Q. How do you prove hyperplane section is in linearly
general position, so the free pencil exists?
A. C in PP^n irreducible curve, not in a hyperplane.
Write phi: C[n] - -> (PP^n)^dual for the rational map taking
a general n-tuple of points to the hyperplane that is their
linear span.
If C is singular, take C[n] the projective closure of the
nth symmetric power of NonsSing C. Since C spans PP^n, a
Zariski open set of (n+1)-tuples of NonSing C consists of
linearly independent points, so a general set of n points
spans a hyperplane.
Now over a field k of characteristic zero, the rational map
phi: C[n] - -> (PP^n)^dual is separable. The same conclusion
holds also if char k > deg(phi), and one sees that
deg(phi) <= (d choose n) with d = deg C.
If phi is separable, then it is a finite unramified cover
over a dense Zariski open set of (PP^n)^dual. For a
hyperplane H of PP^n in this open, every inverse image of H
in C[n] is a set of n points spanning H, therefore
H intersect C is in linearly general position. (Thisnk of
this as a kind of Sard's theorem.)
In small characteristic (and large degree), there are
counterexamples. The main result is due to Pierre Samuel and
treated in Hartshorne, 4.3, p. 310-316. If the general
section of C in PP^n (not in a hyperplane) is not a set of d
points in general position then C is _strange_: the tangent
lines T_{C,P} at all P in NonSing C pass through a common
point Q in PP^n. The example is
C : y^p = x^{p-1}*z in PP^2, image of v |-> (1, v, v^p).
Then projection from Q gives an inseparable map C - ->
PP^{n-1}. For a canonical curve C_{2g-2} in PP^{g-1} this is
impossible for reasons of degree.
I think there exist curves C in PP^n for which every
hyperplane section is a configuration of points with
symmetry by a product (FF_p^+d), so not linearly general.
Exercise: figure this out.
This note still needs to be written out correctly.