OK, I went through all of the comments and we have a brand new version of the Stacks project. Enjoy!

This is just a quick note so if people google this issue they see that one needs to be careful (I think there are some positive things in the literature in special situations).

Let k be a field. Let R be an algebra over k. Let M be an R-module.

For an Artinian local k-algebra (A, m) with residue field k denote F(A) the set of isomorphism classes of pairs (M_A, phi) where M_A is an R \otimes_k A module and phi is an isomorphism of M_A/m M_A with M.

The functor F satisfies condition H1 of Schlessinger’s paper as one can see by using 07RU. Details omitted.

However, the functor F doesn’t satisfy condition H2 in general. As an example, with notation as in Schlessinger‘s Theorem 2.11, take R = k, M = k, A = k, A’ = k[epsilon_1], and A” = k[epsilon_2] where the epsilon have square zero. Then A’ ×_A A” = k[epsilon_1, epsilon_2]/(epsilon_i epsilon_j) and we see that the modules M(c) = A’ ×_A A’/(epsilon_1 – c epsilon_2] for any nonzero c in k map to the same elements of F(A’) and F(A”) for all c. But these modules are all pairwise nonisomorphic.

So of course(!) the functor F in the special case doesn’t have a hull.

I wanted to just present an explicit example of a nonvanishing higher de Rham cohomology group of the spectrum of an Artinian finite dimensional C-algebra.

Consider an element f of C[x, y] where C is the complex numbers. Let ω be a 1-form in x, y such that

d(ω) = f d(x) ∧ d(y)

Such a form always exists by the Poincare lemma for C[x, y]. The form ω will give a nonzero cohomology class in the de Rham complex of A = C[x, y]/(f) unless we can write

ω = d(h) + gd(f) + f η

for some h, g in C[x,y] and 1 form η. Taking d of this relation we find that one needs to have a g and η such that

f d(x) ∧ d(y) = d(g) ∧ d(f) + f d(η) + d(f) ∧ η

This means that with θ = η – d(g) we have

f d(x) ∧ d(y) = f d(θ) + d(f) ∧ θ = d(fθ)

If we write θ = a d(x) + b d(y) then this gives

(*) f = – ∂(fa)/∂ y + ∂(fb)/∂ x

Now we consider an example due to Reiffen. It is carefully written out in the second appendix of 2505.03978 that (*) doesn’t have a solution if f = x^4 + y^5 + x y^4 (see proof of B.8). In fact, the proof shows that there cannot even be a, b in C[x, y] such that (*) holds modulo the maximal ideal (x, y) to the power 6.

Artinian Example. Let B = C[x, y]/(x^4 + y^5 + x y^4, x^100, y^100). Then the de Rham complex Ω^*_{B/C} has cohomology in degree 1. Namely, take the form ω above. If it maps to zero in H^1(Ω^*_{B/C}) then the reader goes through the arguments above and shows that one gets a solution to (*) modulo (x, y)^6 which is a contradiction.

I would welcome a reference for examples of this type (please email me; I will edit the post and put it here). We already have some references to related material in Infinite dimensional de Rham cohomology.

Enjoy!

Recording 2 examples here.

The first is to consider for n > 1 the map

A^n —> P^n, (x_1, …, x_n) maps to (x_1x_2…x_n : x_1 – 1 : … : x_n – 1)

This map is quasi-finite and flat, but it is not surjective as the points (1:1:0…0), (1:0:1:0…0), …, (1:0…0:1) are missing in the image. If we take as homogeneous coordinates on P^n the variables T_0, …, T_n then the inverse image of T_1 + … + T_n = 0 is the hyperplane x_1 + … + x_n = n in A^n. Thus we see

There is a surjective quasi-finite flat morphism A^{n – 1} —> P^{n – 1}.

The map we constructed has degree n and that is also the minimum possible.

The second example is to consider for n > 1 the map

A^n —> A^n – {0}, (x_1,…,x_n) maps to (x_1, …, x_{n – 2}, x_{n – 1}x_n – 1, f)

where

f = x_1x_{n – 1}^{n – 1} + … + x_{n – 3}x_{n – 1}^3 + x_{n – 2}x_{n – 1}^2 + x_{n – 1}(x_{n – 1}x_n – 1) + x_n

This map is surjective, quasi-finite flat of degree n (and again that’s minimal).

Enjoy!

Let A be a ring and let I be an ideal of A. Let (X, O_X) be a space with a sheaf of A-algebras (or X could be a site). Let F be a sheaf of O_X-modules. Set F_n = F/I^{n + 1}F. Then we can ask whether the theorem on formal functions holds in the naive form that

lim H^p(X, F)/I^nH^p(X, F) = lim H^p(X, F_n)

As is shown in EGA this holds if X is a proper scheme over A, the ring A is Noetherian, and F is coherent. Let me give a proof using derived completion for those who’ve not encountered this before.

STEP 1: Cohomology commutes with derived completion. If I is finitely generated, then we have

RΓ(X, F)^ = RΓ(X, F^)

where the wedge means derived I-adic completion on both sides, see Tag 0BLX

STEP 2: Derived completion versus “usual” completion. Provided that X has enough opens that look like spectra of Noetherian rings and F is coherent, then we have that F^ = Rlim F_n. See for example Tag 0A0K. Similarly, if A is Noetherian and the cohomology modules H^p(X, F) are finite A-modules, then the cohomology modules of RΓ(X, F)^ are the usual naive I-adic completions of the H^p(X, F), see Tag 0BKH.

STEP 3: RΓ(X, -) commutes with Rlim. See for example Tag 08U1 (actually we’ve already used this result in the first step).

STEP 4: The pth cohomology of Rlim RΓ(X, F_n) is lim H^p(X, F_n). To see this it suffices to show that the inverse system of modules H^p(X, F_n) has Mittag-Leffler (for all p). By Tag 0GYQ it suffices to show that

H^p(X, F_1) ⊕ H^p(X, IF_2) ⊕ H^p(X, I^2F_3) ⊕ …

satisfies the ascending chain condition as a graded module over

Gr_I(A) = A/I ⊕ I/I^2 ⊕ I^2/I^3 ⊕ …

for all p. This holds as soon as X is proper over A Noetherian and F_0 is a coherent O_X-module (by finiteness of cohomology of coherent modules over proper schemes).

———————

Note that if A is Noetherian and complete wrt I and X is a formal scheme proper over Spf(A) and if we have an inverse system (F_n) of coherent O_X-modules with F_{n – 1} = F_n/I^nF_n giving rise to the coherent O_X-module F = lim F_n then we obtain the same statement with the same proof (in this case completion on the left hand side of the theorem is unnecessary). Of course, it is debatable whether in this case one should really even be interested in the “usual” cohomology groups H^p(X, F)…

OK, I went through all of the comments and we have a brand new version of the Stacks project. The last time we did this was on June 6, 2024. We now have more than 650 contributors. Enjoy!

Let X be an affine variety over a field k of characteristic 0. Then we have the algebraic de Rham complex Omega* of X over k. The algebraic de Rham cohomology H^*_{dR}(X/k) of X over k is then the cohomology of the complex of global sections of Omega* (here we are using that X is affine). So if X is the spectrum of the finite type k-algebra A (which is a domain as X is a variety), then we’re looking at the cohomology of the de Rham complex of A over k.

If X is smooth over k, then Grothendieck proved that each H^i_{dR}(X/k) has finite dimension. The point of this blog post is to clearly state that this doesn’t hold for all singular affine varieties X over k. I could not find a paper literally stating this fact with an explicit example, so I decided to find one myself and present it to you. If someone emails me a reference I would be thankful and would add that here.

A slightly different issue is that if k = C is the complex numbers and X is smooth, then H^i_{dR}(X/C) computes the cohomology of the manifold X(C) with C-coefficients and this is finite dimensional. Googling one easily finds examples where this is false for singular X. Indeed from the work in the references given below it becomes clear that there are many (explicit) examples.

Our example. Let k be any characteristic 0 field, let n > 6 be an integer, and let A be the k-algebra
A = k[x, y, s, t, 1/(st - 1)]/(x^n + y^n + sx^{n - 2}y + t xy^{n - 2})
Let us denote f the polynomial we’re dividing by.

The idea for this came from an example by Brüske mentioned at the very end of Bloom and Herrera, De Rham cohomology of an analytic space, Invent. Math. 7 (1969), 275–296. The purported example is of a singular analytic space where a stalk of the de Rham cohomology sheaf in the analytic topology is infinite dimensional; I didn’t check the example.

In Reiffen Das Lemma von Poincaré für holomorphe Differential-formen auf komplexen Räumen, Math. Z. 101 (1967), 269–284 we find a method to calculate a quotient (!) of the degree 3 de Rham cohomology of X = Spec(A). This quotient itself is the quotient of the k-vector space (f) = fA by the elements bf in (f) of the form
bf = ∂(g_1f)/∂x + ∂(g_2f)/∂y + ∂(g_3f)/∂s + ∂(g_4f)/∂t.
Here is a link to a pdf with a short, but hopefully readable, proof that this quotient has infinite dimension: infinite-de-rham

OK, I went through all of the comments and we have a brand new version of the Stacks project. The last time we did this was on March 3, 2023. We now have more than 600 contributors. Enjoy!

This is to advertise a 2 year research post-doc position with me here in the math dept at Columbia University. We have a very active and strong department and you will get to work with me (I hope that is a plus for you). We are also close to the CUNY graduate school, NYU Courant institute, Rutgers University, Princeton University, the Institute of Advanced Study, Stony Brook University and the Simons Center, etc. Living in NYC isn’t bad either. So apply on mathjobs here!

Let k be a ring, for example a field. Let R be a k-algebra. Let f_1, …, f_r be a regular sequence in R such that k → R/(f_1, …, f_r) is an isomorphism. Let K be the Koszul complex over R on f_1, …, f_r viewed as a cochain complex sitting in degrees -r, …, 0. See Tag 0621. Then we are interested in the hom-complex

E = Hom_R(K, K)

constructed in Tag 0A8H which we may and do view as a differential graded R-algebra, see for example Tag 0FQ2. This algebra is interesting for many reasons; for example because there is an equivalence

D(E) = D_{QCoh, V(f_1, …, f_r)}(Spec(R))

Here the LHS is the derived category of dg E-modules and the RHS is the derived category of complexes of quasi-coherent modules on Spec(R) supported set theoretically on f_1 = … = f_r = 0. To prove this equivalence, use Tag 09IR to see that K gives a generator of the RHS and argue as in the proof of Tag 09M5.

Recall that the underlying R-module of K is the exterior algebra on the free module of rank r, say with basis e_1, …, e_r. For i = 1, .., r let v_i : K → K be the operator given by contraction by the dual basis element to e_i. Of course v_i has degree 1 and a computation shows that v_i : K → K[1] is a map of complexes. Similarly, the reader shows that v_i ∘ v_j = – v_j ∘ v_i and v_i ∘ v_i = 0. Thus we obtain a map of differential graded k-algebras

k ⟨ v_1, …, v_r ⟩ → E

with target E and source the exterior algebra over k on v_1, …, v_r in degree 1 and vanishing differential. A bit more work shows that this map is a quasi-isomorphism of differential graded k-algebras; this is where we use the assumption that f_1, …, f_r is a regular sequence so the koszul complex is a resolution of R/(f_1, …, f_r) by Tag 062F.

Applying Tag 09S6 we conclude that

D(k ⟨ v_1, …, v_r ⟩) = D_{QCoh, V(f_1, …, f_r)}(Spec(R))

Cheers!