Since fppf maps admit quasi-finite fppf quasi-sections, by considering henselizations we see that any fppf cover admits a refinement that is finite flat over an etale cover. So it is equivalent to ask the same question when $B$ is either finite free as an $A$-module or an etale cover of $A$. Not sure where to go from there. But please be more forthcoming and mention that "your" question is actually posed in the Stacks Project (rather than have that fact be hidden behind the link). In view of your flurry of recent MO questions, I hope this isn't the beginning of some trend of SP questions...

The above is only addressing when the trisected angle can be built by straightedge and compass starting from nothing but unit length. This is sort of the "wrong" viewpoint, as Quas identifies the correct viewpoint: one seeks trisectability given the angle $\theta$, so it becomes an algebraicity question over $\mathbf{Q}(\cos \theta)$, which in turn looks basically hopeless for a nice answer.

Since $\mathbf{Q}(e^{i \alpha})$ has degree 2 over $\mathbf{Q}(\cos(\alpha))$ for $\alpha\not\in \mathbf{Z}\pi$, necessarily $e^{i\theta/3}$ has 2-power degree over $\mathbf{Q}$. But $e^{i z} = e^{(z/\pi)(\pi i)}$ with $\pi i$ a complex log of the algebraic $-1 \ne 0, 1$, so by Gelfond-Schneider it's algebraic iff $z/\pi$ is rational. You seek rational $\alpha$ so that $e^{i \pi \alpha}$ has 2-power degree over $\mathbf{Q}$; the $3\alpha$'s are the $\theta$'s. So the answer is $\theta = a/b$ in reduced form with $b = 2^r \prod p_i$ for distinct Fermat primes $p_i > 3$ (!) (avoiding 6...).

Yes, use the "noetherian" property of coherent sheaves on a complex-analytic space $X$ (see Ch. 5, section 6, of the book "Coherent analytic sheaves"): any directed system $\{F_j\}$ of coherent subsheaves of a coherent $O_X$-module is locally stationary (i.e., $X$ admits an open cover $\{U_i\}$ such that for each $i$ there is some $j(i)$ with $F_j|_{U_i} = F_{j'}|_{U_i}$ for all $j, j' \ge j(i)$). I think this is due to Serre. Exhausting your $\mathcal{I}$ by the coherent ideals arising from finite subsets of the global set, this "noetherian" property does the job.

The Hausdorff condition is a matter of taste. The coherence condition on $\mathscr{I}$ is an essential feature to get a reasonable theory, with coherent structure sheaf, etc. The notion of coherence in holomorphic function theory and its robust properties was one of the greater discoveries of Oka, latter clarified and refined via sheaf-theoretic ideas of Cartan, Serre, et al.; one cannot do anything interesting in the style of complex-analytic geometry without it. Motivation for the objects of study is always a good thing to keep in mind. :)

@user45442: Here's an example illustrating the need for context. For an affine finite type group scheme $G$ over a field $k$, $(G_{\overline{k}})_{\rm{red}}$ is a smooth subgroup scheme of $G_{\overline{k}}$ (usually not normal!) but $(G_{k_s})_{\rm{red}} = (G_{\rm{red}})_{k_s}$ is generally not a subgroup scheme of $G_{k_s}$. Yet for some big $n$ the quotient $G/\ker(F_{G/k,n})$ modulo kernel of the $n$-fold relative Frobenius morphism is smooth, so if quotient by an infinitesimal normal $k$-subgroup is harmless for some purpose then one can pass to smooth groups over $k$ after all...

The geometry and etale sites are the same, but reducedness is a big distinction. One merit of just going up to the separable closure is that you can often use Galois descent to return to the "actual fiber", which fails completely if you go up to the algebraic closure (when $k(y)$ is not perfect). On the other hand, if you want to get a rational point you might need to make an inseparable extension on $k=k(y)$. But such extension can fail to commute with the formation of nilradical or "$k$-unipotent radical", etc. Kestutis' final sentence hits the nail on the head.

@Eric: Since I assume you have read the proof of GAGA, and hence must know the analytic properties of $O(1)$ and the basic theory of coherent-analytic sheaves (without which that proof doesn't work), you just use exactly the same proof as in the algebraic case. There is probably no reference because it is literally the same argument. Just talk with anyone in algebraic geometry or number theory on the faculty at your university if you have further questions about it.

Where is the definition of "etale geometric morphism of ringed toposes" to be found? If one removes the $\infty$'s, is the above Theorem 1.2.1 different from the functorial characterization of etaleness for maps of schemes? (That is, does it tell us something different from EGA, or with a different method?) I do not know what a "proper" geometric map of toposes is (where is the definition given?), but should techniques that address an "etale" version be relevant to properties such as properness? Sorry for the list of questions.

The analytic input into the proof of GAGA provides everything which is required, so GAGA isn't needed and the algebraic proof (appropriately formulated) literally works in the analytic setting. Namely, by the analytic ampleness properties of $O(1)$, any coherent analytic sheaf on a closed complex-analytic subspace $X$ of $\mathbf{CP}^n$ has a resolution by vector bundles. So if $X$ is smooth then by Serre's theorem on finiteness of global dimension of regular local rings (applied to stalks of $O_X$) this resolution has vector bundle kernel at the $(\dim X)$-th step.

@Igor: Yes, that sounds like a realistic possibility. Anyway, I prefer the admissions system using recommendation letters rather than entrance exams. :)

@Igor: Probably whomever posed the question had an erroneous solution in mind. This appears to have been a PhD-entrance exam, right? I've seen other situations in which written qualifying exams for PhD students have questions which stump faculty experts because the person who put the question on the exam didn't write out a complete solution and so didn't realize how hard it is.

Just as an aside, when I first began to learn about representation theory as an undergraduate, I had the exact same question as you raise above. Eventually I came to realize that the process of learning a subject via the "abstract" approach rather than through the natural examples that motivate it (and must have been in the back of the mind of the instructor without ever being discussed in the course) often leads to the wrong impression by a beginner as to what the point of a given mathematical theory is.

A large purpose of representation theory is to use "symmetry" of objects arising in algebra, geometry, and analysis to better understand the structure of those objects; the aim is not to describe a group via matrices as an end unto itself. By the classification of finite simple groups, the most important finite groups are largely matrix groups over finite fields. If you study representations of a finite group $G$ in char. 0 and $G$ has a composition series whose successive quotients are matrix groups over finite fields then those matrices have nothing to do with ones in char. 0.

Parahorics are not algebraic groups, but are "just" totally disconnected topological groups, so please define what you mean by "unipotent radical" of a parahoric. For instance, what is your definition of "unipotent radical" of $I$? Do you mean to ask about what is often called the "pro-unipotent radical"?

@QiaochuYuan: It is generally not a good idea (or at least will tend to lead to confusion) to speak of "reductive" beyond the algebraic group setting; e.g., calling $\mathbf{R}$ a "reductive" Lie group isn't a useful thing to do.

The answer to your Zariski-sheaf question is "yes", and this is an exercise in definitions with maps of ringed spaces, so I prefer to say nothing more and let you figure it out for yourself. The rest is just saying that algebraic spaces which aren't schemes probably provide counterexamples to Q2, but I didn't find a rigorous proof and so just put those comments there in case another reader who is also familiar with algebraic spaces might be able to turn that idea into a justified counterexample.

Q2 fails if not a Zariski sheaf. Less trivial: consider q-c separated algebraic spaces $X$. Via an etale chart $R\rightrightarrows U$ in Aff, $|X| :=|U|/|R|$ and $O:=\ker(q_{\ast}(O_U)\rightrightarrows p_{\ast}(O_R))$ (using $p:|U|\rightarrow|X|$ and $p=q\circ p_i:|R|\rightarrow|X|$) make $X':=(|X|,O)$ an LRS. The map $\underline{X}\rightarrow\underline{X}'$ on Aff is initial for $\underline{X}\rightarrow\underline{Y}$ with LRS $Y$. Via $\underline{X}\simeq\underline{Y}$, $X'\rightarrow Y$ is bijective on spaces and an isomorphism on residue fields. Maybe this violates Q2 for $X$ not a scheme?

#1 is correct, #2 is correct if you remove "compact", #3 is correct, #4 is correct with the parenthetical removed (it is the unique one up to conjugation), #5 is correct with "connected" impose on both sides, #6 is probably not what you meant to ask (as every closed subgroup of every real Lie group is a real Lie subgroup, #7 is ambiguous (there is one conjugacy class when the ambient group has finite component group), #8 is correct if the group is required to be connected and simply connected.