Let $X$ be a normal, projective (complex) variety with at worst rational singularities. Let $\pi:Y \to X$ be the resolution of singularities obtained by blowingup the singular points. Is $R^1 \pi_*\mathbb{Z}=0$? I am mainly interested in the case when $X$ is of dimension $3$.
1 Answer
This should be true. Use the exponential sequence to obtain $$\pi_*\mathcal{O}_Y\to \pi_*\mathcal{O}_Y^*\to R^1\pi_*\mathbb{Z}\to R^1\pi_*\mathcal{O}_Y$$ The last term is zero since you have rational singularities. You can check that the first map is surjective as follows: By normality $$\pi_*\mathcal{O}_Y = \mathcal{O}_X,\quad \pi_*\mathcal{O}_Y^* = \mathcal{O}_X^*$$ Choose a local embedding $X\subset Z$, with $Z$ smooth. Consider the diagram $$\begin{array}{ccc} \mathcal{O}_Z & \xrightarrow{exp}&\mathcal{O}_Z^* \\ \downarrow&&\downarrow\\ \mathcal{O}_X & \xrightarrow{exp}&\mathcal{O}_X^* \end{array} $$ The top arrow $exp$ is surjective, and the vertical arrows are surjective. Therefore the bottom $\exp$ is surjective.

$\begingroup$ I am sorry, I do not quite see the surjectivity of the first map. $\endgroup$ Commented Mar 27, 2021 at 15:41


$\begingroup$ Thank you. I think I understand. $\endgroup$ Commented Mar 27, 2021 at 16:47