Solutions to Hartshorne's Algebraic Geometry/Cech Cohomology

Exercise 4.7 Problem Statement: 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 . Solution: 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 .