Solutions to Hartshorne's Algebraic Geometry/Cech Cohomology

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Exercise 4.7[edit]

Problem Statement:[edit]

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:[edit]

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 .