Another way of distinguishing curves of genus \(g\) is to ask, what is the least degree of a birational plane model with only nodes as singularities (3.11)? Let \(X\) be nonhyperelliptic of genus 4. Then:
if \(X\) has two \(g_3^1\)’s, it can be represented as a plane quintic with two nodes, and conversely;
if \(X\) has one \(g_3^1\), then it can be represented as a plane quintic with a tacnode (I, Ex. 5.14d), but the least degree of a plane representation with only nodes is 6.
Solution
Let \[ C\subseteq\PP^3 \] be the canonical model of the nonhyperelliptic genus-\(4\) curve \(X\). By Exercise IV.5.3, \[ C=Q\cap F_3, \] where \(Q\) is the unique quadric containing \(C\). The curve has two \(g^1_3\)’s when \(Q\) is smooth and one when \(Q\) is a rank-\(3\) quadric cone.
1. For a general point \(P\in C\), projection from \(P\) is the morphismdefined by\[ \boxed{|K-P|:X\longrightarrow\PP^2,} \]and it is birational onto a plane quintic.
Hyperplanes in \(\PP^3\) through \(P\) restrict on the canonical curve to canonical divisors containing \(P\). Removing that fixed point gives the complete linear system \(|K-P|\). Since \[ \deg(K-P)=2g-3=5 \] and \[ \ell(K-P)=g-1=3, \] it gives a morphism to \(\PP^2\); geometrically this is the projection of the canonical curve from \(P\), extended across the centre.
Its image spans \(\PP^2\), so it is not a line. If its generic degree were \(e>1\), then \[ 5=e\deg C' \] for the plane image \(C'\). Since \(5\) is prime, this would force \(\deg C'=1\), a contradiction. Thus the map is birational and \[ \deg C'=5. \]
We can make the required generality explicit. Each ruling gives a degree-\(3\) map \[ C\longrightarrow\PP^1, \] whose ramification locus is finite in characteristic \(0\). Avoiding the ramification loci makes the ruling fibre through \(P\) reduced, hence gives three distinct points \[ P+A+B. \] Also avoid the finite canonical Wronskian locus of Exercise IV.4.6. Then the canonical vanishing sequence at \(P\) is \[ 0,1,2,3. \] After subtracting the base point \(P\), the system \(|K-P|\) has vanishing sequence \[ 0,1,2, \] so its map is immersive at \(P\). The complement of these finite exceptional sets is nonempty, and we choose \(P\) there.
2. A line through \(P\) identifies two distinct points of\(C\setminus\{P\}\) under this projection if and only if it is a ruling lineof \(Q\) through \(P\).
Suppose distinct \[ A,B\in C\setminus\{P\} \] have the same image. Then \[ P,A,B \] are collinear. Their line \(L\) meets the quadric \(Q\) in a length at least \(3\) subscheme. Since a line not contained in a quadric meets it in degree \(2\), Bezout forces \[ L\subseteq Q. \] Thus \(L\) is a ruling line through \(P\).
Conversely, if \(L\subseteq Q\) is a ruling line through \(P\), then \[ L\cdot C = L\cdot F_3 =3. \] For our general \(P\), \[ L\cap C=P+A+B \] with \(A,B\) distinct from each other and from \(P\). Projection from \(P\) contracts \(L\), so it identifies \(A\) and \(B\).
3. Assume \(Q\) is smooth. Then the plane quintic of step 1 has exactlytwo singularities, both nodes.
A smooth quadric \[ Q\cong\PP^1\times\PP^1 \] has exactly two ruling lines through \(P\). By step 2 they give two identified residual pairs \[ A_i,B_i, \qquad i=1,2. \]
We first check that each resulting double point is transverse. Fix one ruling line \(L\) and its residual points \(A,B\). The plane spanned by \[ P\quad\text{and}\quad T_A C \] is the tangent plane \(T_AQ\): both contain \(L\), and \(T_AC\subseteq T_AQ\). Similarly the plane spanned by \(P\) and \(T_BC\) is \(T_BQ\). On a smooth quadric the tangent plane varies nontrivially along a ruling, so \[ T_AQ\ne T_BQ \] for \(A\ne B\). Under projection from \(P\), this says that the two branches at the identified image point have distinct tangent lines. Hence that singularity is a node.
Thus the plane image already has two nodes. A plane quintic has arithmetic genus \[ p_a=\frac{(5-1)(5-2)}2=6, \] whereas its normalization \(X\) has genus \(4\). Therefore the total delta invariant of all singularities is \[ 6-4=2. \] The two nodes contribute \[ 1+1=2. \] Hence there are no further singularities. This proves the forward assertion in part (a).
4. Conversely, the normalization of a plane quintic with two nodes hastwo distinct \(g^1_3\)’s.
Let \[ \nu:X\longrightarrow C'\subseteq\PP^2 \] be the normalization of an integral plane quintic whose only singularities are two nodes \[ N_1,N_2. \] The genus formula gives \[ g(X) = 6-1-1 =4. \]
Write \[ \nu^{-1}(N_i)=\{P_i,Q_i\}, \qquad A_i=P_i+Q_i. \] Let \[ H=\nu^*\OO_{C'}(1). \] The pencil of lines through \(N_i\) pulls back, after removing the two fixed branches above the node, to a base-point-free pencil \[ |H-A_i| \] of degree \[ 5-2=3. \] Thus each node gives a \(g^1_3\).
These two pencils are distinct. If \[ H-A_1\sim H-A_2, \] then \[ A_1\sim A_2. \] The effective divisors \(A_1\) and \(A_2\) are distinct, so this would give \[ \ell(A_1)\ge2, \] that is, a \(g^1_2\) on \(X\). Then \(X\) would be hyperelliptic, contrary to the standing hypothesis. Hence the two degree-\(3\) pencils are distinct. This proves the converse in part (a).
5. Assume \(Q\) is a rank-\(3\) quadric cone. Then the plane quintic ofstep 1 has exactly one singular point, and it is a tacnode.
Through a smooth point \(P\) of a quadric cone there is exactly one ruling line \[ L\subseteq Q. \] For general \(P\) write \[ L\cap C=P+A+B \] with \(A,B\) distinct. Step 2 says that projection identifies precisely this residual pair.
The crucial difference from the smooth-quadric case is that the tangent plane to a quadric cone is constant along a ruling. Thus \[ T_AQ=T_BQ. \] As in step 3, the tangent line to the projected branch coming from \(A\) is obtained by projecting the plane spanned by \(P\) and \(T_AC\), namely \(T_AQ\); similarly the tangent line of the branch coming from \(B\) is obtained from \(T_BQ\). The equality above therefore says that the two smooth branches at the identified image point have the same tangent line.
Two distinct smooth branches with a common tangent have delta invariant at least \(2\). On the other hand the plane image is again a quintic whose normalization has genus \(4\), so the total delta invariant is \[ 6-4=2. \] Consequently this one singularity has delta invariant exactly \(2\), there are no other singularities, and the two branches have intersection multiplicity exactly \(2\). Equivalently, the singularity is an ordinary tacnode, analytically of type \[ y^2=x^4. \] This proves the first assertion of part (b).
6. If \(X\) has only one \(g^1_3\), then it has no birational plane model ofdegree less than \(6\) whose only singularities are nodes.
Let an integral nodal plane model have degree \(d\) and \(r\) nodes. Its normalization has genus \[ 4 = \frac{(d-1)(d-2)}2-r. \] For \[ d\le4 \] the arithmetic genus is at most \(3\), impossible.
If \[ d=5, \] the equation becomes \[ 4=6-r, \] so \[ r=2. \] By step 4, the normalization of such a two-nodal quintic has two distinct \(g^1_3\)’s. This contradicts the hypothesis that \(X\) has only one. Therefore every nodal plane model has degree at least \(6\).
7. Every such \(X\) has a degree-\(6\) plane model whose only singularitiesare nodes.
Use the canonical embedding \[ C\subseteq\PP^3. \] It has degree \[ \deg C=\deg K=6. \] The generic projection theorem for curves says that projection from a general point \[ O\in\PP^3\setminus C \] is birational onto a plane curve whose only singularities are nodes. Because the centre does not lie on \(C\), projection preserves the hyperplane bundle, so the plane image still has degree \(6\).
Combining this with step 6 shows that \[ \boxed{6} \] is the least possible degree of a nodal plane model.
8. Q.E.D.
Steps <1>1–<1>4 prove part (a). Step 5 constructs the tacnodal quintic in the one-\(g^1_3\) case, and steps <1>6–<1>7 prove that the least degree of a plane model having only nodes is \(6\).