suppose the double cover of $\mathbb{P}^2$ is $\mathbb{P}^2 \sqcup \mathbb{P}^2$, the disjoint union of two planes. Then $\pi^*D$ is a disjoint union of two line, take either line, I get a Cartier divisor. Where did I make wrong? $\endgroup$ Commented Nov 30, 2014 at 0:06
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1$\begingroup$ @LiYutong If you double this line, it's by no means equivalent to $\pi^*D$. $\endgroup$ Commented Nov 30, 2014 at 2:20
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$\begingroup$ @LiYutong : in other words, $\mathrm{Pic}(\mathbb{P}^2\sqcup \mathbb{P}^2)=\mathbb{Z}\oplus\mathbb{Z}$, generated by $(\mathrm{line},0)$ and $(0,\mathrm{line})$. $\endgroup$– abxCommented Nov 30, 2014 at 5:58