I think it's well known that if $X\subset\mathbb{P}^3$ is a smooth cubic surface and we take the projection $\pi: X\rightarrow \mathbb{P}^2$ from a point off the surface, then it's branched over a sextic curve with 6 cusps.

Why is this true? In particular, I'm not seeing the 6 cusps.

Example of a statement of the fact: the Problem of Existence of Algebraic Functions by Zariski (1929), pg 320

  • $\begingroup$ An important additional point is that the six cusps lie on the same conic (for instance, this is shown by the argument in abx's answer). $\endgroup$ Sep 15, 2016 at 5:52

2 Answers 2


A slightly simpler way: if you project from $(0,0,0,1)$, after a change of coordinates you can write the equation of your surface as $T^3+PT+Q=0$, where $P$ and $Q$ are forms of degree $2$ and $3$ in $X,Y,Z$. The branch curve is given by $4P^3+27Q^2=0$, and it is fairly easy to see that the 6 points given by $P=Q=0$ are cusps.


If you put the point from which you are projecting at infinity in the $z$ axis, you can write your projection in affine coordinates as $\pi: (x,y,z) \mapsto (x,y)$. If the surface is given by a cubic $f=0$, then the ramification locus of $\pi$ is given by $f = \partial f/\partial z = 0$. Since $\partial f/\partial z$ is quadratic, the ramification locus is a curve of degree $6$ and genus (at most) $4$ in $\mathbb{P}^3$. The projection of that to the $xy$ plane is therefore a curve of degree $6$ and genus at most $4$ and therefore has to have $(6-1)(6-2)/2 -4 = 6$ singular points. To get that these singularities are cusps, you have to look at the local description of the projection to see that there is only one branch at each point.


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