It should be noted that identifying which $p$-divisible groups arise from abelian varieties is a rather subtle question (e.g., over finite fields of size $q$ there is a Weil-number restriction on the $q$-Frobenius, though there are further subtleties since a simple abelian variety may have non-isosimple $p$-divisible group), so it is quite remarkable that for many questions about abelian varieties in char. $p$ (such as for deformation theory) one can nonetheless reduce oneself to results that are valid more generally for $p$-divisible groups.

For any (smooth, finite-dimensional) commutative formal group $G$ over a field of char. 0, the theory of logarithms provides an isomorphism to a power of the formal additive group. But in positive characteristic $p$ the formal group of an abelian variety "is" the identity component of its $p$-divisible group, so this has a rich theory of moduli in dimension $> 1$. In dimension 1 over a separably closed field, a $p$-divisible group with finite height is determined up to isomorphism by its height (e.g., height 1 for an ordinary elliptic curve, height 2 for a supersingular one).

This is very nice, though I wonder: how does local cohomology tell us about flatness properties for the absolute integral closure?

@JasonStarr: I was looking at the distinguished element 1 in $B$ (which is what makes $B$ more tractable than a random $A$-module).

Let $A = \mathbf{Z}$, $B = \prod_p \mathbf{F}_p$. Assume the non-flat locus $Y$ in Spec($B$) is closed. It contains the evident clopen points Spec($\mathbf{F}_p$), so if $J$ is an ideal in $B$ cutting out $Y$ then $J$ has vanishing image in each direct factor ring $\mathbf{F}_p$, so $J=0$. Thus, $B$ would be nowhere flat over $A$, so $B \otimes_A \mathbf{Q}$ would vanish and hence $B[1/N]=0$ for some $N > 0$. But this is false.

Yes. (It seems you mean to assume $K$ has characteristic 0.) Look up "Cohen structure theorem" in books on commutative algebra (e.g., Matsumura's "Commutative ring theory": Theorems 29.3 and 29.8(ii)).

Let $A$ be a reduced finite generated algebra over an algebraically closed field $k$, $K/k$ an extension field. Let $X={\rm{MaxSpec}}(A)$. The map $A \rightarrow \prod_xk$ defined by $a\mapsto (a(x))_x$ is injective, so it remains injective after scalar extension to $K$. For arbitrary $k$-vector spaces, $V \otimes_k(\prod_i W_i) \rightarrow\prod_i (V\otimes_k W_i)$ is injective, even with infinitely many $W_i$'s (use direct limits in $V$ to reduce to $\dim V < \infty$, and then to $\dim_k V = 1$). Setting $V=K$ and $\{W_i\}=\{k\}_{x\in X}$, $K\otimes_kA\rightarrow \prod_x K$ is injective. QED

Perhaps you mean SO$_q$ rather than O$_q$, but regardless, the uniqueness of maximal compact subgroups over $\mathbf{R}$ breaks down completely over non-archimedean local fields $k$ (every compact subgroup of $G(k)$ lies in a maximal one, and the number of conjugacy classes of maximal compact subgroups is finite but generally $> 1$). There is a good notion of Cartan decomposition for the group of points of a connected semisimple group over a non-archimedean local field (see the work of Bruhat--Tits), but it has nothing to do with notions like Cartan involution. See Tits' article in Corvallis.

The definitions of "elliptic" as given in the question are not compatible when $K$ is not perfect because ${\rm{R}}_{L/K}({\rm{GL}}_1)$ is not a $K$-torus when $L$ is not separable over $K$ (so a separability condition has to be manually inserted into a GL$_n$-specific description). More specifically, in the notation used in the answer, $u$ is not semisimple as a point of the algebraic group ${\rm{GL}}(V)$ if $P_{\rm{min}}$ is irreducible but not separable. (In linear algebra, semisimplicity of an endomorphism can be lost by an inseparable ground field extension.)

But statement (1) in the question posed is generally false when the field $F$ is not perfect (e.g., counterexamples arise over every global or local function field over a finite field) because $L$ can fail to be separable over $F$: in such cases "$L^{\ast}$ as an $F$-group", which is to say ${\rm{R}}_{L/F}({\rm{GL}}_1)$, is not an $F$-torus. This is related to the fact that Jordan decomposition of an $F$-point need not be $F$-rational when $F$ is not perfect.

@Joe: There is a hidden subtlety in the formulation of the question which "rules out" twisting counterexamples (which is why I focused on non-maximal orders): the OP is beginning with an abelian variety over $\overline{\mathbf{Q}}$ and seems to allow the option to choose whatever descent to a number field we like best. So your (natural) viewpoint of first fixing a descent and then going up from there is sort of "opposite" to how the question is posed. Though I really have no idea if the OP recognizes the subtlety of this aspect of how the question was posed, and whether it is intended.

What you say in the elliptic curve case is false, since the CM order might not be maximal (so the $j$-invariant might generate over $K$ a much bigger class field).

There are definitely subtleties when turning isogenies of algebraic groups into maps between $F$-points. The commutator of $G(F)$ is the image of the group of $F$-points of the simply connected central cover of the derived group. Already for $G = {\rm{PGL}}_n$ one sees that matters are not so simple (e.g., determinant mod $n$th powers gives a nontrivial commutative quotient $F^{\times}/(F^{\times})^n$ of ${\rm{PGL}}_n(F)$), so what Loeffler has written is not correct.

Artin's result in the case of formal completion at a point doesn't require any smoothness hypothesis. By using Popescu's generalization of Artin approximation, the proof adapts to work etale-locally on $S$ (to get a common pointed etale neighborhood when there's a common point-completion over one on $S$) if $S$ is excellent. But in the smooth case, at least Zariski-locally, can't we just take $U$ to be the fiber square of etale maps of $X$ and $Y$ to an affine space (respecting sections), avoiding Artin approximation entirely?

Example 7.5.1 in the paper "Finiteness theorems for algebraic groups over function fields" in Compositio Math. 148 (2012), which uses just standard methods in global Galois cohomology (duality theorems, etc.) There must be earlier references on this issue as well.

No, once you allow non-empty $S$ it is often (maybe always?) provably infinite, in contrast with the case of linear algebraic groups, for which the "$S$-version" is provably always finite.

To "demystify" a bunch of these issues, try to understand why a map of schemes is smooth, or finite type, or affine, or finite presented, or separated (or...) if it becomes so after an fpqc base change. Toy example: if $K/k$ is a field extension and a $k$-algebra $A$ is such that $K \otimes_k A$ is finitely generated as a $K$-algebra then why is $A$ finitely generated as a $k$-algebra? Mull it over for yourself, and then if stuck either look in EGA IV$_3$, sections 8, 9, ... and/or ask suitable people in your department.

@Jason: I agree, it doesn't matter where questions come from. My "objection" was that since the question is already posed in the Stacks Project, it had looked inappropriate for the OP to write "we ask" rather than "the following question is posed in the Stacks Project". Even if someone wants to go through a document and post many of its open questions on MO, it seems reasonable for this to be done gradually over time, not dumping so many in quick succession. Now that "question_bot" has filled out a profile (not at the time of my initial comment), the context for the flood has became clearer.

@LaurentMoret-Bailly: Oops, I must have had in mind the additional (necessary) condition that the $\overline{k}$-points over $x$ are all in the same $G_{\overline{k}}$-orbit. I should think it through again, but that would appear to take care of your suggested counterexample.