Let \(X\) be the plane quartic curve \(x^3 y+y^3 z+ z^3 x = 0\) over a field of characteristic 3. Show that \(X\) is nonsingular, every point of \(X\) is an inflection point, the dual curve \(X^*\) is isomorphic to \(X\), but the natural map \(X \to X^*\) is purely inseparable.
Solution
Put \[ F(x,y,z)=x^3y+y^3z+z^3x. \]
1. The plane quartic\[ X=V(F)\subset\PP^2 \]is nonsingular.
In characteristic \(3\), \[ F_x=z^3, \qquad F_y=x^3, \qquad F_z=y^3. \] Thus the three partial derivatives vanish simultaneously only when \[ x=y=z=0, \] which is not a point of \(\PP^2\). The projective Jacobian criterion therefore shows that \(X\) is nonsingular.
2. If\[ P=[a:b:c]\in X, \]then the tangent line at \(P\) is\[ T_P(X)=V(c^3x+a^3y+b^3z). \]
For a nonsingular projective plane curve, the tangent line at \(P\) is defined by \[ F_x(P)x+F_y(P)y+F_z(P)z=0. \] Step 1 gives \[ F_x(P)=c^3, \qquad F_y(P)=a^3, \qquad F_z(P)=b^3, \] which yields the displayed equation.
3. Every point of \(X\) is an inflection point.
Fix \(P=[a:b:c]\in X\) and choose a standard affine chart containing \(P\). After fixing the nonzero homogeneous coordinate defining that chart, write a nearby lift as \[ (a+u,b+v,c+w). \] Here the increment of the fixed chart coordinate is \(0\), so only two of \(u,v,w\) are independent local coordinates. Because \((r+s)^3=r^3+s^3\) in characteristic \(3\), direct expansion gives \[ \begin{aligned} F(a+u,b+v,c+w) ={}&F(a,b,c)\\ &+(c^3u+a^3v+b^3w)\\ &+(b u^3+c v^3+a w^3)\\ &+(u^3v+v^3w+w^3u). \end{aligned} \] There are no terms of total degree \(2\) in the local increments.
The constant term vanishes because \(P\in X\), and by step 2 the linear term is an equation for the tangent line. After restricting to \(T_P(X)\), the local equation therefore has order at least \(3\). Hence \[ I_P\bigl(X,T_P(X)\bigr)\ge3, \] so \(P\) is an inflection point. Since \(P\) was arbitrary, every point of \(X\) is an inflection point.
4. In dual homogeneous coordinates \([U:V:W]\), the Gauss map is\[ \gamma:X\longrightarrow(\PP^2)^*, \qquad [a:b:c]\longmapsto[c^3:a^3:b^3], \]and its image is the quartic\[ Y=V(U^3V+V^3W+W^3U). \]
2.For a point in its image,\[ (U,V,W)=(c^3,a^3,b^3), \]and therefore\[ \begin{aligned} U^3V+V^3W+W^3U &=c^9a^3+a^9b^3+b^9c^3\\ &=(c^3a+a^3b+b^3c)^3\\ &=F(a,b,c)^3\\ &=0. \end{aligned} \]Thus \(\gamma(X)\subseteq Y\).
The same Jacobian calculation as in step 1 shows that \(Y\) is nonsingular. Moreover it is geometrically integral: if its defining quartic factored over an algebraic closure, two positive-degree components would meet in \(\PP^2\) by Bézout, producing a singular point. The map \(\gamma\) is nonconstant: if all tangent lines were equal, the nonsingular plane quartic would have one fixed tangent line at every point, which is impossible since that line meets \(X\) in only finitely many points. Hence the image of the projective curve \(X\) is a closed irreducible curve contained in \(Y\), and must equal \(Y\).
By definition the closure of the Gauss image is the dual curve. Therefore \[ X^*=Y. \]
5. The dual curve \(X^*\) is isomorphic to \(X\).
The equation found in step 4, \[ U^3V+V^3W+W^3U=0, \] has exactly the same form as \[ x^3y+y^3z+z^3x=0. \] The projective linear identification \[ [x:y:z]\longmapsto[U:V:W]=[x:y:z] \] therefore restricts to an isomorphism \[ X\xrightarrow{\sim}X^*. \]
6. Under the identification in step 5, the Gauss map is\[ \gamma([x:y:z])=[z^3:x^3:y^3]. \]It is a purely inseparable morphism of degree \(3\).
Let \[ q:X\longrightarrow X, \qquad [x:y:z]\longmapsto[x^3:y^3:z^3], \] and let \[ \rho([x:y:z])=[z:x:y]. \] The cyclic permutation \(\rho\) preserves the equation \(F=0\), so it is an automorphism of \(X\). Step 4 gives \[ \gamma=\rho\circ q. \]
Let \(K=k(X)\). On function fields, \(q^*K\) contains the cube of every element of \(K\). Indeed, if \[ f=\sum_m a_m m \] is a homogeneous polynomial written as a sum of monomials, define \[ f^{[3]}=\sum_m a_m^3m. \] Then \[ q^*f^{[3]}=f^3. \] Consequently, if \(r=f/g\in K\), then \[ r^3=\frac{f^3}{g^3} \] lies in \(q^*K\). Thus every element of \(K\) has third power in \(q^*K\), so \[ K/q^*K \] is purely inseparable. Composing with the automorphism \(\rho\) does not change this property; hence the natural map \[ X\longrightarrow X^* \] is purely inseparable.
It is nontrivial and has degree \(3\). Indeed, \(\gamma\) is a nonconstant map of projective curves, hence finite, and its coordinate functions have degree \(3\). Therefore \[ \gamma^*\OO_{X^*}(1)\cong\OO_X(3). \] Since both \(X\) and \(X^*\) are plane quartics, \[ \deg\OO_X(3)=12, \qquad \deg\OO_{X^*}(1)=4. \] For a finite morphism of curves, \[ \deg\gamma^*L=(\deg\gamma)(\deg L), \] so \[ 12=4\deg\gamma \] and consequently \[ \deg\gamma=3. \]
7. Q.E.D.
Step 1 proves nonsingularity, step 3 proves that every point is an inflection point, steps <1>4–<1>5 identify the dual curve with \(X\), and step 6 proves that the natural map to the dual is purely inseparable.