Past exam of the geometry course of the University of Oxford / 2007 / B / 6

For a plane affine curve given by a reduced equation $f(x,y)=0$, a point is singular when $f=f_x=f_y=0$ there. For a projective curve given by a homogeneous equation $F(x,y,z)=0$, a point is singular when $F=F_x=F_y=F_z=0$ on a nonzero representative. These conditions are invariant under projective rescaling. In characteristic zero, Euler's identity $dF=x{F}_x+y{F}_y+z{F}_z$ makes the equation $F=0$ follow from vanishing of the partials when $d\ne 0$ in the field. Over an arbitrary field one should retain the equation itself, particularly if the characteristic divides the degree.

Bézout's theorem says that two projective plane curves over $\mathbb{C}$ of degrees $d,e$ with no common irreducible component have total intersection multiplicity $de$, so at most $de$ distinct intersection points. Let the irreducible curve have reduced homogeneous equation $F$ of positive degree $d$. In characteristic zero some partial derivative is nonzero. For $d>1$ this derivative has degree $d-1$ and cannot share a component with the irreducible degree-$d$ polynomial $F$. Every singular point belongs to their intersection. Bézout bounds their number by $d(d-1)$, proving there are finitely many singular points. For $d=1$ a partial is a nonzero constant and there are none.

Put $F=y^{2}z-x^3+3x{z}^2-2z^3$. Its partials areAt infinity $z=0$, the curve equation forces $x=0$, giving $[0:1:0]$, where $F_z\ne 0$. In the chart $z=1$, the derivative equations give $y=0$ and $x=\pm 1$. Only $x=1$ lies on the curve and also makes $F_z=0$. Thus the only singular point is $[1:0:1]$. As a check, the affine equation factors as $y^2=(x-1{)}^2(x+2)$, whose quadratic tangent cone there is $y^2-3(x-1{)}^2$, giving two distinct tangent lines.

For $f=y^2-x^3-17$, the partials are $f_x=-3x^2$ and $f_y=2y$. If $p\notin \{2,3\}$ their simultaneous vanishing forces $x=y=0$, and that point lies on the curve exactly when $p=17$. In characteristic two, the equations force $x=0$ and then $y^2=1$, so $y=1$. In characteristic three they force $y=0$ and then $x^3=1$, so $x=1$. Consequentlyand there are no singular points for any other prime.