Push forward a Cartier is still Cartier

Question

Suppose that $X$ is a projective complex variety with rational singularities (i.e. for any resolution $f: Y \to X$, $f_* \mathcal{O}_Y=\mathcal{O}_X, Rf^i_*\mathcal{O}_Y=0~ \forall i \geq 1$). Let $f: Y \to X$ be a birational morphism, and $D$ be a Cartier divisor on $Y$. Suppose that $D$ is numerically trivial over $X$ (one can also assume that $D$ is $\mathbb Q$-linearly trivial over $X$ if it helps).

Then it is claimed that there exists a Cartier divisor $D_X$ such that $D = f^*D_X$. Why this is true?

Answer 1

Below is not a complete answer, but I write what I can find so far:

My friend told me that if $X$ is a surface, then the claim is a well-known result of Artin. I found a statement in Miles Reid's book "Chapter on Algebraic surface" (Lemma in page 94), which says that if $X$ is a surface with $P\in X$ a rational singularity, and $f: Y \to X$ is a resolution, then for any line bundle $L$ nef over $f^{-1}P$, $|L|$ is free near $f^{-1}P$. Applying this to our case, first choose $A \subseteq X$ to be an ample Cartier divisor, then $|D+f^*A|$ free over $X$. Thus $|D+f^*A|$ defines a morphism which is exactly $f$. Hence $D+f^*A = f^*H$ with $H$ to be a Cartier divisor. Hence $D=f^*(H-A)$, and $H-A$ is still Cartier.

For higher dimensional varieties, the closest result I can find is in Noboru Nakayama's book "Zariski-decomposition and Abundance" (2.12 Lemma in page 41), it claims that $D$ is a pull-back of a $\mathbb Q$-Cartier divisor from $X$.