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Q: Is there a simple proof of the fact that the Weil restriction of an abelian scheme along a finite étale morphism is an abelian scheme ?

Details: Let $S$ be a scheme and $f:S'\rightarrow S$ a finite étale morphism. Let $A/S'$ be an abelian scheme. Then the following argument shows that the Weil restriction $\mathfrak{R}_{S'/S}$(A) (see section 7.6 of the book Néron Models by Bosch, Lütkebohmert and Raynaud) which a priori is just an fppf sheaf of abelian groups on $Sch/S$ is representable by an abelian scheme over $S$:

1) Theorem 1.5 of M. Olsson's paper Hom stacks and restriction of scalars (Duke Math. J. 134 (2006), 139-164.) gives a general criterion for a Weil restriction to be representable by an algebraic space, which applies here to show that $\mathfrak{R}_{S'/S}(A)$ is representable by an algebraic space over $S$.

By passing to a Galois cover, $\mathfrak{R}_{S'/S}(A)$ decomposes as a product of abelian schemes, so is representable by an algebraic space over $S$.

2) The arguments used to prove Proposition 7.6.5 (f) and (h) of Néron Models show that $\mathfrak{R}_{S'/S}(A)$ is smooth and proper as an algebraic space. Moreover the formation of the Weil restriction commutes with base change and its fibers are connected by Proposition A.5.9 of the book Pseudo-reductive Groups by Conrad,Gabber and Prasad.

3) By 1) and 2), $\mathfrak{R}_{S'/S}(A)$ is an abelian algebraic space in the sense of Section I.1 of the book Degeneration of abelian varieties by Chai and Faltings. By Theorem 1.9 of loc. cit. (due to Raynaud) it is actually an abelian scheme.

The "problem" with this proof is that 3) is a relatively delicate result. Of course, if $A$ is projective over $S$, the Weil restriction is automatically a scheme and we can do without 1) and 3). However, the question of when an abelian scheme is projective over the base is subtle in general: it is known to hold over a noetherian normal base but the argument is a key part of the proof of 3) anyway.

My motivation is to understand the push-forward functoriality of Deligne 1-motives over a base, and I would like to be able to consider general abelian schemes over general base schemes.

1) Does anyone know a simpler proof ?

2) If one assumes only that $S'/S$ is finite locally free, is $\mathfrak{R}_{S'/S}(A)$ (which might not be proper anymore) still a scheme ?

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Dear Simon, Can you check this by etale descent? Namely, if you let $S''$ over $S$ be a Galois etale cover lying over $S'$, and pull-back the restriction of scalars to $S''$, then it should just a product of the appropriate number of copies of the original $A$. So after this etale pull-back we do get an abelian scheme. Hopefully descent now applies to conclude that the restriction of scalars itself was an ab. scheme. Regards, Matthew –  Emerton Apr 10 '13 at 18:07
@Emerton: this shows that the restriction is an algebraic space, but how do you prove that it is a scheme ? Indeed, I see that 1) is overkill, even in the case where $S'/S$ is only finite locally free. I will edit the question accordingly. –  Simon Pepin Lehalleur Apr 10 '13 at 20:05
@Simon: Cf. Murre. –  Jason Starr Apr 11 '13 at 0:53
Do you seek a proof that avoids algebraic spaces entirely? That seems unlikely to be possible, and if one is going to be using algebraic spaces then the Faltings-Chai proof is an extremely natural one (building off of the normal case in a clever way). For your purposes with Deligne 1-motives, is it a problem to work throughout with algebraic spaces (which are well-suited to questions related to quotients in algebraic geometry)? –  user28172 Apr 14 '13 at 4:52
@Jason Starr: which paper of Murre are you referring to ? Is it the Bourbaki seminar on unramified functors ? @nosr: because of Raynaud's theorem, most of the algebraic spaces (e.g. semi-abelian algebraic spaces of constant rank) I need are schemes. On the other hand, I can work with algebraic spaces, since I am ultimately interested in the complex of sheaves and the associated objects in triangulated categories of mixed motives. It is more a matter of not being familiar with the technology. –  Simon Pepin Lehalleur Apr 15 '13 at 18:21

1 Answer 1

Let $X \to S'$ be a smooth projective morphism, and $S' \to S$ a finite étale morphism.

The Weil restriction $Y \to S$ of $X \to S'$ is a closed subscheme of the scheme of morphisms $S' \to X$ on $S$, which is projective; hence it is projective (this is Grothendieck's original construction).

Let us show that $Y \to S$ is again smooth. Construction of the Weil restriction is étale local on $S$; hence we may assume that $S' = S_1 \sqcup \dots \sqcup S_m$ is a disjoint union of $m$ copies $S_1, \dots, S_m$ of $S$. Call $X_i$ the restriction of $X$ to $S_i$. It follows readily from the definition that $Y \to S$ is the product $X_1 \times_S \dots \times_S X_m$, and this proves smoothness.

If $X \to S'$ is an abelian scheme, the group scheme structure of $X \to S'$ induces a group scheme structure on $Y \to S$, and this completes the proof.

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Doesn't this work only for projective abelian schemes? –  Ricky Apr 10 '13 at 19:49
@Ricky: Cf. Murre. –  Jason Starr Apr 10 '13 at 21:12

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