77

Sorry I didn’t see this earlier. My memory is vague, and probably colored by subsequent events and results, but here’s how I recall things happening.
Since I had read and enjoyed Lazard’s paper on one-dimensional formal group (laws), which dealt with the case of a base field of characteristic $p$, I decided to look at formal groups over $p$-adic rings. For ...

40

The question is subject to interpretation, because the Langlands program was never a clearly delimited set of conjectures to begin with, and moreover many things were added during the following decades, and the answer depends if you want to consider these innovations as part of the Langlands program or not.
I believe nevertheless that essentially everything ...

24

Thank you for calling this problem to my attention.
I computed $K$ en route to AWS (though this year's
topics
are a rather different flavor of number theory...).
After some simplification (gp's $\rm polredabs$), it turns out that
the field $K$ is generated by a root of
$$
f(x) = x^{17} - 2x^{16} + 8x^{13} + 16x^{12} - 16x^{11} + 64x^9 - 32x^8 - 80x^7
$$
$$
\...

answered Mar 14 '15 at 4:01

Noam D. Elkies

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23

Teichmüller's article appeared in 1940 in the wartime journal $\mathfrak{Deutsche}$ $\mathfrak{Mathematik}$ (of which I've seen copies in the Tata Institute library in Bombay). He is trying to extend the Galois correspondence (between subextensions of $E\mid F$ on the one hand, and subgroups of the Galois group $\mathrm{Gal}(E\mid F)$ on the other, where $E\...

answered Sep 24 '13 at 11:29

Chandan Singh Dalawat

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23

Even though some of the main ideas have been pointed out above, I will try to add (in all modesty) a complete answer.
The following outline comes from my master thesis and further develops some of the ideas in David Cox's book Primes of the form $x^2+ny^2$.
$\color{red}{\text{PART 1: Preliminary results on the genus field.}}$
Lemma: Given a number ...

23

Takagi's goal is the following:
show the existence of sufficiently many cyclic extensions defined
by division values of sn $u$ (the lemniscatic sine);
prove that each abelian extension of ${\mathbb Q}(i)$ is contained in
the compositum of these fields.
Step 1 is analogous to the construction of the fields of $p^n$-th roots
of unity (in particular, ...

answered Jul 17 '13 at 18:58

Franz Lemmermeyer

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19

I was just thinking about this the other day! Here is one solution.
As in the original post, let $K = \mathbb{Q}(\sqrt{-3})$ and let $\theta$ and $\pi$ be distinct primary primes. Also, set $L = K(\sqrt[3]{\theta \pi})$ and $M = K(\sqrt[3]{\theta}, \sqrt[3]{\pi})$.
We note that $M/L$ is unramified. The only primes which might ramify are those lying over $(\...

answered Mar 23 '16 at 20:14

David E Speyer

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18

The cyclotomic field $K=\mathbb{Q}(\zeta_{23})$ contains the quadratic field $F=\mathbb{Q}(\sqrt{-23})$, and $F$ has class number $3$ (it is the first quadratic field, when those are ordered by absolute value of the discriminant, that has class number divisible by $3$). Thus, $F$ has an unramified cubic extensions $H$, and since the degree of $K$ is coprime ...

16

As GH has already remarked, the same thing happened a lot later after Taniyama and Shimura asked whether elliptic curves defined over the rationals are modular. To begin with your last question, there were no other candidates for the Artin isomorphism; reciprocity laws at the time were intimately connected to power residue symbols. Actually Euler had ...

answered Jul 3 '16 at 20:13

Franz Lemmermeyer

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16

I mention this as an answer since it is too long for comments. I do not know what the adelic proof assumes. Suppose that all but finitely many primes of $K$ split completely in $L$. Suppose $d$ is the degree of $L$ over $K$. Then the zeta function of $L$ is the $d$-th power of the zeta function of $K$, up to finitely many factors. But the Zeta functions of $...

15

Let me first note that there is a slight ambiguity when one says "ramified only at 2". Strictly speaking, that means that the extension is unramified at every place
of $\mathbb Q$ except 2, including infinity. The latter mean that the extension is totally real. Often, however, "ramified only at 2" means "ramified only possibly at 2 and $\infty$", and it is ...

14

The main reference here is the very useful User's Guide,
C. Batut, K. Belabas, D. Bernardi, H. Cohen, M. Olivier, "User's Guide to PARI / GP" (2003)
Particularly the sections about bnrstark (pp. 108), quadhilbert (pp. 87) and quadray (pp. 88).
For more information and examples you may want to look at Roblot's work, for instance
Xavier-François Roblot, ...

14

Every prime $p$ is represented by at least one of the following quadratic forms: $x^2+y^2$, $x^2+3y^2$, $3x^2-y^2$:
if $p=2$ or $p\equiv 1\pmod{4}$, then $p$ is represented by $x^2+y^2$;
if $p=3$ or $p\equiv 1\pmod{3}$, then $p$ is represented by $x^2+3y^2$;
if $p\equiv 11\pmod{12}$, then $p$ is represented by $3x^2-y^2$.
This follows from Lemma 2.5, ...

13

Your conjectures are correct. So was the "someone else at MSRI [who]
muttered something about norm forms" (mentioned in earlier edits
of the question), except for the part about laughing at you.
As you in effect note, $f(a,b,c)$ is the norm $N_{K/{\bf Q}}(a+bx+cx^2)$,
where $x$ is one of the roots of $x^3-x^2-x-1 = 0$
and $K$ is the cubic number field ${\...

answered Apr 11 '13 at 23:27

Noam D. Elkies

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13

We give a uniform approach to $p \leq 61$ by applying
analytic discriminant bounds to the Hilbert class field.
To be sure this is not entirely "conceptual", but then
some computation is needed even to deal with $p < 36$ using Minkowski.
If $p = 4k+1$ is prime then $K = {\bf Q}(\sqrt{p})$ has odd class number $h$,
so either $h=1$ or $h \geq 3$. If $h \...

answered May 25 '20 at 18:42

Noam D. Elkies

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11

Here is an elementary proof, the basic idea for which is in Selberg's 1949 paper "An elementary proof of Dirichlet's theorem about primes in an arithmetic progression" (Ann. Math., vol 2, 1949, pp. 297-304).
It is enough to prove the unboundedness
$$
\sum_{\substack{ p < X \\ p \equiv 1 \mod{N}}} \frac{\log{p}}{p} \neq O_X(1).
$$
This is just a ...

answered Jan 24 '15 at 23:16

Vesselin Dimitrov

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11

Let $\ell$ be an odd prime and $m$ an integer such that
$$|\{p|m,\ p\equiv1\operatorname{mod} \ell\}|\geq8.$$
Then Y.Furuta proved that $\mathbb Q(\zeta_m)$ admits an infinite unramified $\ell$-class field tower (Nagoya Math. Journal,1972). In fact, I.Shparlinski proved using this result that $\mathbb Q(\zeta_m)$ admits an infinite class field tower for ...

11

Abelian extensions of $\mathbb{Q}$ can be described using torsion points in the multiplicative group. If $K$ is a quadratic imaginary field, and $E$ is an elliptic curve where $\mathcal{O}_K$ acts by CM, then abelian extensions of $K$ can be described using torsion points of $E$. Shimura proved similar results about CM number fields and higher dimensional ...

answered Oct 13 '15 at 15:19

David E Speyer

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11

The following result is a special case of a sideresult in my PhD thesis (1995):
Let $k = {\mathbb Q}(\sqrt{d})$ be a quadratic extension with
discriminant $d$, let $d = d_1d_2d_3d_4$ be a factorization of $d$
into coprime discriminants, and assume that
$$(d_1/p_2) = (d_2/p_1) = (d_3/p_4) = (d_4/p_3) = +1 $$
for all primes $p_j \mid d_j$. Then there exist
$\...

answered Dec 22 '15 at 16:39

Franz Lemmermeyer

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11

Evidently no. In "Generic Polynomials Constructive Aspects of the Inverse Galois Problem" by Jensen, Ledet, Yui (2002): generic dimension (i.e. minimum number of parameters in a generic polynomial) of $\mathbb{Z}/5$ over $\mathbb{Q} = 2$ (p.203) and moreover essential dimension (i.e. minimum transcendence degree of parameters) of $\mathbb{Z}/5$ over $\...

11

All the information on the higher ramification groups can be derived from Theorem II.5.6, which TKe notes.
The Galois group of the ray class field is the group of fractional ideals relatively prime to $\mathfrak p$ modulo the principal ideals with generators $1$ mod $\mathfrak p$. This naturally maps to the class group, and the kernel consists of principal ...

10

I am sorry if I see this question only now, but since no one gave the following answer, it seems worth posting it.
There is a general formula for the number of extensions of degree $d$ of a $p$-adic field $K$ contained inside a fixed algebraic closure $\overline{K}$, which is given by
$$
\# \{ L \mid K \subseteq L \subseteq \overline{K}, \, [L \colon K] = d ...

10

In fact Shimura handled the case of an abelian variety $A$ with complex multiplication by an order $O$ inside the maximal order $O_K$ of the CM field $K$.
A very good modern reference is the following article by Marco Streng:
An explicit version of Shimura's reciprocity law for Siegel modular functions.
I also recommend his PhD thesis.
Both are available ...

10

I cannot give the relation with Galois Theory, or Teichmuller's algebraic background, but the relation you quote crops up in terms of the Homotopy Addition Theorem (or Lemma) for the boundary of a $4$-simplex. You can find this theorem in the book of G.W. Whitehead on "Homotopy Theory", but somewhat confusing in proof, and he does not recognise the algebraic ...

10

I assume you want $K$ to be Galois over $\mathbb{Q}$. More generally, let $L/K$ be a Galois extension of number fields. The the class group $C_K$ of $K$ maps to $C_L^{G_{L/K}}$, the part of $C_L$ fixed by the Galois group of $L/K$, and you seem to be asking what the quotient $C_L^{G_{L/K}}/C_K$ looks like.
Taking cohomology of the exact sequences
$$ 1\to ...

answered Aug 28 '14 at 19:56

Joe Silverman

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10

A more general version of this statement was shown by Kimura in Acta Arith. (2003). His corollary gives $\gg \sqrt{X}/\log X$ such fields ${\Bbb Q}(\sqrt{-d})$ with $d\le X$, and also allows you to add further
splitting conditions. There is an extensive literature on divisibility and indivisibility of class numbers and Kimura's paper has many other ...

10

As Serre seems to be fond of saying, life begins at $H^3$.
The first appearance of a $3$-cocycle seems to be in Teichmüller's article Über die sogenannte nichkommutative Galoissche Theorie... in Deutsche Mathematik 5. He was trying to extend Galois Theory to extensions of a commutative field which are not themselves commutative. Teichmüller's formula for ...

answered Aug 23 '16 at 14:31

Chandan Singh Dalawat

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10

I assume you mean $K_1=\mathbb Q(\sqrt{d_1})$.
1) If by $K$ you mean $\mathbb Q(\sqrt{d_1d_2})$, then there is no simple relation
between $h$ and $h_1$ and $h_2$.
2) If by $K$ you mean the quartic biquadratic field $\mathbb Q(\sqrt{d_1},\sqrt{d_2})$, a theorem of Herglotz says that $h=h_1h_2h_3/2^j$, where
$h_3$ is the class number of $\mathbb Q(\sqrt{...

10

If $K$ is 'everything 1 mod N' for some N, then the canonical model of $\mathbf{Q}^\times_+ \backslash \mathbf{A}^\times_{\mathrm{f}} / K$ is exactly $\mu_N / \mathbf{Q}$, the scheme of $N$-th roots of unity. Any open compact $K$ will contain one of these, so $GL_1 / \mathbf{Q}$ Shimura varieties all look like quotients of $\mu_N$ for some $N$.
Hence the ...

answered Aug 29 '20 at 17:23

David Loeffler

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9

This is how to answer the question (but it's not an answer). (Edit: it and the comments below now form an answer).
Let $K_1$ and $K_2$ be the fields in question. By local class field theory you're asking if $\mathcal{O}_{K_1}^\times\times\widehat{\mathbb{Z}}$ and $\mathcal{O}_{K_2}^\times\times\widehat{\mathbb{Z}}$ are isomorphic (the usual exact sequence ...

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