This semester (Spring 2019) I taught our first year, second semester algebraic geometry course for graduate students as a series of ask me anything sessions. This blog post is a report on how it worked and whether I would advise you to do the same when you teach your course.

Some background info: The enrollment was 10 students (including 4 undergraduates) but the course was attended by a few additional graduate students. Here is a link to the course webpage.

Why did I do this? First of all, I have taught algebraic geometry for graduate students many, many times and I have never taught it the same way twice. When I teach these courses, I am trying to “level up” our graduate students: I am trying to show that there is a well defined language of algebraic geometry with very sound foundations, that you can enjoy building more complicated things on top of this, that you can read about fantastic new methods/results in papers and books, and that you can do all of these things yourself actually. Maybe I can put it simply this way: I want to get them to the point where it would be fun for me (and hopefully others) to have them as graduate students. So what is more fun than a student who comes up with good questions? My second motivation was a more personal one, namely I wanted to have a bit more direct contact than usual with our first year graduate students.

How did it work? Well, I simply went to the lecture room and asked if there were any questions. After an awkward silence, usually lasting not more than one or two minutes, somebody would ask a question. Most of the time I would answer. Then another question, and so on. Often the answers were longish because of the backtracking necessary to give sufficient background in order for the answer to make sense.

Initially, the students were reading along and asked questions which allowed me to more or less develop basics of scheme theory. I tried to stimulate questions from people who preferred to think about only classical varieties, as I think this is a perfectly sound way to do algebraic geometry. Roughly in the middle of the course, in one of the lectures I limited myself to 3 minutes per answer. This lecture I think worked rather well, because I ended up getting more different questions from more different people in the room. Later I introduced another ingredient, namely, 20 questions for schemes! The students surprisingly quickly guessed I was thinking about the degree 5 Fermat surface over the field with 7 elements. I repeated this the next time, where the student, again surprisingly quickly, found that I was thinking of the spectrum of the dual numbers over the complex numbers.

An ingredient of the course which I quite enjoyed was that I was able to talk about the de Rham complex and its cohomology, both in characteristic p and in characteristic 0. Namely, the natural question arose to what extend we can find the “usual cohomology” if you are an algebraic geometer who isn’t willing to use inequalities between real numbers (such as myself). I was then also able to tie this in with coherent duality and the relationship between dualizing sheaves and sheaves of differentials.

For each of the lectures, I wrote, from memory, an account of the mathematics discussed during each meeting. You can find these on the webpage of the course linked above. As you can see, we covered an acceptable range of topics during the lectures. Also, I gave problem sets based on the topics discussed previously. (Since you do not know what the students are going to ask, you cannot give problems foreshadowing the upcoming material.) Finally, I decided to give an oral exam to the graduate students and assigned final papers to the undergraduates.

What worked well?

1. The students were amazing: they asked good questions whose answers they really desired to know.
2. You don’t need to prepare the lectures!
3. You get more buy in from your students.
4. You can cover more diverse topics.
5. Different answers can be at different levels (some of the students may already know more theory and you can occasionally discuss material which is more advanced).
6. You can react to the interests of the students in real time.
7. This structure gives you an opportunity to skip worn out paths and reorder the material drastically.
8. Students will help you prove things as you are working through them on the board.
9. I found that I was doing more examples this way. For example, we discussed the blowing up of the cone over a conic along a line through the vertex to show that blowups mayn’t be what you think they are from nice pictures in books.
10. You can skip over annoying verification without the students noticing. You may think this is a drawback, but more and more I think this is a useful and necessary evil of teaching algebraic geometry.

What didn’t work so well?

1. You don’t always remember the slickest proof of every result you talk about.
2. The order of the arguments isn’t always the right one.
3. Students may never be sure that they have completely grokked all of the material up to a certain point. The material is not presented in a linear fashion. They’ll have to do a lot of work themselves.

The last point is perhaps the most serious drawback. However, IMHO it is impossible to teach algebraic geometry and truly cover all the details needed to get a sound theory. It is necessary for those who intend to work in the field or intend to use AG in a serious manner to sit down and work/read through a good deal of the material by themselves.

Would I recommend teaching a graduate course in Algebraic Geometry this way? Actually, no. First of all, if you are teaching algebraic geometry for the first time, I recommend choosing a good text to work with and sticking fairly close to it; or choose a topic, for example linear algebraic groups and representations, choose a good book for it, and aim the development of your algebraic geometry towards the topic (maybe the book already discusses some algebraic geometry in the first chapter). On the other hand, if you have taught algebraic geometry already multiple times, then you know why it is a difficult thing to do. You probably have already found a best possible method of teaching the course for you, and there is nothing I can say to change your mind!

Also, personally, the next time I will yet again use a completely different method!

Let p be a prime number. Let k be a perfect field of characteristic p. Let U be a smooth variety over k. Choose a compactification U ⊂ X over k such that X is smooth over k and such that the divisor D = X – U is a strict normal crossings divisor D = D_1 ∪ … ∪ D_n. Then we can define the log de Rham complex Ω_X^*(log D) and try to define

H^*_{dR, log}(U) = H^*_{dR}(X, Ω_X^*(log D)

I would like to know is whether there is a published/online proof of the independence of the choice of the compactification provided one has a sufficiently strong form of resolution of singularities (RS). I did the calculation myself on a napkin (see explanation below), but it’d be great if somebody can point to a more honest writeup. Of course I searched the web for a while… also I think one of my students told me this calculation works and somebody else (maybe Illusie himself?) told me a student of Illusie worked on a better version of this a while ago? Does this ring a bell?

What do I mean by RS? Well, I mean that we can go from any compactification to any other one by a sequence of good blowing ups and good blowing downs. A good blowing up of an X as above is one which has an irreducible smooth center Z contained in the boundary (of course) such that for any I ⊂ {1, …, n} the intersection Z ∩ ⋂_{i in I} D_i is either all of Z or empty or a smooth closed subscheme of Z of codimension equal to the number of elements of I. (Aside: I think that embedded resolution of singularities will imply RS.)

Assuming RS the independence claimed above boils down to a local calculation. Think of affine r + s space A^{r + s} as the spectrum of k[x_1,…,x_r,y_1,…,y_s] with divisor D given by x_1…x_r = 0. A good blowing up looks etale locally like the blowing up of A^{r + s} in the ideal generated by x_1,…,x_{r’}, y_1, …, y_{s’} for some 1 ≤ r’ ≤ r and 0 ≤ s’ ≤ s. This blowing up is clearly equal to the blowing up of A^{r’ + s’} times the other factors. By a suitable Kunneth argument for logarithmic complexes this reduces us to the case r = r’ > 0 and s = s’. OK, so denote b : W → A^{r + s} this blowing up with exceptional divisor E isomorphic to P^{r + s – 1}. What I did was compute the cokernels of the maps

b^*Ω^i_{A^{r + 1}}(log D) → Ω^i_W(log b^{-1}D)

for all i. My napkin calculation for i = 1 showed the cokernel to be equal to Q = O_E(-1)^s. For notational convenience set S = O_E^r. Then for i = 2 my calculation gave a cokernel with a filtration having 3 graded pieces, namely

S ⊗ Q, ∧^2(Q)(1), ∧^2(Q).

For i = 3 we get graded pieces

∧^2(S) ⊗ Q, S ⊗ ∧^2(Q)(1), S ⊗ ∧^2(Q), ∧^3(Q)(2), ∧^3(Q)(1), ∧^3(Q).

And so on. If correct (caveat emptor), these cokernels have zero cohomology in all degrees. (Note that Q has rank s which is ≤ dim(E) because r > 0.) Hence the displayed arrow defines an isomorphism on cohomology and we get the desired isomorphism on logarithmic de Rham cohomology because Rb_* b^* = id on locally free coherent modules.

Looking forward to your comments!

Edit 3/16/2020. Both Ofer Gabber and Burt Totaro emailed me to say that the result is in the paper: A. Mokrane. Cohomologie cristalline des varietes ouvertes. Rev. Maghrebine Math. 2 (1993), 161-175.

For whatever reason I really enjoy results that tell us certain schemes are affine. Here is a list of a number of results of this nature in the Stacks project (but only those which deal with schemes — there are analogues of most of these results when we look at algebraic spaces and algebraic stacks):

Tag 02O0 A scheme whose underlying space is finite discrete is affine.

Tag 01PV The nonvanishing locus of a section of a line bundle on an affine scheme is affine.

Tag 0C3A Let Y be a locally closed subscheme of an affine scheme X and assume there is an affine open U of X such that Y ∩ U is affine and such that Y ∖ U is closed in X. Then Y is affine.

Tag 04DE If X → Y is a homeomorphism onto a closed subset of the affine scheme Y then X is affine.

Tag 01XF Vanshing of higher cohomology for quasi-coherent modules implies affine.

Tag 0EBE If X is quasi-affine and H^i(X, O_X) = 0 for i > 0 then X is affine.

Tag 0EBR Suppose you have a reflexive rank 1 module L over a local ring A and a section s of L such that s^n is contained in m_A L^[n]. Then the locus where s doesn’t vanish is affine. This generalizes the case of invertible modules mentioned above.

Tag 05YU If X → Y is surjective and integral (for example finite) and X is affine, then Y is affine.

Tag 09NL If a scheme X is the union of finitely many affine closed subschemes, then X is affine.

Tag 0A28 If X is a curve and not proper, then X is affine.

Tag 0F3R If f : X → Y is a morphism of affine schemes which has a positive weighting w, then the set V of points y of Y such that the total weight over y is maximal is an affine open of Y. For example, if f is etale, then V is the maximal open of Y over which f is finite etale. Other cases where one has a weighting are discussed in Lemmas Tag 0F3D and Tag 0F3E

Tag 0EB7 The complement of a 1 dimensional closed subset of the spectrum of a 2 dimensional normal excellent Noetherian local ring is affine.

Tag 0ECD Let f : X → Y be a finite type morphism of excellent affine schemes over a field with X normal and Y regular. Then the locus V in X where f is etale is affine. (This should be true without assuming Y to be over a field.) This result is a strengthening of purity of ramification locus which itself is a result of Gabber you can find in section Tag 0EA1.

I hope you enjoy this kind of result as well! If you know addtional results of this nature, please leave a comment or send me an email. Thanks!

Yesterday I finished working through the comments that were left on the Stacks project site since late January. Thanks for all of these and please keep them coming (both online and by email)!

In this blog post we (partially?) answer one of the questions posed in this mathoverflow post. Namely, let A be a complete discrete valuation ring with uniformizer t and fraction field K. Let X = P^1_A be the projective line over A. Let (X^h, O^h) be the henselian scheme you get by henselizing along t = 0.

Lemma. If the characteristic of K is zero, then H^1(X^h, O^h) is not zero.

Proof. Recall that the underlying topological space of X^h is the projective line over the residue field of A. Consider the standard open covering of this projective line. Then the first Cech cohomology for O^h with respect to this covering is the cokernel of the map A[x]^h x A[1/x]^h → A[x, 1/x]^h. Here the henselizations are taken with respect to the ideal generated by t. Since for H^1 Cech cohomology always injects into cohomology, it suffices to show that this cokernel is nonzero.

What does this mean? Well, by Artin approximation the ring A[x, 1/x]^h is the set of algebraic elements of the t-adic completion of A[x, 1/x] as defined in the previous blog post. A similar statement holds for A[x]^h and A[1/x]^h. See Tag 0A1W for an explanation. Thus we see immediately that the result of the previous blog post exactly gives the nonvanishing of the cokernel. QED.

What is mildly interesting is that this counter example doesn’t work if the characteristic of K is p > 0. Moreover, in this blog post we proved that one does have theorem B for henselian affine schemes in characteristic p. It could still be true that there is some good theory of henselian schemes and quasi-coherent modules on them in positive characteristic. Let me know if you have one. Thanks!

In this post I discuss a funny observation about algebraic Laurent series. In a future post I will explain why it is also an interesting fact. Thanks to Will Sawin and Raymond Cheng for helping me figure this out! As usual, all mistakes are mine.

Let A be a complete discrete valuation ring with uniformizer t and fraction field K. Let x be a variable. Consider an element f in the t-adic completion of A[x, 1/x]. Then we can write

f = sum a_n xⁿ

where the sum is over all integers n (postive and negative) and where a_n is in A and tends to zero t-adically as the absolute value of n goes to infinity. We say f is algebraic if there exists a relation

sum P_i fⁱ = 0

where the sum is finite and the P_i are rational functions of x not all zero. Of course by clearing denominators we may assume P_i is in A[x] and not all of them equal to zero. Finally, given f we can of course write

f = fplus + fminus

where fplus is the sum of a_n xⁿ for n ≥ 0 and fminus is the sum of a_n xⁿ for n < 0.

Question. If f is algebraic, must fplus and fminus be algebraic?

It turns out that if K has positive characteristic, then the answer is yes and if K has characteristic zero, then the answer isn’t yes in general. Let’s get to work.

Char p. If f is algebraic, then we can find a relation of the form

sum P_i f^pi = 0

for some polynomials P_i not all zero. Then looking at powers of x we observe that

sum P_i fplus^pi

is a polynomial in x! This of course means that fplus is algebraic. QED.

Char 0. This is more difficult. We claim that the square root f of (1 + tx)(1 + t/x) which starts as 1 + 1/2t(x + 1/x) + … is a counter example. To see this we consider the differential operator

L = 2x(1 + tx)(t + x)D – t(x^2 – 1)I

where D = d/dx and I is the identity operator. An easy computation shows that L(f) = 0. Looking at powers of x the reader easily shows that

L(fplus) = a + bx

for some nonzero a, b in A. In fact, a computation shows that we have

a = t * sum_{i = 0, 1, 2, …} binomial(1/2, i)^2 * t^(2*i)

and

b = sum_{i = 1, 2, 3, …} -t^(2*i) * binomial(2*i – 2, i – 1) * 4 * 3 * (3 – 8*i + 4*i^2)^(-1) * 2^(-i*4) * binomial(2*i, i)

(It isn’t necessary to verify this for what we’re going to say next.)

Lemma. If fplus is algebraic, then fplus is the sum of a rational function and a K-multiple of f.

Proof. Observe that the operator L’ = (a + bx)D – bI annihilates a + bx. Hence we see that fplus is a solution of L’L. Choose an embedding of K into the complex numbers C. On some affine open U of P^1_C we see that L defines a local system V of rank 1 with finite monodromy; in fact the monodromy group is of order 2 because L is of order 1 with 4 regular singular points whose residues are each a half integer (you can also see this without computation by considering how we chose f in the first place). On the other hand, L’ defines a local system V’ of rank 1 with trivial monodromy (as it has a solution which is a rational function). Finally, L’L defines an extension W of V’ by V, i.e., we have a short exact sequence 0 → V → W → V’ → 0. The monodromy of W is either finite and then the extension is split, or the monodromy is infinite and the extension is nonsplit. Of course fplus defines a horizontal section of W over some small disc. Since L(fplus) is nonzero, we conclude that fplus isn’t in V. If fplus is algebraic (before embedding into C then of course also after embedding K into C), then we conclude that the monodromy of W is finite and the extension is split over all of U. This means there is a rational function h over C with L’L(h) = 0. By a change of fields argument (omitted), this implies we can find a rational function h over K with L’L(h) = 0. Then any solution, and in particular fplus, is a complex linear combination of h and f. QED

OK, so now we know that we have fplus = P/Q + c f for some constant c and some P, Q in A[x] with Q not zero. Multiplying through by Q we obtain Q * fplus = P + c * Q * f. Looking at very negative powers of x we conclude that c = 0 but looking at very positive powers of x we conclude that c = 1. This contradiction finishes the proof.

This post is a follow-up on this post. There we gave an example of a henselian affine scheme which does not satisfy theorem B.

In this post p will be a prime number and we will show that for an affine henselian scheme in characteristic p we do have theorem B. In fact, it turns out this is almost an immediate consequence of Gabber’s affine analogue of proper base change. I’m a little embarrassed that I didn’t see it earlier. The trick is to use the following trivial lemma which will be added to the Stacks project soonish.

Lemma. Let X be an affine scheme. Let F be an abelian sheaf on the small etale site X_{et} of X. If H^i_{et}(U, F) = 0 for all i > 0 and for every affine object U of X_{et}, then H^i_{Zar}(X, F) = 0 for all i > 0.

Proof. Namely, let U = U_1 ∪ … ∪ U_n be an affine open covering of an affine open U of X. Then all the finite intersections of the U_i are affine too. Hence the Cech to cohomology spectral sequence for F in the etale topology degenerates and we see that the Cech complex is exact in degrees > 0. But by a well know criterion this implies vanishing of higher cohomology groups of F on X_{Zar}. See Tag 01EW. □

OK, now suppose that X = Spec(A) and that A is one half of a henselian pair (A, I) with p = 0 in A. Let Z = Spec(A/I) and denote i : Z → X the inclusion morphism. The corresponding henselian scheme is gotten by taking the underlying topological space of Z and endowing this with a structure sheaf O^h obtained by a process of “henselization” on affine opens. We prefer to do this as described below (it gives the same thing).

For any quasi-coherent module F on X, viewed as a sheaf of O_X-modules on the small etale site of X, we set F^h = (i_{et}^{-1}F)|_{Z_{Zar}}. This is a sheaf of modules over the structure sheaf O^h = (O_X)^h of our henselian affine scheme, in other words on the underlying topological space of Z.

Theorem B. Let (A, I) be a henselian pair with p = 0 in A for some prime number p. Let (Z, O^h) be the henselian affine scheme associated with the pair. Then H^i(Z, F^h) = 0 for i > 0 and any O^h module F^h coming from a quasi-coherent module on Spec(A) as in the construction above.

Proof. We will show that the lemma applies to i_{et}^{-1}F on Z_{et} which will prove that H^i(Z, F^h) = 0 for i > 0 and this will finish the proof of Theorem B. For any affine object V in the site Z_{et} we can find an affine object U in X_{et} such that V = Z x_X U. Denote U’ the henselization of U along the inverse image of Z. Denote F’ the pullback of F to U’. Then we see that the restriction of i_{et}^{-1}F to V_{et} is just the pullback of F’ to V by the closed immersion V → U’ (pullback in the etale topology as before). Hence by Gabber’s result (Tag 09ZI) we see that H^i_{et}(V, i_{et}^{-1}F) = H^i_{et}(U’, F’) = 0 because U’ is affine and F’ is quasi-coherent. We may use Gabber’s theorem exactly because p = 0 in A and hence F’ is a torsion sheaf! □

Enjoy!

In the paper “Varieties of small codimension in projective space” Hartshorne has the following conjecture.

Conjecture 5.1. Let A be a regular local ring of dimension n. Let P be a prime ideal of A such that A/P has an isolated singularity. Let r = dim(A/P) and suppose that r > 1/3(2n – 1). Then A/P is a complete intersection.

We don’t have any positive results on this conjecture (as stated) this except in the case where r = n – 1. Namely, if r = n – 1, then A/P is a hypersurface as a regular local ring is a UFD.

We do have a negative result, namely, the conjecture seems to be wrong for n = 6 and r = 4. Namely, Tango constructed a nonsplit rank 2 vector bundle on P^5 in characteristic 2. A general section of a high twist of this vector bundle will give a codimension 2 smooth subvariety of P^5 which is not a complete intersection. Pulling this back to punctured affine 6-space will give the desired counter example.

This is related to what Hartshorne says just before stating his conjecture, namely that his local version is actually stronger than the original conjecture. If we don’t make the conjecture stronger then in 5.1 we would put the inequality r > 1/3(2n + 1). And Tango’s example no longer gives a counter example. Let’s call this the corrected conjecture.

Let us, as is customary, strengthen the corrected conjecture as follows.

Conjecture E.1. Let A be a regular local ring of dimension n with spectrum S and punctured spectrum U. Let V be a closed subscheme of U which is a local complete intersection, whose closure in S is equidimensional of dimension r. Suppose that r > 1/3(2n + 1). Then there exists a complete intersection Z ⊂ S with V = U ∩ Z scheme theoretically.

Conjecture E.1 again holds for r = n – 1. Work in the projective case suggests that E.1 is not much stronger than the corrected conjecture 5.1.

Lemma 1. Let A be a Noetherian local ring with spectrum S. Let E be a vector bundle on the puncture spectrum U of A. Let Z ⊂ S be a complete intersection of codimension c such that E|_{Z ∩ U} is a trivial vector bundle. If depth(A) > 2 + c, then E is a trivial vector bundle.

Proof. Using induction on c we reduce to the case c = 1. Then Z = Spec(B) where B = A/f for some nonzerodivisor f in A. Hence depth(B) > 2. Thus we have H^1(Z ∩ U, O_Z) = 0. Thus we see that E is trivial on a formal neighbourhood of Z ∩ U in U by a standard deformation argument. But the functor from vector bundles on U to vector bundles on the formal completion of U along Z ∩ U is fully faithful by a Lefschetz type result, see Tag 0EKS, and hence we conclude.

Lemma 2. Let A be a Noetherian local ring with spectrum S. Let E be a vector bundle of rank c on the punctured spectrum U of A. Let Z ⊂ S be a complete intersection such that Z ∩ U is the vanishing scheme of a global section s of E. If depth(A) > 2 + c, then E is a trivial vector bundle.

Proof. Note that E|_{Z ∩ U} is the normal bundle of Z in U. Since Z is a complete intersection we find E|_{Z ∩ U} is a trivial vector bundle. We conclude by Lemma 1.

Lemma 3. Let A be a regular local ring. Let B = A/I be a complete intersection of codimension t. If c < 1/3(dim(B) – 2t – 1) and conjecture E.1 holds, then any vector bundle of rank c on the punctured spectrum of B is trivial.

Proof. Let E be a vector bundle of rank c on the punctured spectrum of B. Choose a finite B module M corresponding to E. Choose a “random” element s of m_B^N M for some N ≫ 0. Denote V inside the punctured spectrum of B the vanishing scheme of s. Observe that dim(V) = dim(B) – c = n – t – c where n is the dimension of A. The inequality n – t – c > 1/3(2n + 1) is equivalent to the inequality of the lemma and hence E.1 tells us that V is the intersection of the punctured spectrum of A with a complete intersection Z in Spec(A). Because we chose N ≫ 0 we find that we may choose the first t generators for the ideal of Z to be the t generators for I. Thus we see that Z is a complete intersection in Spec(B). Then we conclude that E is trivial by Lemma 2.

The lemma tells us that complete intersection rings B should have few interesting low rank vector bundles on their punctured spectra provided E.1 is true. But the appearance of the term -2t in the inequality is annoying. By Grothendieck we know that any invertible module on the punctured spectrum of B is trivial if dim(B) > 3. In other words if 1 < 1/3dim(B) or put another way 1 ≤ 1/3(dim(B) – 1). So let’s ask the following question (where we have dropped the -2t and replaced < by ≤ in the inequality).

Question E.2. Let B be a Noetherian local ring which is a complete intersection. Let E be a vector bundle on the punctured spectrum of B. If rank(E) ≤ 1/3(dim(B) – 1), then is E trivial?

So for example if the rank is 2 the inequality gives 7 ≤ dim(B). If true this would be sharp by what we said above. Anyway, I don’t insist on the exact formula for the inequality; I’m not sure why Hartshorne chose 2/3 as the leading coefficient in his inequality. Really, a much more reasonable question is the following.

Question E.3. Given an integer c does there exists an integer n(c) such that if B is a Noetherian local complete intersection of dimension > n(c), then any E vector bundle of rank c on the punctured spectrum of B is trivial?

For those who prefer projective geometry over local algebra, we ask whether there exist indecomposable rank 2 vector bundles on complete intersection varieties of arbitrarily large dimension… Please let me know if you have interesting examples! Thanks.

This is a write-up of an exercise I did in my office with Alex Perry and Will Sawin. Namely we made an example where you can’t flip a Weil divisor. I couldn’t immediately find one by googling; I hope this helps those who google; all mistakes are mine. For the exact notion of flip, please see below (it may not be the same as your notion of flip).

Let C be an elliptic curve. Let E = L_1 ⊕ L_2 be a direct sum of two invertible modules of degree 1 on C. Let L be a third invertible module of degree 1 on C. We will assume L, L_1, L_2 are Z-lineary independent in Pic(C). Let p : X = P(E) —-> C be the corresponding projective bundle which with my normalization means that p_*O_X(1) = E. Observe that O_X(1) is ample on the surface X because E is an ample vector bundle on C.

Let A = ⨁ H^0(X, O_X(n)). Then Z = Spec(A) is the projective cone on X wrt O_X(1). Thus X gives a nice threefold singularity. Denote U the complement of the vertex in Z. There is a morphism U —> X. The pullback of L via the composition U —> X —> C is of the form O_U(D) for some Weil divisor (class) D on Z. If we take the closure Y of the graph of U —> C in Z x C then we see that D pulls back to a Cartier divisor on Y which is moreover ample on Y (equivalently relatively ample with respect to Y —> Z). Finally, note that the fibre of Y —> Z over the vertex has dimension 1.

Another way to construct Y is to consider the graded A-algebra

B+ = ⨁ _{d ≥ 0} H^0(U, O(dD)) = ⨁_{d, n ≥ 0} H^0(X, O_X(n) ⊗ p^*L^d)

(with grading given by d) and then Y = Proj(B+). Proof omitted.

OK, so now we can ask: can we flip (Y —> Z, D)? What I take this to mean is that we want to find a proper morphism Y’ —> Z which is an isomorphism over U, whose fibre over the vertex has dimension < 2 and such that -D determines a Q-Cartier divisor on Y’ which is ample on Y’. Note the sign in front of D!

It turns out that Y’ exist if and only if the algebra

B- = ⨁ _{d ≥ 0} H^0(U, O(-dD)) = ⨁_{d, n ≥ 0} H^0(X, O_X(n) ⊗ p^*L^-d)

is finitely generated; you can find this in the literature when you google the question. Using p_*O_X(n) = Sym^n(E) this becomes

B- = ⨁_{a, b, d ≥ 0} H^0(C, L_1^a ⊗ L_2^b ⊗ L^-d).

Thus we get a natural Z^3-grading for this algebra. By our choice of L_1, L_2, L above we see that we have nonzero elements in the graded piece with (a, b, d) = (0, 0, 0) and in the graded pieces corresponding to (a, b, d) with a + b – d > 0. Thus B- is not finitely generated, because the elements in degrees (a, b , a + b – 1) are all needed as generators for the algebra B-.

This is just to record informally a counter example which we found in March 2017. Please let me know if such an example is in the literature and I will add a reference.

Let (A, I) be a henselian pair. On Z = Spec(A/I) with the Zariski topology consider the presheaf O^h which associates to the open V = D(f) \cap Z of Z the ring A_f^h where (A_f^h, I_f^h) is the henselization of the pair (A_f, I_f) = (A_f, IA_f). It is easy to see that A_f^h only depends on the Zariski open V of Z.

Then O^h is a sheaf (on the basis of standard opens of Z), but it may have nonvanishing higher cohomology.

You can deduce the sheaf property from Tag 09ZH if you think about it right.

To get an example of where the cohomology is nonzero, start with Z_0 : xy(x+ y – 1) = 0 in the usual affine plane over the complex numbers. Let R be the henselization of C[x, y] at the ideal of Z_0. Let A be the integral closure of R in the algebraic closure K of C(x, y). Then A is a domain and (A, I) is henselian where I is the ideal generated by xy(x + y – 1) in A. Denote Z = V(I).

The reason for going all the way up to A is that A is a normal domain whose fraction field K is algebraically closed. Hence all local rings of A are strictly henselian and moreover affine schemes etale over Spec(A) are just disjoint unions of opens of Spec(A). See Tag 0EZN. Thus O^h = O_{Spec(A)}|_Z in this case (restriction is usual restriction of sheaves in Zariski topology). In particular, the map O_{Spec(A)} —> (constant sheaf value K on Spec(A)) induces a map O^h —> (constant sheaf value K on Z).

Observe that since Z_0 is a triangle, we have H^1(Z_0, Z) = Z. Let g be a generator of this cohomology group. Then you check that g|_Z is still non-torsion. I do this using a limit argument and trace maps for finite maps between normal surfaces. If you have a clever short argument, let me know. You do have to use/prove something because we can “unwind” the triangle topologically, so your argument has to show that this doesn’t happen (in some sense) for the map Z —> Z_0 of topological spaces.

Next we consider the maps of sheaves

Z —> O^h —> (constant sheaf with value K on Z)

Since g|_Z is nontorsion we see that its image in the first cohomology of the last sheaf is nonzero and we conclude H^1(Z, O^h) is nonzero.

Conclusion: no theorem B for henselian affine schemes. Enjoy!