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By Akhil Mathew

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So this basis is stable under intersections, which is nice. 7. Let fi , i ∈ I be a family of elements in A. We have D(fi ) = SpecA if and only if the {fi } generate the unit ideal. Proof. If D(fi ) = SpecA, then every prime ideal must be an element of some D(fi ). In particular, the intersection V ((fi )) is trivial. This intersection is just V of the ideal a generated by the {fi }. As a result, the radical of a must be all of A, so a = A. The converse is proved similarly. 8. SpecA is quasi-compact.

The derived functors Ri Γ of the global section functor on Mod(OX ) and the usual cohomology groups H i are both δ-functors on Mod(OX ) which are naturally isomorphic in dimension zero. But both are effaceable (since we can use injective or flabby OX -modules). Consequently, the sequences of functors are naturally isomorphic by a general theorem of Grothendieck. As a result of this result, if F is an OX -module, the cohomology groups can be given a structure of Γ(X, OX )-module. 4. Cohomology with supports.

We can think of elements of A as “functions” on the space SpecA. Each element f ∈ A defines a map sending each p ∈ SpecA to the image of f in κ(p). 3. The image of f in κ(p) is called the value at p. Of course, the definition means that f will generally take values in very different things according to what p is. In the language of algebraic varieties, the residue field was always the same. Now, the set V (a) corresponds to the set of prime ideals where every element of a has value zero. So V (a) can be thought of as an intersection of zero sets, so it should intuitively be closed.

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