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4 votes

Class numbers in the unramified biquadratic extensions of number fields

Kuroda's class number formula does not require the extension to be unramified. It is not true, as you write, that the class number of $K$ can be determined from the class numbers of the $k_i$; you ...
Franz Lemmermeyer's user avatar
4 votes

Equidistribution on $\mathrm{SU}_2$

It was asked in the comments that I provide some details. I prove slightly more: if $\mu_n$ denotes uniform probability on the sphere of radius $n$ and if $\rho:F_2 \to \mathrm{SU}_2$ is a ...
Mikael de la Salle's user avatar
8 votes
Accepted

Equidistribution on $\mathrm{SU}_2$

In the article "On the spectral gap for finitely-generated subgroups of SU(2)" by Jean Bourgain and Alex Gamburd (Invent. Math. 171, No. 1, 83-121 (2008)), they show that free subgroups of $...
Lucas Kaufmann's user avatar
5 votes
Accepted

Generators of the ideal class group

Let $G$ be the ideal class group, and let $H$ be the subgroup generated by the prime ideals of norm at most $3\log^2(d^2)=12\log^2(d)$. Assume that $H$ is a proper subgroup of $G$. Then there is a ...
GH from MO's user avatar
  • 96.9k
2 votes
Accepted

How to compute the asymptotic constant for the count of $S_3$-sextic number fields?

I will do a worked example for the tame case and show that it agrees with the formula from their paper. You should then be able to adapt this to the case $p=2,3$. Let $p > 3$. I use formula (7) ...
Daniel Loughran's user avatar
5 votes
Accepted

Ramification criteria for Kummer extensions

$K( A^{1/n})$ is unramified if and only if the image of $A$ in $K_{\mathfrak p}^* / (K_{\mathfrak p}^*)^n$ lies in a certain cyclic subgroup of order $n$, which is the cyclic subgroup generated by a ...
Will Sawin's user avatar
  • 133k
3 votes
Accepted

Conductor and local Kronecker–Weber theorem

Let $m$ be the order of $p \in \mathbb Q_p^\times$ in the character of $\mathbb Q_p^\times$ associated to $F$ by local class field theory. Let $\ell$ be the least integer prime to $p$ such that the ...
Will Sawin's user avatar
  • 133k
0 votes

Cohomology of $S$-arithmetic groups with trivial coefficients such as $H^n(\rm{PGL}_2(\mathbb{Z}[1/N]);\mathbb{Z})$

I added this non-answer since it is too long for a comment. EDIT: In the first version I claimed that the action is free, which it is not. There is a very explicit model for the classifying space $...
5 votes
Accepted

Cohomology of $S$-arithmetic groups with trivial coefficients such as $H^n(\rm{PGL}_2(\mathbb{Z}[1/N]);\mathbb{Z})$

This is a hard problem in general. When $N$ is not prime, then even the first homology of $\operatorname{PSL}_2(\mathbf{Z}[\frac{1}{N}])$ is non-trivial to compute (cf. Corollary 4.4 of [1]), but I ...
Carl-Fredrik Nyberg Brodda's user avatar
1 vote

Is the completed tensor product (over a complete dvr) of two reduced complete Noetherian local rings again reduced?

Let $K$ be the fraction field of $\mathcal{O}$. Assuming $A$ and $B$ are $\mathcal{O}$-flat, then also $A \widehat{\otimes} B$ is $\mathcal{O}$-flat (exercise), so it's enough to see that $C:=(A \...
Satan's Minion's user avatar
3 votes

Explicit description of the extension generated by the square root of a fundamental unit of a real quadratic field

This has nothing to do with ramification. Write your unit in the form $\varepsilon = t + u \sqrt{m}$; then $t^2 - mu^{2} = 1$, hence $t^2 - 1 = (t-1)(t+1) = mu^2$. If $t$ is odd, then unique ...
Franz Lemmermeyer's user avatar

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