# Three and a half basic questions on the Weil restriction of scalars

(This is reposted from mathstackexchange, where it received no answer so far.)

I am currently trying to get familiar with the Weil Restriction functor.

For a finite field extension $L|K$ it associates a variety over $K$ to every variety $X$ over $L$ as the representing object of the functor defined via $Res_{L|K}X(A)=X(A\otimes_K L)$ for every $K$-Algebra $A$. There is also an explicit description in terms of defining equations for the variety.

I have the following questions:

$1$) Is there a characterisation of the (essential) image of this functor? i.e. can we say which varieties over $K$ "come from" $L$?

$2$) Is Restriction of scalars "injective" i.e. if a $K$-variety admits the structure of an $L$-variety is this structure unique? (up to isomorphism)

$3$) In the case of $L=\mathbb{C}$ and $K=\mathbb{R}$ there are also analytification functors and we can "restrict scalars" on the analytic side by regarding a complex manifold as a real analytic one. It seems to me, by the explicit construction of the Weil-Restriction, that analytification and restriction of scalars commute. Is this correct?

$3.5$) Can you recommend a reference to learn about this stuff which does not focus only on algebraic groups?

• A complex variety has a complex conjugate (concretely, apply complex conjugation to the coefficients of the defining equations). This is not always isomorphic to the original, but surely the associated real varieties are isomorphic. – Tom Goodwillie Aug 4 '15 at 12:08
• One reference that contains some good examples (not necessarily involving algebraic groups) of the pathologies you can encounter with non-separable extensions is Conrad-Gabber-Prasad, "Pseudo-reductive groups". – S. Carnahan Aug 4 '15 at 13:41

It is best not to use the word (quasi-projective) "variety" in this context because (as alluded to in Scott Carnahan's comment to the question posed) Weil restriction does not preserve geometric integrality or even geometric irreducibility or geometric reducedness or non-emptiness in general (allowing $L/K$ that is possibly not separable), and even if $L/K$ is separable one can encounter some oddities such as ${\rm{R}}_{L/K}(X)$ being non-reduced with $X$ that is reduced (though such $X$ are not geometrically reduced, so the latter oddities are a bit anti-climactic after a moment's thought) . A reference for such counterexamples (giving $X$ geometrically integral over $L$ for which ${\rm{R}}_{L/K}(X)$ is disconnected with a connected component that is everywhere non-reduced) is provided at the end of this answer. So one should speak in terms of "schemes" (as always!).

The explicit description of Weil restriction in terms of equations is usually (though not always) useless:

Example: Assume $(X,e')$ is an $L$-group scheme, and consider the natural isomorphism of $K$-vector spaces $T_e({\rm{R}}_{L/K}(X)) \simeq T_{e'}(X)$ (via evaluating on the dual numbers). It is true that this is an isomorphism of Lie algebras, moreover respecting the $p$-operation in characteristic $p$, but I don't think that obvious by thinking in terms of "equations" (even if $X$ is smooth, which is not necessary to assume).

The answers to #1 and #2 are certainly negative in general (if for #1 one intends for an interesting rather than tautologous answer); it is akin to analogues with induction of representations from a (possibly non-normal) subgroup of finite index. For example, if $L/K$ is finite Galois with Galois group $G$ and $Y$ is a quasi-projective $L$-scheme that is not $L$-isomorphic to its twist $g^{\ast}(Y)$ by some $g \in G$ (elliptic curves provide plenty of examples) then nonetheless ${\rm{R}}_{L/K}(Y) \simeq {\rm{R}}_{L/K}(g^{\ast}(Y))$ due to the description of finite separable Weil restriction as mentioned in Fonarev's answer.

As for #3.5, section 7.6 of the book "Neron Models" gives a very elegant introduction to the topic of Weil restriction (over general rings -- as one really ought to do, since $L \otimes_K K'$ is rarely a field when $K'/K$ is an extension field yet we do want to contemplate how Weil restriction interacts with scalar extension from $K$ to $K'$). That reference goes way beyond the setting of algebraic groups, and addresses good behavior of the Weil-restriction functor (on quasi-projective schemes) with respect to etale morphisms and open immersions and much else. It doesn't include the proof that the output is again quasi-projective in general, but that property -- which lies a bit deeper in the story -- is settled affirmatively in Proposition A.5.8 in the book "Pseudo-reductive groups"; Appendix A.5 of this latter book (recently published in a revised/expanded 2nd edition that I will refer to below) gives an alternative general development of the topic of Weil restriction (also going beyond algebraic groups).

Remark: One input into the proof of quasi-projectivity of ${\rm{R}}_{L/K}(X)$ is knowing in advance that this $K$-scheme is quasi-compact. Such quasi-compactness for general (not necessarily separable) finite extensions $L/K$ is not part of the usual construction, which merely gives something locally of finite type over $K$ (due to failure of Weil restriction to carry Zariski covers to Zariski covers, a posteriori -- see Example A.5.3 of "Pseudo-reductive groups" for simple counterexamples with any nontrivial finite separable extension $L/K$, not such a surprise if you think in terms of pushforwards of sheaves of sets) and separated (by valuative criterion). Fortunately, there always does exist a finite Zariski affine open cover $\{U_i\}$ of $X$ for which the associated affine opens ${\rm{R}}_{L/K}(U_i)$ cover ${\rm{R}}_{L/K}(X)$: see page 504 of "Pseudo-reductive groups".

As for question #3, it seems you intend to ask, for a smooth quasi-projective $\mathbf{C}$-scheme $X$, if the real-analytic manifold associated to the smooth quasi-projective $\mathbf{R}$-scheme ${\rm{R}}_{\mathbf{C}/\mathbf{R}}(X)$ is isomorphic naturally in $X$ to the real-analytic manifold underlying the complex manifold $X^{\rm{an}}$. In other words, is the evident identification of sets $$\varphi_X: ({\rm{R}}_{\mathbf{C}/\mathbf{R}}(X))(\mathbf{R}) \simeq X(\mathbf{C})$$ a real-analytic isomorphism? The case when $X$ is an affine space is clear, and for a clean argument it is tempting to try to reduce the general case to that case by using the Zariski-local descrption of smooth morphisms in terms of etale maps to affine spaces. This basically works, but there is a tiny wrinkle, as we will soon see.

We noted above that although Weil restriction generally does not carry Zariski covers to Zariski covers, there always exists some finite open affine cover $\{U_i\}$ of $X$ such that ${\rm{R}}_{L/K}(X)$ is covered by its open affine subschemes ${\rm{R}}_{L/K}(U_i)$. The construction of such $\{U_i\}$ does not permit us to choose such $\{U_i\}$ to be subordinate to a chosen cover of $X$, so that might seem to doom the attempt to reduce to working with $U_i$'s that are etale over affine spaces.

But actually we don't need this: if $\{U_i\}$ is any Zariski-open cover of the smooth $\mathbf{C}$-scheme $X$ then the open submanifolds $({\rm{R}}_{\mathbf{C}/\mathbf{R}}(U_i))(\mathbf{R})$ of $({\rm{R}}_{\mathbf{C}/\mathbf{R}}(X))(\mathbf{R})$ visibly constitute a cover since $\{U_i(\mathbf{C})\}$ covers $X(\mathbf{C})$ (even though the ${\rm{R}}_{\mathbf{C}/\mathbf{R}}(U_i)$'s may fail to cover ${\rm{R}}_{\mathbf{C}/\mathbf{R}}(X)$, which is to say that $\{U_i(\mathbf{C} \otimes_{\mathbf{R}}\mathbf{C})\}$ may not cover $X(\mathbf{C} \otimes_{\mathbf{R}} \mathbf{C})$). Likewise $\{U_i^{\rm{an}}\}$ is an open cover of $X^{\rm{an}}$. Hence, the problem for $X$ does reduce to the same for each $U_i$ separately after all, so in this way we can reduce to the case when there is an etale map $f:X \rightarrow \mathbf{A}^n_{\mathbf{C}}$. But then ${\rm{R}}_{\mathbf{C}/\mathbf{R}}(f)$ is an etale map between smooth $\mathbf{R}$-schemes, so it induces a local real-analytic isomorphism on $\mathbf{R}$-points by the real-analytic inverse function theorem and the Zariski-local description of etale morphisms. Likewise, $f^{\rm{an}}$ is a local complex-analytic isomorphism. In this manner, the bijective map $\varphi_X$ is a real-analytic isomorphism because of the same for $\varphi_{\mathbf{A}^n_{\mathbf{C}}}$. That settles #3 affirmatively.

Although some answers to #3.5 have been given above, I want to warn you again about the dangerous use of the word "variety" in these settings by providing a reference for the disorienting counterexamples alluded to at the start. To give some context, I first note that if the quasi-projective $L$-scheme $X$ is smooth and geometrically connected over $L$ then the $K$-scheme ${\rm{R}}_{L/K}(X)$ is smooth (easy, by infinitesimal criterion; not by staring at equations) and geometrically connected. One cannot "see" connectedness properties in terms of equations, so this geometric connectivity over $K$ is really not obvious in general: the case that $L/K$ is separable is very easy because one has the $K_s$-isomorphism $${\rm{R}}_{L/K}(X)_{K_s} \simeq {\rm{R}}_{(L \otimes_K K_s)/K_s}(X_{L \otimes_K K_s}) = \prod_{\sigma} \sigma^{\ast}(X)$$ where $\sigma$ varies through the $K$-embeddings of $L$ into $K_s$ (so the smoothness hypothesis isn't needed in such cases), but an entirely different idea is needed to handle the possibility that $L/K$ is not separable. The general case (allowing any finite extension of fields $L/K$) is Proposition A.5.9 in "Pseudo-reductive groups".

To appreciate the optimality of the smoothness hypothesis when $L/K$ is not separable, we finally come to the mind-boggling counterexamples: for any imperfect field $K$ one can make examples of purely inseparable finite extensions $L/K$ and geometrically integral affine plane curves $X$ over $L$ that are smooth away from one point (and even regular at that non-smooth point when $[K:K^p] > p$, with $p = {\rm{char}}(K)$) such that ${\rm{R}}_{L/K}(X)$ is disconnected and has a connected component that is everywhere non-reduced! This is discussed immediately after the statement of Proposition A.5.9 in "Pseudo-reductive groups".

[If we relax geometric integrality to geometric irreducibility then other weird things can happen. For instance, ${\rm{R}}_{L/K}(X)$ can be non-reduced with separable $L/K$ and reduced $X$ that is not geometrically reduced -- not so surprising if you think about it for a minute -- and this Weil restriction can be empty for $X = {\rm{Spec}}(F)$ with a purely inseparable finite extension field $F/L$ and suitable purely inseparable $L/K$.]

• Wow, thanks a lot for your effort in writing this very detailed answer! I will carefully work through it:) – jorst Aug 4 '15 at 16:17
• You're welcome. I f you look in the 1st edition of "Pseudo-reductive groups" you'll find after the statement of Proposition A.5.9 the asserted existence of such crazy curves as near the end of my answer, but no details, but the 2nd edition gives actual examples with justification (over any imperfect field). And the loss of properness under inseparable Weil restriction (in the smooth case with positive dimension) is addressed somewhere in A.5 too; not so surprising, as it gives something like a jet bundle (sort of like Weil restriction through $K[\epsilon]/K$ gives the tangent bundle). – grghxy Aug 4 '15 at 16:34

Let me just try to clarify the definitions a bit as it seems to be the case where Wikipedia is actually poorly written.

The first thing to keep in mind is that, reformulating the functor of points definition, the Weil restriction $R_{K/k}$ is the right adjoint to the extension of scalars functor. $$\mathrm{Hom}_K(Y_K, X) \simeq \mathrm{Hom}_k(Y, R_{K/k}(X)).$$

As for an explicit construction, choose an embedding $K\subset k^{sep}$ and let $L$ be the normal closure of $K$ in $k^{sep}$ over $k$. Then there exist exactly $d=[K:k]$ embeddings $\sigma_1,\ldots,\sigma_d: K\to L$. Each embedding $\sigma_i$ defines a variety $X_{\sigma_i}$. Consider the product $Y'=X_{\sigma_1}\times\ldots\times X_{\sigma_d}$. The group $G=Gal(L/k)$ acts on $Y'$ and defines canonical descent data. In particular, there exists $Y$ defined over $k$ such that $Y_L\simeq Y'$.

Now you should be able to understand what is $R_{K/k}(Y_K)$ (I leave it as an exercise).

Edit: as grghxy points out, I indeed assume that $k \subset K$ is finite separable.

• It should be indicated more explicitly that this answer is only for the case when $K/k$ is separable, as a lot of good behavior in that case (e.g., preservation of geometric connectedness, geometric reducedness, geometric irreducibility, properness, non-emptiness, etc.) breaks down when $K/k$ is not separable. – grghxy Aug 4 '15 at 12:51
• @grghxy thank you for pointing that out – Anton Fonarev Aug 4 '15 at 14:10
• thanks! I agree this is a more conceptual view of what is happening than what is presented on the wikipedia page. – jorst Aug 4 '15 at 16:18