de Rham cohomology of an Artinian ring

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!

Surjective map from affine space

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!

The theorem on formal functions

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)…

Updated Stacks Project

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!

Infinite dimensional de Rham cohomology

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

Postdoc position

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!

Endomorphisms of the Koszul complex

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!

A Jouanolou device

Let X be a scheme or an algebraic space. A Jouanolou device is a morphism Y → X such that Y is an affine scheme and such that Y is a torsor for a vector bundle over X.

A scheme with an ample family of invertible modules has a Jouanolou device (due to Jouanolou and Thomason). This is called the “Jouanolou trick”.

Let X be quasi-compact with affine diagonal. Consider the following conditions

  1. X has the resolution property
  2. X has a Jouanolou device
  3. X is the quotient of an affine scheme by a free action of a group scheme
  4. X is the quotient of a quasi-affine scheme by a free action of GL_n for some n

We know from Thomason, Totaro, and Gros that 1, 3, and 4 are equivalent. It is easy to see that 2 implies 3.

I am going to sketch an argument for 1 ⇒ 2. It may be in the literature; if you have a reference, please email me or leave a comment. [Edit: see end of this post.] Thanks!

The case of schemes is easier so I will explain that first. So, assume X is a quasi-compact scheme with affine diagonal and with the resolution property. By standard methods we reduce to the case where X is also of finite type over the integers, i.e., we may assume X is Noetherian. Let X = U_1 ∪ … ∪ U_n be a finite affine open covering. Let I_i ⊂ O_X be the ideal sheaf of the complement of U_i. By the resolution property, we may choose a finite locally free O_X-module V and a surjection V → I_1 ⊕ … ⊕ I_n. Thus we have

V → I_1 ⊕ … ⊕ I_n → O_X

Let E be the dual of V and let O_X(m_1 D_1 + … + m_n D_n) be short hand for SheafHom_{O_X}(I_1^{m_1} … I_n^{m_n}, O_X). Warning: this is just notation and we do not think of D_i as an actual divisor. Taking the dual sequence we obtain

O_X → O_X(D_1) ⊕ … ⊕ O_X(D_n) → E

If s : O_X → E is the composition, then we can consider

f : Y = RelativeSpec_X(Sym^*(E)/(s – 1)) → X

Since the section s is nowhere zero, this is a torsor for a vector bundle. To finish the proof we have to show that Y is affine. To see this it suffices to show that H^1(Y, G) = 0 for every coherent O_Y-module G. Since f_*G is a direct summand of f_*f^*f_*G it suffices to prove that for every coherent O_X-module F the map H^1(X, F) → H^1(X, f_*f^*F) is zero. This will be the case if for every element ξ in H^1(X, F) there exists an m > 0 such that the image of ξ by the map

F = F ⊗ O_X → F ⊗ Sym^m(E)

is zero. Finally, the key point is that the map O_X → Sym^m(E) factors through the map

O_X → ⨁ O_X(m_1 D_1 + … + m_n D_n)

where the sum is over m_1 + … + m_n = m. Thus it suffices to show that ξ dies in H^1(X, F ⊗ O_X(m_1 D_1 + … + m_n D_n)) provided m is large enough. Since m_i > m/n for at least one i, we see that it suffices to show that ξ dies in H^1(X, F ⊗ O_X(m D_i)) for m large enough. This is true because (U_i → X)_*O_{U_i} is the colimit of the modules O_X(m D_i) and hence we have

colim H^1(X, F ⊗ O_X(m D_i)) = H^1(U_i, F|_{U_i}) = 0

Here we use that the open immersion U_i → X is affine and that U_i is affine. This proves the schemes case of the assertion.

How do we change this argument when X is an algebraic space? We reduce to X of finite type over the integers as before. We replace the affine open covering by a surjective etale morphism U → X where U is an affine scheme. Using Zariski’s main theorem we may assume that U is a dense and schematically dense affine open of an algebraic space Z which comes with a finite morphism π : Z → X. Let I ⊂ O_Z be the ideal sheaf of the complement of U. If π where finite locally free, there would be a trace map tr : π_*O_Z → O_X. The replacement for this is that, after replacing I by a large positive power, there is a map

τ : π_*I → O_X

such that for a point u ∈ U with image x ∈ X the restriction of τ on the summand O^h_{U, u} of (π_*I) ⊗ O^h_{X, x} is given by the trace map for the finite free ring map O^h_{X, x} → O^h_{U, u}. In particular, τ is surjective. Choose a finite locally free O_X-module V and a surjection V → π_*I. Thus we have

V → π_*I → O_X

Let E be the dual of V and let s : O_X → E be the dual section. As before we have to show that Y = RelativeSpec_X(Sym^*(E)/(s – 1)) is affine.

Before we prove this, we can reduce to X integral and even normal. Namely, if X’ → X is finite surjective and if we can show that the base change of Y to X’ is affine, then it follows that Y is affine (see Tag 01ZT). The reader can check that the situation after base change has all the same properties as the situation before base change.

OK, we continue the proof that Y is affine, but now with X normal irreducible. Arguing as in the case of schemes, we reduce to proving that given F coherent on X and ξ ∈ H^1(X, F) then for m large enough ξ maps to zero in H^1(X, F ⊗ Sym^m(E)). Observe that Sym^m(E) is dual to the sheaf Syt^m(V) of symmetric tensors. Thus we consider the map

τ^m : Syt^m(π_*I) → O_X

Then I claim given m_0 for all m ≫ m_0 the map τ^m is a sum of compositions

Syt^m(π_*I) → π_*(I^a) ⊗ Syt^{m – a}(π_*I) → O_X

for a ≥ m_0. Here the first map comes from the multiplication maps and the second uses τ on the first tensor factor and some linear map Syt^{m – a}(π_*I) → O_X on the second tensor factor. It follows that F → F ⊗ Sym^m(E) is a sum of maps which factor through F → SheafHom(π_*(I^a), F) for some a ≥ m_0. Thus it suffices to show that

colim SheafHom(π_*(I^a), F)

is the pushforward of a coherent module on U to get the desired vanishing (argument as in the schemes case). This holds because it is the pushforward of the pullback of F as follows from Deligne’s formula and ω_{U/X} ≅ O_U. This finishes the sketch of the proof.

A word about the claim. To prove it we construct global maps over X and then we prove the equality on stalks at the generic point of X (this is why we reduced to the case where X is normal irreducible). Let us first explain what is happening in the generic point. Say L is a finite separable algebra of degree n over a field K. Let τ : L → K be the trace map. Denote Syt^m(L) the symmetric tensors in the mth tensor power of L over K. For m large enough we will write

τ^m : Syt^m(L) → K

as a sum of maps

Syt^m(L) → L ⊗ Syt^{m – a}(L) → K

for a ≥ m_0 where the first arrow uses the product on the first a tensors and the second map is of the form τ \otimes f_{m – a} for some linear map f_{m – a} : Syt^{m – a}(L) → K. In fact f_{m – a}(alpha) will be a universal polynomial in the coefficients of the characteristic polynomial of alpha over K. Thus to prove this we may reduce to the case where K is algebraically closed. Then L = K x … x K is a product of n copies of K. In this case, the dual statement, in terms of symmetric polynomials, asks the following: we have to write the polynomial

(x_1 + … + x_n)^m

as a sum of polynomials

(x_1^a + … + x_n^a) f_{m – a}

for a ≥ m_0 where f_{m – a} is some element in the Z-algebra S of symmetric polynomials in x_1, …, x_n. Now this is possible because the element x_1 + … + x_n maps to a nilpotent element of

S /(p_a; a ≥ m_0)

as the reader easily verifies. To finish the proof, all that is required is to observe that the polynomials f_{m – a} do indeed produce linear maps Syt^{m – a}(π_*I) → O_X over X as X is normal (details omitted).

Thanks for reading!

Edit 6/25/2023. Burt Totaro emailed to say that one can deduce the existence of the Jouanolou device more directly from his and Gross’s papers. I agree. For example, we may using their papers write X as W/GL_n with W quasi-affine and then use the “equivariant Jouanolou trick” in Section 3 of Burt’s paper to get the Jouanolou device for X. Or one can modify the arguments in Burt’s paper directly and get a simple, short, direct proof. I gave the argument above because for a while I’ve wanted to make the thing about powers of the trace map work but didn’t succeed until the writing of this blog post.