Let $B$ be a variety with torsion canonical sheaf, i.e., $\omega^{\otimes n}_B \cong \mathcal O_B$ for some $n>0$. Then, why does there exist a finite (etale?) morphism $X\to B$ such that $K_X$ is linearly equivalent to zero?
I'm thinking $X\to B$ should be some kind of cyclic covering, but how does one construct such a cyclic covering of $B$ in the first place? And why would it have trivial canonical sheaf?
This is a special case of a more general fact. If we are given a line bundle $L$ on an algebraic variety $X$ and an isomorphism $L^{\otimes n} \cong O_X$, then there is uniquely attached to this data an $n$-fold étale cover $Y \to X$ such that the pullback of $L$ to $Y$ is trivial. Note that the pullback of the canonical sheaf under an étale map is the canonical sheaf.
One way of realizing this is to use that $X$ is isomorphic to its image in the total space of $L^{\otimes n}$ under the unit section obtained from the isomorphism above. Now every point on $X$ has $n$ distinct "roots" in the total space of $L$. I think all this is explained in Griffiths-Harris, for instance.