Curve with no embedding in a toric surface 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.
 A: 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 $2i+7$", 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.) 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", 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.
