{"id":4828,"date":"2023-06-09T15:03:07","date_gmt":"2023-06-09T15:03:07","guid":{"rendered":"https:\/\/www.math.columbia.edu\/~dejong\/wordpress\/?p=4828"},"modified":"2025-11-17T14:45:15","modified_gmt":"2025-11-17T14:45:15","slug":"a-jouanolou-device","status":"publish","type":"post","link":"https:\/\/www.math.columbia.edu\/~dejong\/wordpress\/?p=4828","title":{"rendered":"A Jouanolou device"},"content":{"rendered":"<p>Let X be a scheme or an algebraic space. A <em>Jouanolou device<\/em> is a morphism Y &rarr; X such that Y is an affine scheme and such that Y is a torsor for a vector bundle over X.<\/p>\n<p>A scheme with an ample family of invertible modules has a Jouanolou device (due to Jouanolou and Thomason). This is called the &#8220;Jouanolou trick&#8221;.<\/p>\n<p>Let X be quasi-compact with affine diagonal. Consider the following conditions<\/p>\n<ol>\n<li>X has the resolution property<\/li>\n<li>X has a Jouanolou device<\/li>\n<li>X is the quotient of an affine scheme by a free action of a group scheme<\/li>\n<li>X is the quotient of a quasi-affine scheme by a free action of GL_n for some n<\/li>\n<\/ol>\n<p>We know from Thomason, Totaro, and Grosa that 1, 3, and 4 are equivalent (for some discussion and references see eg <a href=\"https:\/\/arxiv.org\/pdf\/2109.09623\">this paper<\/a>). It is easy to see that 2 implies 3.<\/p>\n<p>I am going to sketch an argument for 1 &rArr; 2. It may be in the literature; if you have a reference, please email me or leave a comment. [<strong>Edit:<\/strong> see end of this post.] Thanks!<\/p>\n<p>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 &cup; &#8230; &cup; U_n be a finite affine open covering. Let I_i &subset; 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 &rarr; I_1 &oplus; &#8230; &oplus; I_n. Thus we have<\/p>\n<blockquote><p>\nV &rarr; I_1 &oplus; &#8230; &oplus; I_n &rarr; O_X\n<\/p><\/blockquote>\n<p>Let E be the dual of V and let O_X(m_1 D_1 + &#8230; + m_n D_n) be short hand for SheafHom_{O_X}(I_1^{m_1} &#8230; 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<\/p>\n<blockquote><p>\nO_X &rarr; O_X(D_1) &oplus; &#8230; &oplus; O_X(D_n) &rarr; E\n<\/p><\/blockquote>\n<p>If s : O_X &rarr; E is the composition, then we can consider<\/p>\n<blockquote><p>\nf : Y = RelativeSpec_X(Sym^*(E)\/(s &#8211; 1)) &rarr; X\n<\/p><\/blockquote>\n<p>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) &rarr; H^1(X, f_*f^*F) is zero. This will be the case if for every element &xi; in H^1(X, F) there exists an m &gt; 0 such that the image of &xi; by the map<\/p>\n<blockquote><p>\nF = F &otimes; O_X &rarr; F &otimes; Sym^m(E)\n<\/p><\/blockquote>\n<p>is zero. Finally, the key point is that the map O_X &rarr; Sym^m(E) factors through the map<\/p>\n<blockquote><p>\nO_X &rarr; &bigoplus; O_X(m_1 D_1 + &#8230; + m_n D_n)\n<\/p><\/blockquote>\n<p>where the sum is over m_1 + &#8230; + m_n = m. Thus it suffices to show that &xi; dies in H^1(X, F &otimes; O_X(m_1 D_1 + &#8230; + 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 &xi; dies in H^1(X, F &otimes; O_X(m D_i)) for m large enough. This is true because (U_i &rarr; X)_*O_{U_i} is the colimit of the modules O_X(m D_i) and hence we have<\/p>\n<blockquote><p>\ncolim H^1(X, F &otimes; O_X(m D_i)) = H^1(U_i, F|_{U_i}) = 0\n<\/p><\/blockquote>\n<p>Here we use that the open immersion U_i &rarr; X is affine and that U_i is affine. This proves the schemes case of the assertion.<\/p>\n<p>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 &rarr; X where U is an affine scheme. Using Zariski&#8217;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 &pi; : Z &rarr; X. Let I &subset; O_Z be the ideal sheaf of the complement of U. If &pi; where finite locally free, there would be a trace map tr : &pi;_*O_Z &rarr; O_X. The replacement for this is that, after replacing I by a large positive power, there is a map<\/p>\n<blockquote><p>\n&tau; : &pi;_*I &rarr; O_X\n<\/p><\/blockquote>\n<p>such that for a point u &in; U with image x &in; X the restriction of &tau; on the summand O^h_{U, u} of (&pi;_*I) &otimes; O^h_{X, x} is given by the trace map for the finite free ring map O^h_{X, x} &rarr; O^h_{U, u}. In particular, &tau; is surjective. Choose a finite locally free O_X-module V and a surjection V &rarr; &pi;_*I. Thus we have<\/p>\n<blockquote><p>\nV &rarr; &pi;_*I &rarr; O_X\n<\/p><\/blockquote>\n<p>Let E be the dual of V and let s : O_X &rarr; E be the dual section. As before we have to show that Y = RelativeSpec_X(Sym^*(E)\/(s &#8211; 1)) is affine.<\/p>\n<p>Before we prove this, we can reduce to X integral and even normal. Namely, if X&#8217; &rarr; X is finite surjective and if we can show that the base change of Y to X&#8217; is affine, then it follows that Y is affine (see <a href=\"https:\/\/stacks.math.columbia.edu\/tag\/01ZT\">Tag 01ZT<\/a>). The reader can check that the situation after base change has all the same properties as the situation before base change.<\/p>\n<p>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 &xi; &in; H^1(X, F) then for m large enough &xi; maps to zero in H^1(X, F &otimes; Sym^m(E)). Observe that Sym^m(E) is dual to the sheaf Syt^m(V) of symmetric tensors. Thus we consider the map<\/p>\n<blockquote><p>\n&tau;^m : Syt^m(&pi;_*I) &rarr; O_X\n<\/p><\/blockquote>\n<p>Then I <strong>claim<\/strong> given m_0 for all m &gg; m_0 the map &tau;^m is a sum of compositions<\/p>\n<blockquote><p> Syt^m(&pi;_*I) &rarr; &pi;_*(I^a) &otimes; Syt^{m &#8211; a}(&pi;_*I) &rarr; O_X<\/p><\/blockquote>\n<p>for a &ge; m_0. Here the first map comes from the multiplication maps and the second uses &tau; on  the first tensor factor and some linear map Syt^{m &#8211; a}(&pi;_*I) &rarr; O_X on the second tensor factor. It follows that F &rarr; F &otimes; Sym^m(E) is a sum of maps which factor through F &rarr; SheafHom(&pi;_*(I^a), F) for some a &ge; m_0. Thus it suffices to show that<\/p>\n<blockquote><p> colim SheafHom(&pi;_*(I^a), F) <\/p><\/blockquote>\n<p>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&#8217;s formula and &omega;_{U\/X} &cong; O_U. This finishes the sketch of the proof.<\/p>\n<p>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 &tau; : L &rarr; 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<\/p>\n<blockquote><p>\n&tau;^m : Syt^m(L) &rarr; K\n<\/p><\/blockquote>\n<p>as a sum of maps<\/p>\n<blockquote><p>\nSyt^m(L) &rarr; L &otimes; Syt^{m &#8211; a}(L) &rarr; K\n<\/p><\/blockquote>\n<p>for a &ge; m_0 where the first arrow uses the product on the first a tensors and the second map is of the form &tau; \\otimes f_{m &#8211; a} for some linear map f_{m &#8211; a} : Syt^{m &#8211; a}(L) &rarr; K. In fact f_{m &#8211; 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 &#8230; 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<\/p>\n<blockquote><p> (x_1 + &#8230; + x_n)^m <\/p><\/blockquote>\n<p>as a sum of polynomials<\/p>\n<blockquote><p> (x_1^a + &#8230; + x_n^a) f_{m &#8211; a} <\/p><\/blockquote>\n<p>for a &ge; m_0 where f_{m &#8211; a} is some element in the <strong>Z<\/strong>-algebra S of symmetric polynomials in x_1, &#8230;, x_n. Now this is possible because the element x_1 + &#8230; + x_n maps to a nilpotent element of<\/p>\n<blockquote><p> S \/(p_a; a &ge; m_0)<\/p><\/blockquote>\n<p>as the reader easily verifies. To finish the proof, all that is required is to observe that the polynomials f_{m &#8211; a} do indeed produce linear maps Syt^{m &#8211; a}(&pi;_*I) &rarr; O_X over X as X is normal (details omitted).<\/p>\n<p>Thanks for reading!<\/p>\n<p><strong>Edit 6\/25\/2023.<\/strong> Burt Totaro emailed to say that one can deduce the existence of the Jouanolou device more directly from his and Gross&#8217;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 &#8220;equivariant Jouanolou trick&#8221; in Section 3 of Burt&#8217;s paper to get the Jouanolou device for X. Or one can modify the arguments in Burt&#8217;s paper directly and get a simple, short, direct proof. I gave the argument above because for a while I&#8217;ve wanted to make the thing about powers of the trace map work but didn&#8217;t succeed until the writing of this blog post.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Let X be a scheme or an algebraic space. A Jouanolou device is a morphism Y &rarr; X such that Y is an affine scheme and such that Y is a torsor for a vector bundle over X. A scheme &hellip; <a href=\"https:\/\/www.math.columbia.edu\/~dejong\/wordpress\/?p=4828\">Continue reading <span class=\"meta-nav\">&rarr;<\/span><\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-4828","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/www.math.columbia.edu\/~dejong\/wordpress\/index.php?rest_route=\/wp\/v2\/posts\/4828","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.math.columbia.edu\/~dejong\/wordpress\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.math.columbia.edu\/~dejong\/wordpress\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.math.columbia.edu\/~dejong\/wordpress\/index.php?rest_route=\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/www.math.columbia.edu\/~dejong\/wordpress\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=4828"}],"version-history":[{"count":75,"href":"https:\/\/www.math.columbia.edu\/~dejong\/wordpress\/index.php?rest_route=\/wp\/v2\/posts\/4828\/revisions"}],"predecessor-version":[{"id":5008,"href":"https:\/\/www.math.columbia.edu\/~dejong\/wordpress\/index.php?rest_route=\/wp\/v2\/posts\/4828\/revisions\/5008"}],"wp:attachment":[{"href":"https:\/\/www.math.columbia.edu\/~dejong\/wordpress\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4828"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.math.columbia.edu\/~dejong\/wordpress\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=4828"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.math.columbia.edu\/~dejong\/wordpress\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=4828"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}