Q: How can I prove that a given scheme is not affine?

A: One way is to figure out what its global sections are, identify the resulting affine scheme, and somehow distinguish it from your given scheme. For example, if you start with Spec k[x,y] and remove the origin (the point corresponding to the ideal (0,0)), then any global section must restrict to a global section in Spec k[x,y,x^{-1}] and in Spec k[x,y,y^{-1}], and so must actually belong to Spec k[x,y]. But the induced map from this scheme to Spec k[x,y] is not surjective, so our scheme is not affine. For a more general version of this argument, see exercise II.2.17.

Q: Is any open affine subscheme of an affine scheme Spec A of the form Spec A_f for some f in A?

A: Not in general. I don't have a concrete counterexample on the top of my head, but here is a general construction that should produce one: if P is a non-principal minimal ideal of A, then I think the complement of V(P) is always affine (and this should follow from exercise II.2.17), but it will only be of the form Spec A_f if there exists some f such that Rad((f)) = P. What is true that any open affine subscheme of Spec A is a finite union of opens of the form Spec A_f (that plus the fact that the intersection of Spec A_f and Spec A_g is Spec A_{fg} means they form a basis for the Zariski topology); that is enough for all "sheafy" purposes. [Note added 26 Feb 09: the complement of V(P) need NOT be affine. Example suggested by David Speyer: A = k[w,x,y,z]/(wz-xy), P = (w,x).]

Concrete answer (suggested by Bhargav Bhatt): let E be a smooth plane cubic over the complex numbers, let O be the point at infinity, and let P be a point such that (P)-(O) is not torsion in Pic^0(E). (Such points exist because the group is isomorphic to a complex torus C/L, for L a lattice in C, and that has lots of non-torsion points.) Then X = E - O is affine (any curve minus a point is affine, thanks to Riemann-Roch!), Y = E - O - P is an open affine, but Y is not a localization of X. (If it were, then X would have a function having only zeroes or poles at E, so there would be a rational function f on X with (f) = c(P) - c(O) for some nonzero integer c, contrary to the non-torsion hypothesis.)

Q: Is the intersection of two affine open subschemes of a scheme necessarily affine?

A: Yes if the intersection is taking place inside a separated scheme; this is exercise II.4.3 (and I believe the same statement for varieties came up last semester). Otherwise not: glue two copies of Spec k[x,y] together along the complement of the point (x,y). Then these intersect in that complement, which is not affine.

Q: If U = Spec A, V = Spec B are open affines inside some scheme (with nonempty intersection), can I find some open affine W in their intersection such that W = Spec A_f for some f in A and W = Spec B_g for some g in B?

A: Yes! First, since U cap V is open in U, I can find inside it some (nonempty) open of the form Spec A_a. That is an open in an open in V, so inside Spec A_a I can find some Spec B_b. Now think of b as a global section of the structure sheaf O on V, and restrict it to Spec A_a. You get a global section of Spec A_a, i.e., an element of A_a, which you may write as c/a^m for some c in A and some nonnegative integer m. So now you have Spec B_b = Spec A_{ac} inside U cap V.

Q: Why is an affine scheme irreducible if and only if it it has a generic point?

A: On one hand, any topological space with a generic point is irreducible, since by definition the generic point is absent from every proper closed subset (and irreducible means precisely that you can't cover the space with two proper closed subsets). On the other hand, if X = Spec A is irreducible, let N be the nilradical of A. Then N is the intersection of the minimal prime ideals of A; if P1 and P2 are two such minimal primes, then V(P1) and V(P2) are closed subsets of X which cover X, and neither of which is proper (since V(P1) does not contain P2, and vice versa). That contradicts irreducibility, so there must be only one minimal prime of A, namely N itself. The corresponding point of X is a generic point.

Q: Why is a general scheme irreducible if and only if it has a generic point?

A (from Genya): If X has a generic point, it is irreducible by the argument above. Conversely, if X is irreducible, then so is any open subscheme (since irreducibility of a topological space passes to open subspaces); in particular, we may pick any one open affine U in X, and it will be irreducible. Now U is affine and irreducible, so it has a generic point x. This point must also be generic in X; otherwise if Z is a closed subset of X not containing x, then Z and (X-U) form a cover of X by two proper closed subsets, contradiction.

Q: Is the generic point of an irreducible scheme unique?

A: Yes. For X = Spec A affine, this follows because as above, the generic point must correspond to the nilradical of A. For X general, a generic point must lie in any open affine subset, so again we may argue by passing to a single affine.

Q: Is Hartshorne's notion of a "variety" the same as an "abstract algebraic variety" from last semester?

A: No! Hartshorne's varieties are all separated and of finite type over the base field.

Q: What does it mean for a property of morphisms of schemes to be "local on the base"?

A: The property is local on the base if for any morphism f: Y -> X and any open cover {U_i} of X, the morphism f has the property if and only if each induced morphism f^{-1}(U_i) -> U_i has the property. For example, most of the properties we are interested in (separated, proper, quasi-compact, finite, of finite type/presentation...) are local on the base. Note that Hartshorne's definition of projective is broken precisely because it fails to be local on the base when the base is "too big".

Q: (PS 1, #12(a); Hartshorne II.2.14) If S is a graded ring and Proj S is empty, why must S_+ be a nilpotent ideal (in other words, every element of S_+ is nilpotent)?

A: Maybe the best way to think of this is to recall that for f in S_+, the open subscheme D_+(f) of Proj S (consisting of those homogeneous prime ideals not containing f) is isomorphic to Spec (S[f^{-1}])_0 (Proposition II.2.5). If f is not nilpotent, then S[f^{-1}] is not the zero ring, so (S[f^{-1}])_0 isn't either. Hence Spec (S[f^{-1}])_0 is nonempty, as then is Proj S. (Word to the wise: rather than work this out myself, I simply looked this up in EGA II.2!)

Q: (PS 1, #12(c); Hartshorne II.2.14) If S -> T is a homomorphism of graded rings which is an isomorphism in sufficiently high degrees, why is the induced map Proj T -> Proj S a bijection?

A: Say the isomorphism is in degrees d >= d_0. Note that every distinguished open can be written as D_+(f) for some homogeneous f of degree >= d_0 (because D_+(f) = D_+(f^n) for any positive integer n). Given such an f, viewed inside both S and T, the two rings S[f^{-1}]_0 and T[f^{-1}]_0 are the same because any element of either can be written as (something of degree >= d_0)/(power of f). So you can view the map Proj T -> Proj S as being obtained by glueing together isomorphisms between two systems of affine schemes, hence it's an isomorphism.

Q: Under what conditions is the dimension of a scheme necessarily finite?

A: It's a bit hard to say in general. Of course you need the underlying topological space to be noetherian. But that's not enough; one can have a noetherian scheme (even an affine one) in which each descending chain of closed subspaces has finite length, but there is no uniform bound on the lengths of these chains.

On the positive side, the dimension of any scheme of finite type over Spec Z, or over a field, is finite. I think more generally you can take a scheme of finite type over what is called an "excellent ring".

Q: According to exercise II.3.11, if f: X -> Y is a morphism of schemes with X reduced, the underlying set of the scheme-theoretic image of f is the closure of the set f(X). Is that true for general X?

A: No! The scheme-theoretic image can behave in extremely counterintuitive ways (thus making it not a very useful concept in general). x=0, but the scheme-theoretic image is all of Spec k[x].

It may be possible to force better behavior by making stronger assumptions (the morphism is of finite type, X and Y are noetherian, etc.) but I haven't looked into this.

Q: How does the degree of a complete linear system relate to the degree of the corresponding map to projective space?

A: The degree of the complete linear system associated to a divisor D on a projective variety X is, by definition, the degree of the divisor D (or indeed of any other divisor in the system). It is equal to the "degree" of the induced map f: X -> P^n (which I haven't defined). However, it is not in general the same as the degree of the map X -> f(X)! In fact, it is the product of that degree with the degree of the image variety f(X) (i.e., the number of intersection points with a typical hyperplane of complementary dimension). For instance, if X is a curve and n = 1, then the latter is always 1 as soon as the map is nonconstant.

Q: Given a projective variety X and a complete linear system |D| on X, how do I tell whether the induced map X -> P^n is a closed immersion?

A: You have to first check that |D| separates points (that is, given any two distinct closed points P and Q, you can find a section in L(D) vanishing at P but not at Q) and that |D| separates tangent vectors (that is, given any point P, the sections in L(D) vanishing at P span the tangent space at P). See Proposition II.7.3 for the proof.

Q: What are some examples of quasi-finite morphisms that are not finite (in re II.3.5, from PS3)?

A: For starters, if R is an integral domain, the map Spec K -> Spec R is quasi-finite but typically not finite. If you want surjective and finite type, as in II.3.5(c), the usual nonseparated example works: take two affine lines glued away from zero, mapping to one affine line. That's quasi-finite but not finite (because finite implies affine implies separated).

If you want your example to also be separated, you should take an actual finite map and merely shrink the source. Simplest example: take the disjoint union of Spec k[x] and Spec k[x,x^{-1}], mapping to Spec k[x].

Q: Is Omega^1_{A/B} the same thing as Omega_{A/B}?

A: Yes. In general, Omega^i_{A/B} denotes the i-th exterior power of Omega_{A/B} over A.

Q: How do I show that a smooth plane curve of degree 5 has no g^1_3 (IV.5.6, PS 7)?

A: Sketch: first verify that the curve has genus 6. Then note that the canonical divisor is equivalent to two lines (in general, the canonical divisor of a smooth plane curve of degree d is equivalent to d-3 lines, as can be seen by calculating that the divisor of, say, dx looks like the zero locus of a polynomial of degree d-1 minus twice the line at infinity); that means that the canonical embedding of your curve factors as your original plane embedding followed by the Veronese map (the 2-uple embedding of P^2 into P^5). Finally, use Riemann-Roch to show that if E was an effective divisor of degree 3 generating a g^1_3, then the three points of E would map to collinear points on the Veronese surface. That's a contradiction because any line meets the Veronese in at most two points (because the degree of the Veronese map is 2).

Q: In II.6.10(c), the hint explains how you can check that if a coherent sheaf F on X has support in Y, then its class in K(X) is in the image of K(Y). But in fact you have to show that if you have any class in K(X) (not necessarily represented by a single coherent sheaf), and its image in K(X-Y) is the zero class, then it comes from a class in K(Y). How do you do this?

A: I'm still working on this...

Q: If I is an ideal sheaf on the scheme X, over which points of X is the blowup at I an isomorphism?

A: At precisely those points where I is locally principal. That is, if x is a point of X, then the blowup is an isomorphism at x if and only if the stalk of I at x is a principal ideal in the local ring O_{X,x}. (Necessity: the inverse image sheaf is locally principal, so if I isn't locally principal at x, the blowup can't be an isomorphism there. Sufficiency: clear from the form of relative Proj.) Corollary: multiplying by an invertible ideal has no effect on the blowup (exercise II.7.11(b)).

Q: So what does the previous question/answer have to do with II.7.11(c)?

A: By (b) and the fact that X is regular, you can choose the closed subscheme Y so that each irreducible component of its support has codimension at least 2. (If you had a codimension 1 component, i.e., a Weil divisor, it would give rise to a Cartier divisor since X is regular, and the ideal sheaf of said divisor is invertible. So you can get rid of it without changing the blowup.) Now note that if Y has codimension at least 2, then the blowup in Y fails to be an isomorphism precisely at Y, because the ideal sheaf corresponding to Y fails to be locally principal everywhere at Y. (If it were locally principal somewhere, that would force Y to have a codimension 1 component.) So Y = X-U as desired.

Q: Exercise II.5.7(a) (PS 5) asks: if X is a noetherian scheme, F is a coherent sheaf on X, and x is a point at which the stack F_x is free over O_x, then F is freely generated by sections on some neighborhood U of x. Is it enough to take free generators of F_x over O_x and simply note that they come from sections over some U?

A: No, you still have to show that your sections s_1, ..., s_m generate F(V) *freely* for some V. Namely, consider the map O_U^m -> F defined by (a_1, ..., a_m) -> a_1s_1 + ... + a_m s_m, and let G be the kernel of this map. By hypothesis the stalk G_x is zero; hence for any section t, I can find f in O(U) not vanishing at x such that ft = 0. Since U is noetherian, G is coherent, so I can find a finite set of generators t_1, ..., t_n of G; now pick f in O(U) annihilating t_1, ..., t_n, and put V = D(f) in U. Now s_1, ..., s_m really do freely generate F(V).

Q: How do you check Exercise II.5.8(c) (PS 5)?

A: Let x be a point in X. As in 5.7(a), on some open affine U = Spec A, you can find sections s_1, ..., s_m of F which freely generate the stalk F_u. Again look at the kernel G of O_U^m -> F (given by the s_i). Suppose this kernel has a nonzero section, i.e., there is a relation a_1 s_1 + ... + a_m s_m with not all a_i zero. WLOG a_1 is nonzero; since X is reduced, that means a_1 fails to belong to some prime ideal of A, corresponding to some point y in U. But now phi(y) < m because over k(y), I can eliminate s_1 in terms of s_2, ..., s_m; contradiction. So G has zero sections over an open affine and so is the zero coherent sheaf; hence F is locally free.

Q: Is there something wrong with problem 12 on PS 9?

A: Yes, sorry. I should have said that there is a spectral sequence with E_2^{p,q} = H-check^p(X, H^q(F)) (where H^q(F) is the presheaf associating to each U the sheaf cohomology group H^q(U,F)) abutting to H^n(X,F). (This is the same spectral sequence that "explains" why you can compute Cech cohomology on a "good cover", i.e., one such that your sheaf is acyclic on each finite intersection of opens in the cover.) Then you proceed by induction: once you know that H^q(U,F) vanishes for each U in your basis for 0 < q < n, you can deduce that the map H-check^n(X,F) -> H^n(X,F) is an isomorphism; since the former is zero, so is the latter. (Reference: section 3.8 of Grothendieck's Tohoku paper "Sur quelques points d'algèbre homologique".)

Q: In the last GAGA problem on PS 10, how do you know that the map G(-1) -> G is injective?

A: Actually, you don't; that is a typo. There should be a kernel there, but the point is that you know it is supported on the hyperplane, so by the induction hypothesis, it (like the cokernel) must come from an algebraic coherent sheaf. Split this four-term sequence into two three-term sequences, twist, take cohomology, and see what happens.

Q: In the first GAGA problem on PS 10, do you really need the Dolbeaut theorem in full force for part (a)?

A: No. It suffices to check that the analytic structure sheaf is acyclic for each intersection in the standard cover of projective space by affines. Note that each intersection is isomorphic to a product of copies of C and copies of C^*; moreover, by the Kunneth formula, you can write H^m(X x Y, O) as the direct sum of H^p(X, O) * H^q(Y, O) over all p+q = m. Thus all you need to show is that H^1(C, O) = H^1(C^*, O) = 0. The former is trivial because C is contractible, so all sheaves are acyclic. So in fact you only need the Dolbeaut theorem on C^*! (You might be able to prove this case "with bare hands", but the only way I was able to do it was to actually trace through the proof of the Dolbeaut theorem using the de Rham theorem and the d-bar Poincare lemma. See Chapter 0 of Griffiths and Harris, Principles of Algebraic Geometry.)

Q: In the last GAGA problem on PS 10, do I need any analytic input?

A: It appears there is no way around assuming that H^1 of any coherent analytic sheaf on P^n is finite dimensional. This is a special case of a theorem of Cartan, proved with a little functional analysis as follows: exhibit two finite good covers of P^n (i.e., with contractible intersections) with the property that each element of one cover is an interior subspace of some element of the other cover (that if, given U in one cover, its closure is contained in some V in the other cover). Then the Cech cohomology of either computes H^1, so the comparison map between the two is an isomorphism. On the other hand, that interior setup forces the map to actually be a completely continuous map of topological vector spaces (because the restriction map from analytic functions on any open V to an open whose closure is contained in V is completely continuous), and an isomorphism can only be completely continuous if it is between two finite dimensional spaces.