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Just to tack something on about your final question: There exists a rather concrete "yes" answer to your last question in the form of Arakelov theory. One defines an Arakelov divisor as a finite for …

Kolyvagin's "On the First Case of the Fermat Theorem for Cyclotomic Fields" is worth mentioning here. The main point of this paper is that while various case-by-case eliminations (e.g., Kummer's reg …

No, I'm pretty sure not. In general, the theory is much more developed for the maximal pro-p-quotient of the groups you're asking about, and even in this more explored setting, not a single explicit …

Two things: 1) Yes, certainly. By class field theory and the finiteness of the class group, the maximal abelian unramified extension of any number field is of finite degree. Thus any infinite un …

Yes, there are many such fields. (Edit: Let me put up here that "many" is still finite, and only in the hundreds if my list below represents most of the known examples. Note that "state-of-the-art" …

Yes, you get some interesting dynamics out. The derivative with all $\alpha_p=1$ goes by the name "the arithmetic derivative," and there are a few references around (Ufnarovski's being the most compl …

The question itself is certainly still open. Mostly as an exercise for myself, I'll coalesce my comments above into an answer, and add in some details about where various pieces of the philosophy com …

Yes. Take $$ \alpha=\sqrt{2-\sqrt{2}}+i\sqrt{\sqrt{2}-1}. $$ Neither of the conjugates $$ \sqrt{2+\sqrt{2}}\pm \sqrt{\sqrt{2}+1} $$ have absolute value 1. It is impossible, however, if $\mathbb{Q} …

I believe the question is: "Does there exist a constant $A$ such that there exists infinitely many totally real number fields with regulator less then $A$?" Ignore this response if that's incorrect. …

Here is an explicit construction$^*$. Since there exists such an unramified $p$-extension, by class field theory the $p$-part of the class group of $\mathbb{Q}(\mu_p)$ is non-trivial. Further, speci …

It is very likely that every (positive) odd number is covered by a sum of this type. As Robin Chapman points out, this is equivalent to asking, for a given odd number $h$, whether there exists an odd …

To add on to Pete's answer, let me comment that the differences are even more pronounced if we look at the maximal pro-$p$ quotient $\pi_1(\operatorname{Spec}(\mathbb{Z}\left[\frac{1}{p_1p_2\cdots p_r …

The short answer to your question is basically no, there's essentially no connection between the prime powers $q^i$ dividing $h_p$ and $h_{-p}$. It's true that there's a general relationship betwee …

This result is pretty shy of needing the full Hasse-Minkowski Theorem. Indeed, since Fermat already knew which integers were a sum of two integer squares, it would suffice for him to show that those …

Hopefully I've read all your notation correctly. If so, by playing (very) fast and loose with heuristics, I think your friend is right that the answer is 0. Your function $Log(n)$ is the additive fu …