Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hello Everyone,

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 $R_{L/K}X$?

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?


share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

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.)

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).

To be precise, the construction in that book for affine $S$ and $S'$ (which is the essential case) is initially done using all 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 ${\rm{R}}_{S'/S}(U') \rightarrow {\rm{R}}_{S'/S}(X')$ 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!).

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 $\{U_i\}$ of affine opens such that any ordered $d$-tuple in $X'$ (allowing repetitions, 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 $\{U_i\}$ 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 $\{U_i\}$ that is finite.

Assertion (2) is point-set topology without Hausdorffness, as follows. Pick some $\{U_i\}$ as above but perhaps infinite. The topological product ${X'}^d$ of all ordered $d$-tuples (allowing repetitions!) is quasi-compact since $X'$ is quasi-compact, and the open 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, $\{U_{i_1},\dots,U_{i_r}\}$ answers (2) affirmatively.

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 ${\rm{R}}_{S'/S}(X')$ corresponds to the image of a map $x:{\rm{Spec}}(F) \rightarrow {\rm{R}}_{S'/S}(X')$ 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 ${\rm{R}}_{S'/S}(U_i)$.

share|improve this answer
Excellent answer! A possible dumb question: why is Weil restriction of an open immersion etale? –  minimax Nov 20 '12 at 5:02
@ylin2: Given where you indicate you are located (I presume as a graduate student), I think you can find classmates with whom to discuss this question -- but first please give yourself more time to figure it out on your own (hint: how do you prove anything is etale?). –  user28172 Nov 20 '12 at 6:52
@nosr: I think I figured this out, by infinitesimal criterion for etaleness: a morphism $f: X\to Y$ of locally of finite presentation iff for any $Y$-scheme $T$ and closed subscheme $T_0$ of $T$ defined by a square zero ideal, the map of sets $\Hom_Y(X,T)\to\Hom_Y(X,T_0)$ is bijective. Now etaleness follows if we use the universal property defining Weil restriction and the above criterion.... –  minimax Nov 20 '12 at 7:52
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.