New answers tagged


No, because almost all numbers have at least $4$ distinct prime factors, making the divisibility graph contain a hypercube and thus be nonplanar.


Assume $\mathcal{L}$ is associated with an effective Cartier divisor $D$. Let $D'$ be another Cartier divisor such that $D + D'$ is divisible by 2 in $\mathrm{Pic}(X)$. Let $$ g \colon X' \to X $$ be the double covering branched at $D + D'$. Then $g^{-1}(D) = 2R$ for a Cartier divisor $R$ on $X'$, hence $g^*\mathcal{L} \cong \mathcal{O}_{X'}(2R)$ has a ...

Top 50 recent answers are included