How Fine One Must Choose an Affine Cover to get Weil Restriction? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T04:44:59Z http://mathoverflow.net/feeds/question/113866 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/113866/how-fine-one-must-choose-an-affine-cover-to-get-weil-restriction How Fine One Must Choose an Affine Cover to get Weil Restriction? minimax 2012-11-19T20:24:49Z 2012-11-20T02:22:11Z <p>Hello Everyone,</p> <p>It is well-known that Weil restriction does not commute with the formation of affine open covers, so I am wondering how fine one must choose an affine cover to recover the Weil restriction. More precisely, let $L/K$ be a finite separable extension, $X$ be a variety over $L$. Let ${U_i}$ be an affine open cover of $X$. For each $i$, one can Weil restrict to get the Weil restriction $V_i$ which is a variety over $K$. The gluing data for $U_i$ descends to gluing data for $V_i$ so we can glue the $V_i$'s together. But it is not true that for any open cover ${U_i}$, the $V_i$'s glue to get $R_{L/k}X$. So the question is, under what condition does those $V_i$'s glue to get $R_{L/K}X$?</p> <p>Upon reading the book Neron model, in particular the last paragraph of the proof of Theorem 4 on page 195, I thought that it is sufficient if any $d$ points of $X$ lie in some $U_i$, where $d=[L:K]$ is the degree of field extension (reason is that, for any $K$-scheme $T$, let $T'=T\times_KL$, then any fiber in the projection $T'\to T$ contains at most $d$ points, now follow the proof of that Theorem). Is this correct?</p> <p>Thanks! </p> http://mathoverflow.net/questions/113866/how-fine-one-must-choose-an-affine-cover-to-get-weil-restriction/113891#113891 Answer by nosr for How Fine One Must Choose an Affine Cover to get Weil Restriction? nosr 2012-11-20T02:22:11Z 2012-11-20T02:22:11Z <p>You are correct, and something similar holds more generally for Weil restriction for a quasi-projective $X' \rightarrow S'$ through a map $f:S' \rightarrow S$ that is finite locally free of constant rank $d$. (So in the field case your separability hypothesis is unnecessary.)</p> <p>In fact this is related to an apparent technical gap in the discussion in that book: they don't address why the Weil restriction is again quasi-compact (and hence of finite type) over the base, and its proof seems to require your observation (generalized to the relative base case).</p> <p>To be precise, the construction in that book for affine $S$ and $S'$ (which is the essential case) is initially done using <em>all</em> open affines in $X'$, so a-priori it is a just locally of finite type over $S$. Note that existence as a locally finite type $S$-scheme is sufficient for knowing that <code>${\rm{R}}_{S'/S}(U') \rightarrow {\rm{R}}_{S'/S}(X')$</code> is an open immersion for any open subscheme $U'$ of $X'$ because open immersions are the same thing as etale monomorphisms (without any quasi-compactness hypotheses on morphisms!). </p> <p>Since $X' \rightarrow S'$ is quasi-projective over an affine $S'$, any finite subset of $X'$ lies in an affine open subset. Thus, there does exist a (perhaps infinite) collection <code>$\{U_i\}$</code> of affine opens such that any ordered $d$-tuple in $X'$ (<em>allowing repetitions</em>, for a technical reason to be seen later!) lies in some $U_i$. Now there are two things to be done: (1) prove that for any such <code>$\{U_i\}$</code> the collection of open subschemes ${\rm{R}}_{S'/S}(U_i)$ covers ${\rm{R}}_{S'/S}(X')$ (a generalization of your guess), (2) show that there is such a collection <code>$\{U_i\}$</code> that is <em>finite</em>. </p> <p>Assertion (2) is point-set topology without Hausdorffness, as follows. Pick some <code>$\{U_i\}$</code> as above but perhaps infinite. The <em>topological</em> product ${X'}^d$ of all ordered $d$-tuples (allowing repetitions!) is quasi-compact since $X'$ is quasi-compact, and the <em>open</em> subsets $(U_i)^d$ do cover it (by the quasi-projectivity, as indicated above), so there is a finite subcover, say $(U_{i_1})^d, \dots, (U_{i_r})^d$ for some $i_1,\dots, i_r$. Thus, <code>$\{U_{i_1},\dots,U_{i_r}\}$</code> answers (2) affirmatively.</p> <p>It remains to prove (1), which in effect is the problem you're asking about and for which you have already identified the argument. To be pedantic, here is your argument. A point of a scheme is the image of a morphism from a field, so a point of <code>${\rm{R}}_{S'/S}(X')$</code> corresponds to the image of a map <code>$x:{\rm{Spec}}(F) \rightarrow {\rm{R}}_{S'/S}(X')$</code> for a field $F$. Composing $x$ with the structure map to $S$ defines an $F$-valued point $s: {\rm{Spec}}(F) \rightarrow S$, and by the functorial meaning of Weil restriction we see that $x$ viewed as an $S$-morphism (using $s$) corresponds to an $S'$-morphism $x':S'_s \rightarrow X'$. But $S'_s$ is an $F$-scheme of rank $d$, so it consists of at most $d$ physical points, and hence lands inside one of the open subschemas $U_i$ (because we allow our ordered $d$-tuples to contain repetitions). Now we can run the calculation in reverse with that $U_i$ in the role of $X'$ to deduce that $x$ factors through <code>${\rm{R}}_{S'/S}(U_i)$</code>.</p>