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!