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14h

comment 
Connection between Haar measure of locally compact group G and Haar measure compact subgroup of it
It depends on what you mean by the word "connection." In the case of $\mathbf C^{\times}$, Lubin pointed out that the compact subgroup could have measure $0$, and thus its Haar measure is certainly not the restriction of the Haar measure on $G$ (even up to scaling). But at the same time the Haar measure on the circle group is a "factor" of the Haar measure on $\mathbf C^{\times}$ since $dz/z^2 = dr/r \times d\theta$. So there is a "connection" between them, but not in the sense of restriction. Can you make your question more precise as far as what you intend by "connection"? 
2d

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Trivial zeroes of the Riemann Zeta function are simple
In section 6, chapter 16 of Ireland & Rosen's book they show how to extend the zetafunction to the halfplane $\{s : {\rm Re}(s) > k\}$ for $k = 0, 1, 2, \dots$ step by step. No functional equation is used or even arises in this method. (Whether this method can be carried out more carefully to tell you that a trivial zero is a simple zero is not something I checked.) When Gerry writes that you must use the functional equation to define the zetafunction on all of $\mathbf C$ he is incorrect. 
2d

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What are the restrictions in the ramification behavior of a Galois extension of number fields imposed by the Galois group of the extension?
You should think not about how ramification in two Galois extensions $K/E$ and $L/E$ are related when their Galois groups are isomorphic, but how ramification in $F/E$ and $FK/K$ are related where $F/E$ is a finite Galois extension (not necessarily the Hilbert class field of $E$). 
2d

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What are the restrictions in the ramification behavior of a Galois extension of number fields imposed by the Galois group of the extension?
If $L$ and $K$ are two quadratic extensions of $\mathbf Q$ then they have isomorphic Galois groups over $\mathbf Q$, but you certainly can't read off the ramification in $L$ just from knowing it in $K$. 
Aug
27 
comment 
BatemanHorn, continued even further
@kantelope, from the way your wrote your comment, do you mean you don't think in principle that Cohen does anything that Moree and Kurokawa don't do? 
Aug
27 
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BatemanHorn, continued even further
I put more references to Kurokawa's paper as an answer below. It is more directly applicable than the work by Moree, as you noted. 
Aug
27 
answered  BatemanHorn, continued even further 
Aug
27 
comment 
BatemanHorn, continued even further
Nobushige Kurokawa wrote a series of papers in the 1970s and 1980s, including a paper "Special Values of Euler Products and HardyLittlewood Constants" (Proc. Japan Acad. 62 Ser. A (1986), 2528), where he cites his earlier papers for how to write them as more rapidly convergent products. 
Aug
27 
comment 
BatemanHorn, continued even further
Replace the role of the zetafunction there with the zetafunction of the number field $K$ cut out by a root of $f$ or some of its "irreducible parts" (Dirichlet $L$functions or more general Artin $L$functions). For instance, the linear part of the $p$factor is $(n_f(p)1)/p^s$, which is the linear part of $\zeta_K(s)/\zeta(s)$. So multiply $L_f(s)$ by this ratio and its reciprocal, and insert the $p$factors for one of them into the Euler product for $L_f(s)$ to speed up convergence at $s=1$ (if you can compute $\zeta_K(s)/\zeta(s)$ rapidly by other methods). Does that make sense? 
Aug
27 
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BatemanHorn, continued even further
Look at the page guests.mpimbonn.mpg.de/moree/Moree.en.html and the links there. 
Aug
27 
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BatemanHorn, continued even further
By "evaluate" do you mean get more rapidly convergent expressions for them? 
Aug
20 
comment 
Springer GTM Reprints in China?
@DimaPasechnik, the line is not "Not for sale in..." but rather "... for sale in the Mainland China only and not for export therefrom." Hmm, not surprised it has an English error, like the chopstick instructions. 
Aug
20 
comment 
Springer GTM Reprints in China?
When I was in Beijing I found a bookstore near my hotel with a lot of these Springer books (Chinese writing added to front cover, English unchanged). It is totally legit. The copy of Griffiths & Harris I saw had the table of contents in Chinese, but other English math books did not translate that part. When I returned to the US I was advised by several Chinese people that the books easily fall apart under regular use, so I have treated mine extra carefully. 
Aug
18 
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Important formulas in Combinatorics
How is this a formula in combinatorics? 
Aug
17 
comment 
Writing papers in preLaTeX era?
There is at least one other answer here (by lhf) mentioning ChiWriter. 
Aug
16 
comment 
Finding the inertia group
Knowing how $h(x) \bmod 2$ factors does not by itself tell you the residue field degree for the splitting field of $h(x)$ over $\mathbf Q_2$. Consider the polynomials $F(x) = x^2+1$ and $G(x) = x^2 + 3$. Both factor mod 2 as $(x+1)^2$, but their respective splitting fields over $\mathbf Q_2$ are $\mathbf Q_2(i)$ and $\mathbf Q_2(\sqrt{3}) = \mathbf Q_2(\zeta_3)$. As extensions of $\mathbf Q_2$ the first of these is ramified and the second is unramified. 
Aug
16 
comment 
Finding the inertia group
As a first step, have you been able to compute the Galois group of the splitting field of $h$ over $\mathbf Q_2$? I worked out that it is $D_4$ (group of symmetries of a square). 
Aug
16 
comment 
Looking for the name of a mathematical symbol that looks remotely like 1 (answer: indicator function)
Use the detexify webpage and you find code for it. 
Aug
2 
comment 
When did people know that all real polynomials of degree greater than 2 are reducible?
This question would be more suitable on the History of Science and Mathematics stackexchange site. You are essentially asking who first proved the fundamental theorem of algebra. That is generally attributed to Gauss, in his doctoral thesis, although I believe his argument had some topological gaps. 
Jul
30 
revised 
Cases where the number field case and the function field (with positive characteristic) are different
added 358 characters in body 