Seventeenth post: Flatness (finished), and proof of Serre duality

July 21, 2011

Seventeenth post: Flatness (finished), and proof of Serre duality

Posted by ravivakil under Actual notes
[43] Comments

The seventeenth version of the notes is the July 21 version in the usual place. This version has a complete exposition (i.e. everything I currently intend to say) of flatness (chapter 25), and a proof of Serre duality (chapter 28). Some content is added earlier (e.g. the Artin-Rees Lemma). The next post may appear in August, depending on baby constraints.

$FlatnessIsGod$

Status report.

There are only three more content chapters still to come, one on smooth/etale/unramified morphisms; one on formal functions and related issues (Zariski’s main theorem, Stein factorization, etc.); and one on regular sequences and related issues (local complete intersections, Cohen-Macaulayness, etc.). I’m 100% sure they will appear, but I’m not sure when (again, due to oncoming family constraints). Of course, a lot of work remains to be done to fill in holes and patch problems in the rest of the notes (and responding to old comments), so I may spend some time doing that.

I also want to take this opportunity to thank Sándor Kovács for advice throughout this project (and before), both technical and otherwise.

For learners.

Flatness is confusing the first time you see it. Also the second and the third. But with each iteration, you will digest and master more aspects of flatness. With most parts of algebraic geometry, when you learn a concept, you get used to one strange thing and then you’re good to go. That’s not true with flatness — when thinking over this chapter, I realized that there are many different types of results and arguments that come up. I’ve done my best to organize them, and to discuss no more than I find brutally necessary. (Many important flatness facts are left unproved or unstated, but hopefully by the end you will know enough to be able to read what you need elsewhere.)

The structure of the chapter is described in 25.1.1, so I won’t repeat it here.

I hope some of you read the proof of cohomology and base change — if you do, please let me know, and please let me know what is most confusing!

Here as always are some suggested problems. Of course, try every exercise marked “easy”. If it isn’t easy, let me know!

Here are twelve problems on flatness: 25.2.E (transitivity of flatnes); 25.2.G (relating flatness in algebra with flatness in scheme theory); 25.2.L (explicit examples), 25.2.M (cohomology commutes with flat base change — this looks hard but isn’t), 25.3.A (explicit and important Tor calculation), 25.3.F (practice with Tor), 25.4.D (flat = torsion-free for a PID), 25.4.F (“finite flat morphisms have locally constant degree”), 25.4.I (an explicit example that will come up later, involving two planes meet at a point), 25.5.D (going-down for flat morphisms), 25.5.F (fibers of flat morphisms have the “expected dimension”), 25.7.B (important! invariance of many important numbers in flat families), some 25.8.A-F (using cohomology and base change),

If you want to work through some pleasant explicit examples, I’d recommend 25.4.8 onwards, on flat limits. Another fun discussion that will help you see if you understand flatness well enough to do something is Hironaka’s example of a proper nonprojective nonsingular threefold, 25.8.6. If you get stuck, please let me know.

If you read 25.10 on flatness and completions, please let me know how it went.

Chapter 28 is a starred proof of Serre duality. I hope some of you try to read it — it is not double-starred, which means that I intend for this to be readable, and not just an indication that a proof exists. As with flatness, I try to prove no more than I really need to given this stage of the notes/course. If you read this, some good problems to try are 28.3.C (relations among Ext, sheaf Ext, and H^i), 28.3.H (Ext and vector bundles), and 28.3.J (the local-to-global spectral sequence for Ext).

For experts (and general discussion)

The Artin-Rees Lemma.

Greg Brumfiel explained the Artin-Rees Lemma to me in a way that made it very natural — enough so that I can no longer forget the proof. I’d never understood it well before. I hope I’ve gotten it across with some semblance of Greg’s clarity (13.6 and parts of 25.10).

Question: suppose $A$ is a Noetherian ring, and $I$ an ideal. Suppose $0 \rightarrow M \rightarrow N \rightarrow P \rightarrow 0$ is an exact sequence of $A$ -modules. I find it entertaining that it remains exact when tensoring by $\hat{A}$ , but that $0 \rightarrow \hat{M} \rightarrow \hat{N} \rightarrow \hat{P} \rightarrow 0$ need not be exact (without coherence hypotheses on the modules — perhaps just on $P$ ?). Does anyone have a( reference to a)n example where exactness does not hold? (And as a consequence, we’ll see an example where completion is not the same as tensoring with $\hat{A}$ .) Update August 17, 2011: An answer is given in the section “Completion is not exact” in the “Examples” chapter of the Stacks Project. I’m not sure how to find out the tag. But I’ll add this example, in the version to be released around the end of August.

Flatness.

I found the flatness notes of Brian Lehmann (available on his webpage) very nice. Andrew Critch’s enlightening a postiori explanation of how to think about flatness is incorporated into
Remark 25.4.2, just before the equational criterion. (Side remark: we don’t use the equational criterion for anything.)

I hope someone looks closely at my exposition (and proof) of Cohomology and Base Change. That’s a topic where I think I learned the right perspective only by talking to people, and part of my goal is to translate some of the folklore into writing. If you have never bothered fully understanding the proof, and want to, please take a look and let me know where the exposition confuses you. (It is now divided up into some general facts about cohomology of complexes, and a very short argument for the theorem itself.)

Max Lieblich told me that he first figured out Cohomology and Base Change by translating to local rings, and working there, which has the advantange that you can make short exact sequences split. I could imagine that this would yield an even faster exposition. (One worry: the statement I want of cohomology and base change involves an honest Zariski open set, see part (i) of my statement. But I bet Max’s approach would give that too.) Partially because I’d already written this, I haven’t tried to piece together how Max’s argument should go. But if someone does, or someone thinks I should because it would make things more transparent, please let me know.

The flatness chapter is disappointingly long — and I even didn’t prove (for example) that the flat locus is open (under reasonable hypotheses). I didn’t prove Grothendieck’s generic freeness lemma because I didn’t use it (but I stated it). I didn’t prove the fibral flatness theorem, but stated it. Are there things that I really should include? Are there things I’ve included that you think could reasonably be tossed in a first course? (You’ll noticed that lots of the chapter is already starred or double-starred.) One fact that isn’t there but will be (in a later chapter) is what Brian Conrad calls “miracle flatness”, about a morphism $\pi: X \rightarrow Y$ to a nonsingular scheme, and relating the flatness of $\pi$ , the equidimensionality of the fibers, and the Cohen-Macaulayness of $X$ .

Serre duality

I’m going to upset some people here, by not proving the “right” statement. My goal, given that this discussion comes at the end of a long set of notes, and at the end of a long course, is to prove just enough to justify the statements made earlier in the notes.

Here’s what gets used:
(i) we need a perfect pairing (28.1.1.1) $H^i(X, \mathcal{F}) \times H^{n-i}(X, \mathcal{F}^{\vee} \otimes \omega_x) \rightarrow k$ in good circumstances.
(ii) We need the dualizing sheaf to be the determinant of the cotangent bundle if $X$ is smooth.
(iii) Perfect pairing (i) often arises from something better (which I call Strong Serre duality) which is an isomorphism $\rm{Ext}^i_X(\mathcal{F}, \omega_X) \rightarrow H^{n-i}(X, \mathcal{F})^{\vee}$ .
I show (iii), but don’t show that these maps are well-behaved in any way at all — for what we do, we don’t need the perfect pairing (28.1.1.1) to be “natural” in any way — we just need dimensions. The reason I can’t show any sort of naturality (in a pedagogically easy way) is that it isn’t worth the trouble of defining the natural maps $\rm{Ext}^i_X( \mathcal{F},\mathcal{G}) \times H^j(X, \mathcal{F}) \rightarrow H^{i+j}(X, \mathcal{G})$ . In 28.3.4, I do mention where these maps come from, and outline the Yoneda cup product for Ext’s (following Grothendieck’s Theoreme de dualite pour les faiscaux algebriauqes coherents — apologies for lack of accents).

The advantage of my approach is that we can prove the statement we actually use relatively easily (although not so easily that I’d remove the star from the chapter). Keep in mind that we are the end of the course; I want to prove what we use as easily as possible.

(The disadvantage is that we clearly prove the wrong statement!)

Side remark: in an earlier version of the course, I proved Serre duality via duality for finite flat morphisms. This results in a proof which is much easier and shorter. (To apply it, we need the “miracle flatness theorem” I mentioned above; but that will be included.) The serious downside of this approach was that I was unable to prove (ii). So instead I decided to go with the current exposition, which requires more work.

Random questions for experts

1. I’ve proved uppersemicontinuity of fiber dimension on the target (for a projective morphism). But I haven’t proved uppersemicontinuity of fiber dimension on the source (for locally finite type morphisms to locally Noetherian schemes; or if you really care, for locally finitely presented morphisms in general, but that’s just an easy generalization once you’ve got the hard part). I don’t know an easy proof (i.e. short given what is already done in the notes). Does anyone know one (or have a reference)? It seems to be surprisingly hard work. (I also asked for a trick solution here.)

2. A reference questions about the category of O-modules on a scheme. I have heard that they don’t have enough projectives. (I asked a variant of this question here.) Does anyone have a reference (ideally with a proof)? I’ve heard that locally free sheaves on a scheme are not necessarily projective in the category of O-modules. Reference with proof? (Update August 2, 2011: see David Speyer’s comment below.)

3. (unimportant; maybe better suited to mathoverflow) It was in grad school that I first heard about the Lefschetz principle, allowing you to reduce all statements over an algebraically closed field of char 0 to $\mathbf{C}$ . Even now I’m not sure precisely what this principle is supposed to be (except in a rather baby case, where it is basically elimination of quantifiers). Is there a reference somewhere? Here is an interesting article complaining about it. (Warning: you need access to jstor to access it. Bibliographic info: A. Seidenberg, Comments on Lefschetz’s Principle, The American Mathematical Monthly, Vol. 65, No. 9 (Nov., 1958), pp. 685-690.) Here is a possible reference (that I’ve not read): Frey, Gerhard and Rück, Hans-Georg, The strong Lefschetz principle in algebraic geometry, Manuscripta Math. 55 (1986), no. 3-4, 385–401.

43 Responses to “Seventeenth post: Flatness (finished), and proof of Serre duality”

Allen Knutson Says:

July 21, 2011 at 8:16 pm
One benefit of establishing Serre duality for finite flat morphisms is that that statement is needed in order to correspond near-splittings of a scheme over F_p with sections of omega^{1-p}. (See p21 of the [Brion-Kumar] book on Frobenius splitting.) It’s one of only two references in that book to [Hartshorne '66].

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1. ravivakil Says:
  
  August 2, 2011 at 8:30 am
  That’s certainly pretty! I’m not yet convinced that it’s of the level that requires mention (even at the level of statement), as opposed to, say, the valuative criterion for flatness (see David’s comment below) or generic freeness.
  
  Reply
David Speyer Says:

July 22, 2011 at 10:39 am
I’m a big fan of the valuative criterion of flatness, EGA IV_3.11.8: If S is reduced and noetherian and f: X –> S is locally of finite presentation, then f is flat if and only if, for every map of Spec of a D to S, the pull back $X \times_S \mathrm{Spec}\ D \to \mathrm{Spec}\ D$ is flat.

Since flatness over a dvr just means no associated primes over the special point, I find this easy to visualize. It says that for any s in S, and any path gamma through s, all of f^{-1}(s) can be obtained as limits along paths in X lying over gamma. That’s my mental image of flatness (at least over reduced bases.)

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1. David Speyer Says:
  
  July 31, 2011 at 3:58 pm
  “Spec of a D” should read “Spec of a dvr, D”.
  
  Reply
2. ravivakil Says:
  
  August 2, 2011 at 8:44 am
  I agree — this is helpful to visualize flatness. I’ve now added this (the statement without proof, and a pointer to EGA) just after they learn how to visualize flatness over a DVR. Thanks!
  
  Reply
David Speyer Says:

July 22, 2011 at 10:49 am
A proof that locally free sheaves need not be projective: Look at $\mathbb{P}^1$ . We have a surjection

$\mathcal{O}(k-1)^{\oplus 2} \to \mathcal{O}(k)$

If $\mathcal{O}(k)$ were a projective object, this surjection would split, and it doesn’t.

A modification of this argument shows that there are no non-zero projective objects in the category of coherent $\mathcal{O}$ -modules on $\mathbb{P}^1$ . I’m not sure about quasi-coherent.

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1. ravivakil Says:
  
  August 2, 2011 at 9:18 am
  What’s the easiest way of seeing that the surjection doesn’t split?
  
  Reply
  1. David Speyer Says:
    
    August 2, 2011 at 10:22 am
    The kernel is $\mathcal{O}(k-2)$ , so you are asking how to check that $\mathcal{O}(k-1)^{\oplus 2} \not \cong \mathcal{O}(k) \oplus \mathcal{O}(k-2)$ . Probably the easiest argument is to twist by $\mathcal{O}(-k)$ , so we are trying to show that $\mathcal{O}(-1)^{\oplus 2} \not \cong \mathcal{O}(0) \oplus \mathcal{O}(0-2)$ . The RIHS has global sections, and the LHS doesn’t.
    
    Reply
  2. ravivakil Says:
    
    August 2, 2011 at 10:25 am
    Yes, of course — nice!
    
    Reply
2. ravivakil Says:
  
  August 2, 2011 at 9:35 am
  See also David’s link to mathoverflow here.
  
  Reply
3. ravivakil Says:
  
  August 2, 2011 at 9:43 am
  And by “modification to this argument” do you mean, if $\mathcal{F}$ is coherent sheaf on $\mathbb{P}^1$ that is a projective object (in the category of coherent sheaves, or quasicoherent sheaves, or O-modules, it doesn’t matter), then choose a surjection from a direct sum of line bundles, and do something like the above? How does the argument work?
  
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4. ravivakil Says:
  
  August 7, 2011 at 1:11 pm
  Max Lieblich sent me the following great discussion of when there are projectives in the category of coherent sheaves.
  
  “Btw, I see some discussion of projective things on your site, and it reminds me of a result I worked out in grad school: a separated noetherian scheme with the resolution property in characteristic 0 (surely this is not necessary) has enough projectives in Coh(X) iff it has a single projective of positive rank iff it is affine. Proof of non-stupid direction: say V is a projective coherent sheaf of rank r. Since V admits a surjection from a locally free sheaf, it is a summand hence is itself locally free. But then the sheaf End(V) is projective. Since the trace map End(V) -> O splits the inclusion (char 0 being used here) the structure sheaf is projective. But then Serre’s criterion shows that X is affine (because O being projective says that Gamma is exact!).
  
  There is undoubtedly a way to do this without the trace map (hence without the char 0 hypothesis).”
  
  Reply
David Speyer Says:

July 22, 2011 at 11:16 am
Potential typo:

p. 546, middle of the page “For example, cohomology commutes with flat base change, Theorem 25.2.8, so the result holds if g is flat.” There is no map called g. Should that be f?

I really like the organization of 25.8 and 25.9. I’m afraid I don’t have time to actually check the proofs.

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1. ravivakil Says:
  
  August 2, 2011 at 8:51 am
  Indeed the g should be f; now fixed.
  
  Thanks for the comments about 25.8 and 25.9 — certainly I wouldn’t expect you to have time to check the proofs! Already you are a huge help.
  
  Reply
Charles Staats Says:

July 29, 2011 at 7:53 am
For a discussion in which Brian Conrad gives specific references for versions of the Lefschetz Principle based on flatness, see the remarks after http://mathoverflow.net/questions/44758#44758.

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1. ravivakil Says:
  
  August 2, 2011 at 9:28 am
  Thanks Charles!
  
  Reply
Charles Staats Says:

August 2, 2011 at 11:05 am
Ravi,

Based on certain comments you’ve made on your blog, I get the impression that you are trying to avoid the inclusion of the generic freeness lemma in your notes. I am going to make a case that you should include it. My main points are these:

1) In addition to allowing a proof of semicontinuity of fiber dimension on the source, generic freeness can make a lot of arguments easier. For instance, the “hard” part of Chevalley’s theorem (showing that the image of a dominant morphism contains an open set) is taken care of immediately.
2) Generic freeness is a great illustration of the general principle that “if something is true over the generic point, then it should be true over a dense open set.”
3) Although the theorem seems to have a reputation for “hardness”, there is at least one proof that is not too long and entirely elementary.

Best,

Charles

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1. ravivakil Says:
  
  August 16, 2011 at 9:53 am
  Charles, you are right that I was trying to avoid the inclusion of generic freeness in these notes. My reasoning was that it is a nice result, but avoidable, and thus a suitable sacrificial victim to the brutal goal of excising as much as possible so the rest could fit into a single year. You have convinced me otherwise, and I’ll think about how to best include this. All three of your points are good — the third is one that others might overlook.
  
  To everyone else: Charles also gave me a short and pretty exposition of the argument, which you can find here. He says he heard it from Nori, although from Charles’ earlier posts, I find it easy to believe that he contributed a great deal to the explanation.
  
  Update later the same day: my first stab at distilling Charles’ argument is here. (The page number does not reflect where it will eventually sit.)
  
  Reply
  1. Charles Staats Says:
    
    August 16, 2011 at 10:12 am
    I’ve asked Nori where he got this from, and he said it came from Matsumura’s Commutative Ring Theory, verbally via another mathematician (Pavaman Murthy). Having glanced at Matsumura’s proof, I have a feeling that all three of us made significant contributions to the explanation. No doubt if Ravi chooses to use this argument, his version will be nicer still.
    
    One other note–one limitation of this argument is that it seems to rely heavily on the Noetherian hypothesis.
    
    Reply
  2. ravivakil Says:
    
    August 16, 2011 at 11:01 am
    I also looked at Matsumura’s exposition, and your explanation is much simpler. (Before anyone complains: I agree that logically they are the same argument.) I love how mathematics can be successively distilled through a sequence of mathematicians. (I have had a number of interesting discussions on this point with a number of people: once someone has given the first prove of a result, to what extent is the story over? I gather my own view lies more toward one extreme — that there is a great deal of value added by this distillation process.)
    
    On another note: the Noetherian hypotheses seem necessary to me, so they don’t bother me. (Is there some more general version of this result? And I mean substantively more general — nothing that is just a base change of the Noetherian statement.)
    
    Reply
    1. Johan Says:
      
      August 16, 2011 at 1:35 pm
      Actually, you can improve generic flatness in two ways: (a) Noetherian hypotheses aren’t necessary (this is in EGA), and (b) the base need only be reduced. This is a key step in proving the existence of residual gerbes for algebraic stacks with quasi-compact inertia in the stacks project. Take a look at Proposition Tag 052B of the stacks project to see one possible statement of generic flatness with minimal assumptions. (As usual there is also a commutative algebra version, a version for morphisms of algebraic spaces, and a version for morphisms of algebraic stacks — this last one is not yet coded up.)
      
      Reply
    2. ravivakil Says:
      
      August 17, 2011 at 6:19 am
      Johan, this is very nice! (I was thinking of applications of just generic freeness, but of course a central use is generic flatness.) I think this is important to know, so I (currently) intend to state this without proof, and refer to tag 052B. I remember wondering about this when first learning about generic flatness, and vaguely assuming that something like this was too good to be true, under the assumption that if it were true, someone would have told me. And now someone has!
      
      Reply
      1. Johan Says:
        
        August 18, 2011 at 3:25 am
        David Eisenbud points out in his book on commutative algebra (page 307) that for many applications the freeness is essential. In the application to “global” schemes (i.e., to X —> S not affine) it is hard to formulate what “freeness” would mean. (There is something you can say; it is related to the morphism being “pure” but it takes a lot of work to formulate properly.) What is easy to say and useful for some (very technical) applications is that starting with a finite type morphism the resulting morphism over the open is of finite presentation (as well as flat).
        This is a slight improvement of the nonnoetherian version in EGA IV, Section 11.1-11.3
        
        Just a caution about tag 052B: the dense open you get can be “bad”, for example not quasi-compact (even if the base S starts out as an affine). So be careful! If the base is irreducible you can replace it by any affine open contained in it (because this will still be dense) — in this case I suggest referencing the easier to prove Proposition Tag 052A.
        
        Reply
        
        Charles Staats Says:
        
        August 18, 2011 at 7:07 am
        In the global case, you can say that above the dense open set, the coherent sheaf in question is either zero or faithfully flat.
        
        Reply
        
        Johan Says:
        
        August 19, 2011 at 9:31 am
        No!
        
        Reply
        
        Charles Staats Says:
        
        August 19, 2011 at 11:56 am
        Okay, I just thought more carefully about my argument for this, and realized it relies on the inverse image of $S$ being sufficiently nice (e.g., quasicompact and irreducible). Thus, it might not have been an appropriate remark in a discussion on extending to the non-Noetherian case. But just to check–if you do have such niceness hypotheses, this is an appropriate way to use generic freeness to get a stronger global statement than generic flatness, right? Or am I totally confused?
        
        Another thought: I also think that even in the general case, generic freeness should imply that you can restrict to a dense open subset $U$ of $S$ such that whenever $x \mapsto s \in U$ , then the stalk $\mathcal{F}_x$ is free (hence either zero or faithfully flat) over the stalk $\mathcal{O}_S,s$ . Is this valid, or am I totally confused?
        
        Reply
2. ravivakil Says:
  
  December 8, 2011 at 8:44 am
  Update much later: I’ve tried this out in the course (with admittedly very strong students), in the section on Chevalley’s theorem (section 8.4 in the most recent version posted) and this was something lots of people enjoyed. Before, the Chevalley’s section was uniformly something people strongly disliked. So you are completely right — this is the way to go. Thanks!
  
  [And I am again behind in responding to others' comments. After term ends this week I will begin to catch up.]
  
  Reply
Charles Staats Says:

August 19, 2011 at 11:58 am
[Note: The following comment is a re-post of a reply I made above, and then realized was too narrow for readability.]

Okay, I just thought more carefully about my argument for this, and realized it relies on the inverse image of $S$ being sufficiently nice (e.g., quasicompact and irreducible). Thus, it might not have been an appropriate remark in a discussion on extending to the non-Noetherian case. But just to check–if you do have such niceness hypotheses, this is an appropriate way to use generic freeness to get a stronger global statement than generic flatness, right? Or am I totally confused?

Another thought: I also think that even in the general case, generic freeness should imply that you can restrict to a dense open subset $U$ of $S$ such that whenever $x \mapsto s \in U$ , then the stalk $\mathcal{F}_x$ is free (hence either zero or faithfully flat) over the stalk $\mathcal{O}_S,s$ . Is this valid, or am I totally confused?

Reply
1. Johan Says:
  
  August 20, 2011 at 8:28 am
  [Too bad I was hoping to see if we could get it down to one letter per line:)] I don’t think you’re totally confused at all, but I think it is very hard to get used to how bad things can be! Also the stalks F_x aren’t free over O_{S,s} because when you take the stalk at x you are localizing _upstairs_… In my comment I just meant to say that it is better to switch to the algebra side of the picture when you want to use the freeness part of generic flatness/freeness. That’s all.
  
  Reply
  1. Charles Staats Says:
    
    August 20, 2011 at 9:02 am
    Of course! Somehow, it did not occur to me that localizing _upstairs_ need not preserve freeness or faithful flatness, but you’re absolutely right: If you have a morphism of rings $A \to B$ , and $B$ is faithfully flat over $A$, then $\mathrm{Spec } B \to \mathrm{Spec } A$ is surjective. In a large number of cases, we can dispose of this surjectivity property (and, hence, of faithful flatness, and hence freeness if we had it) by inverting an element of B. Since “move to an affine cover upstairs” was part of my argument in the global case, I was wholly mistaken.
    
    Reply
Georges Elencwajg Says:

August 20, 2011 at 2:34 am
Dear Ravi,
I’m desperate: I can’t download your July 21 notes. Whenever I click on the red link “Actual notes” above, the system brings me back to this very page.
Am I the only one in this unfortunate situation?

Reply
1. ravivakil Says:
  
  August 20, 2011 at 3:57 am
  Dear Georges,
  My apologies! Click on “the usual place”, and it will take you there. More directly:
  http://math.stanford.edu/~vakil/216blog/FOAGjul2111public.pdf
  (The various posts are classified under a very few subheadings, and this is under the subheading of “Actual Notes”.) Please let me know if you have further troubles!
  
  Reply
Georges Elencwajg Says:

August 20, 2011 at 12:31 pm
Dear Ravi,
Hurrah! I followed your instructions and successfully downloaded the July 21 notes: a wonderful week-end in perspective!
Best wishes to you and your growing family,
Georges.

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Lalit Says:

August 26, 2011 at 1:45 pm
Hi,

I had a quick question about the statement of theorem 18.4.5. Currently you have no requirements on f in the statement of the theorem but I was not sure if it is necessary that f be a dominant(surjective?) map in the statement of the theorem? For exercise 18.4.C to make sense it is certainly needed. Also in the discussion following 25.4.J (about how 18.4.5 can be deduced from flatness) you certainly need it since you need the generic point of C(assumed to be irreducible) to map to the generic point of C’.

I wasn’t sure if we can automatically assume that f is dominant for some reason. By 25.4.G for f.O_C to be free we would need f to be flat and then 25.4.J would give that the map was dominant. Would it be possible to shine some light on this? Thanks.

Lalit

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1. ravivakil Says:
  
  August 29, 2011 at 7:05 am
  That’s a good question. Dominance (indeed surjectivity) is indeed forced in the situations you have in mind. Hypotheses need to be added to make it work (e.g. if C is empty, or if C is irreducible, but C’ has many components, then you won’t have surjectivity for silly reasons). The result you want is this: if $\pi$ is as in the statement of 18.4.5, and C is nonempty, and C’ is irreducible, then $\pi$ is surjective. Hint for that: because finite morphisms are closed, the main issue is to make sure that the image of C can’t be dimension 0. If $\pi: X \rightarrow Y$ is a dominant morphism of irreducible varieties, then $\rm{dim} \; X \geq \rm{dim} \; Y$ . (The condition of working in the category of varieties can be weakened but not removed: witness the map of schemes corresponding to $k(x) \rightarrow k[x]$ . So that’s a hint of what is needed to show this: the transcendence interpretation of dimension for varieties.)
  
  I’ve added a couple of sentences to try to help the reader avoid this confusion. If I should say more, please let me know!
  
  Reply
ravivakil Says:

September 2, 2011 at 9:29 am
Regarding the Artin-Rees section (13.6, 13.7): Jakub Byszewski sent me a number of insightful comments on many parts of the notes, and I’ve dealt with them all except those dealing with completions. I’m recording them here so I can remember to deal with them later (and so others won’t be confused). These refer to the July 21 2011 notes.

13.6.1. “Any I-filtered module is an A_{\bullet}(I)-module” should probably be “\oplus M_n is a graded A_{\bullet}(I)-module”. Also, you probably want to define “M_{\bullet}(I):=\oplus M_n” rather than “M_{\bullet}(I):=\oplus I^n M”. Otherwise, the subsequent proposition makes no sense.

13.7.B. The problem with this exercise is that you have already defined the completion of J (since you’ve defined the completion of any module), as lim J/I^nJ. However it seems that in this exercise you would like to redefine it as lim J+I^n/I^n, or equivalently, the image of the completion of J in the completion of A. Now, if A is noetherian (which you don’t assume yet) this is the same by Artin-Rees Lemma, but not in general.

13.7.H. This is just not true. Take p=0, A – the local ring of a nodal curve. For this reason, the next exercise makes little sense. Perhaps you should state these exercises assuming A is regular.

[***Note to self: not dealt with. - R]

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Preston Wake Says:

September 6, 2011 at 2:39 pm
Hi Ravi,

In September of last year, I started a seminar at the University of Chicago for beginning graduate students to learn algebraic geometry, following your notes. Your excellent notes are the reason that many students are not afraid to volunteer to give talks, so we have you to thank for the success of our seminar.

We are now working on the proof of Serre Duality, and I thought this would be a good time to go over spectral sequences in detail. I’ve noted some comments I had while preparing my lectures. I hope they are helpful.

Best,
Preston

Notes:

23.3.D: [He meant 24.3.D. Yuncheng Lin noticed this too. -- R.]

The hint suggests that we use 24.3.3, but I think 24.3.3 is not true (if it is true, please give a more elaborate hint, as I cannot figure it out). Here is why I think it is not true:

Suppose 24.3.3 were true, then take A \to I^* and injective resolution, and do 24.3.3 (dual) to F(A) \to F(I^*) to get a injective complex K^{*,*} with exact rows and columns. Then look at the spectral sequence of E^{p,q}=G(K^{p,q}). The vertical E_2 will give you R^p(GF)(A) if q=0 and 0 else, as usual. But the rightward E_2 will give you R^pG(F(A)) if q=0 and 0 else – you will never get R^pG (R^qF(A)) as the result requires.

The statement in 24.3.3 can be weakened so that it’s easy to prove and is enough to prove the Grothendieck spectral sequence. For example, the existence of ‘totally injective resolution’ (although this is more complicated to state).

24.4.A:

This is false, I think. Proof: Let X= Spec\Z/2, Y= Spec\Z, and f:X \to Y be the natural map. Let \F be the sheaf \Z/2 on X. Then \F is injective because exact sequences of vector spaces split. However, f_*(\F) is the skyscraper sheaf of \Z/2 at the point P=(2) – this is not injective: let \G be the skyscraper sheaf of \Z/4 at the point P=(2); the usual injection f_*(\F) \to \G does not split.

The obvious categorical proof doesn’t work:
Let \F be injective, then the functor Hom(-,f_*(\F))=Hom(f^*(-),\F) is the composition of the right-exact functor f^* with the exact functor Hom(-,\F), therefore it is right-exact.

This is wrong because of some contravariance vs covariance trickery with the words ‘right-exact’ and ‘left-exact’. Try to show that Hom(-,f_*(\F)) takes injections to surjections. Let \G’ \to \G be an injection. Then f^*(\G’) \to f^*(\G) is not necessarily injective, so we can’t conclude that Hom(f^*(\G),\F) \to Hom(f^*(\G’),\F) is surjective.

In order for right adjoints to take injectives to injectives, the corresponding left adjoint has to be exact; thus flat pushforwards of injectives are injective (which is enough for 24.4.B), but not generally. What is true is that pushforwards of flasque modules are flasque (easily) and this is enough for 24.4.E (the Leray spectral sequence).

28.4.5:

The proof shows that H^0(X, \mathcal{E}xt^r(\O_X, \omega_\P)=Ext^r(\O_X, \omega_\P), but this is not the result we are trying to prove. At least, I don’t see how this implies the result.

I think that one needs to show an adjuction. To be more precise, let j: X \to \P be a closed immersion. Let F be the functor F:Mod_X \to Mod_X,
F(\G)=j^{-1} \mathcal{H}om_{\P}(j_* \O_X, \G)). [which, as you point out, has a natural module structure]
One wants to show that Hom_{\P} (j_*\F, \G) = Hom_{X}(\F, F(\G)). I don’t know if this is true, but if it is then the proof follows by 28.4.4 and some work.

Less specifically, I think that some confusion is caused in this section by leaving out notation. For example, when we write Ext^i_{\P}(\O_X,\omega), what we really mean is Ext^i_{\P}(j_*(\O_X),\omega), right? Usually the ‘pushfoward by a closed immersion’ is superfluous because there is only one thing we could mean by Ext^i_{\P}(\O_X,\omega) – it is analogous to using notation for the forgetful functor. However, in this case, I think the main difficultly of the argument is keeping track of what scheme each sheaf is on and where each Hom is taking place, so this notation is helpful.

[*** Note to self: this thread is not yet dealt with (but might be soon). --- R]

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1. ravivakil Says:
  
  September 7, 2011 at 2:38 pm
  Thanks — this is very helpful! I’m going to think about this as soon as I can, although I think I won’t manage to patch anything before the next posting (and I’m about to have a big delay in responding for family reasons).
  
  I’m impressed you managed to get to the proof of Serre duality in a single year — you must have been going at a ferocious pace. From your comments (and others at Chicago), you’re obviously deep in the guts of the subject.
  
  I’ll write again before too long (I hope)!
  
  Reply
  1. Preston Wake Says:
    
    September 7, 2011 at 6:33 pm
    Thank you for reading my comments. I think I figured out 28.4.5:
    
    Indeed, the functor F:Mod_{\P} \to Mod_X,
    
    F(\G)=j^{-1} \mathcal{H}om_{\P}(j_* \O_X, \G))
    
    is a right adjoint to j_*, when j is a closed immersion. Thus j_* is exact and F takes injectives to injectives. The proof then follows from the following 2 facts:
    
    - if I^{\bullet} is an injective resolution of \omega_{\P}, then the cohomology for J^{\bullet}=F(I^{\bullet}) vanishes for i<r. (this is prop 28.4.4)
    
    -if J^{\bullet} is an injective complex who's cohomology vanishes for i<r, then for any right-exact additive functor G, G(H^r(J^{\bullet}))=H^r(G(J^{\bullet}))
    
    Reply
2. Yifei Zhu Says:
  
  November 12, 2011 at 12:56 pm
  Hi Preston,
  
  I’ve just noticed that I had similar questions for 24.3.D and 24.4.A.
  
  For 24.3.3, I think it is true. The key property needed is that direct sums of projectives are projective. For example, P_{0,0} is in fact the direct sum of two projective modules, one for P_0 and one for A_0. But I got something similar as you did for 24.3.D. (I sent Professor Vakil an email about that a while ago which he decided to post at http://math216.wordpress.com/2011/05/14/fifteenth-post/#comment-1734.) In particular, by “totally injective resolution”, did you mean something like a Cartan-Eilenberg resolution?
  
  For 24.4.A, I was not able to prove it due to lack of left-exactness of f^*. But I did not try to come up with a counterexample. Your example and explanations of how things should go in general are very helpful. Thank you.
  
  Yifei
  
  [*** Still to deal with (see also Matt below). -R]
  
  Reply
3. Matthew Emerton Says:
  
  November 14, 2011 at 10:06 am
  Regarding 24.4A:
  
  What is true is that pi_* of injective abelian sheaves are again injective (just as abelian sheaves). The point is that in the setting of general abelian sheaves (i.e. if we don’t impose auxiliary sheaves of rings onto the situation) then pi_* is right adjoint to pi^{-1}, and the latter functor is exact.
  
  This is one reason for defining sheaf cohomology (even if the sheaves are sheaves of O_X modules for some sheaf of rings O_X) to be the cohomology of the underlying sheaves of abelian groups; i.e. to compute sheaf cohomology via injective resolutions in the category of sheaves of abelian groups. (This is done e.g. in Hartshorne, where he then immediately checks that one would get the same answer using injective resolutions by sheaves of O_X modules.)
  
  Reply
zlbb Says:

November 12, 2011 at 2:13 pm
Dear professor Vakil!
I think in 13.7.L the answer should be m instead of n-m (and there’s also trouble with parentheses).

Reply
1. ravivakil Says:
  
  December 21, 2011 at 10:42 am
  Thanks, fixed (in the next version to be posted)! If you were thinking about this final problem in 13.7, I’m hoping that you did many of the earlier exercises in that section — please let me know (e-mail is fine) if they were fine, or if there were problematic ones!
  
  Reply