# 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).