For every degree \(d > 0\), and every \(p = 0\) or a prime number, give the equation of a nonsingular curve of degree \(d\) in \(\PP^2\) over a field \(k\) of characteristic \(p\).
Solution
Let \([x:y:z]\) be homogeneous coordinates on \(\PP_k^2\). We give equations whose Jacobian has no common projective zero even after extending the ground field, so the resulting curves are geometrically nonsingular.
1. If \(p=0\) or \(p\nmid d\), the Fermat equation\[ \boxed{x^d+y^d+z^d=0} \]defines a nonsingular plane curve of degree \(d\).
Put \[ F=x^d+y^d+z^d. \] Its partial derivatives are \[ F_x=d x^{d-1},\qquad F_y=d y^{d-1},\qquad F_z=d z^{d-1}. \] Under the stated hypothesis \(d\) is nonzero in \(k\). Thus simultaneous vanishing of the three partials forces \[ x=y=z=0, \] which is not a projective point. Hence the projective hypersurface has no singular point by the Jacobian criterion [@Har10a, Chapter I, §5]. The same argument works over every extension field of \(k\).
2. If \(p>0\) and \(p\mid d\), the equation\[ \boxed{x^{d-1}y+y^{d-1}z+z^d=0} \]defines a nonsingular plane curve of degree \(d\).
Now \(d=0\) in \(k\), while \(d-1=-1\) is nonzero. For \[ G=x^{d-1}y+y^{d-1}z+z^d \] the partial derivatives are \[ \begin{aligned} G_x&=(d-1)x^{d-2}y,\\ G_y&=x^{d-1}+(d-1)y^{d-2}z,\\ G_z&=y^{d-1}. \end{aligned} \] Suppose all three vanish at a projective point of \(G=0\). The equation \(G_z=0\) gives \(y=0\). Then \(G_y=x^{d-1}=0\), so \(x=0\). At such a point the defining equation becomes \[ G(0,0,z)=z^d, \] forcing \(z=0\). Again this is not a projective point. Thus no singular point exists. This calculation also remains valid after any field extension. For the smallest possible case \(d=p=2\), the same formulas read \[ G=xy+yz+z^2,\qquad (G_x,G_y,G_z)=(y,x+z,y), \] and give the same conclusion.
3. In either case the nonsingular hypersurface is a curve, i.e. a geometrically irreducible one-dimensional projective variety.
5.At a point of intersection, every first partial derivative\[ \partial(AB)=A\,\partial B+B\,\partial A \]vanish.That would contradict steps 1 or
2.Thus the hypersurface is geometrically irreducible and nonsingular, hence is a plane curve of degree \(d\) in the sense of the statement.
4. Q.E.D.
Step 1 covers all characteristics not dividing \(d\), and step 2 covers the complementary positive-characteristic case. Step 3 verifies that the resulting smooth hypersurfaces are curves.