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Answer by nosr for How Fine One Must Choose an Affine Cover to get Weil Restriction?

2012-11-20T02:22:11Z

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 ${\rm{R}}_{S'/S}(U_i)$ .

Answer by nosr for Groups becoming algebraic groups

2012-11-17T23:39:36Z

I predict that in whatever is the situation of motivating interest, you know more: for any algebraically closed field $K/k$ you likewise have a group structure on $G(K)$ functorially in $K/k$ making the translations and inversions by $G(K)$ also be morphisms on $G_K$. Under this additional condition the answer is always affirmative, by using a trick that I believe goes back to Hasse, namely base change to the function field of $G$ (or rather, in our situation, an algebraic closure thereof). This is a Weil-style way of getting at Yoneda's Lemma.

The inversion hypothesis implies that left-translations are also morphisms, and we will use this condition instead of the inversion hypothesis (and deduce at the end that inversion is a morphism too). In fact, in char. 0 we won't even need this condition with the left-translations (or inversion) at all, but we are using a stronger assumption on the existence of group laws on $K$-points for lots of $K/k$. This stronger initial assumption across many $K/k$ bypasses Bondarko's examples in char. 0. The near-miss examples avoiding the inversion condition in char. $p$ in the comments are not ruled out by our stronger hypothesis across all $K/k$, but they are ruled out by the left-translations being morphisms (!), so matters will be more delicate in positive characteristic. We need the left-translation condition in positive characteristic to circumvent some inseparability issues with function fields of $k$-varieties.

Before we begin the actual arguments, note that since $k$ is algebraically closed, so $G(k)$ is Zariski-dense in $G$, for any field extension $F/k$ the subset $G(k)$ inside $G(F) = G_F(F)$ is Zariski-dense $G_F$. Hence, $F$-morphisms among $G_F$, $G_F \times G_F$, etc. are determined by their effect on $k$-points promoted to $F$-points. I mention this at the outset for peace of mind in some later discussions.

The "translation is a morphism" hypothesis ensures that $G$ is smooth, so its connected components are irreducible. I will assume that the variety $G$ is connected, a harmless assumption since (i) for $g \in G^0(k)$ the right-translation morphism by $g$ preserves $G^0$ because it carries $e$ to $g$, (ii) inversion preserves $G^0$ since it fixes $e$, (iii) the morphism property on $G^0$ clearly implies the general case by using the "morphism" property for a few translations by points in the various connected components of $G$. Thus, now $G$ is irreducible and so has a "function field" $k(G)$.

By applying the functoriality in $K/k$ to $k$-automorphisms of $K$, we see via Galois descent that the initial hypothesis for algebraically closed extensions is actually valid for all perfect extensions of $k$ as well. Let $K$ be the perfect closure of $k(G)$, and let $\eta \in G(k(G)) \subset G(K)$ correspond to the generic point. This induces a right-translation morphism $\rho_K:G_K \simeq G_K$ (computing right-translation by $\eta_K$). Since $G$ is finite type over $k$, by expressing $K$ as a directed union of finite purely inseparable extensions of $k(G)$ we obtain a finite purely inseparable extension $F$ of $k(G)$ and an $F$-morphism $\rho_F:G_F \simeq G_F$ that descends $\rho_K$ and clearly computes right-translation by $\eta_F$. The normalization $X$ of $G$ in the finite extension $F/k(G)$ is a $k$-variety finite radiciel over $G$, and $X_{\eta} = {\rm{Spec}}(F)$ over $k(G)$. Thus, by expressing $k(G)$ as the directed union of the coordinate rings of affine opens around $\eta$ in $G$, we find such an open $U$ so that over its preimage $U'$ in $X$ (which is finite radiciel over $U$) there is a $U'$-morphism $\rho_{U'}:G_{U'} \simeq G_{U'}$ extending $\rho_F$.

The map $U'(k) \rightarrow U(k)$ is bijective, so for each $g \in U(k)$ let $g' \in U'(k)$ be the unique point over $g$. Consider the specialization $\rho_{g'}:G \simeq G$ of $\rho_{U'}$ over $g'$. I claim that this is the right-translation by $g$. To see this, pick $h \in G(k)$ and consider the morphism of left-translation $\ell_h:G \simeq G$ by $h$. This carries $\eta$ to another $k(G)$-point of $G$ (over $k$) that "spreads out" to a $U$-point of $G$ whose specialization at $g \in U(k)$ is $\ell_h(g) = hg$ (so if we work instead with the $U'$-point over this via the canonical $U' \rightarrow U$ then its specialization at $g' \in U'(k)$ is also $\ell_h(g) = hg$). Now inside the group $G(K)$ we have $\ell_h(\eta_K) = \rho_K(h_K)$, so $\rho_{U'}(h_{U'})$ is the $U'$-point obtained from applying $\ell_h$ to the canonical $U'$-point of $G$ (spreading out $\eta_F$) because this comparison of $U'$-points may be checked by working at the generic point of $U'$ (and this generic point is dominated by the canonical $K$-point, over which we have the equality $\ell_h(\eta_K) = \rho_K(h_K)$). Specializing this equality of $U'$-points at the point $g' \in U'(k)$ over $g \in U(k) \subset G(k)$ gives the equality of $k$-points $\rho_{g'}(h) = \ell_h(g) = hg$, as desired.

To summarize, we have constructed a $U'$-morphism $\rho:G_{U'} \simeq G_{U'}$ whose specialization at any $g' \in U'(k)$ is right-translation on $G$ by the corresponding point $g \in U(k)$. Suppose ${\rm{char}}(k) = 0$, so $U' = U$ and the composite morphism $${\rm{pr}}_1 \circ \rho:G \times U \rightarrow G$$ on $k$-points is the group law restricted to $U(k)$ in the 2nd variable. For any $g \in G(k)$ the right translation morphism $\rho_g$ carries $U$ to another open subscheme $\rho_g(U)$, and I claim that these opens cover $G$. It suffices to check the covering property on $k$-points, so for $h \in G(k)$ we seek $g \in G(k)$ such that $h \in \rho_g(U)$, or equivalently $hg^{-1} \in U(k)$. Pick any $u \in U(k)$ and let $g = u^{-1}h$. This proves the covering property, and the multiplication map $G(k) \times U(k)g \rightarrow G(k)$ arises from a morphism, namely the composition of ${\rm{pr}}_1 \circ \rho$ and the morphism $\rho_g$. We conclude (in char. 0) that the group law on $G(k)$ is induced by a morphism $G \times G \rightarrow G$.

Now we can deduce in char. 0 that inversion is a morphism (so we have a group variety), even though we never used the left-translation or inversion conditions. Indeed, the "universal right-translation" $G \times G \rightarrow G \times G$ defined by $(x,y) \mapsto (xy,y)$ between fppf $G$-schemes (via ${\rm{pr}}_2$) is a scheme isomorphism between fibers over all points in $G(k)$ and therefore is a scheme isomorphism (by fibral isomorphism criteria, adapted to the peculiarities of $k$-points when $k$ is alg. closed). This yields the morphism property for inversion for free! This was a char-free argument, but it required the composition law to be a morphism. Anyway, char. 0 is now settled.

Assume ${\rm{char}}(k) = p > 0$, so for sufficiently large $n \ge 0$ the finite flat $n$-fold relative Frobenius morphism $G^{(1/p^n)} \rightarrow G$ of $G^{(1/p^n)}$ dominates $U'$ over $U$ (namely, pick $n \ge 0$ so that the finite purely inseparable extension $F/k(G)$ is contained inside $k(G)^{1/p^n}$). Since the initial choice of $F$ could be replaced with a finite purely inseparable extension at the outset if we wish, we may therefore assume that $U'$ is open inside $G^{(1/p^n)}$. Using a covering and translation argument similar to characteristic 0, we arrive at a morphism $$m_r:G \times G^{(1/p^n)} \rightarrow G$$ that recovers the given group law on $k$-points via the natural identification of $G^{(1/p^n)}(k)$ with $G(k)$ (for some $n \ge 0$). Our problem is precisely to show that $m_r$ factors (in the sense of morphisms of varieties) through the $n$-fold relative Frobenius morphism in the 2nd variable.

Now we shall use that left-translations are morphisms too. At the cost of increasing our $n$ if necessary, we can run through the same arguments with "left" instead of "right" to arrive at another morphism $$m_{\ell}:G^{(1/p^n)} \times G \rightarrow G$$ which recovers the given group law on $k$-points (with the same $n$).

Returning to our task of checking that $m_r$ factors through the appropriate iterated Frobenius in the 2nd variable, since that iterated Frobenius is fppf (as $G$ is smooth!) we conclude via fppf descent that it is harmless to check the existence of such a factorization after precomposing $m_r$ with an fppf morphism in the first variable. So let's compose with the same iterated Frobenius in the first variable, arriving at a morphism $$G^{(1/p^n)} \times G^{(1/p^n)} \rightarrow G$$ that recovers the composition law of $G(k)$ on $k$-points. It suffices to show that this latter map factors through the $n$-fold Frobenius in its 2nd variable, but this factorization is clear: it is the composition of $m_{\ell}$ with that iterated Frobenius (as we may check by computing on $k$-points, since we're working with reduced $k$-schemes of finite type). QED

Answer by nosr for Is the n-torsion of an extension of an abelian variety by a torus, finite and flat?

2012-11-17T21:18:14Z

It is an exercise with descent theory and the snake lemma for fppf abelian group sheaves to deduce the result for $G[n]$ from the cases of $T[n]$ and $A[n]$.

In more detail, by the snake lemma $G[n]$ is an extension of $A[n]$ by $T[n]$ in the sense of such abelian sheaves. Since $A[n]$ and $T[n]$ are each finite fppf over $S$, the same then holds for $G[n]$. Indeed, rather generally, if $$1 \rightarrow G' \rightarrow G \rightarrow G'' \rightarrow 1$$ is a complex of $S$-group schemes with $G'$ affine fppf over $S$ and the diagram is short exact for the fppf topology (so $G'$ is the scheme-theoretic kernel of $G \rightarrow G''$) then the functor of points of $G$ as a $G''$-scheme is a $G'$-torsor for the fppf topology on $G''$, so the $G''$-scheme $G \rightarrow G''$ becomes isomorphic fppf-locally on $G''$ to $G'$ (over the base) as a scheme. Hence, by fppf descent for properties of morphisms, $G \rightarrow G''$ inherits many "nice" properties that may be satisfied by $G' \rightarrow S$, such as: proper, flat, smooth, etale, finite, etc. In particular, $G$ is fppf over $G''$ and if $G'$ is finite over $S$ then so is $G \rightarrow G''$ (and hence so is $G$ if $G''$ is also finite over $S$).

See Oort's LNM book on commutative group schemes for generalizations with the fpqc topology (around section 18, IIRC).

Answer by nosr for Adelic description of moduli of $G$-bundles on a curve

2012-11-17T06:03:31Z

The one-sentence answer to this question is: use fpqc descent theory (and an "answer" which doesn't address the role of fpqc descent -- sometimes presented in the form of a reference to a paper of Beauville and Laszlo -- is missing the key technical issue in the rigorous proof when working with general $G$, as far as I know).

We make two "necessary" hypotheses (as noted in the comments to Sawin's answer) for our smooth connected affine $k$-group $G$: ${\rm{H}}^1(K,G) = 1$ for $K = k(X)$ and ${\rm{H}}^1(k',G) = 1$ for all finite extensions $k'/k$. For example, if $k$ is finite then this holds for any simply connected semisimple (connected) $G$ by the theorems of Harder and Lang respectively. If instead $k$ is algebraically closed of char. 0 then it holds for any $G$ by theorems of Tsen and Springer. If $k$ is algebraically closed of positive characteristic then it holds for any connected reductive $k$-group $G$, but Springer's theorem doesn't literally apply; see Remark 2(b) of the Drinfeld-Simpson paper mentioned in the comments to Sawin's answer.

We shall allow $X$ to be any 1-dimensional reduced and irreducible $k$-scheme of finite type, not assumed to be proper or even normal. This way we incorporate Chervov's observations about using singular curves to build in more level structure.

We construct the desired bijection as follows. Consider a left $G$-torsor $E \rightarrow X$ (local triviality equivalent for the fpqc and etale topologies due to the smoothness of $G$; the equivalence will be crucial later on and is false in general if we try using the Zariski topology, and officially we work with the etale topology in the definition as is traditionally done). Since ${\rm{H}}^1(K,G) = 1$, the generic fiber $E_{\eta}$ has a $K$-point and this spreads out over a dense open $U$ in $X$. Fix such a $U$ and trivialization $\xi \in E(U)$.

Let $X^0$ be the set of closed points of $X$. Consider the pullback of $E$ over the completion $O^{\wedge}_x$ at $x \in X^0$. This pullback is a smooth $O^{\wedge}_x$-scheme whose special fiber is a $G$-torsor over the finite extension $k(x)$ of $k$ and so has a $k(x)$-point due to the other vanishing hypothesis. By smoothness (!) of $G$ (and the henselian property of $O^{\wedge}_x$) this lifts to a point $\xi_x \in E(O^{\wedge}_x)$.

For each $x \in X^0$ consider the pullbacks of $\xi$ and $\xi_x$ over $U \times_X {\rm{Spec}}(O^{\wedge}_x) = {\rm{Spec}}(K_x)$ where $K_x$ is the total ring of fractions (product of finitely many fields) of the 1-dimensional reduced complete local ring $O^{\wedge}_x$. These two $K_x$-points of a common $G$-torsor are related through the action of a unique $g_x \in G(K_x)$; to be precise, $\xi = g_x \xi_x$ in $E(K_x)$. If $x \in U^0$ then clearly $g_x \in G(O^{\wedge}_x)$, so $(g_x) \in G(\mathbf{A}_X)$.

If we change $\xi$ then we multiply every $g_x$ on the left by some common $g \in G(U)$, and if we change the various $\xi_x$'s then we multiply each $g_x$ on the right by an element of $G(O^{\wedge}_x)$ . Finally, taking into account that we may shrink $U$ (and thereby enlarge $X - U$), we obtain an element $$(g_x) \in G(K)\backslash G(\mathbf{A}_X)/G(O^{\wedge})$$ (where $O^{\wedge} = \prod_{x \in X^0} O^{\wedge}_x$) that depends only on the isomorphism class of $E$ over $X$.

Our problem is to show that (i) this adelic double coset determines the isomorphism class of $E$ and (ii) all double cosets arise in this way.

The assertion (i) is proved as follows. Assume $E$ and $E'$ give rise to the same double coset, so for a Zariski-dense open $U$ in $X$ trivializing $E$ and $E'$ we have $(g'_x) = \gamma (g_x) h$ for some $\gamma \in G(K)$ and $h \in G(O^{\wedge})$ with $g'_x, g_x \in G(O^{\wedge}_x)$ for all closed points $x$ of $U$. By shrinking $U$ we may assume $\gamma \in G(U)$. We may replace $\xi_x$ with $h_x\xi_x$ for all $x \in X^0$ and replace $\xi$ with $\gamma \xi$ so that $g'_x = g_x$ for all $x$. In other words, $E_U$ and $E'_U$ are each identified with the trivial $G_U$-torsor and has corresponding trivial $K_x$-fiber identified with the generic fiber of $G_{O^{\wedge}_x}$ via $g_x$-translation. Working one point of $X - U$ at a time, we just have to check:

${\mathbf{Claim}}$: The category of $G$-torsors over $O_x$ is equivalent to the category of $G$-torsors over $O^{\wedge}_x$ equipped with a $K$-descent on its generic fiber over $K_x$.

The categorical aspect of this Claim is essential (i.e., we do not just consider sets of isomorphism classes).

Proof: By fpqc descent theory (the "Beauville-Laszlo step", though for us all we need was provided by Grothendieck) applied to the fpqc cover $${\rm{Spec}}(K) \coprod {\rm{Spec}}(O^{\wedge}_x) \rightarrow {\rm{Spec}}(O_x)$$ whose fiber square is ${\rm{Spec}}(K_x)$, the category of affine $O_x$-schemes is equivalent to the category of affine $O_x^{\wedge}$-schemes equipped with a $K$-structure on the generic fiber over $K_x$. Since the notion of $G$-torsor is well-behaved for the fpqc topology (and recall that fpqc $G$-torsors are automatically etale-topology torsors, due to the smoothness of $G$!!!), this equivalence specializes to the case of $G$-torsors. QED

Now we run the game in reverse. Pick a class in $G(K)\backslash G(\mathbf{A}_X)/G(O^{\wedge})$ represented by some $(g_x) \in G(\mathbf{A}_X)$. Since $g_x \in G(O^{\wedge}_x)$ for all but finitely many $x \in X^0$, we can pick a Zariski-dense open $U$ in $X$ such that $g_x \in G(O^{\wedge}_x)$ for all $x \in U^0$, so we may change our representative to satisfy $g_x = 1$ for all $x \in U^0$. Applying the above Claim then enables us to extend the trivial $G_U$-torsor to a $G$-torsor $E$ over $X$ by fpqc-gluing using the elements $g_x$ for each $x \in X - U$ one at a time, and by design this $E$ gives rise to the chosen adelic double coset (if we are careful not to mix up $g_x$ and $g_x^{-1}$ for $x \in X - U$).

Comment by

2013-05-18T22:29:22Z

In the non-compact case there are problems caused by unipotent elements. For example, all non-trivial unipotent elements in ${\rm{PGL}}_2(\mathbf{R})$ are conjugate to each other, so a continuous class function would have the same value on all such elements, hence also on the identity, as that is a limit of such elements. Hence, such class functions cannot separate the identity from that other conjugacy class.

Comment by

2013-05-17T13:30:13Z

No. The case of quadratic fields is misleading, or rather has a special property that fails in higher degree: the ring of integers is monogenic over $\mathbf{Z}$. Once you drop that property, the invertibility can fail (and all orders in rings of integers of number fields are Cohen-Macaulay). I think there are counterexamples given in Shimura's introductory book on modular forms, around where he discusses the invertibility in the quadratic case.

Comment by

2013-05-17T13:22:56Z

Yes; see Illusie's paper that surveys Grothendieck's work on the deformation theory of p-divisible groups. Early in there he gives Dieudonne-module arguments of Gabber and Ekedahl to handle the extension problem over any perfect field.

Comment by

2013-05-17T00:00:43Z

Another proof: $\Omega^1_{K(X)/k}$ has $K)X)$-dimension $d$ by computing after extension to the perfect closure of $k$ (preserves topology -- irreducibility -- and reducedness by hypothesis), using existence of separating transcendence bases over perfect fields. Since $\Omega^1_{K(X)/k}$ is $K(X)$-spanned by elements ${\rm{d}}f$ for $f \in K(X)$, there is $\{f_j\}$ so ${\rm{d}}f_j$'s are a $K(X)$-basis. The $f_j$'s are algebraically independent over the perfect $k$ (differentiate a potential irreducible relation) and $L=k(t_1,\dots,t_d)\rightarrow K(X)$ has $\Omega^1_{K(X)/L}=0$. QED

Comment by

2013-05-07T20:09:56Z

Yes. Let $U$ be the open affine curve away from $S$ over the constant field $k$. For any elliptic curve $E \rightarrow U$ there's a finite etale cover $U' \rightarrow U$ of universally bounded degree over which $E$ acquires a point of order 1728. Since $\pi_1(U)$ has only finitely many quotients of any given size, there are only finitely many possibilities for $U'$ up to $U$-isomorphism. Each connected component $U'_i$ of $U'$ has a specified map to $Y_1(1728)$ and it is dominant if $j(E) \not\in k$. There are only finitely many such maps since $X_1(1728)$ has genus $> 1$. QED

Comment by

2013-05-07T19:52:19Z

Aside from the fact that the phrase "doesn't have to do with torsion data" is vague, if you consider a $\mathbf{Z}[1/N]$-schemes $S$ that is sufficiently disconnected then the set of level-$N$ structures on a fixed elliptic curve $E$ over $S$ can be arbitrarily large and in particular not finite (akin to global sections of a constant sheaf on a disconnected space). So do you mean just that for elliptic curves over algebraically closed fields the associated set should be finite? If so, then how about assigning to any $E \rightarrow S$ the automorphism group?

Comment by

2013-05-02T00:04:05Z

It's the same reason that one develops a general theory of finitely generated modules over noetherian rings rather than focus exclusively on locally free modules (akin to vector bundles): we get more robust notions of kernel and cokernel, and a wider framework in which to prove results about ideals of holomorphic functions by viewing them as instance of coherent sheaves (for which many operations are available which cannot be expressed purely within the setting of ideal sheaves: tensor product, Hom-sheaves, etc.). And one can perform normalization of coherent sheaves of algebras, etc., etc.

Comment by

2013-05-01T10:00:58Z

Stripping away the geometric terminology, it sounds like you seek a commutative ring $A$ and ideal $I$ such $I_{\mathfrak{p}}$ is invertible as an $A_{\mathfrak{p}}$-module for all prime ideals $\mathfrak{p}$ of $A$ but $I$ is not invertible as an $A$-module (equivalently, $I$ is not finitely presented as an $A$-module). Is that correct?

Comment by

2013-05-01T02:33:43Z

@Jason: For another example, $X_0(N)$ (appropriately defined as a proper flat Artin stack over $\mathbf{Z}$) is not Deligne-Mumford in characteristic $p$ when $p^2|N$.

Comment by

2013-04-30T23:24:34Z

It would be a bit more accurate to say "finite etale automorphism schemes" (though since offered just as an "expectation", perhaps one cannot insist on too much precision).

Comment by

2013-04-30T15:37:13Z

[I assume when you wrote "maximal (split) torus" you mean that $T$ is a split maximal $E$-torus of $G$ that is also maximal as an $E$-torus.]

Comment by

2013-04-30T15:35:24Z

They're equal. This can be seen in multiple ways. For example, the evident isomorphism $S'_E \rightarrow T$ induced by $G_{E'} \twoheadrightarrow G$ gives an identification ${\rm{X}}_F(S') = {\rm{X}}_F(T)$ under which $\Phi(G',S')$ is carried isomorphically onto $\Phi(G,T)$ (using the equality of $\mathfrak{g}'$ with the underlying $F$-vector space of $\mathfrak{g}$ to match root spaces), and the Weyl group of this common root system is naturally identified with the finite constant groups $W$ and $W'$. This respects the induced map $W'_E \rightarrow W$, so the latter is an isomorphism.

Comment by

2013-04-23T01:49:13Z

An abstract group of points over an algebraically closed field is not an algebraic group. One has to specify the algebro-geometric structure over the ground field. Do you mean for $G$ to be the linear algebraic group ${\rm{GL}}_n$ over $k = \mathbf{F}_p$, or in other words is your implicit Galois action of Frobenius given by the usual $p$-power on matrix entries? If so, then since $T$ as a $k$-group is a split torus, the answer to your question is negative since every element of $W(\overline{k})$ arises from a point in $N_G(T)(k)$ modulo right translation by a point in $T(\overline{k})$.

Comment by

2013-04-22T22:31:21Z

Dear Robert: For a non-degenerate quadratic space $(V,q)$ over a field $k$, usually ${\rm{SO}}(q)$ denotes the algebraic $k$-group classifying automorphisms of $(V,q)$ (over extensions of $k$). If $q$ is the standard split quadratic form $q_n$ on $k^n$ ($x_1 x_2 + x_3 x_4 + \dots + x_{n-1}x_n$ for even $n$, $x_0^2 + q_{n-1}$ for odd $n > 1$), it is common for algebraists to write ${\rm{SO}}_n$ to denote ${\rm{SO}}(q_n)$. So for $k = \mathbf{R}$, the Lie group ${\rm{SO}}_n(\mathbf{R})$ is not the same as ${\rm{SO}}(n)$. And ${\rm{PGL}}_2 = {\rm{SO}}_3$ as algebraic groups (over any $k$)!

Comment by

2013-04-18T05:45:52Z

The formation of the Neron model over henselian discrete valuation rings commutes with scalar extension to the maximal unramified extension and its completion, so for any "table" of reduction types it is often sufficient to consider only separably closed residue fields. Hence, to the extent the residue field is perfect, it usually may as well be algebraically closed. (It isn't clear if Tate's algorithm works for imperfect residue field $k$ of char. 2 or 3, due to the existence of non-smooth regular Weierstrass cubics over such $k$.) For much more, read 10.2 in Qing Liu's textbook.