Q. I'm interested in knowing the full proof of Max
Noether's theorem on canonical curve

==
Max Noether's theorem

C a nonhyperelliptic curve embedded in PP^{g-1} by the
RR space L(C,KC).

Then forms of every degree d >= on PP^{g-1} map
surjectively to L(C,dKC).

The forms of degree d are linear combinations of
products of elements s1^d, s1^{d-1}s2, .. sg^d,
where s1,.. sg bases L(C,KC). The multiplication
takes place in k(C).

The basic case to work with is d=2. Choose a basis s1,..
sg of L(C,KC). The product si*sj is in L(C,2KC) because
div(si*sj) = div s1 + div s2. Required to prove that
the products si*sj include 3g-3 linearly independent
elements. The key point is the Castelnuovo free pencil
trick.

Part 3, 8.3 of the notes proved that KC is very ample.
It can be shown that a general hyperplane H in PP^{g-1}
cuts the canonical image C in 2g-2 points that are
linearly in general position. (I omit this because it
is quite tricky and tedious.) Choose g general points
P1,.. Pg of C. Because the points are general, each
L(D-Pi) in L(D) has codimension exactly one. Choose the
basis { si } of L(C,KC) so that each si vanishes on the
g-1 points Pj (j <> i).

Write A = P3+.. Pg. Then s1, s2 vanish on A, the g-2
points P3,..Pg, and do not have any other common point.
(This is where the general position assumption is
used.) It follows that s1,s2 in L(C, KC-A) form a free
pencil: that is, KC-A + div s1 = D1 (all the points of
the hyperplane s1 = 0 except for P3..Pg) and
KC-A + div s2 = D2, where D1 and D2 are effective
divisors having no common points.

The products s1*L(KC), s2*L(KC) are both vector
subspaces of L(C, 2KC-A). The free pencil trick implies
that they intersect in L(C,A). Now dim L(C,A) = 1. In
fact by RR, L(A) - L(KC-A) = 1-g + g-2 = -1, but since
the points are general, the usual induction shows that
L(KC-A) has dimension g - (g-2).

Since each of s1*L(KC) and s2*L(KC) has dimension g
and they intersect only in the 1-dimensional space,
it follows that they span a subspace of dimension
2g-1, that is the whole of L(2KC-A).

Now to go from L(2KC-A) of dimension 2g-1 up to
L(2KC) of dimension 3g-3, take the squares si^2 of
the remaining elements s3,.. sg. These are linearly
independent modulo L(2KC-A) because each si is
nonzero at Pi, and zero at every Pj.

The argument for d >= 3 is similar, but slightly
easier. With the same basis s1,.. sg, consider
the two subspaces s1*L((d-1)KC) and s2*L((d-1)KC)
in L(dK-A). The Castelnuovo free pencil trick
shows that their intersection is exactly
L((d-2)KC+A). The numerics to complete the proof
are a straightforward exercise. Q.E.D.