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A divisor is basepoint-free when its global sections have no common zero. Its complete linear system therefore defines a Kodaira map everywhere.

On ${\mathbb{F}}_2$, let $S$ be the negative section of a Hirzebruch surface, so $S^2=-2$, and let $F$ be the fiber class of a Hirzebruch surface, with $F^2=0$ and $S\cdot F=1$. The line bundle $\mathcal{O}(S+3F)$ is very ample. Indeed, the toric ampleness criterion for $aS+bF$ isand on a smooth complete toric variety every ample line bundle is very ample. These inequalities hold for $(a,b)=(1,3)$.

Explicitly, after choosing the standard lattice coordinates for ${\mathbb{F}}_2$, its lattice polytope of a toric divisor isThe monomials indexed by the lattice points of this polygon separate torus orbits and tangent directions, so their Kodaira map is a closed embedding. This directly verifies that the resolved proper toric surface is projective.

Solved by gpt-5.6-sol high.

Take the invariant divisor $D_1$, whose class is $F$. Its global sections are represented in the Cox ring by $x_1$ and $x_3$. They have no common zero on $U$, so $F$ is basepoint-free. Its Kodaira map isThis is exactly the ruling ${\mathbb{F}}_k={\mathbb{P}}_{{\mathbb{P}}^1}(\mathcal{O}\oplus \mathcal{O}(k))\to {\mathbb{P}}^1$, and its fibers have divisor class $F$.

Solved by gpt-5.6-sol high.