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\beginsection{ MATHEMATICAL BACKGROUND}\par

\def\P{\bf P} The gonality of a curve is defined to be the smallest degree of a morphism from the curve to the projective line $\P^1$. It is known that a curve $C$ of genus $g$ admits a map to $\P^1$ of degree at most $[(g+3)/2]$. Further, if $C$ is $d$-gonal, then in its canonical embedding $C$ lies on a rational normal scroll of dimension $d-1$, and the free resolution of the homogeneous coordinate ring of the scroll is a subcomplex of the free resolution of the homogeneous coordinate ring of $C$. Thus for example the $2$-linear part of that resolution has length at least $g-d$, and ``Green's Conjecture'' states that if one computes Clifford index instead of gonality, a slight refinement, then equality holds. For example, Green's conjecture predicts that the resolution of the homogeneous coordinate ring of a general curve of genus 7 and gonality 4 is:

       --     total: 1 10 25 25 10 1
       --         0: 1 .  .  .  .  .
       --         1: . 10 16 9  .  .
       --         2: . .  9  16 10 .
       --         3: . .  .  .  .  1

We will make these computations for three types of plane sextic curves (of genus seven):

{\tt C1} will have 3 ordinary nodes;

{\tt C2} will have one ordinary triple point;

{\tt C3} will have one tacnode and one ordinary node. \beginsection{ Computation}\par We take {\tt C1} to be a curve of degree $6$ having $3$ ordinary double points:

We define the ideals of the points. We could write

but the following shortcut is faster!

For a curve to be double at the $3$ points, its equation must lie in the ideal

The matrix with the generators of {\tt icurves1} as its entries is obtained by

We find the equation {\tt F1} of a general curve of degree $6$ with these double points by composing a random matrix of forms having the correct degree with the matrix of generators of {\tt icurves1}.

We now look for the equation {\tt F2} of {\tt C2}, a curve with an ordinary triple point at {\tt point1}. It must lie in the cube of the ideal {\tt ipoint1}.

Finally, the equation of a curve with a tacnode at $a=b=0$ having tangent line $a-b=0$ there must lie in the ideal

and adding a node at {\tt point3} we get

so

It is evident from the discussion above that {\tt C1} and {\tt C3} have gonality $\leq 5$ (indeed, every curve of genus $7$ has gonality $\leq 5$) and that {\tt C2} has gonality $\leq 4$. We can establish lower bounds for the gonalities by looking at the canonical embeddings. The canonical series of {\tt C1} is cut out by

Some explanation regarding the {\tt basis} command is needed here. {\tt can1} is a matrix whose target is the ideal of the intersection of these three points:

and whose source is a free module over the coefficient ring:

For our purposes, there are two problems with this. The first is that we want a map where both the source and target have the base ring $R$. This can be accomplished by tensoring with $R$:

The second problem is that the image of a basis element is not obviously in the ideal: it is represented in terms of the generators of $I$. This can be alleviated by applying {\tt super}: this takes a homomorphism $f : M \rightarrow N$, where $N$ is a submodule of a quotient module $F/I$, and returns the homomorphism $f : M \rightarrow F/I$.

similarly, for {\tt C2} and {\tt C3}:

These should all give embeddings of the curves in $\P^6$, so we expect them to be linear series of vector space dimension $7$. Here's how to check:

To compute the homogeneous coordinate rings of the canonical curves, we must find the algebraic relations among the generators of {\tt cani} modulo {\tt Fi}.

The coordinate ring of $\P^6$

Find the canonical ideal {\tt IC1} of {\tt C1}, that is, the kernel of the map $S \rightarrow R/(F1)$ defined by the canonical series.

and similarly for {\tt C2, C3}

We now find the $2$-linear part of the free resolution of {\tt IC1} and its betti numbers.

From this computation, and the easy implication of Green's conjecture explained above, we see that the gonality of {\tt C1} is exactly 4, the gonality of the linear series obtained by projection from any one of the three double points. We now do the same for {\tt IC2} and {\tt IC3}:

and we find that in the tacnodal case the gonality is still 4, while in the triple point case the gonality is 3. Note that we could have made the computation faster, as in the following example. In these cases the resolution is so fast that the speedup is not noticeable, but in larger cases it would be worthwhile. First clear the info computed in {\tt IC1}

Now redo the resolution, this time bounding the degree to which the computation is carried.

Instead of computing the canonical model of {\tt C1} directly, we could have treated the structure sheaf of {\tt C1} as a sheaf on the projective plane, and compute its pushForward under the map to $\P^6$ given by {\tt can1} (the image of the plane under this map is a Del Pezzo surface on which the canonical curve lies.) This is done as follows:

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