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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 $smooth$.

Now we have

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

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.

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.

For a $general$ flat triple cover $f \colon X \to Y$ the situation is as follows.

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 $smooth$.

Now we have

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

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.

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 $smooth$.

Now we have

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

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.

For a $general$ triple cover $f \colon X \to Y$ the situation is as follows.

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 $smooth$.

Now we have

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

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 $B$ (and not from the ramification divisor $R$, which 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.

For a $general$ flat triple cover $f \colon X \to Y$ the situation is as follows.

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 $smooth$.

Now we have

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

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.