Yes if $S$ is reduced, and no otherwise. The case of a field is classical (ultimately because $k[t,1/t]^{\times} = k^{\times}\cdot t^{\mathbf{Z}}$ for fields $k$), and I assume you are familiar with ...

[This answer is an elaboration on what KConrad mentions in his comments below, posted while I was first writing this.] At finite level one has a perfectly good theory of Frobenius elements attached ...

This is an application of the geometric method of constructing highest-weight representations via line bundles on the "variety of Borel subgroups", together with the theory of split semisimple ...

I assume you meant for the fiberwise rank of $E$ to be constant, say $r>0$ (or at least uniformly bounded above). The answer is "yes" when the Stein space has finite dimension (equivalently, when ...

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 ...

The general technique goes as follows. Let $\{A_i\}$ be a directed system of rings with limit $A$; e.g., $A=k$ and $\{A_i\}$ the set of finitely generated $k_0$-subalgebras of $k$. Let $X$ and $Y$ ...

Are you trying to justify some of the assertions about the Tate curve in Katz-Mazur? (Would help to know the motivation for the question.) Anyway, the affirmative answer is part of the "standard" ...

Suppose $k$ is finitely generated over a perfect field $k_0$ of characteristic $p$ (e.g. $k_0 = \mathbf{F}_p$), with transcendence degree $d > 0$. We'll make infinitely many such pairwise non-...

Smoothness of $X$ is not needed (neither for the comparison isomorphism nor for the result in question). Let $X$ be any quasi-separated scheme over a separably closed field $k$, equipped with an ...

Yes, and you can make it split away from whatever even finite set $S$ of places of $\mathbf{Q}$ you wish that contains the archimedean place, and such a form of ${\rm{Sp}}_{2n}$ is uniquely determined ...

Yes. The task is to show that $X$ is a scheme (as then Lichtenbaum's result may be applied). By standard "spreading out" arguments, we may assume $S = {\rm{Spec}}(R)$ for a discrete valuation ring $R$,...

There's more than enough information: the answer is that $S$ is never projective when it isn't "obviously" projective (i.e., never happens when the leading coefficient of $f$ is a non-unit). This is ...

For any finite separable extension of fields $k'/k$ and geometrically connected smooth affine $k'$-scheme $X'$, the affine finite type Weil restriction $X := {\rm{R}}_{k'/k}(X')$ over $k$ is smooth ...

The answer is affirmative with $R$ any henselian local ring and $G$ any smooth $R$-group scheme; this is a special case of Lemma 11.4 in Grothendieck's article "Le Groupe de Brauer III: Examples et ...

For a scheme $S$, every proper finitely presented map $f:X \rightarrow S$ from an algebraic space $X$ admitting a line bundle $L$ that is ample on each geometric fiber (which of course forces such ...

One can also give explicit high-rank counterexamples using special orthogonal groups of quadratic lattices over $R$ with non-degenerate reduction, and we can also arrange that $R$ has $K$ as its ...

The question has some minor misstatements. You didn't really want to mention $R$ anywhere: the base ring is $A$, over which $A'$ should be assumed to be "finite locally free" (equivalently, "finite ...

There are really two separate assertions here: one concerns the generic fiber $A = \mathbf{Q} \otimes_{\mathbf{Z}} R$, and the other concerns orders in semisimple $\mathbf{Q}$-algebras. To clarify ...

Yes. In fact, if $B \subset C$ is a module-finite extension of noetherian rings with $C$ local and complete then $B$ is local and complete. Indeed, by standard prime-lifting stuff with module-finite ...

As desired in the question, the action of $A \times B$ on $G$ is the key point. This is a transitive action, and it is a general fact in the theory of actions of (separable) Lie groups on manifolds (...

It seems easier to work this out than to dig up a literature reference. This is an application of the basic structure of connected reductive groups and the dictionary between connected reductive ...

The answer is sort of maximally negative. For a (smooth) connected linear algebraic group $G$ over any field $k$ whatsoever, if $G$ admits a Borel $k$-subgroup (as happens for finite $k$) then all ...

Gauss' Disquisitiones gives a completely algorithmic solution to this question in his development of reduction theory for ternary quadratic forms over the integers (via artful systematic use of his ...