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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.

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2 Answers 2

up vote 4 down vote accepted

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.

Moreover

  • $R$ 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.

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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.

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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? –  Makhalan Duff Aug 31 '10 at 21:08
    
The answer is still no, $R_1$ is smooth for a generic projection. See Polizzi's answer. –  jvp Sep 1 '10 at 11:44

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