On a multiplication map of divisorial sheaf

Question

It maybe a very foolish question, but I can't prove or disprove this.

Suppose $X$ is a normal variety, and $L$ a $\mathbb{Q}$-line bundle on $X$. I.e., there is a $\mathbb{Q}$-Cartier Weil divisor $D$ on $X$ such that $L=\mathcal{O}_X(D)$. Let $L^{[i]}$ be the $i$th reflexive tensoring of $L$.

Then, can we prove that for any $i$, the multiplication map $$ \underbrace{L\otimes \cdots \otimes L}_{i\text{ times }}\to L^{[i]}$$ is a surjection? In my eyes, Kollár used the fact to prove $L_i=L^{[i]}(D_i)$ for some Weil divisor $D_i$ on $X$ in 2.44 and Corollary 2.46 in [Kol13], if we use the Kollár's notations. So I tried to prove this, but it seems to be very hard, that I can't prove the torsion-free part of the left hand side is reflexive. How to prove it? Or, is it false in general and there is another argument to prove $L_i=L^{[i]}(D_i)$?

[Kol13]: J. Kollár, "Singularities of the minimal model program", Cambridge Tracts in Mathematics, vol. 200, Cambridge University Press, Cambridge, (2013)

Answer 1

The surjectivity is not true in general. For instance, let $X$ be the affine quadratic cone $$ \{xz - y^2 = 0\} \subset \mathbb{A}^3 $$ and let $L$ be the ideal of a ruling $R \subset X$. Then the multiplication map factors as $$ L \otimes L \cong I_{R_1} \otimes I_{R_2} \twoheadrightarrow I_P \subset \mathcal{O}_X = L^{[2]}, $$ where $P$ is the vertex.