The thirteenth post is the March 31 version in the usual place. It covers up to 21.4 (although more is included). The next post should appear in three weeks, on April 23.

We have arrived in the promised land. We have now learned so much that we can reap the benefits and think cleanly about lots of things.

**For learners**

Read 20.4, 20.5, 20.7, the unstarred part of 20.8, and 21.1-21.4.

20.4 is on Riemann-Roch, degrees of coherent sheaves, arithmetic genus, and Serre duality. You will discover that Riemann-Roch is not hard (given what we know). We will later see that Serre duality *is* hard.

20.5 is on Hilbert polynomials and functions, and the notion of genus.

Skip 20.6 on Serre’s cohomological characterization of ampleness (which we mostly won’t use) unless you feel like reading it.

20.7 is on higher direct image sheaves. Like much of what we’re doing, you may find it surprisingly easy.

20.8 is on the fact that proper pushforward of coherent sheaves is coherent (in reasonable situations): Grothendieck’s coherence theorem. Read everything up until the double-starred proof of Chow’s lemma. Spectral sequences make an appearance here at the very end (of the proof of Grothendieck’s coherence theorem), and I wonder if this entire section should be starred. (At this point, I’m intending to star it in the next version.)

I hope someone tries to read the proof of Chow’s Lemma, and lets me know what can be improved! I tried to isolate the hard parts (which in some expositions can be somewhat elided, e.g. Hartshorne Exercise II.4.10).

Chapter 21 is on curves, and is a possible punchline for a year-long course. Or it can be done earlier, taking things as black boxes.

21.1 is a criterion for a morphism to be a closed immersion. It has the hardest argument (for me) in the chapter, so don’t be deterred.

21.2 collects what we know into some useful tools we will repeatedly use.

In 21.3, we apply them to curves of genus 0.

In 21.4, we apply them to hyperelliptic curves.

(If you read ahead, you’ll read about curves of genus up to 5, except for genus 1. The genus 1 — elliptic curve — case will be in the next posting, or so I hope.)

If you are interested in curves over fields that are not necessarily algebraically closed, I’d like to know if the extension to that case is readily comprehensible, or if there are steps that are harder than they should be. In particular, if you have trouble with 21.2.D, I would like to know, because the fault will be mine, not yours.

Recommended exercises: 20.4.A, 20.4.B (Riemann-Roch!), 20.4.C, 20.4.E, 20.4.F, 20.4.H, 20.5.C, 20.5.M (Bezout!), 20.7.B, 20.7.D, 20.7.E, 21.2.A, 21.2.B.

Exercises involving explicit examples:

20.5.A, 20.5.F, 20.5.S, 20.5.T, 20.5.U, 21.1.A, 21.1.B, 21.4.A.

**For experts**

The pushforward of coherent sheaves under proper morphisms of locally Noetherian schemes are coherent. I’m idly curious: is there a counterexample in the wacky non-Noetherian world without Noetherian hypotheses?

In Exercise 21.7.B, I want to mention the fact that there is no number N such that every genus 1 curve over a field k has a point of degree at most N over k. Is this actually known? Does anyone know a reference? On a related note, is there an easy example of a genus 1 curve with no degree 1 points (easy = they can prove it themselves)? Or failing that, is there a reasonably canonical reference? (I should presumably browse through Silverman.) [Update May 4, 2011: both are now answered, in these comments and the 14th notes comments.]

**For everyone**

I may have the definition of Hilbert function wrong, and at some point I will check. Currently for any coherent sheaf F, I define it as h^0(F(n)). It’s possible it should be defined only for the structure sheaf of a projective variety X, and then it shouldn’t be h^0(O_X(n)), but the dimension in the image of H^0(O_X(n)) of H^0(O_{P^N}(n)). I will check. (Or if someone can immediately point me to some reference, that would be great…)

I have some notes on defining the cup product in Cech cohomology. This is a bilinear map H^i(F) x H^j(G) –> H^{i+j}(F \otimes G) that is associative, and “symmetric” (up to a sign). This is something that I’d not otherwise learned. Does anyone feel this is worth including? Is the cup product in cohomology something that people use often enough to make this worthwhile? The fact that I haven’t included it in this draft shows that my default answer is “no”: I’m trying to find excuses not to include things.

April 8, 2011 at 1:58 am

Hi, Ravi,

personally – my perspective is that of a master’s student at a German university – I would like to see Kunneth included. I had some need for it in a small way in my master’s thesis and found Kempf’s treatment in his “Algebraic Varieties” (Prop. 9.2.4) too condensed and EGA too general and scary.

April 17, 2011 at 11:12 am

Thanks! I’ve added it to my “possibly add” list, which means it will happen a bit later (as I don’t have exposition ready to go). Thanks also for possible references. My ideal way of getting this across would be in an exercise (or series of exercises) in which the strategy is laid out clearly, but where readers can be enlightened by thinking through the right details. (Others should write in too in support, if they would strongly like Kunneth.)

April 8, 2011 at 2:06 am

By the way, I think your notes are fantastic. I took courses in scheme theory some time ago and, being of a more geometric and less algebraic bent, missed the motivation and geometry behind many concepts. Unsatisfied, I subsequently focused on learning smooth manifolds and complex geometry, but reading your first chapters has made me want to (re)learn schemes properly.

April 17, 2011 at 11:13 am

Thanks very much for the kind words! They give me energy to do more writing and editing.

April 8, 2011 at 8:36 am

Actually for no good reason but curiosity I’d also be interested in seeing the cup-product in Cech cohomology.

April 17, 2011 at 11:14 am

I’ve largely thought through that exposition, so that may happen. (Or I may write something short that I’ll link to…)

April 22, 2011 at 12:12 pm

I wasn’t able to get the right answer for exercise 20.5.A. I assumed the sheaf in question to be the pushforward of the structure sheaf of m distinct closed points, which would be the direct sum of skyscraper sheaves. Twisting skyscraper sheaves does not change the sheaf. Hence the Hilbert polynomial should be constant (not eventually constant) and independent of the position of the points.

April 23, 2011 at 9:54 am

Thanks Daniel! You’re right, and that’s because I have the wrong definition of the Hilbert polynomial, and I haven’t yet figured out what is correct, and what works best for learning. What may be right is the following. The Hilbert polynomial is defined only for a closed subscheme X of projective space P^m, and it is defined as the image of h^0(P^m, O(n)) in h^0(X, O(n)). It is indeed eventually polynomial, for reasons you will see. With the correct definition of Hilbert polynomial, the exercise is right. What I will likely have to do is the following: deal both with the Hilbert polynomial, and with h^0(X, F(n)), both of which are eventually polynomial. The latter I guess doesn’t have a “name”.

Warning to other readers: this is not yet fixed!

And question for experts: is this revision “right”? I’m hoping someone will confirm it, and that’s why I’m not making this change immediately.

April 23, 2011 at 8:54 am

You ask for an easy example of a genus 1 curve with no rational points. Consider the plane curve Ax^2+By^4+C=0. For various A,B,C, the smooth completion of this curve has no points over some completion of Q. eg, A=2, B=3, C=-1 has no 3-adic points or A,B,C>0 has no real points.

April 23, 2011 at 8:57 am

Nice! I’d come up with a similar but different example (while I think sitting in a number theory seminar that you were at too): x^3 + 2 y^3 + 4 z^3 = 0. I like your “no real points” one the best, because it is essentially instantaneous (with the briefest of thoughts about infinity).

April 25, 2011 at 5:11 pm

A few typos:

Exercise 20.6.C, there is a bogus closing parenthesis ) on the second line.

Exercise 20.6.D, the second sentence has to many “if”s. [Minor note: the sentence "Perhaps first reduce to the case where X is reduced," while perfectly clear and grammatical, is a little bit peculiar on account of the two uses of the word "reduce" with unrelated meanings.]

A minor point: given how many of the other exercises use exercise 20.6.A, it seems odd to me that this exercise is not upgraded to at least a corollary (with its proof, perhaps, as an exercise).

A final remark: Another interesting fact that can be proven from Exercise 20.6.A is that any normal affine variety can be embedded in affine space in such a way that its projective closure is also normal. (The basic idea is to take any embedding in affine space, take the projective closure, pull back the hyperplane divisor to the normalization and note that it is still ample. I learned this proof from Donu Arapura via mathoverflow.net/questions/39172 )

April 26, 2011 at 7:21 am

Thanks Charles! I’ve fixed 20.6.C and D. About the minor note: that is indeed unfortunate (and I’m sure comes up repeatedly), and I can’t think of a better wording. Somehow “show it suffices to consider the case where X is reduced” isn’t so great. 20.6.A is now “Important Exercise (used repeatedly)”. If it isn’t sufficient straightforward that essentially every reader (following well) will be able to solve it, then I might upgrade it further, giving a proof (or a giveaway hint).

I’ve kept your note about 20.6.A in my “notes to myself” embedded in the tex file, so I can tell more people about it.

As usual, any further advice is appreciated. (Also, I certainly have earlier comments of yours still to respond to. As of a few days ago, I’ve “tagged” all comments that I still want to respond to.)

April 26, 2011 at 9:59 am

Maybe, “Perhaps first reduce to the case where $X = X^{red}$”?

April 28, 2011 at 4:18 pm

Thanks, good idea — done!

April 26, 2011 at 9:54 am

Additional typos (I think):

(minor) In the next to last paragraph before 20.6.F, in the next to last line, there is a space missing after the third word.

In the last paragraph before 20.6.F, L^N should be L^n, and F^n should be F \otimes L^n. (I’m omitting some \otimes signs for the sake of readability sans tex.)

April 28, 2011 at 4:18 pm

Thanks for catching those — now fixed!

April 29, 2011 at 8:49 am

One additional note, that you may or may not want to do anything about: I’ve recently seen a reasonable proof of 20.6.A that does not seem to require any cohomology or other remotely high-brow stuff. See http://mathoverflow.net/questions/63053#63053. [Edited by Ravi so that the link is clickable.]

April 30, 2011 at 1:28 pm

Very nice! I want to point out something before discussing this argument: a key thing is that you asked the right question. Some might argue that it doesn’t matter; if you know something is true by any means, then that’s enough and you should move on. I think it can help if you pay attention to how hard it was, and if you have to work harder than necessary, there is insight in finding a better way of thinking about the problem. So nicely done in asking the right question.

About the argument: I think it gives a stronger result (and thus is better than the cohomological argument even to those who think purely instrumentally). I think it shows the following facts (and please correct me if I’m wrong).

(a) for any finite morphism of schemes *at all*, the pullback of an ample line bundle is ample. Reason: Saying a line bundle is ample implies that the underlying space is quasicompact and quasiseparated (the first by definition, see 16.3.14, and the second by Thm 16.3.15). Noetherian hypotheses are not used: the pushforward of a finite type quasicoherent sheaf (which is what comes up in the definition of ampleness, not coherence) is finite type quasicoherent cheaply, and that is what is used.

(b) The same thing makes sense relatively: if X –> Y is a finite morphism of S-schemes, and L is an S-ample line bundle on X, then the pullback of L to X is also S-ample. Proof: it suffices to check this over Spec A. Then apply the argument of (a) without change.

You(=Charles) may not care (and in fact I(=Ravi) may never in my life use this (although I might!)), but I think your question led to the right statement.

(Anyone) please let me know if I’m doing something foolish. I intend to add this before long otherwise.

(And a shout-out to K. Halb for that nice argument.)

April 30, 2011 at 1:33 pm

Further follow-up: I put a link to here from mathoverflow, and when I did that, I noticed that Torsten Wedhorn said that EGA II.5.1.12 gives a noncohomological proof of (a) above where “finite” is replaced by “quasiaffine”. I haven’t looked it up. The argument above doesn’t generalize (the pushforward of finite type isn’t necessarily finite type). But the version you asked in your MO question is the one that gets used lots, so I see no need to give a more general version at the cost of losing transparency.

July 25, 2011 at 6:27 pm

I’ve now added this argument, as described above.

June 4, 2011 at 8:56 am

I just realized that I don’t actually understand example 2 from 20.5, in which you calculate the hilbert polynomial of a degree-d hypersuface H in projective space. I think I am missing some statement like “If H is a hypersurface and i: H -> P^n is the closed immersion into projective space then i_*(O_H(m)) = (i_*O_H)(m)”. But possibly I am not understanding the argument.

Some elucidation: The proof works by taking the closed subscheme exact sequence

0 -> O_P(-d) -> O_P -> O_H -> 0

and then twisting with O_P(m) and applying additivity of the Euler characterisitic. However, the last entry of the short exact sequence really ought to be i_*O_H where i: H -> P^n is the inclusion.

And it is not clear to me why the Euler characteristic of O_H(m) is the same as that of i_*O_H(m). Pushforwards preserve Euler characteristic so it is clear that chi(O_H) = chi(i_*O_H). But it is not obvious that (i_*O_H)(m) IS the pushforward of (O_H(m)), so it is not obvious that they have the same Euler characteristic.

June 9, 2011 at 11:15 am

That’s a very helpful question. The reason this is true is because of the projection formula: if is any quasicompact quasiseparated morphism (in this case, just the closed immersion ), and is any quasicoherent sheaf on , and is any

locally freesheaf on , then we have a natural isomorphism . (If is only quasicoherent, then we still have a map from left to right.) This is an important fact, called theprojection formula.And because of your question, I realized that I somehow didn’t state the projection formula earlier! (It is a special case of 20.7.E, but that’s no help, as 20.7.E is after the part of the notes you asked about.) I’ve now added it as Exercise 17.3.H. (If you want to do it now, you can do it as follows. First show that in the case of general , you get the map as described above, by the FernbaHnHoF (FHHF if you prefer) theorem. Then show that if is free, the map is an isomorphism. Then deal with the locally free case by working locally on the base.)

I’ve also answered your question in the text as well, referring back to the (newly added) projection formula.

Please let me know if I should say more (either here, or in the notes)!

June 4, 2011 at 9:03 am

Oops, meant to say that “pushforwards along affine morphisms preserve Euler characteristic”, not “pushforwards preserve Euler caharacteristic”. Don’t want to muddy the waters.

June 9, 2011 at 2:08 pm

Dear Ravi,

Fernbahnhof is spelled without a p. Its official abbreviation seems to be Fbf by the way (according to Wikipedia http://en.wikipedia.org/wiki/Railway_station_types_in_Germany )

June 10, 2011 at 8:24 am

Thanks, I’ve fixed the spelling. (

Christian Liedtkewill confirm that my German could use some improvement.)I’m disappointed that Fbf is the official abbreviation. Perhaps I could email Angela Merkel and see if that could be changed. (For better or worse, there is no German analogue of the Academie Francaise.) Or I could appeal to a higher authority, and try to change it on wikipedia.

Or I could try to find another mnemonic to remember that it is FHHF and not HFFH.

There are a few optionsout there, but they aren’t great. (I agree with Allen Knutson’s philosophy — which admittedly I’ve never heard him say explicitly — that naming things well is very important.)Added July 26, 2011: I didn’t intend a train theme on this site, but I’ve added a couple of FHHF pictures to the

category theory post.September 19, 2011 at 7:04 am

[...] I remain confused on the right definition of Hilbert function. It is one of two things. (i) If M is a finitely generated module over a graded ring, it can be the dimension of the various graded pieces. Special case: the graded ring of a projective variety. (ii) If X is a projective variety (with embedding into projective space), it can be the dimension of the restriction of degree n polynomials (in the projective coordinates) to X. I want to use (i), but fear that (ii) may be right, at least for some people. (The definitions disagree: consider 3 distinct points on a line, where the value of the first polynomial is 3, and the value of the second polynomial is 2.) Can anyone give an opinion, informed or otherwise? (This was discussed earlier here.) [...]

April 26, 2012 at 4:32 am

Dear Ravi,

21.2.5.1) seems incorrect to me. Shouldn’t it be

h^0(C,L) = deg L – g +1, instead of , as it is now, h^0(C,L)=deg L – g – 1?

April 26, 2012 at 10:26 am

Dear Magnus,

You’re absolutely right — thanks for catching that! It is now fixed (in the next version to be posted). If you catch anything else like that, please let me know!