REMARKS: corrections, supplements, etc.
[09.18.2012] After constructing the sheaf of rings, I should have stated O(Spec A)=A. Here I'm omitting the subscript "Spec A" for O. (Thanks for asking this after class.) This follows from: O(Spec A)=O'^+(Spec A)=O'(Spec A)=A, where O' denotes the sheaf defined only on the basic open sets and O'^+ is the sheafification. Note that 1st/2nd/3rd equalities follow from definition of O/sheaf is isomorphic to its sheafification/definition of O', respectively. It's no harder to show O(D(f))=A_f.
[09.27.2012] Here are my answers to the questions during and after class.
Let X be a scheme and Z_{top} be a closed topological subspace of X_{top}. Write Z for the reduced induced closed subscheme associated to Z_{top}. On the other hand, let Z' be (any) closed subscheme of X such that Z'_{top}=Z_{top}. Let Z'_{red} denote the reduced subscheme associated to Z' as in Exercise II.2.3. (Note that Z, Z'_{red} and Z', as closed subschemes, are equipped with morphisms into X.) QUESTION: Is Z canonically isomorphic to Z'_{red}? ANSWER: Yes. By the universal property of Z, there's a unique morphism from Z to Z' such that a natural diagram commutes. Then the universal property for Z'_{red} (see Exercise II.2.3.(c)) tells us that there's a unique morphism from Z to Z'_{red} such that another natural diagram commutes. The question is whether the canonical morphism from Z to Z'_{red} is an isomorphism. This can be checked locally: For each open affine subscheme U=Spec A of X, is Z\cap U isomorphic to Z'_{red}\cap U (which are closed subschemes of U)? So we assume X=Spec A without loss of generality. Then clearly Z=Spec(A/a) for some a which is its own radical, and Z'=Spec(A/a') for some a' whose radical is a. But then it is not hard to show Z'_{red}=Spec((A/a')_{red})=Spec(A/a). From here one sees that the canonical morphism from Z to Z'_{red} is nothing but the obvious isomorphism between Spec(A/a) and Spec(A/a).
Let i:Y->X be an immersion. I stated as a fact that i factors through a closed immersion Y->U followed by an open immersion U->X. QUESTION: Is this factoring unique up to a unique isomorphism (in a suitable sense)? ANSWER: No. Stupid examples can be constructed using disconnected X. Concretely let Y=Spec k and X=disjoint union of Spec k and Spec k. Then the map Y->X sending Y to the first copy of Spec k is an (open and closed) immersion. It can factor through either U=Spec k (first copy) or U=X; in either case Y->U and U->X are closed and open, resp.
We haven't got to quasi-coherent sheaves and ideals yet, but there was QUESTION: When Z is a closed subscheme of X, is the kernel of the surjection O_X->i_*O_Z always a quasi-coherent ideal of O_X? ANSWER: Yes, see Prop II.5.9 of Hartshorne.
[10.24.2012] I was imprecise in defining a direct sum of O_X-modules. A correct definition should sheafify the direct sum of O_X-modules (for an infinite direct sum). Namely the direct sum of O_X-modules {F_i} is defined to be the sheafification of the presheaf U -> \oplus_{i} F_i(U). Alternatively we can say that the direct sum O_X-modules is the one satisfying U -> \oplus_{i} F_i(U) on every *affine* open U. In particular the direct sum of \tilde{M_i} is canonically isomorphic to \tilde{\oplus_i M_i}. The direct product of O_X-modules does not require sheafification. It would be instructive to find a counterexample where U -> \oplus_{i} F_i(U) is *not* a sheaf. (For instance, let G_i be sheaves of O_{X_i}-modules (or abelian groups) for an infinite index set, and then consider the disjoint union X of X_i's and F_i's where each G_i is trivially extended from F_i outside X_i.)
For an A-module M, we defined \tilde{M} to be the sheaf such that \Gamma(D(f),\tilde{M})=M_f=M\otimes_A A_f. A question during class was whether (#) \Gamma(U,\tilde{M})=M\otimes_A \Gamma(U,O_{Spec A}) for arbitrary open sets U. The answer is no. Example: A=k[x,y], M=k[x,y]/(x,y), U=complement of (0,0). Nevertheless there is also a positive result that if U is *affine* then (#) holds true.
