I am looking for a smooth proper curve $C$ such that there does not exist any closed embedding $C \to S$ where $S$ is a (normal projective) toric surface.


Since $C$ is smooth I believe it suffices to consider smooth projective toric surfaces $S$ since we may always perform a toric resolution of singularities and the strict transform of $C$ will be isomorphic to $C$ since $C$ is smooth.


Using the result on p.25 of Harris Mumford, On the kodaira dimension of the moduli space of curves, I can conclude that a very general curve cannot have any such embedding.


However, I am not able to write down an explicit example. Does anyone know such an example or what sort of obstruction might work to check this in particular examples.



Curve with no embedding in a toric surface
A generic curve of genus $5$ is not a hypersurface in a toric surface. This argument is going to use conceptual ideas from Haase and Schicho's paper ["Lattice polygons and the number $2i+7$"](https://arxiv.org/abs/math/0406224v3), plus a bunch of case analysis.


Let's start with generalities about a curve $C$ in a toric surface $S$, other than one of the boundary divisors. The divisor $C$ gives a nef line bundle $\mathcal{O}(C)$ on $S$, which gives a lattice polygon $P$ in $\mathbb{Z}^2$. (See, for example, [Lecture 4 here](https://dacox.people.amherst.edu/lectures/coxcimpa.pdf).) The number of interior lattice points of $P$ will be the genus of $C$.


Let $Q$ be the convex hull of the interior lattice points. So, in our genus $5$ case, $Q$ will be a lattice polytope with $5$ lattice points. Such a polytope either (Case 0) has no interior lattice points (Case 1) has one interior lattice point (Case 2) is a triangle with two interior lattice points or (Case 3) is a line segment of length $4$.


**Case 0** Polytopes with no interior lattice points are trapezoids with vertices of the form $(0,0)$, $(a,0)$, $(0,1)$, $(b,1)$. If we want $5$ lattice points, our options are
$$\mbox{polygon 0a} = \begin{matrix} 
\bullet&&& \\ \bullet&\bullet&\bullet&\bullet \\
\end{matrix} \qquad 
\mbox{polygon 0b} = \begin{matrix} 
\bullet&\bullet& \\ \bullet&\bullet&\bullet \\
\end{matrix}$$


**Case 1** There are $12$ polytopes with $1$ interior lattice point, which you can see in Figure 6 of Haase and Schicho. There are $3$ of these with $5$ lattice points:
$$
\mbox{polygon 1a} =\begin{matrix} &\bullet& \\ \bullet&\bullet&\bullet \\ &\bullet& \\ \end{matrix} \qquad
\mbox{polygon 1b} =\begin{matrix} \bullet&& \\ \bullet&\bullet&\bullet \\ &\bullet& \\ \end{matrix} \qquad
\mbox{polygon 1c} =\begin{matrix} \bullet&& \\ \bullet&\bullet&\bullet \\ \bullet&& \\ \end{matrix}.$$


**Case 2** Wei and Ding, "[Lattice polygons with two interior lattice points](https://doi.org/10.1134/S0001434612050343)", list all lattice polytopes with $2$ interior lattice points. Only one of them has $5$ lattice points:
$$
\mbox{polygon 2} = \begin{matrix} &&\bullet& \\ \bullet&\bullet&\bullet& \\ &&&\bullet \\ \end{matrix}
$$


**Case 3** Finally, here is the line segment:
$$\mbox{polygon 3}=\begin{matrix} \bullet&\bullet&\bullet&\bullet&\bullet \end{matrix}$$


So $Q$ must be one of the $7$ polygons above.


Haase and Schicho show that not every polytope can occur as $Q$. Hasse and Schicho make the following definition: Let $v$ be a vertex of a lattice polytope $Q$, and let $x$ and $y$ be the minimal lattice vectors along the edges incident to $v$. Hasse and Schicho say that $v$ is a "good corner" if $(\vec{x}, \vec{y})$ is $SL\_2(\mathbb{Z})$ equivalent to $((1,0), (-1,k))$ for some $k$. Haase and Schicho, Lemma 9, show that $Q$ is always either the empty set, a line segment, or a polytope with good corners. Polygons 0a and 2 do not have good corners, so we can eliminate them now.


We now look at Haase and Schicho, Lemma 8. This lemma says: Let $a\_1$, $a\_2$ and $b$ be integers with $GCD(a\_1, a\_2)=1$. Suppose that $a\_1 x\_1 + a\_2 x\_2 \leq b$ on $Q$, and that equality occurs at at least two lattice points. Then $a\_1 x\_1 + a\_2 x\_2 \leq b+1$ on $P$.


We will apply this result to eliminate polygons 0b and 3. Take $(a\_1, a\_2) = (0,1)$, so $a\_1 x\_1 + a\_2 x\_2$ is the second coordinate. Then Hasse and Schicho's lemma show that the polytope P is contained in a horizontal strip of width $3$ when $Q$ is polygon 0b, and width $2$ in polygon 3. This means that the corresponding curve is trigonal or hyperelliptic, respectively. A generic genus $5$ curve is not trigonal or hyperelliptic.


This leaves cases 1a, 1b and 1c. Haase and Schicho define a polygon they call $Q^{(-1)}$, whose definition I will let you read in their paper, and show that $P \subseteq Q^{(-1)}$. In the figure below, I have redrawn polygons 1a, 1b and 1c with solid dots $\bullet$, and drawn the additional vertices of $Q^{(-1)}$ as hollow circles $\circ$.


$$
\begin{matrix} &&\circ&& \\ &\circ&\bullet&\circ& \\ \circ&\bullet&\bullet&\bullet&\circ \\ &\circ&\bullet&\circ& \\ &&\circ&& \\ \end{matrix} \qquad
\begin{matrix} \circ&&&& \\ \circ&\bullet&\circ&&& \\ \circ&\bullet&\bullet&\bullet&\circ \\ &\circ&\bullet&\circ& \\ &&\circ&& \end{matrix} \qquad
\begin{matrix} \circ&&&& \\ \circ&\bullet&\circ&&& \\ \circ&\bullet&\bullet&\bullet&\circ \\ \circ&\bullet&\circ&&& \\ \circ&&&& \\ \end{matrix}.$$


In each case, there are $13$ lattice points in $Q^{(-1)}$, and thus $\leq 13$ lattice points in $P$. The curve $C$ is the zero locus of a polynomial with this Newton polytope, so it depends on $13$ parameters. But rescaling either of the variables, or rescaling the whole polynomial, does not change the isomorphism class of the curve, so we really only have $10$ parameters. The moduli space of curves of genus $5$ has dimension $12$, so the generic curve of genus $5$ is not a zero locus of a polynomial of any of the above forms. This concludes our case analysis.




---


Writing down an explicit genus $5$ curve which works still seems interesting. We can avoid cases 0b and 3 by just not making our curve hyperelliptic or trigonal. It isn't clear to me what explicit criterion avoids cases 1a, 1b or 1c.


