**The sixth post is the November 5 version here. The next reading is 8.1-9.2.** Sections 9.3 and 10.1 are bonuses. I had pondered including 9.3 in this batch of required reading, but I wanted to have only one “hard” section per batch if possible.

**About my slowness.** As promised, this fall is much busier than the rest of the academic year will be, so I continue to be very very slow on responses. I promise to start catching up by the end of the academic year, and intend to respond to essentially everything! (In the last while I’ve been working through emailed comments.) In order to accommodate my slowness, the next posting will be in three weeks (target: Nov. 27), and the posting after that will be three weeks later (very tentative target: Dec. 18). I intend to then continue posting fortnightly.

**For learners.**

*8.1 (open immersions)* is short and straightforward.

*8.2 (algebraic interlude)* has one pretty algebraic trick which can be used in lots of circumstances. You should make friends with Nakayama’s lemma, and in particular try to get some sense of when it can be used.

*8.3 (finiteness conditions on morphisms)* has a number of definitions (many taking advantage of 8.2), but isn’t too bad.

*8.4 (Chevalley’s theorem and elimination theory)* is supposed to be pretty, but people have found earlier versions of this section harder than I would like. How do you find it? (Feel free to email me privately.) Can anything help make it better (e.g. a particular extra sentence in some proof, or a particular exercise)?

*Chapter 9: * closed immersions are surprisingly tricky, and much of the trickiness is encoded in a particular exercise (9.1.F).

*Exercises: * Here are some good ones, although as usual you should try a random sampling of others that catch your eye. 8.1.A, 8.2.D, 8.2.H, 8.3.G and 8.3.O (practice with the Affine Communication Lemma), 8.3.H, 8.3.K, 8.3.Q, 9.1.F (hard but key), 9.1.G, 9.2.C, 9.2.N.

For worthwhile problems involving getting your hands dirty with examples (very much worth doing!), you could try 8.4.C, 9.2.B, 9.2.F, 9.2.J, 9.2.M, and 9.2.N (yes, that last one is a repeat from the above list).

I have some questions for you too, about other exercises you may try.

Can you do the Exercise 8.3.H on classifying finite morphisms to Spec k by bare hands, without having seen it before? Or would more of a hint help?

If someone (likely arithmetic or non-Noetherian) tries Exercise 8.3.R (affine local nature of locally finitely presented morphisms), please let me know. I haven’t had a chance to think it through, and I don’t have a sense of how hard it is (as an exercise). If it is too hard, I’ll just give a reference. Or if it could be easy with an outline/hint, I’d prefer to do that instead. (If you solve it, please send me your solution!)

Do you have opinions on the do-ability of Exercise 8.4.C on elimination of quantifiers over an algebraically closed field? In an earlier incarnation, people didn’t like it, but I’d botched the exposition, and I think it is now repaired.

**For experts.**

It seems to me that the fact that the composition of two quasiseparated morphisms is quasiseparated is from the (“wrong”) definition (that the preimage of any affine is a quasiseparated scheme) is surprisingly difficult. If true, this is another good sales pitch for thinking of things in terms of the diagonal map (coming soon, after fiber product).

My definition of quasifinite (after 8.3.Q) includes a finite type hypothesis, which I believe agrees with everyone but Hartshorne.

I use the phrase “the reduced subscheme structure on a closed subset” rather than the “reduced induced subscheme structure”. If this is terrible, please let me know.

I prove that finite morphisms are closed (8.4.4) using elimination theory. One can also prove this with the Going-Up Theorem, but I find that it takes a little care, and that the elimination theory argument is kind of pretty (using just that finite morphisms are projective, which is useful later too). *[Update Nov. 15 2010: useful comments below changed my opinion on this.]*

I wonder if I should mention the fact that you can check surjectivity of maps of varieties by just checking closed points, which is a nice application of Chevalley for people thinking mainly of varieties (most people). But perhaps they are already sold on how it is a geometrically important fact. (I should say that I think that Chevalley is currently undervalued in the way we often learn the subject. I only learned to appreciate it from Matthew Emerton, long after I should have.)

**For everyone.**

A visiting topologist asked the reasonable question: where are all the examples? That’s a fair question. In some sense we have lots of examples (we have all varieties! all schemes!), but we don’t have the ability to do too much with that. Hopefully by this posting, examples have started to appear (both geometric and arithmetic). But if any “obvious” or particularly enlightening examples are missing, please let me know!

November 6, 2010 at 5:12 am

Typo: In equation 8.3.4.1 (just after “localization commutes with \emph{finite} products”) I assume the last subscript should be $s_{ijk}$.

Exercise 8.3.F seems hard to me. I’m curious how many of your students got it.

November 6, 2010 at 5:52 pm

Dear David,

Were you using the full strength of affine locality in your argument? All the difficulty should be encoded in that result.

Regards,

Matt

November 15, 2010 at 6:24 am

Thanks David! (8.3.4.1) is now fixed. Regarding 8.3.F: it is true that without affine locality, it is a remarkably hard fact (at least the way I think about it). With affine locality, I want it to be easier. But I can’t put myself in the head of someone learning this for the first time. Can someone learning this for the first time please try this exercise and let me know how hard it is for you? Feel free to email me directly if you feel shy. (And “I couldn’t see how to do it” is a completely acceptable unembarrassing answer — hopefully you see that people like David or Matt or myself are very comfortable saying this, and the greatest danger in a developing mathematician is when they are *afraid* of saying such a thing.)

November 6, 2010 at 5:57 pm

Dear Ravi,

Amusingly (to me), I just proved that finite morphisms (between quasi-projective algebraic sets) are projective by showing that they are proper (using Going-Up) because it seemed a little tedious to write down the homogeneous equations directly. Having looked at how you do it, I now appreciate the elegance of allowing fairly general graded rings in the treatment of projective schemes.

Best wishes,

Matt

November 7, 2010 at 1:05 pm

“I just proved … ” refers to the algebraic geometry class that I am teaching this quarter.

November 13, 2010 at 6:34 am

Thanks Matt! After reading your comment, I realize that the alternate argument is worth mentioning, because it is just as short, more direct, and gives the reader more practice with and appreciation for Going-Up. So I’ve put it back in.

Text:

You can also prove Corollary 8.4.4 using the Going Up

Theorem 8.2.5. Here is a sketch. Reduced to the affine case.

If $f^*: B \rightarrow A$ is a ring map, inducing finite $f: \Spec A

\rightarrow \Spec B$, then suppose $I \subset A$ cuts out a closed set

of $\Spec A$, and $J = (f^*)^{-1}(I)$, then note that $B/J \subset

A/I$, and apply the Going Up Theorem 8.2.5 here.

November 15, 2010 at 5:53 am

After further thought, I’m now convinced by Matt’s implicit point. Using going up shows a stronger result — that *integral* morphisms are closed. Plus the argument is short, motivated, and can be done earlier. So I have now done this. I still remark that the fact that finite morphisms are projective, and projective morphisms are closed gives an enlightening *second* proof that finite morphisms are closed. (But this proof is now relegated to 9.2, because as D.H. points out below, we need to know that Proj of this ring can be interpreted as a closed subset of P^n.)

Side comment: I think the reason I didn’t appreciate this approach was better was because (back in the triassic) I didn’t have a long proper commutative algebra class before learning algebraic geometry. This drives home in my mind the fact that it’s best to try to learn things the “right” way (not necessarily the most general way or the most abstract way) the first time you learn a subject seriously. I’m glad this project provides a communal way of trying to figure out candidates for what this “right” way is.

November 15, 2010 at 8:29 pm

Dear Ravi,

I now feel a little guilty that the winds of generality and commutative algebra have pushed things so that the elimination theory argument is now a second proof rather than the primary one. I do see why you are making the change (it is closely related to why I chose this route rather than a more explicit projective geometry argument) — to really write down the homogeneous equations seems to be tedious, however you go about it.

On the other hand, I like the elimination theory argument because it is concrete, whereas the argument via Going Up has something of a black magic feel.

November 22, 2010 at 7:35 pm

Don’t feel guilty — I still state the other argument later (as it is pretty, and gives different enlightenment).

November 7, 2010 at 2:01 am

Dear Ravi,

perhaps you might name 8.2.5. the Lying Over Theorem (a fairly standard denomination, which your drawing clearly illustrates), reserving the terminology Going-Up Theorem for the statement in Exercise 8.2.F and thus eliminating the parenthetical remark “…sometimes also called going-up”.

Friendly,

Georges.

November 11, 2010 at 5:12 am

I second this!

November 11, 2010 at 5:13 am

Moreover, this version of “going-up” (properly formulated) holds for any integral ring map.

November 13, 2010 at 6:48 am

Georges and Johan, thanks!

(1) Change made! I’m not hyphenating “Lying Over Theorem” but I *am* hyphenating “Going-Up Theorem”. Someone please complain if I should do something differently.

(2) I oscillated in this section between writing f: B –> A and phi: B –> A. I’ve made it consistent, using phi.

(3) Johan, good point; I’ve changed the Going-Up Exercise to integral extensions. (Text: Suppose $\phi: B \rightarrow A$ is an integral {\em homomorphism} (not

necessarily an integral extension). Show that if $\fq_1 \subset \fq_2 \subset

\cdots \subset \fq_n$ is a chain of prime ideals of $B$, and $\fp_1

\subset \cdots \subset \fp_m$ is a chain of prime ideals of $A$ such

that $\fp_i$ “lies over” $\fq_i$ (and $m < n$), then the second

chain can be extended to $\fp_1 \subset \cdots \subset \fp_n$ so that

this remains true.)

November 7, 2010 at 4:27 am

Dear professor Vakil,

Here are some typos + some questions as always.

4.4.H. It might be good to state about the (non-)relationship between topology of SpecS^-1B and topology of SpecB.

p.114 l.6 again : Did you define f^*? I pointed it out last time since I cannot see the definition of f*.

Prop 6.3.3 again : At the last line of p.131 of this version.

p.137 (4) a zero divisor -> a zero-divisor

6.5.1 again : I just pointed out that ‘the set is’ should be ‘the set of’.

6.5.D. It is not do-able at least for me(but take into account that I am an undergraduate.)

6.5.E. Should I use Zorn’s lemma? Then why I need the Noetherian assumption?

7.3.J. pi and f are not consistent.

p.153 line 20 : There should be space next to ‘to the generic point of Y.’

line 30 : we need Spec’s both for k[x,1/x] and k(x).

7.5.D. Show there -> Show that there

p.155. => part : (-2)x -> (-2m^2)x

y=-2m/(m^2+1) (two times)

7.5.G. do you mean 6.4.6, not 9.2.5?

7.5.7. see describe

7.6.G. translate -> translates

7.6.M. group invertible functions -> group of invertible functions

right before 8.2 : we will see …??

Three weeks are also great for me because I am getting busy preparing applying for graduate program in US including Stanford!

See you three weeks later.

All the best,

PhilSang

November 13, 2010 at 7:11 am

Thanks! I’ve dealt with all but the following. (I’ll gradually work through these too.)

p.114 l.6 again : Did you define f^*? I pointed it out last time since I cannot see the definition of f*.

Prop 6.3.3 again : At the last line of p.131 of this version.

6.5.D. It is not do-able at least for me(but take into account that I am an undergraduate.)

6.5.E. Should I use Zorn’s lemma? Then why I need the Noetherian assumption?

7.5.G. do you mean 6.4.6, not 9.2.5?

November 13, 2010 at 7:13 am

Important question: I don’t understand your comment on 4.4.H. The topology on Spec S^{-1} B is the subspace topology induced by inclusion in Spec B. Is it not clear that that is what is meant? (He responded here.)

September 2, 2011 at 11:31 am

Thanks again! I’m sorry for how long it took to get to these last few points.

You wrote:

p.114 l.6 again : Did you define f^*? I pointed it out last time since I cannot see the definition of f*.

My response: Thanks, now fixed.

You wrote:

Prop 6.3.3 again : At the last line of p.131 of this version.

My response: I still don’t see a period-less sentence that you referred to in a comment on the previous notes.

You wrote:

6.5.D. It is not do-able at least for me(but take into account that I am an undergraduate.)

My response: Can you say which parts you managed to do, and where you got stuck? It’s probably too long since you were stuck. But I’ll keep an eye out for how students do with this question this fall.

You wrote:

6.5.E. Should I use Zorn’s lemma? Then why I need the Noetherian assumption?

My response:

No, you don’t need Zorn’s lemma. I’m not sure where you want to use it, but I suspect that if you see where you use it, you’ll be able to use Noetherianness instead.

You wrote:

7.5.G. do you mean 6.4.6, not 9.2.5?

My response: Yes, thanks, now fixed!

I hope you are starting graduate school very soon (maybe even around now)!

November 7, 2010 at 6:24 pm

Dear Ravi: Perhaps the non-word “P-ness” in the Intro part of chapter 8 should be replaced with something else, depending on whether the similarity in sound to a certain actual word is something you had in mind.

Just above 8.3.10, when you tell them the punchline of Zariski’s Main Theorem (in the EGA formulation), why not clue them in to the fact that you just told them the content of an extremely deep and important theorem (i.e., that it is totally not obvious that all quasi-finite separated maps between qcqs schemes are open immersion followed by finite map).

On the issue of using Chevalley to show the sufficiency of checking something at closed points when working with finite type schemes over a field, this should be a kind of exercise which comes up over and over and over: for every new concept, when appropriate ask them to prove the sufficiency of checking whatever at closed points (since it is ubiquitous when passing between schemes and classical varieties, etc.)

Finally, concerning the preservation of quasi-separatedness under compositions, the issue is to prove that if f:X –> Y is quasi-separated and Y is quasi-separated then so is X. This seems not too bad to prove (with small hints?) even using your “wrong” definition (but the diagonal-based proof is obviously far superior). Cover X by q-c open U_i such that f(U_i) is contained in q-c open V_i of Y, and want each U_i \cap U_j is q-c. But f^{-1}(V_i) is q-septd, so enough that f^{-1}(V_i) \cap U_j is q-c (as its open in q-septd f^{-1}(V_i) which contains U_i as q-c open). Thus, want open immersion f^{-1}(V) –> X to be q-c for any q-c open V in Y. Suffices to test against a q-c opens that cover X, so consider q-c open U in f^{-1}(V’) for affine open V’ in Y. Then U meets f^{-1}(V) where it meets f^{-1}(V \cap V’), so by q-septdness of f^{-1}(V’) (in which U is q-c open and f^{-1}(V \cap V’) is open) we win. QED

April 24, 2011 at 11:08 am

Par 1: that made me laugh, enough so that I now don’t want to change it!

Par 2: Thanks, I’ve added that comment, and intend to refer back to here from the (as yet unwritten) ZMT discussion.

Par 3: Thanks again! Do you (and others) have favorite examples? The only examples I currently have remembered to keep track of (with the intend of making them a list somewhere) are, in no particular order: (i) functions on varieties (by which I mean reduced separated finite type schemes over a field) are determined by their values at closed points. (ii) Maps of varieties over algebraically closed fields are determined by what they do set-theoretically on the level of closed points.

I have another short list of things that can be checked on geometric points, but it is also incomplete.

And par. 4: thanks yet again, that is enlightening, and I’ve kept it so I won’t forget it!

November 10, 2010 at 1:48 am

8.4.2:

In your proof you say that it suffices to show that pi(X) is contructible before you reduce to f: Spec B -> Spec A. How does this work?

The only way I see is to reduce to affine schemes first and then a locally closed subset X has a finite type morphism X -> Spec B which is an inclusion of topological spaces.

8.3.J: What is the “obvious multiplicative structure”?

Thank you,

D.H.

November 12, 2010 at 3:00 am

I have now understood 8.3.J. But now in 8.4.4 X is not a projective A-scheme that can be directly plugged into the fundamental elimination theorem, because it does not have the form P^n_A.

With the next section about closed immersions it seems to be possible to do

X -> P^n_A -> Spec A, and then it would work, is this correct?

And using the going up theorem did not seem so bad to me, use

Spec A -> Spec f#B -> Spec B

the first morphism is surjective and show that the second one sends f#B to a closed subset, is this correct?

November 15, 2010 at 6:16 am

These comments are very helpful!

8.4.2: Good question. Now added: “(Reason: If $X’ \subset X$

is a constructible subset of $X$, then by

Exercise~8.4.A,

we can write $X’$ as a finite union $X_1 \cup \cdots \cup X_n$

of locally closed subsets of $X$. It suffices to show that the image

of each $X_i$ is constructible.)”

8.3.J Even though you figured it out, that’s a good sign I should haven’t used the loaded word “obvious”. The text is now improved a bit there.

8.4.4: Good point! (See also my response to Matt Emerton above.) I’ve now relegated it to the next section, and the going up proof is now the “main” one of this fact (in the earlier section on integral and finite morphisms).

November 10, 2010 at 2:32 pm

This is going back a bit, but we found a minor mistake in exercise 6.3.I. Setting S_{\bullet} = k[x_0, x_1, x_2, …] (countably many variables) is a counterexample, since then Proj wouldn’t be quasicompact. I’m guessing we need the additional assumption that S_\bullet is a finitely generated graded ring.

November 15, 2010 at 6:18 am

Thanks Silas! Hypothesis explicitly added.

November 11, 2010 at 5:26 am

In the proof of 8.3.3 you use 8.3.4 but you forgot to check that X is quasi-compact and quasi-separated which is an assumption of 8.3.4. This requires the result of Exercise 8.3.C.

[Thanks, Johan, fixed! – R.]November 15, 2010 at 9:57 am

I’m having a few difficulties with the proof of Chevalley’s Theorem (8.4.2, proved on page 178).

1) You seem to be reducing to the case that X is a closed subset of affine n-space. But this does not necessarily imply that the image of X in (n-1)-space is closed, so in fact, you can still only assume X is a locally closed subset of $A^{1}_A$.

2) This may or may not be a minor point, but depending on your choice of the f_i, as I understand it, $\overline{X}$ may include more than the closure of $X$. It would be nice to have this clarified.

3) Main difficulty: I cannot follow most of the last paragraph. Clearly Z is constructible, but why is it locally closed? Why is the image of X’ the only part of the image of X we are missing? Most problematically, why is X’ a closed subset of X? It looks open to me.

I may be able to figure out what is going on here with some more work, but I thought the fact that I and other members of my reading group are having difficulty here would in itself be valuable information.

November 23, 2010 at 11:02 am

The Chevalley proof was indeed botched. A revised version is here (and is part of the next update later this week). Would you mind taking a look and letting me know if the argument is now reasonable and readable?

November 23, 2010 at 12:23 pm

Thanks for the clarification.

I don’t immediately see how to reduce from (e) to (f), specifically in the case that Z = S – T, where T \subset S are closed subsets, S is irreducible, and the image of T is all of Spec B.

I have not really put the time into 8.4.F that a “subtle exercise” should require. Based on its similarity to an argument in the proof sketched below, into which I did put a fair amount of time, I would expect to be able to prove it using dimension theory, but would not know how to go about it otherwise.

[Note: since this is likely to inform how I am interpreting your exercises, I am going to sketch the version of the proof that I finally came up with to present to my reading group. Reduce to a locally closed, irreducible subset X of P^1_B, where B is an integral domain and the image of X is dense. Set X = X’ – X”, where X’ and X” are closed and X” is a subset of X’. Consider two cases: the image of X” is a proper subset of Spec B (use Noetherian induction) and the image of X” is all of Spec B (use dimension theory to show that X’ is all of P^1_B, and then the image of X is precisely Spec B minus the (closed) set of points whose fibers are contained in X”).]

November 24, 2010 at 6:04 am

I also have problems reducing from (e) to (f). Was the earlier version of reducing to (f) not correct?

(That was: after reducing to Spec A -> Spec B, show that it is sufficient to show that the image of Spec A is constructible, then do the induction step. Then you can reduce from a closed subset in A^1 to an irreducible closed subset.)

November 24, 2010 at 2:51 pm

D.H., that’s indeed confusing, so I’ve rewritten this again (see the Nov. 24 post). If you are still game to try it, please take a look, and let me know where any issues remain. I always find that when I rewrite something and immediately post it, there are numerous issues, so I apologize in advance.

And Charles, your approach is a good one (even if not the one I was aiming for). One reason I like Chevalley’s theorem is that it is not easy, yet there are several roads there, which all highlight different interesting points.

November 27, 2010 at 5:30 am

Thank you for your answers.

Now I am stuck with Exercise 8.4.D. Is there any hint you can give us?

Maybe I have not been thinking about it long enough, but since you said that now “all the steps [are] handed to the reader on a silver platter”, I assumed it should be doable quickly.

November 27, 2010 at 2:18 pm

I’m glad you asked, because I indeed want it to be something you can quickly see, given how much you’ve thought about the problem. Here’s the idea. Say Spec B – F is covered by U_1, …, U_n. Then the image of pi restricted to U_i (call that image V_i) is constructible, and hence the image of pi, being the union of the V_i and F, is also constructible. Do you buy that?

November 27, 2010 at 3:14 pm

Sorry, I think I meant another exercise than you. I meant the completely new version, where Exercise 8.4.D is algebra: “Prove the statement of Exercise 8.4.C in the case C = B[t]…”

November 28, 2010 at 3:01 am

I think I did 8.4.D, but 8.4.E seems not to be doable.

Assume C = Z[t]/(t^2 + t) and B = Z. Then t is not zero in C

and a prime element, so (t) is a point in V(t) in C. But then we have maps

B=Z -> C -> C/(t)=Z

and we find a prime ideal (the preimage of zero of the last map) in C which is not zero but whose preimage is zero, therefore pi sends that point to the generic point in B, which you said was not the case.

Did I make any mistake here?

November 28, 2010 at 8:57 am

For 8.4.E, try long division of an arbitrary element in C_{b_n} by the given polynomial in I, with b_n inverted.

For your example, Spec C should be irreducible.

December 5, 2010 at 3:40 am

Thank you! I had missed the integrality. I now have everything.

November 15, 2010 at 11:42 am

I’m transcribing an email I received (which I’ll leave anonymous because I didn’t ask permission to post) relating to the proof of Chevalley. Along with Charles’ comment above (which I haven’t yet had a chance to digest), clearly I need to think about this proof. I’ll respond as soon as I can on this issue!

[begin quote]

Hello,

I am following your notes on algebraic geometry and I am having problems with your proof of Chevalley’s

theorem (Proof of 8.4.2 on page 178 of your last version).

1. Why is the locus of points of Spec A whose entire fibres are contained in X exactly the vanishing locus of

the coefficients of the homogeneous polynomials f_1, …, f_m? (What you call F.)

2. What is the set F needed for?

3. Is your definition of X’ correct? Why are you using “-” and “\”? What do they mean?

4. Why does pi(X’) lie in the closure of Z in Y and why is this fact important?

5. Why does pi(X’) lie in (closure of Z) – Z?

6. Why is X’ strictly smaller than X? (Maybe this follows from 5. if Z is nonempty?)

Thank you,

[end quote]

November 23, 2010 at 11:02 am

My response to Charles immediately above now applies.

November 16, 2010 at 6:23 am

In the definition of quasi-separated morphism you really have to look at inverse images of affines and not arbitrary quasi-compact opens!

November 22, 2010 at 4:23 pm

Whoops! Now fixed.

November 16, 2010 at 6:46 am

What is really an “extension” of rings? I think this terminology is potentially confusing. In the definition of integral morphisms you use this word. Now closed immersions are integral morphisms, but would you call C[x, y] —> C[x, y]/(x^2 – y^2) an integral ring extension? Some people certainly would, but earlier in your manuscript it appeared as if “extension” means “subring”.

November 16, 2010 at 9:40 pm

Definitely “extension” has to mean “subring”; anything else will be totally confusing.

November 22, 2010 at 10:30 am

Thanks, that was a “typo” rather than anything intentional. In my definition of integral morphism, I said that affine-locally it is an integral *extension* of rings when I meant to say integral *homomorphism*. That is now fixed. But if there was some other mistake, or if this error propagated without my realizing it, please stop me!

November 17, 2010 at 7:32 am

In 9.1.K. can you give us the correct definition? My problem is the the result should somehow not depend on the factorization X -> Z -> Y of X -> Y.

November 18, 2010 at 11:21 am

Is the intersection of two closed immersions defined earlier the fibered product? In that case I think it can be shown that the result of 9.1.K is also the fibered product which would the be unique.

November 22, 2010 at 10:25 am

That is an excellent point which I didn’t see. Rather than worrying about showing that it is independent of factorization, I now have this problem in the fibered product section.

November 18, 2010 at 5:11 am

Comment on constructible subsets. There is a lot of possibility for confusion here. (1) Between the two versions of EGA I the definitions differ, (2) in the first edition of EGA I a constructible set constructed by starting with retrocompact opens, taking complements, finite intersections and finite unions, (3) you cannot check for constructibility on affine opens, hence 1st ed of EGA I introduces the notion of a locally constructible set, (4) in general an open is not locally constructible, (5) in a Noetherian topological space we have: constructible = locally constructible = finite union of locally closed subsets.

My suggestion: Do not formally introduce constructible sets and formulate 8.4.2 (Chevalley’s theorem for finite type morphisms of Noetherian schemes) purely in terms of finite unions of locally closed sets. Then have separate remarks for non-Noetherian schemes.

November 18, 2010 at 11:37 am

I have found the naive notion of constructible sets helpful in learning and remembering Chevalley’s Lemma. I have also never before heard of the more complicated versions in EGA, although I am hardly an expert. My thought here is that as long as this naive notion is kept, without introducing all the complicated notions in EGA (except possibly in a starred or double-starred remark), it is not likely to cause much confusion. But then again, I’ve never tried to read a paper dealing with constructible sets, so perhaps I am mistaken.

November 23, 2010 at 10:14 am

I’m swayed by Charles’ comment. So to assuage Johan’s concerns, I’m going to change “Recall from topology that a **constructible** subset of a topological space” to “A **constructible** subset of a *Noetherian* topological space”. Here **=bold, *=italics. There is still a comment later about the more serious definition in EGA (later). I removed “Recall from topology” because I realize that this notion may not be at all common in topology, and may have just been introduced by Grothendieck. (Please feel free to debate!)

November 23, 2010 at 5:35 pm

Ravi, maybe the noetherian reader (who doesn’t care about non-noetherian details) should at least be warned right away in a Remark that the correct generalization of “constructible” for arbitrary (perhaps non-noetherian) schemes is more subtle than one may first guess based on the definition in the noetherian case, and that many theorems about constructibility (such as Chevalley’s theorem) remain valid in general when suitable definitions are made.

November 24, 2010 at 2:49 pm

Done (in the Nov 24 version now posted). I also realized that in the (double-starred) exercise on Chevalley for locally finitely presented morphisms, I used the word “constructible”, and I now just use “finite union of locally closed subsets.

November 19, 2010 at 4:03 am

In Example 9.2.1, it seems that you have to show that the closed subschemes of the open sets can be glued together. This seems to require some work.

Is there an easier way to understand this Example?

November 22, 2010 at 10:11 am

I agree that one must show that the closed subschemes can be glued together. I think the second equation display does this. If I’m wrong, or if there is something I can say to make it clearer, please let me know!

November 19, 2010 at 6:33 am

In 9.3.3, I think that you mean B_g instead of B_f (you wrote B_f twice).

And the coproduct sign here seems to be simply a union of sets. Is this correct?

November 22, 2010 at 10:09 am

About B_g: thanks, fixed!

And about the coproduct: I’d meant just disjoint union, but now that I’ve made a big deal about “coproduct”, I realize this is confusing, so I’ve turned it into “union”.

February 2, 2011 at 9:52 am

Hi,

In Edinburgh we are a bit behind, still working through 8-9, mainly due to the fact that we have not met in a month due to trips. I think I have found a typo in the definition of integral extension of rings.

We have \pi^{-1}(Spec B)=Spec A. It reads that ‘the induced map A \ra B is an extension of rings’.

Shouldn’t the induced map go B \ra A? I understand Spec is a contravariant functor, right? This is still there in the jan 29 version now in 8.3.8

I browsed through the coments above and it seems no one said it. Forgive me if I read too fast.

February 8, 2011 at 5:08 am

Dear Jesus,

Many thanks — you are right, and I’ve now fixed it (Feb. 8 2011). It’s important to catch these, as small typos can really throw off a reader (as I know from experience).

Also, there are a lot of comments already posted, so don’t worry about looking through them all before making suggestions or reporting answers. I won’t mind if there are repeats, or if something is reported that has already been corrected. It would be worse if people didn’t post things because the effort of checking was too much. (Also, there are many valuable earlier comments I still haven’t replied to.)

Thanks for catching that issue!

March 8, 2011 at 2:26 pm

Hi Ravi

Are you missing some hypotheses in exercise 8.2.B(b), showing that if B -> A is an integral extension then B/J -> A/JA is too?

Here’s the simplest counterexample I could find: If k is a field, the kernel of the natural map

k[x,y] -> k[x,y,z]/(z^2,xz-y)

is (y^2), so we get an integral extension

B:=k[x,y]/(y^2) -> k[x,y,z]/(z^2,xz-y)=:A

But if J=Bx, then JA contains y, so B/J -> A/JA is not an inclusion. Or have I made a mistake somewhere?

Another question: The only way I can think of solving exercise 8.2.A uses the fact that the sum of two integral elements is integral, but this depends on lemma 8.2.1, which comes later. Is there another way of doing this?

[Update March 14, 2011: Pieter came up with a solution to 8.2.A before I had a chance to respond. I asked him for it, and it was indeed very nice. — R.]March 24, 2011 at 9:44 am

About your counterexample to 8.2.B(b): it looks right to me! (It is beautiful and geometric, and clearly didn’t come from nowhere — can you say how you came up with it? Can anyone else think of a simpler version?) I checked to see where this was used, and unfortunately it was indeed used, but fortunately only once — in the hint to 12.1.A, which is an important exercise showing that if Spec A –> Spec B corresponds to an integral extension, then dim Spec A = dim Spec B. I will have to patch this at some point! (Or maybe someone has a quick patch.)

Update August 23, 2011:this is now patched (in the next version t be posted), thanks to Pieter’s advice. In fact, the property of f being an integral extension is indeed preserved by quotients by prime ideals (and more generally, by those ideals that are restrictions of ideals of A).March 24, 2011 at 11:50 am

My counterexample didn’t come from any geometric insight, I just tried to find an ideal of B that would generate a sufficiently large ideal of A, and this was largely a matter of trial and error.

8.2.B(b) does hold if you assume J \subset B is the preimage of an ideal I of A, since then B/J -> A/I is injective, and B/J -> A/JA factors through this. In exercise 12.A.1, this is the situation you’re dealing with, so all is good.

August 23, 2011 at 8:51 am

Even though your counterexample didn’t come from geometry in your head, people may enjoy trying to picture it, and see “why” (in geometric terms) it works. (Also, as mentioned above, this exposition is now fixed in the notes.)

September 10, 2011 at 10:15 am

Dear professor Vakil!

Right after 9.2.Q you write that V(T) is locally principal in Proj(S[T]). Could you please give a hint as to why this is true if S is not generated in degree one? I feel that something like \mathbb{P}(1,2,3) as a cone over \mathbb{P}(2,3) might be a counterexample.

September 29, 2011 at 7:38 am

Good point! Your example kills the imprecise idea I had in my head. I’ve removed this paragraph. I want to check (out of curiosity and interest) if your potential counterexample really is a counterexample (by looking at the Zariski tangent space at a point), but haven’t yet had a chance. If someone else does, please let me know! Or if I do, I’ll comment here.

November 15, 2011 at 12:48 pm

Aaron Mazel-Gee from Berkeley points out that the notation (while common) could be improved to . I like this suggestion. I’ve never seen it before, but there is no doubt what it means (like the best notation). I’m leaning toward using it. Any complaints or comments?

December 1, 2011 at 11:42 am

Response to this suggestion is mixed, although I weight Georges’ opinion below highly, so I remain leaning toward doing it. I’m hoping more people will express an opinion, to make my decision clearer!

November 17, 2011 at 2:04 pm

Dear Ravi,

it is a wonderful notation, exactly for the reason you say: it is self-explanatory.

I can even imagine its twin brother for the normal bundle of a subvariety.

December 21, 2011 at 10:46 am

Here is further justification for this notation, which applies especially for : I see (here) as being equivalent to , where is a closed immersion where the points of are taken to be a subset of points of . So in particular, the notation is also defined, which is useful too (and used).

March 27, 2012 at 9:05 pm

In the end, I’ve not moved forward with this new notation, due to insufficient enthusiasm (except from Georges and myself).