Is the direct limit of Weil restriction of an elliptic curve a scheme? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T13:43:06Z http://mathoverflow.net/feeds/question/44722 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44722/is-the-direct-limit-of-weil-restriction-of-an-elliptic-curve-a-scheme Is the direct limit of Weil restriction of an elliptic curve a scheme? Dror Speiser 2010-11-03T20:33:51Z 2010-11-04T03:21:29Z <p>In a discussion today on the Shafarevich-Tate group of an elliptic curve, the following structure and question came up. I will abuse many notations and be very vague about some things, but am very open to suggestions for clarification. Not to mention that some, if not all, of the following is incorrect.</p> <p>A key ingredient in the rich structure of number fields is Hilbert 90, asserting that the first galois cohomology of the multiplicative group is trivial. Elliptic curves, on the other hand, have no analogy (as far as I know). The discussion was about looking for a natural object to inject an elliptic curve into, that has trivial first galois cohomology.</p> <p>Given an elliptic curve, an interesting looking (well, maybe not) object is the direct limit of Weil restriction of the curve, going over all the galois extensions of the base field and the morphisms are the natural inclusions. As a $G_K$ module the Weil restriction is the induced module $Ind_{G_L}^{G_K} E$, so by Shapiro's lemma the cohomology is isomorphic to $H^1(G_L,E)$. Applying the direct limit, we indeed see that the first cohomology of the direct limit of the Weil restriction is trivial (since $G_L$ keeps getting smaller).</p> <p>And so many questions are raised regarding this direct limit. The first one is in the title, and is my question.</p> <blockquote> <p>Is the direct limit of Weil restrictions, going over all galois extensions of the base field, of an elliptic curve a scheme?</p> </blockquote> http://mathoverflow.net/questions/44722/is-the-direct-limit-of-weil-restriction-of-an-elliptic-curve-a-scheme/44776#44776 Answer by BCnrd for Is the direct limit of Weil restriction of an elliptic curve a scheme? BCnrd 2010-11-04T03:21:29Z 2010-11-04T03:21:29Z <p>I agree with Adam Smith that the question seems a bit misguided, but let me show anyway that the answer is negative away from certain silly cases. Well, first to make a more well-posed question, one first has to adjust the definition of the functor so that it is at least a Zariski-sheaf (ideally without changing the "value" on affines). So let's first do that. Let $E$ be an elliptic curve over a field $k$, and for the directed system of finite extensions $k_i/k$ inside of a fixed separable closure $k_s/k$ let $E_i = {\rm{Res}}_{k_i/k}(E_{k_i})$ denote the corresponding abelian varieties arising by Weil restriction. Define the functor $G = \injlim E_i$ on the category of $k$-schemes. This is an fpqc sheaf on the category of affine $k$-schemes, since we can explicitly compute it: for any $k$-algebra $R$, $G(R) = \injlim E(R_{k_i}) = E(R_{k_s})$, the final equality using that the functor of $E$ commutes with the formation of direct limits in $k$-algebras (thanks to Grothendieck's awe-inspiring necessary and sufficient functorial characterization of locally finitely presented morphisms of schemes, from EGA IV$_3$, 8.14, which we will use in a more impressive converse direction shortly). </p> <p>By inspection, the functor $G$ satisfies the fpqc sheaf axioms on affines. Thus, if we Zariski-sheafify $G$ on the category of $k$-schemes then the resulting functor $G'$ has the same value on affines and so is easily seen to be an fpqc sheaf. So the "right" question is whether $G'$ is representable. I will now prove the "expected" negative answer whenever $k$ is not separably closed or real-closed, which is to say (by the Artin-Schreier theorem) that $k_s/k$ has infinite degree. (Obviously need to rule out the cases when $[k_s:k]$ is finite!)</p> <p>Suppose $G'$ is represented by a $k$-scheme (which we will also denote by $G'$), so by Grothendieck's functorial criterion we see that $G'$ is locally of finite type over $k$. Thus, the identity component ${G'}^0$ makes sense and it <em>finite type</em> (as for any locally finite type group scheme over a field; see SGA3, VI$_{\rm{A}}$, 2.4). As such, it contains the connected $E_i$ as closed $k$-subgroup schemes since any monomorphic homomorphism between finite type group schemes over a field is a closed immersion (proved in SGA3, VI$_{\rm{B}}$, 1.4.2, for example). But $E_i$ has dimension $[k_i:k]$, which grows without bound since $[k_s:k]$ is infinite by hypothesis. Hence, ${G'}^0$ contains closed subschemes of arbitrarily large dimension, an absurdity since ${G'}^0$ is finite type over a field. QED</p>