Let be an equation cutting out a degree d curve in . Suppose that doesn't contain the point . Use \v{C}ech cohomology to calculate the dimensions of of .
The degree d curve is the vanishing locus of , so we have a short exact sequence:
where without further decoration denotes the structure sheaf of . Specifically, the map on the left is multiplication by our polynomial f, which is a degree d map , but a degree 0 map . This is an injective map, and we are quotienting out precisely by its image, so it's equivalent to the usual short exact sequence associated to a closed subscheme.
Then apply the H functor to get a long exact sequence:
Which vanishes in higher degrees by dimensional vanishing.
Now to figure out what these things are:
for in projective space .
This gives us that . Furthermore, assuming degrees must be positive .
actually vanishes again by dimensional vanishing. , either by general knowledge (constants are the only globally defined homogeneous polynomials with degree zero on any of the standard open affines) or by the fact that in general; when e = 0, this gives dimension 1 over k. ().
Our last trick we shall use is Serre duality (here just for projective space):
, where represents the dual.
Since the dimension of a vector space (these H's are vector spaces in this context because of III 5.2 in Hartshorne, pg 228) is the same as its dual, . Moreover, , which has dimension , so it's 0. Hence .
Moreover, , and by the same trick (Serre duality), , which has well-known dimension (e.g., Vakil 14.1.c) of .
Combining all of the above results, we get two short exact sequences:
So we have and .