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)!

# Coh proper henselian schemes

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!

# Algebraic Laurent series

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}x^{n}

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}f^{i}= 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} x^{n} for n ≥ 0 and fminus is the sum of a_{n} x^{n} 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.

# Thm B for hens sch in pos char

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!

# Hartshorne on local algebra

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.

# Nonexistence of flip

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-.

# No Theorem B for henselian affine schemes

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!

# Up to date

This is just a short post letting you know that I have updated the Stacks project by working through all your comments once more. The last time I did this was on May 19 of this year. I try to respond to mathematical errors very quickly, but sometimes I do not realize a comment is pointing out an error until I really sit down and look carefully at the comment. If you’ve left a comment pointing out a mathematical error, feel free to also email. I remind you that it is very helpful to me if you suggest a fix or if you have a counter example to the statement you are objecting to or more generally if you discuss what you think went wrong.

Enjoy!

# Question about links in proofs

Yesterday I got an email asking about links between tags in proofs. Here is the question:

————————–

On the page 00KD in the proof of Lemma 00KK the reader will find the strings “If (00KN)” or “Assume (00KP)”, provided you view the page in “tags” mode. However, these strings are not found in the pdf. You have to click on them to see what they mean, and it turns out to be the items (1), (2), … of Lemma 00KK. How does this work?

Related but different question: on the page 00KA shouldn’t there be a hyperlink to the definition of χ_{M}?

—————————-

The explanation is that in Lemma 00KK there is an itemized list in the underlying latex file. Then I decided to give latex labels to the items so I could refer to them in the proof. If you have a pdf reader which can deal with (internal) links, then you can click on the occurences of (1), (2), … in the proof of 00KK and you’ll be thrown to the corresponding statements in the proof of 00KK.

There are several other places in the Stacks project where the items of an itemized list have latex labels. But most of the time we don’t do this. (It turns out to work to be best for the structure of the Stacks project if each lemma has a single conclusion.) I would appreciate feedback on whether readers think it works well in the cases where we do do this.

I hope that this addresses the first question somewhat.

Second question. Ideally, mathematicians who help out with the Stacks project should only have to worry about helping make the pdfs readable and mathematically correct, etc. The underlying website code will hopefully display the mathematics in such a way that the experience is similar (or better than) reading the pdf. So if you want a “link” back to the definition of χ_{M} in Definition 00KA, then try to change or suggest a change in the latex so that it looks good in the pdf.

Specifically in Definition 00KA we could change the latex code to read

`$\chi_M$ as defined in Definition \ref{definition-chi}.`

except that then we would also need to insert, earlier, a definition environment where we define χ_{M} and φ_{M}. Perhaps this isn’t a bad idea.

Technically speaking (please ignore this): I do not want to add another layer to the underlying latex files. However, in the future we could have a “hover” functionality where hovering would show you the definitions of defined terms… I would want this to be implemented in such a way that no changes need to be made inside the latex files, but perhaps a separate file would be added (similar to the tags file being maintained outside the latex).

# Kerodon

Kerodon is a site modeled after the Stacks project maintained by Jacob Lurie.

1. It uses the tags system for stable references as originally devised for the Stacks project by Cathy O’Neil.

This means that if you reference a tag in the Stacks project or Kerodon, you should make sure to specify which of these two you are referring to (if you use the “cite” links in either project and copy-paste from there this will work fine). All of the references I’ve seen to the Stacks project already do this, so I am not worried.

2. It is running the Gerby website infrastructure with some additional work by Opus Design for a distinct look.

Some of the work done for this by Pieter Belmans will also benefit the Stacks project (this will probably be mostly invisible to the user though).

3. It has a comment system, so please go over there and leave mathematical comments!

4. The mathematics in Kerodon is written and copyrighted by Jacob Lurie.

This is a difference in philosophy: (a) the contributors to the Stacks project collaboratively own the Stacks project and (b) you can directly access the underlying latex files to make changes to submit to the maintainer (me).

5. Currently the pdf version of Kerodon has 85 pages.

The plan is to add more over time. As you can see, the current material covers a tiny fraction of Jacob’s book entitled “Higher Topos Theory”. If you want to know more about what is planned in Kerodon, you’ll have to ask him. But for now, I am kind of taking this opportunity to read Kerodon the way you would read a webcomic. As new material gets added I will head over there and read it. I hope you will enjoy it as well.