**The fourth post is the October 9 version here. This fortnight’s reading is 5.3-6.4.** At the end of it, you will begin to have some experience with schemes. If you are comfortable with it, you may wish to read ahead, so I’ve included 6.5 (on associated points — as suggested in the first paragraph of 6.5, just consider the case where A is reduced), and 7.1 (an introduction to morphisms).

As I feared, I am falling behind in responding to comments. I’ve chosen to post notes and move forward rather than wait to respond to earlier comments, but I promise to respond to comments before too long. (The fall of 2010 will be the worst time time of the academic year for me. Last week was one of the worst weeks of the fall. When the season of recommendation-writing ends, I expect to be better.)

**For learners.**

I’m not happy with my description of projective schemes (5.5); it seems much harder than it needs to be. If you have any suggestions (or even references where the development seems easier or more natural — in the same generality of course, where the ring is not necessarily generated in degree 1), please let me know! The key idea is that one wants to understand this in terms of covers of affines. I hope this is clear.

If I had to pick the eleven most important exercises, they would be 5.3.A, 5.3.B, 5.3.F, 5.4.A, 5.4.B, 5.4.H, 5.5.B, 6.1.E, 6.2.E, 6.4.F, and 6.4.H. But these are necessarily notably harder than many of the other problems. The best exercises to get your hands dirty with explicit examples are 5.4.B, 5.4.E, 5.4.F, 6.2.A, and 6.4. H.

I’d also like you to ask you for some advice about four exercises.

Exercise 5.5.B is important and tricky. Is it too hard? Is the hint sufficient?

Exercise 5.4.A is essential. Are the hints sufficient for you to figure it out (after a lot of thought)? This is often done explicitly (I think for example in Eisenbud and Harris), but I find that it is something that you have to do yourself to understand. If this exercise is far too hard, I should merge it into the text. If it is only a little too hard, I shouldn’t.

Is Exercise 6.4.K do-able (notably the non-UFD-ness)? I haven’t thought this part through.

**For experts.**

See my plea about the discussion of projective schemes above.

I call a ring graded by the nonnegative integers, with S_0 = A and the irrelevant ideal finitely generated over A, *a finitely generated graded ring over A*. That is slightly unwieldy, but I can’t think of anything better.

The important and tricky exercise 5.5.B is to identify a bijection between the primes of ((S_.)_f)_0 and the homogeneous prime ideals of (S_.)_f. This seems quite tricky if S_. is not generated in degree 1. (Hartshorne says that “the properties of localization” show that this is a bijection, but I don’t understand this.) If there is a better way to proceed than my hint, I’d be interested in hearing it.

Very unimportant point: Exercise 6.2.I says that a locally Noetherian scheme is integral if and only if it is connected, and all stalks (of the structure sheaf) are integral domains. Is there a counterexample if the Noetherian hypothesis is removed? (Update July 25 2011: Here is a link to an answer.) Also, I was very surprised to hear that “locally Noetherian” is not a stalk-local condition — Joe Rabinoff gave me a short neat example of a scheme that is not locally Noetherian, yet has Noetherian stalks.

I’d like to give a name (other than “Nike’s Lemma”) to the proposition that the intersection of 2 affine opens can be covered by opens that are simultaneously distinguished in both of the big affine opens (Prop. 6.3.1).

Unimportant question (cf. Ex. 6.4.G): is there a reference that if R is integrally closed, so is R[x]?

What’s the easiest example to show that factoriality is not affine-local? (Of course, I don’t just want an example stated. That’s not hard, and one is given in 6.4.K. I want an example that is easy to prove with bare hands.)

**For everyone.**

I’ve gradually come around to the idea that when learning about some category for the first time, the notion of *isomorphism* is *pedagogically* prior to the notion of morphism. I won’t argue about whether it is logically prior; that’s not my point. When learning groups, students first propose the notion of isomorphism (as they figure out what they mean by the intuition of two groups being the same) before the notion of how you map from one group to another. With schemes too isomorphisms come first.

In mathematics notation, we have a symbol for “is isomorphic to” (\cong). But this is the wrong notion in general: we need a symbol for “a map that is an isomorphism”. We already have a reasonable answer: a right arrow with a \sim on top. But might it be nicer to make it look slightly different, and have a symbol that looks like a \cong, where the bottom is a rightarrow? I like this idea because although it is (slightly) new notation, it is patently clear what it means, and also useful.

When dealing with projective space P^n, with “projective coordinates” x_0, …, x_n, we often use the n+1 “standard” affine patches U_0, …, U_n, and it is convenient giving names to the coordinates on the affine patches that helps avoid confusion. I’ve used “x_{i/j}” on patch U_j (with convention/definition x_{j/j}=1).

October 10, 2010 at 6:58 pm

Concerning 6.2.I, you ask for an example of a connected (non-empty!) scheme whose local rings are domains but which is not integral. I don’t know what example Rabinoff mentioned to you, but the usual suspect $R = \prod_{n \ge 1} \mathbf{F}_2$ once again leads to an example. Namely, since the reduced $X = {\rm{Spec}}(R)$ is quasi-compact and infinite, it has some connected component that is not a point. That is closed, so it has the form ${\rm{Spec}}(A)$ where all elements of $A$ are idempotent and hence all stalks of $A$ are $\mathbf{F}_2$. But it sure isn’t a domain, since $A \ne \mathbf{F}_2$. So there’s an “abstract” example.

October 18, 2010 at 5:05 am

Fantastic! It is currently embedded as a “to add” comment, so it will be done at some point. This is one of the small list of “universal examples” as it serves as an example of so many things. Do you have a comprehensive list (or semi-comprehensive list) of things you pull out this scheme for?

Also, Joe Rabinoff’s example was of a non-locally-Noetherian scheme that still has Noetherian stalks, not of this.

October 18, 2010 at 11:29 am

Additional comment from Brian (to add later): “note that since all of its elements are idempotent,

its local rings are all F_2. So it is a non-noetherian ring whose stalks

are all noetherian.”

October 19, 2010 at 3:33 am

Whoops, the above is wrong. It’s true that there must be a connected component that is not an *open* point (as all connected components are closed, and a quasi-compact infinite space cannot be a disjoint union of clopen points), but this doesn’t rule out that they’re all (closed) points. In fact, the p-adic integers form a compact space whose connected components are all points. And likewise for the above alleged example: any connected component has the form Spec(A) for a quotient ring A of $\prod_{n \ge 1} \mathbf{F}_2$, and all elements of A are idempotent. Thus, by connectedness $A = \mathbf{F}_2$, so Spec(A) is a 1-point space.

[ 😦 — R.]October 10, 2010 at 7:24 pm

Concerning the definitions of “projective” and “quasi-projective”, your definition of projectivity over a ring is not consistent with the definition in EGA; see EGA II, 5.5.1 (and 5.5.2) with $Y = {\rm{Spec}}(A)$ there. For instance, EGA requires generation in degree 1! Likewise, your definition of “quasi-projective” is not consistent with the definition in EGA (see EGA II, 5.3.1 and especially 5.3.2); the error is that you do not require the scheme to be quasi-compact (equivalently, finite type) over the base. Maybe some of the content of EGA II, 2.9.2(ii) should also be taken into account, but that may not be compatible with the development of the material in the notes up to this point.

I now wonder: how do the notes define “quasi-affine”, and do they address different ways of characterizing “quasi-affine of finite type”? (See EGA II, 5.1, especially 5.1.2 and 5.1.9.)

The “very rarely” should be removed from the text discussion about finite generation conditions on graded algebras when forming Proj, since even in the presence of noetherian conditions it is perfectly convenient to be able to take Proj of graded algebras such as $\oplus_{n \ge 0} \Gamma(X, L^{\otimes n})$ prior to knowing their finite generation properties.

October 11, 2010 at 5:29 am

I think you can defend Ravi’s definition of what it means to be a projective scheme over a ring as follows.

In EGA a projective morphism X —> Y is one which factors as X —> P(E) —> Y where the first morphism is a closed immersion, and E is a quasi-coherent, finite type O_Y-module. In EGA a quasi-projective morphism is a morphism of finite type such that there exists a relatively ample invertible sheaf on the source.

But since in Ravi’s notes the definition of a projective A-scheme comes before the notion of a morphism of schemes, there is no way he can use those definitions. In fact, the definition of “scheme over A” comes on page 130. I think it would be very useful to point the reader to the definition of scheme over A

beforedefining what is a projective scheme over A.If S is a graded A-algebra then Proj(S) is quasi-compact if and only if S_+ is contained in the radical of an ideal generated by finitely many elements. If S is finitely generated as an S_0-algebra then this always holds. Moreover, if S is finitely generated as an S_0-algebra and S_0 is finite over A, then Proj(S) —> Spec(A) is projective.

Perhaps a good way to clarify all of this in Ravi’s notes, is to define weighted projective spaces before defining projective A-schemes, and to show that they can always be embedded into projective space. From Ravi’s definition of a projective scheme over A as Proj(S) where S is a quotient of a graded polynomial ring over A, it is morally clear that every such scheme is a closed subscheme of a weighted projective space, thereby removing any doubt as to what it means to be projective over A. Hmm?

Of course, my inclination is to define morphisms of schemes before doing any of this at all, so maybe another solution is to move projective schemes later? After all a key point in all of this is to study the Serre twists of the structure sheaf, i.e., one has to know about invertible sheaves.

All of this is in the stacks project, but it has to be cleaned up and improved. For example, to piece together the things I just said above about Proj of graded rings requires quite a bit of looking around. Partly the problem seems to be that if you try to be very precise and match the construction of the relative Spec with the construction of the relative Proj, then there are many many details to check.

The exposition of Proj of a graded ring in Hartshorne is a miraculous feat of exposition. Every time I read it I am completely convinced, but whenever I am not actually reading it I say to myself: “How could he possibly explain all of this in < 2 pages?"

October 12, 2010 at 2:33 pm

Johan, I agree that Ravi’s order of presentation forces him to have to break with the definitions in EGA. On the other hand, when it comes to the concept of quasi-affine then one can wonder if he will also be inconsistent, and I wanted to raise that issue for him to contemplate. Also, I think it is worth at least warning the reader in the notes when there is given a definition of a basic concept in a manner which is not consistent with EGA. Otherwise one creates the same confusion that has arisen among many students from Hartshorne’s definition of coherent sheaf. (Hartshorne notes this in his textbook for “projective” and “very ample”, but should not have buried the remark where he did and should have added a few more sentences to clarify the distinction.)

October 12, 2010 at 3:33 pm

Yes, I agree with what you say.

October 20, 2010 at 9:50 am

I don’t understand why the definition of “projective over a ring” in these notes is not consistent with that in EGA. It seems to me that in this case, Vakil’s definition differs from EGA II,5.5.1(b) only in that it does not require generation in degree one. This condition is unnecessary in this case, since if $S$ is finitely generated graded over $A$, then we can get an isomorphic scheme that is generated in degree one by replacing $S$ by the graded ring whose $d$th component is $S_{nd}$, for suitable (fixed) $n$.

October 21, 2010 at 2:38 pm

Charles, after some theory is developed one can relate the two definitions under some noetherian hypotheses, but a-priori the two definitions have a very different “flavor”. They also suggest possibly inconsistent ways to generalize to the case of a non-affine base. So perhaps I should have used a less loaded word than “inconsistent”.

March 17, 2011 at 9:47 am

Hi everyone,

Here is a much-delayed response.

I think in the current (March) version, all of the issues have been sorted out except for the following two issues I have not sorted out in my own mind (and I need to). When I define Proj, do I want to define it only for graded rings (and in the analogous way for sheaves) that are (i) finitely generated, and (ii) generated in degree 1? I need to make a decision, and be consistent. The current version may now be consistent, but Brian is now convincing me to relax both (i) and (ii). The annoying consequence is that often one wants to just deal with Proj of things are nice in one or both of these ways. (For example, we will definitely want projective schemes to be finite type!) I’m willing to put up with that annoyance, but it will just take me some time to go carefully through the notes.

Small comments: “very rarely” is now fixed. And Johan, I agree with you that I’d prefer to defer the definition of projective scheme to after morphisms. But in teaching the class, students were reasonably surprised that such a classical notion had to wait so long for a definition. So this section can be be happily deferred until later, but is there so people can read it right away if they want to. (I remain surprised that projective geometry of this sort is so much harder, at least for me, than the seemingly forbidding machinery built up in the 1960’s.)

If any issues remain that I forgot to address, don’t hesitate to remind me!

October 10, 2010 at 7:32 pm

Finally, concerning Exercise 5.5B, even in EGA the proof (see II, 2.3.6 in surjectivity aspect) is not so simple. They appeal to a slightly exotic side lemma (II, 2.1.9) that axiomatizes when certain “graded constructions” yield a prime ideal.

[Thanks! Note to myself now embedded in text. — R.]October 10, 2010 at 9:00 pm

OK, one more thing. Concerning Exercise 6.4K, any Dedekind domain which is not a UFD provides an example, so $\mathbf{Z}[\sqrt{-5}]$ is a pretty simple example (unit group is $\pm 1$, so the usual identity $2 \cdot 3 = (1 + \sqrt{-5})(1 – \sqrt{-5})$ shows it is not a UFD). Here too, the class group has order 2, but that’s not relevant to the example.

October 12, 2010 at 2:11 pm

(Thanks Brian!)

BCnrd and I had a brief follow-up discussion on email about this. The proof of factoriality can reasonably be punted forward to when we talk about DVRs in 13.3. One thing I find interesting about this example (which I intend to add

[update Oct. 11, 2011: added at some point in the past]) and the example I currently have in the notes (which I will likely subtract) is that in one case the non-factoriality of global sections is easy and the factoriality of the scheme is hard and in the other case the reverse is true. (Here “hard” = “can be done by hand by a novice with no machinery and just hints”.)Another possible example: take our old warhorse the cone over the quadric surface k[w,x,y,z]/(wz-xy), remove the cone point (to get something which is factorial because it can be covered by open subsets of A^3’s), and show that the global sections are not a UFD (using wz=xy). This could be slick; but is there a quick fast *proof* that x,y,w,z are irreducible?

October 18, 2010 at 5:41 am

I should have been clearer in the above: this question is for everyone, not just Brian: is there a short proof that k[w,x,y,z]/(wz-xy) is not a UFD? (This may have a short fast answer; I’ve not yet thought about it, but I’m hoping someone will settle it quickly.)

October 20, 2010 at 11:29 am

Showing that this is not a UFD is equivalent to showing that some height-one prime is not principal. A sketch of this is given in the “for the experts among the experts” note after Example 5.4.10.

October 20, 2010 at 11:50 am

Is this easy enough – the ring is graded, and we know the degree 0 and 1 pieces?

March 17, 2011 at 9:48 am

Tom, you’re right. Now stated clearly in the notes.

October 28, 2011 at 3:27 pm

Much later update: I realize now what I wanted in that example: I wanted a *local ring* that was normal but not a UFD. This would give (provably) a normal non-factorial scheme. The global argument currently there (with Tom G’s proof) wouldn’t give this to you. I can’t see how to make Tom G’s proof work for the local ring. Charles’ argument (showing a height 1 prime that is not principal) does it, so I can at least refer forward. But it would be great if there were a direct argument. (If anyone has one, please let me know…)

October 29, 2011 at 5:20 am

Since we have wx=yz in the local ring, it is enough to show that (1) w, x, y and z are primes and (2) No two of them have unit ratio.

Let m be the maximal ideal.

Proof of (1): w, x,y and z are in m, so they are not units. If w=w_1 w_2, with w_i each not units, then w would be in m^2, and it is not. Similarly for x, y and z.

Proof of (2): Suppose for contradiction that u were a unit with wu=x. Reducing modulo m, we get that the images of w and x in m/m^2 are proportional. This is not true, so we have a contradiction. Similarly, none of the other pairwise ratios can be units.

October 28, 2011 at 3:47 pm

It’s true that you can cover by two distinguished open sets that are Spec’s of UFD’s, i.e. there are with and and are both UFD’s. What are examples? 2 and 3? (Any examples with a short proof?)

October 29, 2011 at 5:26 am

2 and 3 work. If R is a noetherian ring, and S a multiplicative set, then every ideal of S^{-1} R is of the form s^{-1} I for I an ideal of R and s in S. Therefore, the map from the class group of R to that of S^{-1} R is surjective.

The class group of has one nontrivial element: . If is invertible, then this is obviously principal. If is invertible, then , so this ideal is again principal.

I don’t see any proof that doesn’t rely on knowing the class group of $\mathbb{Z}[\sqrt{-5}]$ in advance.

October 11, 2010 at 8:17 am

In exercise 5.5.B

1) There is a sentence, “Showing that homogeneous a is in […] a^2 in P;”, which does not make sense. Do you mean “Show” instead of “Showing”?

2) Why do you have to show that P is an ideal, if it is already defined to be the ideal generated by some elements?

January 10, 2011 at 4:02 pm

I thought about this again, and still do not get your hint. However it seems that the Proof of Lemma 7.49.2 of the Stacks project can be used here.

March 17, 2011 at 10:05 am

Hi D.H,

You’ve long forgotten this question, but I’m only now beginning to respond. (I’ve fixed the first error just now.) Can you tell me the current number of the lemma in the stacks project you mean? (It’s easiest if you just give me the tag, as that will stay fixed forever, but I can get the tag quickly.)

March 18, 2011 at 2:20 am

It is still 7.49.2 and the tag seems to be 00JO. I think there might be an error in the statement of the lemma itself, because I see no localization there, but in the proof they divide by f.

[Hereis a link to the tag in the stacks project. — R.]October 11, 2011 at 11:05 am

This is another long-delayed response, so very likely you (=D.H.) have long stopped thinking about this, but if anyone else is stuck (or if I can say more, or if this response is helpful, or if this response is not helpful) please let me know.

I’ve just worked through the hint myself, and it still seems to work. Can you say at which sentence you got stuck? About your second question: P is only described as a *subset*, not an ideal; as it is initially defined, it is not even clear that it is an abelian subgroup! I’ve been slightly more explicit: the second last stage of the hint is “show that P is a homogeneous prime ideal of “.

October 11, 2010 at 8:36 am

Dear professor Vakil,

I found out some typos in solving more exercises. Some of them are not fixed yet + some points I do not understand + many remaining typos this time:

2.6.1. Definition : maybe too many unnecessary space?

3.2.F. Exercise, 3.2.G. Exercise : reference to Example 3.2.D(b) should be Exercise 3.2.D(b)

3.3.I. Exercise : unnecessary weird space in short exact sequence.

3.4.E. Exercise : I don’t quite understand what you mean in this exercise. Is isomorphism of sheaves a categorical one(existence of inverse morphism)?

3.4.K. Exercise : I don’t understand the last remark – sheafification is left-adjoint but left-adjoint commutes with colimit and kernel is limit .. so what?? maybe I am missing some point ..

3.5.3. Exercise : the section functor -> the global section functor &

in the second exact sequence, U -> X

3.6.G. (need a word ‘Exercise’) : line 3, ringed spaces (missing period) &

line -2, FG -> F,G

3.7.1. Theorem : in the definition of F(U), there is some unnecessary space.

4.2.D. Exercise : line 4, k(y)[x] -> k(x)[y] &

line 6, show that you can -> you can show that

4.2.6. : how prime behave (?)

4.2.K. Exercise : it is somewhat confusing, I think interchaging x and y for compatibility of notation with 4.2.L. might be better and I don’t understand the picturing procedure. How can I understand it well?

right before 4.4 : (x) in Speck[x] -> speck[x]_(x)

right before 4.5.A. : the (Zariski) topology (missing period or colon?)

5.1.3. Remark : (f_1, …, f_r) = R or A? (Is there any reason for this non-consistency in one paragraph?)

5.2. line 1 : set of schemes,adding (no space after ,)

5.4.2 last line : there is ) with no (

5.4.A. Exercise : the cocycle condition is wrong. U -> X

5.4.7. Proposition line -3, in in -> in

I hope what I found would be helpful.

Thank you.

October 18, 2010 at 5:31 am

Thanks Phil-Sang! I’ve made all the changes with the possible exceptions of the below.

2.6.1 In general, I’m not worrying about spacing issues right now (as they can be fixed later, and may be specific to some document class or some such that may change), but sometimes they are easy to change, as in this case and 3.3.I and 3.7.1, so thanks.

3.4.E: That’s right, isomorphism means what the word means in all categories. I’ve not made any change here.

3.4.K I’ve edited it a little bit, but it’s phrased in a way that the reader shouldn’t necessarily expect to understand the comment (and I don’t want to distract most readers with this). But here’s a clue for you (also now in the text): the fact that the presheaf kernel (of a map of sheaves) is also a kernel of a map of sheaves can now be seen purely in this language.

3.5.3 Edited in a different way: I *want* the functor of sections over any U, not just over all of X (the latter is the “global section functor”).

3.6.G Thanks! I suspect you are the first to tackle this exercise, and it can certainly be useful.

4.2.K: I need to improve this and 4.2.L (see other comments); I owe you a response on this one.

5.1.3: The only reason for continual inconsistency between R and A is that I usually use R for rings, but in these notes I’m trying to follow the convention of using A.

October 2, 2011 at 3:28 pm

Long-delayed response to 4.2.K: I’ve now changed the wording. For various reasons I resisted switching the x and y, but I’m now convinced (by you, Tom Church, and others). There should be a new picture there (when I have access to xfig soon), which will be in the version next appearing.

October 11, 2010 at 11:48 am

In 5.5.L we have to define a closed subscheme, but I think we do not know yet what a closed subscheme is. At least I have not noticed it, and the table of contents also suggests that closed subchemes are only introduced later.

October 17, 2011 at 1:13 pm

This is finally patched in the new version, likely to be posted in about a week.

October 11, 2010 at 11:12 am

Concerning projective schemes:

– in 5.5.C must I really be a subset of S+ or is that an error? Is I anywhere required to be a subset of S+ at all?

– do the sets D(f) form a base with f running over all homogeneous elements or only those from S+?

– In 5.5.E, is f intended to be from S+?

– In 5.5.G, is f intended to be from S+?

– If D(f) for f in S+ is not a base, it seems that we have to show that those D(f) do at least cover Proj S, is that correct?

March 17, 2011 at 10:33 am

Sorry for the terrible delay! I understand why you were asking the questions, and I’ve clarified the exposition. (Just let me know if you’d like me to send it to you before the next posting.) In 5.5.C, I must be a subset of S_+, and an ideal of S_*. We really want f to only be in S_+. An idea of why: we want D(f) to be an *affine* cover, and if f is degree 0, it might not be. (Example once we’ve completed the construction: let S_*=k[x,y], so Proj S_* = P^1, and let f = 1.) So everywhere we indeed want f to be in S_+.

October 12, 2010 at 12:38 am

Dear Ravi, the theorem that a domain R is integrally closed if and only if R[X] is integrally closed can be found in

Bourbaki, Algèbre Commutative, Chapitre 5, §1, n°3, Corollaire 2 (page 16).

Godspeed to this great project of yours.

[Thank you very much Georges! I’ve added a comment to let the reader know about this. — R]

October 12, 2010 at 5:48 am

Thanks for the update!

I like the updated section on the Yoneda lemma. Now I have a question regarding this section, it’s something I have been brooding on for a while.

In 2.3.X you write that Yoneda’s lemma gives a more general statement than what has been proved in the exercise, implying that 2.3.X can be deduced from 2.3.Y. However, I fail to see the connection. Yoneda’s lemma gives us a fully faithful embedding (the injectivity on the objects still needs proof – it’s almost immediate though) of a category into it’s functor category. The way I understand it, 2.3.X gives us another property of this functor, unrelated to the ones secured by the lemma:

If there is a natural isomorphism between h^A and h^B, 2.3.X says that then A and B are already isomorphic.

At the moment I can’t see how this could follow from fullness or faithfulness of the functor…please do tell if it does!

In general I would like to know if there is a notion for a functor that behaves like the Yoneda in the sense of 2.3.X. I immediately thought that this should be “essentially injective” but I couldn’t find this notion in connection with category theory. Isn’t such a functor, additionally given faithfulness, a “better” notion of embedding, i.e. better than to demand injectivity on objects?

Many thanks, Felix

November 17, 2010 at 8:59 am

I just quickly want to add this:

Fully faithful functors do reflect isomorphisms!

This only needs an elementary proof, I could now easily see – having rested the case for a while. I can’t judge if it’s worth mentioning, but it would be a nice exercise…

(Show that for a fully faithful functor F two objects F(X),F(X’) are isomorphic if and only if X and X’ are isomorphic.)

With this knowledge everything else I have written is, of course, also obsolete.

January 4, 2011 at 1:50 pm

Great!

October 12, 2010 at 2:07 pm

Unimportant question I’m curious about: is there a locally Noetherian scheme where the closed points aren’t dense? I asked Karl Schwede if the examples in his paper are of this form, and he said they aren’t (it isn’t locally Noetherian, although it is locally of finite Krull dimension), but he suspects there should be such examples. He suggested looking at Liu’s wonderful book “Algebraic geometry and arithmetic curves”, and I haven’t yet.

October 12, 2010 at 2:29 pm

The spectrum of any local noetherian ring of positive dimension is an example; e.g., Spec(dvr). Basically, once you leave the world of Jacobson schemes (a concept which deserves to be more widely known!) then it becomes possible to get rid of the horrible psychological grip that closed points have on too many students.

[Oops! Warning to this effect now added. — R.]October 12, 2010 at 3:43 pm

A fun exercise is to show that every locally Noetherian scheme has a closed point! (The standard proof uses dimension theory of Noetherian local rings, hence for experts only. But there might be an argument avoiding this…)

October 12, 2010 at 5:30 pm

Non-empty…

October 12, 2010 at 3:55 pm

Another remark on this topic. Let X be a scheme. Let X_{ft} = {x ∈ X | x is a closed point of some open neighbourhood of x} = {x ∈ X | x is a closed point of some affine open of X}. Then X_{ft} is large, in the sense that for any locally closed subset T of X the intersection T ∩ X_{ft} is dense. Sometimes X_{ft} can be used as a replacement for the set of closed points of X. See section entitled “Points of finite type and Jacobson schemes” of the chapter on morphisms of the stacks project.

October 16, 2010 at 10:41 am

I am having problems with exercise 6.3.H, because if I write 1 = Σ c_ik*(f_i)^k, then the c_i are not necessarily in B. And since the exponent k can get arbitrarily high I need infinitely many c_ik.

October 17, 2010 at 3:26 am

Okay, I solved this. The coefficients c_ik can actually be taken from (Σ c_i*f_i)^K for K big enough, so they are polynomials in c_i,f_i. However your set of generators stated in the hint will then not work. You need to add the c_i to that, if you are using my solution.

Is there a solution that really does not need the c_i?

October 26, 2010 at 8:38 am

D.H., you are absolutely right! Now patched.

October 27, 2010 at 7:35 am

You write “Our graded rings are indexed by . One can define more general graded rings, but we won’t need them.” On the next page, you refer to “the graded ring “. This has terms in negative degree.

In my experience, it is better to allow graded rings to have negative parts. You can certainly say that you want to focus on positive graded examples, but there are too many places where you will want some auxiliary ring with negative grading to build this into your definition. For example, surely you want to be able to talk about the section ring for a line bundle which you have not yet proved to be ample.

October 11, 2011 at 11:09 am

Long-delayed response: Good point! I was going to deal with this, and found that I appear to have fixed this (no longer emphasizing rings with nonnegative grading) at some point, and don’t recall doing it.

October 20, 2011 at 11:34 am

This is causing me more headaches, so I’ve re-patched this. For these notes, “graded ring” will mean “non-negatively graded ring”. When we want a Z-graded ring, I intend to say that explicitly. This will be patched in the next version to be posted — hopefully tomorrow (Oct 21 2011), and at worst in the next week.

October 27, 2010 at 7:44 am

Here is a “big picture” point which took me far too long to learn. Given a ring , specifying a grading on is equivalent to specifying an action of on . It might be nice if you could at least fit this in as a remark.

October 27, 2010 at 7:59 pm

It probably clarifies the ideas to focus on actions on modules rather than rings: fix a base ring $R$ (e.g., $\mathbf{Z}$) and identify linear $\mathbf{G}_m$-actions on an $R$-module $M$ with linear $\mathbf{Z}$-gradings on $M$. That being said, a pre-exercise is to make a suitable *definition* for what it means to make an $R$-group scheme $G$ “act” linearly on an $R$-module $M$ (and to identify what is the “universal case”, and to relate it to $R$-homomorphisms $G \rightarrow {\rm{GL}}(M)$ when $M$ is a finite free $R$-module).

October 27, 2010 at 8:05 pm

Oh whoops, if ${\rm{Spec}}(R)$ is disconnected then my comment above is not true; the grading is really by locally constant $\mathbf{Z}$-valued functions on ${\rm{Spec}}(R)$ (i.e., the character group of $\mathbf{G}_m$ over $R$ is the global sections of the constant sheaf $\mathbf{Z}$). OK, so best to stick with $R$ having connected spectrum, such as the implicit $R = \mathbf{Z}$ in David’s example…all right, this is getting cumbersome, so perhaps ignore my comments on this.

October 11, 2011 at 11:29 am

Good idea! I’ve added this as Exercise 7.6.O(b) (part (a) is for them to define group scheme action). It is in a double-stared section (7.6.3), but one which should attract a lot of readers (of a certain kind).

October 27, 2010 at 7:48 am

(I assume it is easier for you if I leave separte thoughts as separate comments.)

[Yes, thanks! — R.]I’m still thinking about whether you can improve the presentation of Exercise 5.5.B. I think I would find the following statement slightly easier to think about: “Let be a graded ring with a unit in positive degree. Then there is a bijection between prime ideals of and homogenous prime ideals of .” All I’ve done is renamed to , but I find it a little simpler.

I wonder whether it would be a good idea to include the “Veronese is an isomorphism” lemma as a (challenging) exercise. That is to say: “Let be a graded ring and a positive integer. Define the ring . Then (1) there is a bijection between homogenous primes of and and (2) this bijection extends to an isomorphism of the corresponding projective schemes.

October 27, 2010 at 4:32 pm

Dear David,

I agree. (This renaming was my first step when thinking about this problem!)

Best wishes,

Matt

October 11, 2011 at 11:22 am

David, your first suggestion is excellent, and I’ve just implemented it. And the second suggestion is already done; it is 7.4.D in the current version.

March 9, 2011 at 2:10 pm

I presume the hint for exercise 6.4.M should be

(xy/(x^2+y^2))^2 = (x^2/(x^2+y^2))(y^2/(x^2+y^2))

Or am I missing something?

March 10, 2011 at 10:48 am

You’re right, thanks. And if you solve the problem, please let me know. The part I fear will be too hard is proving that it is not a unique factorization domain.

March 10, 2011 at 11:56 am

I solved the problem without too much trouble, so perhaps I’ve overlooked something. Here’s the idea: any element of the form f/(x^2+y^2) is irreducible (f not a multiple of x^2+y^2), in particular xy/(x^2+y^2) is, but your hint shows it isn’t prime.

March 10, 2011 at 12:06 pm

Perfect!

July 25, 2011 at 7:57 am

In the post, I asked: “Exercise 6.2.I says that a locally Noetherian scheme is integral if and only if it is connected, and all stalks (of the structure sheaf) are integral domains. Is there a counterexample if the Noetherian hypothesis is removed?” In the responses to this MathOverflow question, you will find a very beautiful (one-dimensional) counterexample by t3suji, and a link to a family of counterexamples by Mel Hochster, from Georges Elencwajg.

December 21, 2011 at 10:52 am

[Peter Johnson sent me a compiled list of his comments in an email to me on March 21, 2012. I’ll work from that email rather than the blog comments. — R.]More comments from Peter Johnson:

Comments on Ch. 6, based on draft of Dec 2 2011.

6.1 2nd para: is 4.5–4.6.*10* (until those parts are reorganized).

6.3.2 Worth comment: open subschemes inherit all affine-local properties.

6.3.B Probably best to delete first part; is only weak hint for the second.

6.3.C Helpful to remind about 6.2.F. [end: Could even say a scheme has

cover of irred. opens iff conn. cpts are irred. clopens.]

6.3.6 Maybe put k after fifth word; k is used later and in 6.3.7 without

comment about what it is. A global assumption could profitably be

stated earlier: k denotes a field, except where it clearly stands

for a (nonnegative?) integer. Could also have default notation

k bar for alg. closed fields. Would improve (make tighter) MANY

future phrases, at your discretion. Perhaps list most notational

assumptions in 1.2, with forward refs to where restated later.

Notion of A-scheme extends (it seems without conflict?) earlier def.

of projective A-scheme. For any B -> A whatsoever, A-schemes are

B-schemes. But may need care for projective/finite concepts.

p143 Do you really need to use the overloaded phrase “a locally of

finite type A-scheme”? Say X, as an A-scheme, is loc. of finite type.

6.3.7 Why such an indirect def. of affine k-variety? 6.3.3(c) will imply

a simple concrete one. line 4: or an <– or a

6.3.D First ref. to 6.1.E is OK, second may have been meant to point a

different Ex. — 6.2.D?

6.3.E Seems OK, intuitively, but the continuous map like that of 4.2.G

should go the other way. Maybe (if I understand right) clearer

with a brief remark/ref. on implicit function theory. Analytic is

already OK over R; add holomorphic and say what more that implies.

6.3.F "This" argument means (i) and (ii) of (a) ["one direction" is (ii)],

so best to WRITE "(i) and (ii) of (a)". Only finish later, with (c).

A bit odd to interrupt proof with Ex.

6.3.9 Right after def. of J_j maybe add: the image _of r_ in ..

6.3.G May mislead to refer to 5.1.2 for partition of unity. Here need a

high power of 1 = … (line 1).

6.3.I "Easy" <– not especially. Is another place with A = S_0 — as always,

or as extra assumption? line 4: quasiprojective A-scheme (best here not

to drop A from the notation). 2nd half: "is q.compact and hence" <–

clearer if deleted; opens doubts whether def. in 5.5.8 of q.proj

was as intended. Shouldn't "q.proj .." be "open subscheme" (3 x)?

E.g., on l 4: "any q.proj scheme is" <– "its open subschemes are..".

End could also be clearer.

6.4.C Cruel not to give some hint.

6.4.D At end: define clearly what I is; cf. other I (of s) after (6.4.2.1).

Just after: "soon" <– Wow, pace is really fast!

6.4.E Before and after, repetitive. Do you need to quote 6.4.N thrice?

6.4.5 Tiny ambiguity just before: "one can" <– at times

6.4.H Say f is in A even before (a). Assume root is in K(B) but not in A.

Prudent to define the F bar in (a). Probably need ref. to Gauss' Lemma.

6.4.I (b) And m \leq n. Easy to avoid alg. closed in (c), w/o 6.4.M.

6.4.J OK, presumably readers should know def. and basics of quadratic forms.

6.4.M OK, recently improved. But, at end, "Show … " can be false (A = l,

l not k). The main idea doesn't even need this, works for ANY field

extension. May want to give a better example – cf. my rmk on 6.4.I.

Maybe reorder: leave no hint that Hint is about version of converse.

6.4.N In 2nd hint, "each" between "first" and "t = y/x" is a misstatement.

Say more about Z[sqrt(-5)]: what affine opens work? Maybe give ref.

6.5 1st para.: "important" 3 times – could rewrite.

6.5.2 Unfortunate: Associated points "as" components. ("representing"?)

6.5.A Where will you explain "no coincidence"? Connection with non-reduced?

6.5.B-H Imp. note: fix "(C)to" on l4. Of course, still use algebra selectively.

But see below. Maybe remove first "geometric axioms" to avoid repeating.

6.5.C OK for f=0. Supp uses germs, D(f) doesn't. Needs work (unlike 6.5.D).

Does NOT seem to be a good illustration of geometric ideas. Bad hint;

relies on algebra deeper than needed: to exclude points in Supp(f), not

in D(f) bar, must know no minimal primes can contain both f and Ann(f).

Equivalent: all elements of minimal primes are zerodivisors [AM Ch3 Ex9].

I did not see how doing (i) first would help. Another way is below.

6.5.2.1 With (A), p maximal in the list suffice; also in 6.5.2.2, Fig 6.2.

6.5.E To get the locus of nonreduced points closed, not just a union of

closures of associated points, do need something Noetherian-like.

Sufficient (with more natural stronger and/or local versions): all

h with h^2 = 0 lie in some finitely generated ideal of nilpotents.

Hints: Study p in/not in the union of V(Ann h), h^2 = 0.

When f^2.g = 0, let h = f.g.

all 6.5 Assoc. primes interest you, so here's my opinion: Avoid Noetherian

assumptions until needed. I did not like the organization of your

[E]-based approach, so rethought all, in the following way I believe

is shorter and easier to follow. Idea of using more geometry/less

algebra seems appealing. Is it convincing? Algebra gets initial

results. What nontrivial arguments use little algebra?

In a set U of points of a scheme, with U open or closed or even

constructible, every point of U lies above a U-minimal one, by Zorn.

Such points determine the irreducible components of U. These ideas

are not essential, but show that finiteness/Noetherian assumptions

are irrelevant for 6.5.D.

Now work in an affine open Spec A. What makes a section m tilde

for M vanish at a point p?: f.m = 0 for some f in A, f not in p.

Use the ideal Ann m of such elements f, or its radical. The support

of m tilde is clearly V(Ann m). In the case A=M, m=f, it's better to

use the ideal A_f of elements that annihilate some power of f. One

sees that primes minimal above A_f do not contain f and consequently

(think of f^n.g = 0) are minimal primes of A. At other primes p not

in V(A_f), a power f^n tilde_p is 0, and p could be embedded — but

clearly not if A is reduced. [Is there a nice weaker assumption

forcing some f^n to have support V(A_f)? ]

Optional: Exactly one of the following holds for f in A and its

section f tilde on Spec A:

(i) f is a zerodivisor and some minimal p contains f with

zero germ f tilde_p.

(ii) Rad A contains Ann f; f has no zero germs;

f is not in any associated p.

Last by [AM] as above, so a bit deeper. Good to know; not central.

The key assumption to go on is weaker than Noetherian. For m in M:

(*) Finitely many primes of A are minimal subject to containing Ann m.

It's easy to see that these are the primes p_1,..,p_n minimal among

those at which the section m tilde does not vanish; i.e., the points

of (A). To get an element m_1 with Ann m_1 = p_1, choose elements

f_i in p_i but not in any other p_j and let m_1 be f.m, where f is

the product of all f_i except f_1.

Thus primes of (A) are annihilator ideals that happen to be prime: (D).

The slightly dubious geometric/algebraic distinction has disappeared.

[For incomparable assoc. p_i, sum of m_i has Ann=intersection of p_i.]

Can M have submodules like A/p_i for infinitely many primes p_i?

Annihilators show that a sum of these with all p_i different must

be direct, so such an M cannot be Noetherian. If more is wanted here

(I think not), maybe focus on indecomposable M.

Next we show that the union of all these primes, as m ranges over M,

is the set of zero-divisors (with 0) of A: algebraic form of (C).

[In (C), "vanishing" in _sectional_ sense is my (i) and (ii) above.]

Given nonzero f,g in A with f.g = 0, let p be a prime minimal above

Ann g. Then p is an associated prime in which f lies. On the

other hand, given an f in an associated prime p, we know p = Ann g

for some g. Then f.g = 0 with g nonzero.

On (non)reduced points: would be good to make clearer their relevance.

Study of filtrations of M is no longer needed, at least for some time.

6.5.F The defn <– not yet given; should say defn in (A). Obvious by germs.

p152 Sentence (ii) needs fixing — remove beginning, etc.

Fig 6.2 Slightly odd: curve as drawn seems to stop when it reaches the plane.

6.5.I Ex. no longer so important. Needs "nonzero" after "annihilators of"

6.5.J Implied by 6.5.I. After: "It also implies"?? <– not "it" but 6.5.I.

6.5.K In Hint, = should be isom (\cong).

6.5.M Don't use Hom — too imprecise for desired result. Use fractions.

6.5.7 Vague or inaccurate use of "appearing": OK if you say prime decomp.

of radical. I see [E] treats associated primes BEFORE getting to

3.3: primary decomposition.

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