Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 21 / 3

Over ${\overline{\mathbb{F}}}_3$, take the projective cubicIn characteristic three its partial derivatives are $F_X=Z^2$, $F_Y=2YZ$ and $F_Z=Y^2+2XZ$. If they vanished simultaneously, $Z=0$ and then $Y=0$. The equation of $C$ would force $X=0$, impossible for a projective point. Thus the Jacobian criterion proves smoothness everywhere, including infinity.

It is an irreducible cubic: on the affine chart $Z=1$, the polynomial is $y^2-(x^3-x)$, and $x^3-x$ is not a square in ${\overline{\mathbb{F}}}_3(x)$ because its pole at infinity has odd order three. A quadratic in $y$ would factor over that field only if its right side were a square. Gauss lemma for polynomials then gives affine irreducibility, and homogenization introduces no factor at infinity because $Z$ does not divide $F$. This supplies the smooth plane cubic in characteristic three. In particular, choosing a Fermat cubic here would fail, because its cubic partial derivatives vanish in characteristic three.

Take the projective hypersurfaceAs a polynomial in $X_0$ over $\mathbb{C}[X_1,X_2,X_3]$, it is primitive because $X_1$ and $X_2^2$ are coprime. It is linear and irreducible over the fraction field, so Gauss lemma for polynomials proves irreducibility. Thus it is an irreducible surface.

Its gradient is $(X_1,X_0,-2X_2,0)$. The simultaneous zero set on $S$ is exactlyThe Jacobian criterion proves every other point is smooth. At that vertex the affine equation in the chart $X_3=1$ has no linear term, so the tangent space has dimension three whereas the surface has dimension two; the vertex is genuinely singular. This is a projective quadric cone with one singular vertex.

Take the plane quartic with exactly two nodesPut $d=X^2-Z^2$. Its partial derivatives areOn $Y=0$, the equation gives $X^2=Z^2$, producing exactly $P_{+}=[1:0:1]$ and $P_{-}=[-1:0:1]$, and their gradients vanish.

There are no singular points with $Y\ne 0$. Normalize $Y=1$. The derivative equations give $X=4dZ$ and $(16d^2+1)Z=0$. If $Z=0$, then $X=0$ and $F_Y=4\ne 0$, so $Z\ne 0$ and $d^2=-1/16$. The remaining equation gives $XZ=-2$, hence $Z^2=-1/(2d)$. But $d=X^2-Z^2=-2Z^2=1/d$, forcing $d^2=1$, a contradiction. Thus the singular set is exactly $\{P_{+},P_{-}\}$.

In the chart $Z=1$ near $P_{\epsilon }$, put $X=\epsilon +u$, $Y=v$ with $\epsilon =\pm 1$. The lowest-degree part is $4u^2+\epsilon v^2$, which splits into two distinct complex linear factors. Each singularity is therefore an ordinary double point, with two smooth transverse branches.

It remains to prove irreducibility. A repeated polynomial factor would make every point on that component singular, contradicting the finite singular set, so $F$ is reduced. If it were reducible, partition its irreducible factors into two coprime factors of degrees $r$ and $4-r$, $1\le r\le 3$. Every intersection of the two components would be singular on $C$, hence one of the two ordinary double points. At an ordinary double point shared by them, the two branches are transverse, so its intersection multiplicity is one. Their total intersection multiplicity would therefore be at most two. The Bézout theorem instead makes it $r(4-r)\ge 3$. This contradiction proves the displayed quartic is irreducible and has exactly two singular points.