The question can be generalized, but we might as well restrict to this case.

Let $X \rightarrow Y$ be a morphism between nonsingular surfaces (say over $\mathbb{C}$). Let $R_1$ be an irreducible component of the ramification divisor (in $X$). Let $n$ be how much $R_1$ ramifies generically, and let $S$ be the finite set of points of $R_1$ that ramify to a degree which is not $n$. Is it true that the set $S$ is contained in the set of the points where $R_1$ intersects the other irreducible components of the ramification divisor (and maybe where $R_1$ is singular)?

The intuitive answer is yes, but I'm still somewhat skeptical.


2 Answers 2


Let me expand jvp's answer, giving a picture of the situation in the case of a $general$ flat triple cover $f \colon X \to Y$.

Let $R \subset Y$ be the ramification divisor and $B \subset Y$ the branch divisor, that is $B = f(R)$. Then $R$, $B$ are both reduced and irreducible, and $B$ has only a finite number of ordinary cusps $q_1, \ldots, q_t$ as singularities. These cusps are exactly the points over which $f$ is $totally$ $ramified$. Moreover $R$ is isomorphic to the normalization of $B$, in particular it is $smooth$.

One has the equality of divisors

$f^*(B)=2R + R'$,

where $R'$ is another irreducible curve, isomorphic to $R$, which meets $R$ in a finite number of points $p_1, \ldots, p_t$. Notice that $R'$ is $not$ a component of the ramification locus, since the latter consists of $R$ alone.


  • $R$ and $R'$ are tangent at $p_1, \ldots, p_t$;
  • $p_1, \ldots ,p_t$ are the preimages of the cusps $q_1, \ldots, q_t$.

Summing up, in this case your $S$ is the set whose elements are the points $p_1, \ldots ,p_t$. They correspond to the points where the ramification divisor $R$ meets the curve $R'=f^*(B) \setminus R$. In other words, they come from the singular points of the branch divisor $B$ (whereas the ramification divisor $R$ is smooth).

This is easy to see; a good reference is Miranda's paper "Triple covers in algebraic geometry".

Anyway, the crucial fact here is that a general triple cover is not a Galois cover, so over the branch locus $B$ there are both points where $f$ is ramified (the curve $R$) and points where it is not (the curve $R'$).

If you consider instead any Galois cover, say with group $G$, then every preimage of a branch point is a ramification point (and the stabilizers of points lying on the same fibre are conjugated in $G$). In this case there are formulae relating the ramification number of a point on $X$ with the ramification numbers of the components of the ramification locus passing through it.

See Pardini's paper "Abelian covers of algebraic varieties" for more details.


The answer is no. For instance the generic linear projection of a surface of degree at least three on $\mathbb P^3$ to $\mathbb P^2$ has irreducible ramification and cusps singularities.

  • 1
    $\begingroup$ How about if I change the set S to be the points in R_1 which are in the singular locus of the ramification divisor? $\endgroup$ Aug 31, 2010 at 21:08
  • $\begingroup$ The answer is still no, $R_1$ is smooth for a generic projection. See Polizzi's answer. $\endgroup$ Sep 1, 2010 at 11:44

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