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May be you can consult Latimer's paper On ideals in generalized quaternion algebras and Hermitian forms., Trans. Amer. Math. Soc. 38 (1935), no. 3, 436–446. The main theorem (Theorem 3) establishes …

By Theorem 13.31 in L. Washington Introduction to Cyclotomic Fields, Second Edition, GTM 83 we know that $\mathcal{X}\sim \Lambda^{r_2}\oplus(\Lambda-$torsion), where $r_2$ is the number of complex em …

Yes, there is and is called $t$-expansion rather than $q$-expansion - may be, though, you might want to replace your $\Omega$ by $$ \widehat{\bar{K_\infty}}\setminus K_\infty $$ where I denote by $K_\ …

Well, if you consider the definition of cohomology, you see that $H^0(N_0,V)$ are the fixed point, namely $v\in V$ such that $gv=v$ for all $g\in N_0$ (which, I suppose, is your group). If the group w …

My answer is just a detailed version of David Loeffler's comment. First of all, I would observe that $N\cong \mathbb{Q}_p$ via $$ \left( \begin{array}{cc} 1&a\\ 0&1 \end{array}\right)\mapsto a $$ Mor …

Well, I would say that this crucially depends on what you define to be "quick". If you admit global class field theory, at least in its idelic formulation, the fact that all primes in $K$ are split in …

Well, let me try in an elementary way. Pick any field $K$ of characteristic prime to $p$ and suppose it does not contain any $p^n$-th root of unity. Let $a\in K^\times\setminus (K^\times)^p$. Then, th …

This is done in Chapter 7.3 of Ramakrishnan and Valenza's Fourier Analysis on Number Fields (GTM 186) which, despite the title, describes in some details also the situation over function fields in one …

I come later than GH and paul Monsky and with an intricate and way too long answer, but since I had fun in solving the exercice, I post it anyhow. Most probably (as paul Monsky suggested) there are ea …

Following Felipe's answer, I would like to suggest a slightly different strategy which makes deep use of your very peculiar setting (and probably boils down to Felipe's advise). Warning: At the very e …

I think your cup-product is always zero independently of Leopoldt (at least if $\chi$ is of finite order). Consider $\mathbb{Z}_p$-coefficients instead (enough by Neukirch-Schmidt-Wingberg, 2.3.10). I …

As KConrad says, the answer to the first question is "not yet known". For Iwasawa theory, you could consider the extension $L_\infty=\mathbb{Q}(\sqrt[p^\infty]{2},\zeta_{p^\infty})$ which is Galois ov …

The answer is yes. Let $K=\mathbb{Q}(\zeta)$ and suppose that there is an element $u\in\mathbb{Z}[\frac{1}{2},\zeta]$ such that for some embedding $\sigma_0\colon K\hookrightarrow \mathbb{C}$ the abso …

I think you can look at page 532 of (the first version of) J. Neukirch, A. Schmidt, K. Wingberg, Cohomology of Number Fields, Springer, 1999. They explicitly write that we do not even know precisely w …

The von Staudt-Clausen theorem indeed asserts that the denominators of Bernoulli numbers have special behaviour, but from a modern perspective what is even deeper are the so-called Kummer's congruenc …