# Smoothness of ramification divisor

Let $f:X \to Y$ be a finite morphism of smooth projective surfaces over a field $k$ of characteristic $0$. Is the branch divisor $B \subset Y$ a smooth curve?

The answer is negative if $\mathrm{cha}(k) = 2$. Consider the del Pezzo surface of degree $2$ given by $$X: \quad w^2 + xyw = f(x,y,z) \subset \mathbb{P}(1,1,1,2),$$ where $f$ is homogeneous of degree $4$. Then the projection $(x,y,z,w) \mapsto (x,y,z)$ is a separable double cover branched over the reducible plane curve $xy = 0 \subset \mathbb{P}^2$ when $\mathrm{cha}(k) = 2$. In particular it is singular curve, even though $X$ is smooth over $k$ when $f$ is general.

• You might find useful results in Edixhoven-de Jong- Schepers article Covers of surfaces with fixed branch locus Jul 12 '18 at 20:56
• I do not understand this question. For a smooth surface $X$ in $\mathbb{P}^n$, for a general linear projection from $X$ to $Y=\mathbb{P}^2$, the branch divisor is almost never smooth. Classically, the double points, cusps, etc., of the branch divisor were the invariants that were studied in trying to understand the geometry of polarized surfaces. For instance, for a hypersurface in $\mathbb{P}^3$, the double points correspond to bitangent lines that contains a specified general point of $\mathbb{P}^3$. These numbers were stand-ins for Chern numbers prior to Chern. Jul 12 '18 at 21:41
• I see, I had no idea of this theory. I'm just familiar with the case of double covers, where I believe the branch curve is always smooth. Jul 12 '18 at 21:55
• Here is a classical example when the branching divisor is not smooth. Take $\mathbb C^2\to \mathbb C^2$ with $(x,y)\to (x^2,y^2)$. This extends to a map $\mathbb CP^2\to \mathbb CP^2$. Jul 12 '18 at 23:06
• Even that is too optimistic. It might help to take a look at this paper in combination with Jason's comment above arxiv.org/abs/0811.0467 Jul 14 '18 at 9:18

Exercise: Consider the action of $$S_3$$ on the plane $$x_1+x_2+x_3=0$$ in $$\mathbb C^3$$ by permutations. Then the quotient $$\mathbb C^2/S_3$$ is again $$\mathbb C^2$$ and the branch divisor is the cuspidal cubic $$x^2=y^3$$ in $$\mathbb C^2$$.
If you want to make the surfaces projective, take an elliptic curve $$E$$ with a line bundle $$L$$ of degree $$3$$ then the projectivisation of the space of non-zero sections of $$E$$ is $$\mathbb CP^2$$. On the other hand this $$\mathbb CP^2$$ is a quotient of a complex $$2$$-torus (an Abelian surface) by $$S_3$$. The branch divisor in $$\mathbb CP^2$$ is a curve of degree $$6$$ with $$9$$ cusps - it is dual to a smooth cubic curve in $$\mathbb CP^2$$.