The eighth post is the December 17 version here. The next reading is 11.1-12.2. Section 12.3 is included in case you want to read ahead, and the double-starred Section 12.4 (proof of Krull) is included just so the chapter is complete. As always, you should skip the starred bits (unless you are moved to read them), and skip the double-starred bits (unless you are really really moved to read them).

As promised last time, I expect the next post to be on or about January 8, and I then hope to return to a fortnightly schedule. Many pages of changes from Daniel Murphy; Ezra Miller and the Duke group; and Brian Osserman and the Davis group are incorporated, but there are more to go. Thanks to them, and to many others!

**For learners.**

*11.1* is mainly about separatedness, and is surprisingly straightforward. (This is one of the many ideas in algebraic geometry that inspired a musical tribute.) The Cancellation Theorem 11.1.19 is fun, has a fun proof, and is surprisingly useful.

*11.2* is about rational maps. The key (easy) result is the fact that two morphisms from a reduced scheme to a separated scheme are determined by their behavior on a dense open set. I find Proposition 11.2.3 strangely confusing, and the hardest part of the chapter (see below)

*11.3* is on proper morphisms. Again, it is surprisingly easy.

*12.1* is on the general theory of (co)dimension.

*12.2* connects dimension to transcendence theory, via Noether normalization.

(*12.3 and 12.4* are about two ways in which codimension 1 behaves miraculously well.)

Eleven problems worth doing (including four involving explicit examples): 11.1.B, 11.1.C, 11.1.J, 11.2.A, 11.3.A, 12.1.A, 12.1.C, 12.1.E, 12.2.B, 12.2.E, 12.2.H.

Comments on three more problems:

- If you aren’t very comfortable with transcendence theory, you should do 12.2.A. Please tell me where you got stuck — I want to include enough hints so the problem is as do-able as possible.
- I want to add a hint to 12.2.D, so don’t try it unless you want a challenge.
- Can you do 12.2.F, the exercise on “a first example of the utility of dimension theory”? This is very enlightening, but will likely be a fair bit of work. If someone manages it (or has interesting comments while failing), I’d be interested in hearing about it.

**For experts.**

Radiciel morphisms are now added (10.4.1, 10.4.5), but are optional.

Reference question: what’s the right reference for the Noether-Lefschetz theorem implying that the only curves on a very general surface of degree more than 3 are complete intersections?

Notation question: Ezra Miller proposes that I use colons rather than semicolons for points of projective space. He makes a convincing case. Opinions?

Terminology question: “Constructible” or “constructable”? I’m currently using “constructible”, following Bjorn Poonen (and my instincts), but my spellchecker disagrees (and some readers).

Random question: Mumford (in the red book?) mentions the following result: if f: X –> Y is surjective, and h: Y –> Z is separated and finite type, and h \circ f is proper, then h is proper. I’m happy to mention this if it is useful, but I’ve never used it. Is this anything that anyone finds important?

The proof of Prop 11.2.3 (that two birational integral separated schemes must have isomorphic open sets) is not transparent. Is there a better explanation? (Bjorn sent me an improved version just over a month ago, and I just looked at it, and I’ve just asked him about some details. So this may soon be fixed anyway.)

**For everyone.**

I find Theorem 11.1.19 (the “cancellation theorem for a property of morphisms”) remarkably and unexpectedly useful, in much the same way as the magic diagram. I hope I’ve made this clear. I want to give it a memorable name, and this is the best I could come up with. Does anyone have a better name?

Can someone (e.g. someone who already knows it) read the proof of Algebraic Hartogs’ Lemma (12.3.11) and Krull (12.4), and point out the parts that need renovation? I want these proofs to be short but readable (with work), so people won’t need to read them, but feel reassured by their presence.

Question: can you give a short clean complete proof that quasifinite morphisms to (Spec of) a field are finite? (Not acceptable: “Use the Nullstellensatz.”)

December 17, 2010 at 8:43 pm

Ravi, some thoughts.

For the discussion of geometric points, you should distinguish an important special case, namely the “algebraic” geometric points, meaning that is algebraic over the residue field at the physical image point (i.e., is an alg. closure of ). It is really important/useful to observe that for checking a property “on geometric fibers”, in all reasonable situations it suffices to check at the algebraic geometric points (all this amounts to is that the property should satisfy the axiom that it cannot be gained or destroyed by scalar extension between alg. closed ground fields), even though in real life we also want to use the non-algebraic ones (such as complex points of a variety over Q). This also takes care of the obvious set-theoretic quantification issue. So clarify that 10.4.J is well-designed to handle the sufficiency of “algebraic” geometric points.

[I’ve now rewritten and expanded this discussion, with the intent of emphasizing the point you are making. Based on feedback in the 2011-12 course, I think people appreciate this discussion. – R.]In 10.4.5 you say that radiciel (not radicial) is “more useful” in positive char, but more accurate to say it is “more subtle”. The concept is equally useful without regard to characteristic (e.g., normalization of cusp is radiciel, and integral radiciel surjections are the universal homeomorphisms, etale radiciel maps are open immersion, etc.).

[Radiciel morphisms will indeed be added. It is on my to-do list, which is currently shrinking. — R, March 27, 2012]10.4.4 should be “Hard Exercise”, not “Hard Fact”. Just warn that it requires knowledge of the Nullstellensatz.

[Now done, with hints. -R]Speaking of the Nullstellensatz, I still don’t understand the rationale of developing up to this point of the theory and yet resisting having developed the Nullstellensatz earlier. (Hilbert would probably have a heart attack if he found out.) It seems a bit bizarre.Response from Ravi Aug 23 2011: Nullstellensatz now done with Chevalley — you and Matthew and others have convinced me that this is by far the best and most obvious thing to do.

Is Noether normalization also illegal? (There’s a pretty easy argument using that.) Honestly, “quasi-finite = finite” for finite type schemes over a field *should* be a simple exercise, and if it’s not then to me it says there’s something wrong with how the development has been set up, not that we should find an ad hoc proof avoiding basic facts from commutative algebra.

Response from Ravi Aug 23 2011: Now this is a simple exercise, right after Chevalley, following Matthew’s advice.Or put another way, the Nullstellensatz seems like such a basic source of geometric insight into finite type schemes over a field, so why wouldn’t one want to have it available pretty early in the theory? Or why can’t 12.1 and 12.2 appear *much* earlier in the story? The subject is “algebraic” geometry, and that bit of algebra seems very geometrically compelling.

Response from Ravi Aug 23 2011: now done in chapter 8.I agree with Bjorn: ignore your spell-checker on “constructible”.

From Ravi Aug 23 2011: done.As for the result you mention from Mumford, that sort of thing is definitely used a lot in arguments with Chow’s Lemma (e.g., the good behavior of the condition “proper” in the direct limit game, the proof that properness interacts well with analytification for finite type morphisms, etc.). Try to prove that a map of finite type C-schemes is proper when the analytification is proper, and I guarantee you will want this fact.

Rather than semi-colons or colons for points of a projective space, why not simply use commas? The square brackets seem entirely adequate to indicate “homogenous coordinates”.

Ravi on Aug 23 2011: I’m convinced. It will be done at some later point. Ravi on March 27, 2012: done some time ago.For 11.2.B, replace “fully faithful” with “faithful”.

(Ravi Aug 23 ’11: done.)Augment the exercise by asking them to find a counterexample whenever k is not separably closed.For 11.2.3, probably the novice reader would gain much insight by first seeing a direct proof in the special case when everything is finite type over a field, where it can be proved by hands-on methods (which have the virtue of explaining why the concept was isolated in the first place and why birationality is a very natural geometric concept; it must seem quite weird from the super-general point of view without some classical awareness, but as Mumford wrote, it just falls into your lap when thinking about quasi-projective varieties — in contrast with everywhere-defined morphisms, that only seem natural from the viewpoint of schemes).

December 19, 2010 at 2:37 pm

A proper response to this (and later) messages will come in due course.

(Update Aug 23 11: partial responses are now embedded in your original comment.)A quick comment: after a brief e-mail discussion with Brian Conrad and Johan de Jong (in which Ofer Gabber was invoked), I think the consensus is for “radicial” rather than “radiciel”.A response/question for Brian on “radicial in characteristic 0”: perhaps “useful” was not the right word. What I’d like to get across is my sense that “characteristic 0 people” (whoever they are) might have less use for the notion. I’ve never needed this notion. I’d like to point out the kind of person who might run into this, and to make sure that everyone else sees it as a fun aside. (Because already there are a huge number of definitions. I’d like to be careful to clearly mark the ones which are “optional”.) So this leads to a question for the broader public: who uses the notion of “radicial morphism”? This is likely a good question for Math Overflow, so I have asked it there.

December 20, 2010 at 10:24 pm

Ravi, I agree that this notion is not as important as others being introduced, but there is something deeply satisfying about understand that radicial = universally injective. For instance, it completely de-mystifies those examples where injectivity is destroyed by base change (contrast with surjectivity being always preserved by base change).

In the same spirit (and I know I am beating a dead horse), why not make the exercise that for a map between finite-type schemes over an alg. closed field $k$ (any char.), it is injective on $k$-points iff it is radicial in the scheme sense. That illuminates the scheme-theoretic significance of classical injectivity.

For example, if a map between finite-type schemes over Qbar is injective on Qbar-pts then it is injective on C-pts. (See, this can be interesting in char. 0…) True, one doesn’t actually need to introduce the concept of “radicial” to prove this, but the argument one goes through in its proof is exactly the same as the sol’n to the exercise I just proposed. So maybe you have used ideas around radiciality (is that a word?) without recognizing it explicitly…

December 21, 2010 at 5:48 am

Dear Ravi,

I had originally been unsure about the wisdom of including radicial morphisms, for similar reasons to you: there are many, many definitions already, and so one wants to avoid adding more if at all possible.

However, on reflection, I think I am inclined to agree with Brian. The thing that has me most sold on this is the idea that most (maybe even all) of the properties of radicial morphisms can be developed in a series of exercises, which are of strong pedagogical value.

To wit: suppose that one wants to determine whether a morhpism is universally injective. One gets to apply the following concepts in turn: a map of sets is injectivce if and only if each fibre contains at most one point; the underlying set of the scheme-theoretic fibre is the fibre of the map on underlying sets; the fibre of a base-change is the base-change of the fibre. With these three ideas, the question turns into one about a question of classical field theory, which is interesting and not too difficult (and having the students think about field extensions is necessary at some point anyway, since it an important aspect of dimension theory).

This also suggests a place to include it: in the exercise section of the discussion of fibre products.

Best wishes,

Matt

August 23, 2011 at 9:53 am

Partial responses are embedded in the original comment. Some are still in process!

December 17, 2010 at 9:22 pm

Dear Ravi,

Suppose that A is a finite type algebra over a field k that is not finite. Then there is an injection k[x] –> A (choose a generator of A over k that is not algebraic over k, the existence of which is the practical meaning of non-finite in the context of a finite type k algebra), giving a dominant map Spec A –> affine line. By Chevalley’s theorem, the image is open, and in particular Spec A contains infinitely many points, and so is not quasi-finite over Spec k.

This is a typical foundational application of Chevalley, and the same kind of argument in fact proves the Nullstellensatz. (This is the way it is done in the Mumford–Oda notes, for example.) This is more or less the same as the usual Noether normalization proof, but Chevalley is a little softer than Noether normalization, so I prefer this argument. (For more musings on possible proofs of the Nullstellensatz, see this MO post.

(Apologies in advance if this doesn’t actually make a link; I’m not sure how to put links in wordpress.)

I agree with Brian that the Nullstellensatz should have appeared by now; the nice thing is that, with your choice of development, it logically fits immediately after proving Chevalley, which seems like a good place for it to be anyway. (You probably don’t need it before then, since you are deep in the foundations, but if you do it there, you will then have it available as you move into more explicit geometry.)

By the way, I also think you should include Noether normalization, but you should give the geometric proof (take projective closures, project from a point in the hyperplane at infinity, and use the fact that projective plus quasi-finite implies finite) rather than the more common algebraic one. That way you get to prove a useful result and at the same time introduce a number of memorable geometric techniques.

Now let me apologize: this comment is more opinionated than usual, but I guess these are points in the foundations that I’ve pondered quite a bit, and have strong opinions on!

Best wishes,

Matt

December 18, 2010 at 4:36 pm

Oops, I just realized, after rereading Brian’s comment and the actual notes, that you *do* prove Noether normalization, but more or less algebraically rather than geometrically. Let me register a minor protest at this.

Best wishes,

Matt

August 23, 2011 at 9:57 am

Hi Matt,

Thanks for your opinionated comments — I am completely convinced by everything but your 2nd last paragraph, on the proof of Noether normalization (see also your follow-up). And I’m basically in agreement with the spirit of your 2nd last paragraph. But the tools to make this precise don’t come until much later (the fact that projective plus quasifinite implies finite). I interpret Nagata’s proof as actually *being* geometric, and I have a paragraph in the notes making the case. My current intent (not yet implemented — but it should happen with ~60% certainty) is to give the argument you describe much later, and to refer forward to it from where I give Nagata’s proof.

All the other changes are now implemented, and the notes are much better because my own understanding is much better, thanks to your advice.

December 17, 2010 at 10:09 pm

Dear Ravi,

Regarding Ex. 12.3.C, why *don’t* you prove the general statement about intersections in projective space. You are basically just one step away: you only need to add the “form the product and intersect with the diagonal, which is linear” argument to go from the linear to the general case. Do you plan to do this later? (I always thought this was a really nice reduction step, which is why I bring it up.)

January 8, 2011 at 5:10 am

Dear Matt,

This is certainly worth including, as it’s beautiful, natural, believable, and introduces a useful technique. This is (literally) on my to-do list. I first need to resolve one important issue: how to show that the product of two varieties doesn’t have any components of smaller than expected dimension. In the forthcoming “regular sequences” chapter, I’ll also do this for intersections of things on smooth varieties.

December 18, 2010 at 6:24 am

In my class I proved the Hilbert Nullstellensatz right after doing Chevalley’s Theorem as Matt mentions above. Then I had the students do the easy exercise (see Brian’s comment) that a quasi-finite scheme over a field is finite over that field. Moreover, I then had them prove that a finite type morphism of affines f : X –> Y is finite in the neighbourhood of any generic point y of Y such that f^{-1}(y) is finite.

This is useful when you discuss what it means for a morphism of schemes to be “generically finite”, see the section “Generically finite morphisms” in the chapter on morphisms of the stacks project. Note that you don’t need ZMT for this.

[Changes in this spirit made some time ago – R, March 27, 2012]August 23, 2011 at 9:59 am

Thanks Johan, this (Nullstellensatz right after Chevalley, then the easy exercise) is absolutely the right way to go. Change now made.

December 19, 2010 at 5:38 am

Dear professor Vakil,

I was busy applying for graduate program so I missed the previous note, but I will be catching it up gradually.

your last reply : 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?

-> It is clear what you mean, but I just wanted to point out that Spec S^{-1} B is in general neither open nor closed in Spec B because at first I suspect that there might be some relation as the quotient case does. Never mind.

[My responses are in square brackets. Now addressed in 4.4.I(a), in September 2011 version.]6.2.I : I think you did not define Noetherian scheme yet.

[Thanks, moved forward.]6.3.2 line 12 : where an affine scheme has property R if and only if and only if … I think this sentence is weird.

[It is weird, and I’ve now broken it up into two, but can’t see how to fix it further.]p.130, 149, 174, 198, 218, 225 : finite-type -> finite type

[I’m using the hypenated version for the adjectival version.]7.3.6. line 8 : were -> where

[fixed, thanks]7.5 line 3 : We will later see that in good situations that … ??

[fixed, thanks!]7.5.8. line 1 : We now describe two examples of curves C such that do not admit a … ??

[fixed, thanks!]7.5.L. line 2 : show that x(t) .. -> show that if x(t) …

[fixed, thanks!](a) line 4 : P, P-Q, P-lambdaQ -> we need Q

[fixed, thanks!]line 7 : p and q -> P and Q

[fixed, thanks!](b) line 3 : q^3|sr -> q^3|s^2

[fixed, thanks!]7.6.C. line 3 : represented -> representable

[It was the earlier definition that was wrong; now fixed. We say a functor is “represented by” something else.]chapter 8 line 9 : every affine open U in X -> U in Y

[fixed, thanks!](8.3.4.2) : Gamma(X, O_X)_s -> Gamma(“D(s)”, O_X)

[Nicely done! I can’t believe no one caught that earlier!]8.3.5. : does the notation s_i mean s_i, f^#s_i, h^#s_i simultaneously? I got confused.

Example 2 (p.172 line 1) : A/I is generated as a A]-module …

[fixed, thanks!]Example 3 : D(t^2 − 1)) -> D(t^2 – 1)

[fixed, thanks!]8.3.H. : What is A?

[fixed, thanks!]8.3.K. : then we can throw out everything in A outside y by modding out by p; you can show that the preimage is A/p. … I do not understand this hint. What do you mean?

line -3 : Do you intentionally use (Ap)/pAp, not Ap/pAp?

line -2 : that ?

[This problem is now edited significantly.]8.3.M. line 3 : Show that A is surjective -> Show that f is surjective

[Thanks — this exercise is wrong, and now removed.]p.174 line -4 : a scheme X is of finite type over k means … ?

[Now fixed, thanks!]8.4.1. line 6 : The image is the plane, with the x-axis removed, -> y-axis

[Now fixed — I can’t believe no one caught that earlier!]pf of Them 8.4.5. line 5 : If we could show this for all points of pi(Z) .. pi(Z) -> Spec A

[Good point — fixed!]line 15 : by multiplying by -> by multiplying

[I think that’s okay as it stands]a sufficiently high power x_i^n -> x_i^N

[fixed, thanks!]p.181 line 7 : f1,…, fa are annihilate the generators h1,…, ha …

[Ithinkthis is okay as it stands.]p.183 line -2 : when the information … determine … ?

[now fixed!]9.1.G.(d) : we need V in front of y^2-x^2 and y.

[Now fixed, thanks!]Figure 9.1. : we need Spec in front of k[x,y]

[Now fixed, thanks! Amazingly missed by everyone else.]9.2.B. : don’t we need algebraic closedness?

[The phrase “make sense of this” gets around this issue.]What is H?

[Fixed, thanks]line 5 : line H -> line L ?

[Fixed, thanks]9.3.3. line 15 : cover if -> cover

[Fixed, thanks]10.1.3. fancy remark : products of schemes are not

necessarily are products .. ?

[Fixed, thanks]p.208 line -8 : it would be better to remove Spec.

[Fixed, thanks — again, I’m surprised no one else caught this]p.212 line 4 : Trivially (Ra) implies (Rb) implies (Rc) … ?

[I think that’s okay. Is if it isn’t clear that this means (Ra) implies(Rb), and (Rb) implies (Rc), just let me know.]

pf 2 of 11.1.5 : x_0^2 and x_0x_0 are used in different places. Is it done intentionally?

[It was, but I realize it was more confusing than enlightening, so it is now undone.]11.1.13 pf line 1 : two arrows are different. Is it done intentionally?

[No, now fixed.]right before 11.1.19 : I find this result unexpected useful and ubiquitous … ?

[Now fixed.]I hope that my effort is helpful for you.

Best,

PhilSang

April 24, 2011 at 11:29 am

It is indeed, and I will indeed respond! I’m beginning to get caught up, and right now I’m just going back to find out what things from long ago I didn’t respond to yet.

September 2, 2011 at 12:49 pm

I have now (finally!) responded to your very helpful comments — thanks once again! My responses are in-line (italicized) above. There is one comment I have left:

8.3.5. : does the notation s_i mean s_i, f^#s_i, h^#s_i simultaneously? I got confused.

March 27, 2012 at 8:21 pm

A very late response: all of these notations refer to pullbacks of the s_i to different spaces. That portion is slightly rewritten, so hopefully it is a bit better.

December 21, 2010 at 2:02 pm

I have a few questions/notes for chapter 12:

The one that interests me most is marked with a star (*).

12.3.E: Do we not have to additionally assume that f is a nonzerodivisor or something?

[No, it is fine as is. But I’ve edited the wording slightly to try to be clearer. “Unit” has become “invertible”. A hint is added. — R.]12.3.5: Why do we need the first paragraph of the proof at all? Is it not true that an element that generates a prime ideal is prime itself and therefore irreducible?

[That’s right; but I’m proving it here in case people hadn’t seen this before. — R.]12.3.11: In the second half of the proof, I think you meant y IN q^{n-1} – I, but you omitted the “IN”-symbol.

[Thanks, fixed. — R.]*12.3.11: I do not understand the last sentence of this proof, namely the first and second inequality (and where Nakayama’s lemma is used). — But could this last sentence not be replaced by an invocation of Krull’s principal ideal theorem instead?

[I’ve now rewritten that proof (following a suggestion of Charles Staats). — R.]12.4.A-12.4.C: Sometimes it is not clear to me what exactly is meant with m.

[I now repeat explicitly that m is a maximal ideal. — R.]Also the statement between 12.4.A and 12.4.B is strange, does it belong into 12.4.1?

[That’s now moved. — R.]Thank you,

D.H.

March 27, 2012 at 8:38 pm

Long-delayed response: thanks for the comments! Many were dealt with earlier, but it still sparked some further changes. Responses are added to your comment, in italics.

December 31, 2010 at 7:42 pm

I do not know whether I should post here or at the old entry, but I went back to 2.5.1.

2.5.1: In the third line from the bottom I think that

eta F * F epsilon = id

should be

epsilon F * F eta = id

instead.

January 4, 2011 at 12:54 pm

Thanks, fixed!

January 8, 2011 at 4:56 am

This is related to Matt’s commment above, but raises an issue I hadn’t considered. It is an important fact that the fibered product (over k) of k-varieties (indeed finite type k-schemes) plays well with dimension (the dimension of the product is the sum of the dimensions). I’ve just added this as an exercise (12.2.I in the version to appear ~Jan. 8 or 9). My best first idea is to use Noether normalization (use X -> A^m and Y -> A^n to construct X x Y -> A^{m+n}, where all maps are dominant and finite).

But this doesn’t show that if X is pure dimensional, and Y is pure dimensional, then their product is pure dimensional. For example, we haven’t shown that you can’t take the product of two curves, and get a surface union some stray points.

This seems a bit delicate, for the same reason that I haven’t discussed yet the fact that the product of two irreducible varieties over an algebraically closed field is also irreducible (which I want and need to add at some point). They haven’t seen any flatness.

Less important side point: One might hope that this is true for fibered products of more general k-schemes, but I can’t see any argument, but I don’t see anything natural. (And it’s probably not too important. The most important case of this sort is extension of field, e.g. X = Spec of a much bigger field, but we can use Noether normalization for Y as well.) Or perhaps there is some crazy but entertaining counterexample (likely due to Nagata)?

January 8, 2011 at 7:40 am

Given the lack of opinions otherwise, I will (at some point) change notation to projective space from semicolons to colons.

January 8, 2011 at 3:23 pm

Au contraire, as I think I said above, I don’t see what’s wrong with using commas instead of semicolons or colons. The square brackets already distinguish it from affine space, and the colons are more unpleasant than commas when writing by hand. As for pure-dimensional stuff, one can reduce to the irreducible case using the finite surjection from disjoint union of irreducible components, or alternatively using dense open affine with irreducible connected components (noting that tensor product over a field preserves injectivity of linear maps, hence dominance on the geometric side; no need for flatness generalities to see this).

[Update March 27, 2012: the pure-dimensional stuff is all fixed, in some earlier edit. — R.]January 25, 2011 at 9:26 am

Okay, the colon vs. comma vote is deadlocked at one vote apiece, meaning that the semicolons temporarily remain by default.

March 3, 2011 at 12:00 pm

Although I at first liked your reasoning for using colons rather than commas, after actually using this notation a few times while working things out, I found myself in agreement with Brian Conrad: colons are more unpleasant than commas when writing by hand.

March 8, 2011 at 10:28 am

Charles, your timing was perfect. Before reading your comment, I was going to sit down and start changing semicolons to colons. I’ll at least hold off; but at the very least semicolons seem to be ranked below colons and commas, and I’m now leaning toward commas.

August 23, 2011 at 10:01 am

Update: I will change semicolons to commas, gradually.

January 25, 2011 at 9:25 am

Johan pointed out that it makes sense to prove that the product of two irreducible varieties over an algebraically closed field is an irreducible variety directly using the Nullstellensatz, rather than making it a consequence of much harder general discussion. That is now added (in the version to appear after January 25), as exercises 10.4.E (well hinted) and 11.1.E.

March 3, 2011 at 9:55 am

In mid-February, Johan de Jong wrote to me: “I think exercise 12.2.D [now 12.2.E: the fact that codimension is the difference of dimensions for varieties] is too hard. The easiest way I know how to do it is to argue as in Mumford’s red book (what Mumford calls an argument of Tate using a trick with norms).”

I agree it is too hard, and I want to fix it.

Update August 23, 2011: proposed fix is in my response below.Junsoo Ha gave a very nice short explanation (following Eisenbud’s “Commutative Algebra”) in answer to this homework problem, which I’d saved (by scanning).

Update August 23, 2011:I’ve posted it here (with his permission).August 23, 2011 at 10:10 am

This problem (that codimension is the difference of dimensions for varieties; or basically, that polynomial rings over a field are catenary) was brought up by many people. I intend to add the following short argument. I’m posting it here so people can stop me if I’m being foolish (or can confirm, maybe privately by email, if they buy it). This may look a bit different, but it follows Dima Trushin’s argument here. (Comment to readers wondering about Mumford’s argument from Tate: he proves something I might call Krull’s principal ideal theorem for the transcendence version of dimension. But I find Trushin’s argument fits in well with the Noether normalization theme.)

We need the Going-Down Theorem (which needs to be added).

This can be stated better in the language of rings, but I prefer to write it here geometrically.

First, reduce the problem to the following. If is an irreducible affine k-variety, and is a closed irreducible subvariety, that is maximal among those properly contained in X, then dim Z = dim X – 1. Let dim X = d for convenience.

Next: by Noether Normalization, we have a map corresponding to a finite extension of rings. Then is a closed subvariety of . It suffices to show that is a hypersurface, as we know the dimension of any hypersurface is .

But if it isn’t a hypersurface, then it is contained in an irreducible hypersurface , so then by the Going-Down Theorem, is contained some irreducible subvariety of mapping to , contradicting the maximality of .

September 1, 2011 at 1:18 pm

Very minor typo: p. 262 of the July 21 version, second line should be “then the ideal of denominators *of* in is “; the second “of” was omitted.

September 1, 2011 at 1:26 pm

Thanks, now fixed!

September 1, 2011 at 6:16 pm

I would like to disagree with the implicit suggestion that the proof of the algebraic Hartogs’ lemma is an algebraic trick that provides little geometric insight (although this is a much stronger statement than you actually make). As I see it, the steps of the proof are something like this:

Let be a rational function over , where is an integrally closed Noetherian domain. Let be the maximal open set over which is regular. The goal is to show that either or $Y := latex X \smallsetminus U$ has codimension one. [Note that set-theoretically, is precisely , where is the ideal of denominators of .]

Restrict (or if perhaps expand) attention to the functions defined locally about an irreducible component of . [Disclaimer: when I say something is defined “locally about , I really mean it is defined in a neighborhood of the generic point of . It need not be defined at every point of .]

Show that, replacing by something else if necessary, we may assume that if we multiply by anything that vanishes along , we get something regular in a neighborhood of .

At this point, we have two cases. In the first case, there exists some , vanishing on , such that does not vanish on . Thus, is a unit (locally about ), and one shows that is a uniformizing parameter for .

If no such exists, then a bit of algebraic magic shows that must have been regular to begin with. This is the only really non-geometric part of the proof.

September 2, 2011 at 10:00 am

Hi Charles,

I agree with the premise, and also like your explanation. As you noticed, I didn’t state that, although I did indeed implicitly suggest it. The algebraic magic in that last paragraph is really short (one sentence in the second-last paragraph of 12.3.11 in the July 21 notes), I find your explanation a much more geometrically motivated way of describing the proof of Algebraic Hartogs, and this is worth doing. I’ve edited the explanation following your line of reasoning here, and upgraded it from two stars to one star (I still won’t present the proof in class, but people shouldn’t be frightened of it). It has four parts (the last in two pieces), each of which I find quite digestible. And this is the first time I feel I’ve suitably understood this proof, although I imagine that I’ll understand it even better after digesting it more (or more likely when someone tells me something else enlightening).

September 3, 2011 at 6:01 am

Ravi,

After looking over your revised version, I somehow got the impression that the geometric additions “missed the point”, which led me to think more carefully about what exactly the point was. Here is what I came up with.

First, the key insight that allowed me even to try thinking about this proof geometrically was that the ideal of denominators of gives precisely the closed subset of on which is not defined. Thus, I’d kind of like to suggest you mention this. It could also be useful in other contexts, since it looks to me like the ideal of denominators gives a natural closed subscheme structure to the set of “poles” of a rational function, over very general schemes (e.g., non-integral). This also ties in with the fact that the ideal of denominators behaves well under localization.

Second note: The proof seems to proceed by giving two different notions of the “order of vanishing of along ,” and then showing that these two notions agree. I have not thought about this carefully enough know whether the “codimension-one-ness” is a consequence of this agreement, or just a byproduct of its proof; but I’m fairly sure it’s one or the other.

September 4, 2011 at 12:46 pm

Hi Charles,

I certainly also believe that the geometric interpretation of the ideal of denominators is central to understanding it, so I’ve tried to make this clearer. I don’t yet understand your second note, and look forward to hearing more!

September 4, 2011 at 1:01 pm

On the one hand, it’s easy (although a bad idea in general) to define the order of the pole of along to be the minimum integer such that . On the other hand, we would like there to be some , vanishing to order exactly along (in the sense that ), such that has neither a pole nor a zero along .

The proof shows that such a exists, and in the process constructs a uniformizing parameter for $\mathfrak{q}$. (More precisely, it replaces by something that with a “pole of order one” in the first sense; then the for this new is precisely a uniformizing parameter for .

September 8, 2011 at 1:10 pm

Hi Charles,

I buy what you say, and currently see it as an after-the-fact comment. Do you see it as clarifying the argument for someone seeing it for the first time? I’m not yet convinced, but am open to be convinced in the future.

September 8, 2011 at 4:46 pm

Ravi,

One thing that has occurred to me is that the best presentation might be to give the brief, rigorous, completely algebraic proof you started with, followed by a less rigorous geometric explanation. That way, readers can choose which versions and “cross-references” they prefer. (Part of the problem with the geometric language in this particular proof is that when you put out the necessary steps to make it rigorous, it really interrupts the “flow” of the proof.)

I’ll let you know if I have any further thoughts that I think are worth sharing.

September 29, 2011 at 7:34 am

Great, I’ll look forward to hearing more (if you have more ideas).

February 16, 2012 at 9:54 pm

Hi Ravi,

It seems that the two places (that I’ve found) where you state instances of “Going Down” have problems. When you first state it as Theorem 12.2.12, you actually state “Going Up.” In the other instance, which is Exercise 25.5.D, the second should be . If I’ve got things mixed up myself, I apologize.

March 25, 2012 at 4:31 pm

Thanks! It is impressive that I screwed up Going Down twice! Both are now fixed. (Thanks also to Iurie Boreico for catching the first…)

March 5, 2013 at 7:13 pm

There are two well-known ways to show that projective morphisms are proper: elimination theory and the valuative criterion for properness. Kiran Kedlaya now has a third way, which is great. You can get a copy of it

here. (It is part of the algebraic geometry course he is teaching this spring.)The fundamental theorem of elimination theory is (in my mind) not fantastically motivated. The valuative criterion (if you actually prove it) is quite hard and not easy to remember (again, in my mind). Kiran’s approach is genuinely a third way. It is elementary. And if you read it, your mind is prepared to understand how and why valuations are important.

It is rare to get a new take on something so fundamental. If anyone else has seen this argument, please let him know — perhaps most easily by just letting me know, so you don’t waste his time.

March 8, 2013 at 9:42 pm

I don’t know why proofs of the fundamental theorem of

elimination theory deserve much fuss. For what it’s worth,

here’s another, in notation like that of 7.4.8, but with

A = R_0 in any ring R graded by nonnegative integers.

Let Z be a closed set of Proj R, defined from a homogeneous

ideal I of R, with \pi: Proj R -> Spec A as usual. Suppose

p in Spec A is not in \pi(Z). In the localization R_S of R

at the complement S of p in A, the ideal generated by p and I

must be the whole ring. In R, some sum i+x lies in S, where

i \in I, x \in pS. Let f be the 0-degree part of i. Then,

in Spec A, p lies in D(f), which is disjoint from \pi(Z).

[Also, Proj R can be regarded as a subspace of Spec R, and \pi

is the restriction of another closed map Spec R -> Spec A.]

March 9, 2013 at 4:09 am

Oops – last 2 lines were a hasty addition; that idea at end fails.

March 16, 2013 at 6:10 pm

My last posts had ideas too incomplete to be comprehensible.

After more analysis, for me the most satisfying way to remove

all mystery about what is going on in the fundamental theorem

of Elimination is nothing slick, but a return to basics:

Recall (4.5), the construction of Proj for a ring S graded

over the nonnegative integers, where localizations by

homogeneous elements f put us in rings graded over Z, and

Ex. 4.5.E provides a crucial bijection. Given a subset I

of S, let V(I)_f = V(I) \cap D(f) in Proj S, or harmlessly

the set of homogeneous prime ideals of S_f that contain I.

Let k = deg f. The corresponding subset of Spec (S_f)_0

is V(J), J the sum of all I_{ki}/f^i, the “I_0” in S_f,

after replacing I by the homogeneous ideal it generates.

Version of Elimination: Assuming S generated over A=S_0

by finitely many f, say homogeneous, and given I as

before, V(I) has a finite cover by sets V(I)_f. We now

only need to see that Spec (S_f)_0 -> Spec A is closed.

One can show that the first ring is f.g. over A. Then

reduce to showing Spec A[X_1, ..,X_n] -> Spec A, say

with A an integral domain, is closed. That is easy.