User chandan singh dalawat - MathOverflow

Answer by Chandan Singh Dalawat for Retracted Mathematics Papers

2013-01-18T11:49:00Z

A general existence theorem is proved :

1933 : W. Grunwald, Ein allgemeines Existenztheorem für algebraische Zahlkörper, J. reine angew. Math. 169 (1933), 103–107.

and reproved :

1942: G. Whaples, Non-analytic class field theory and Grünwald's theorem. Duke Math. J. 9, (1942). 455–473.

A counter-example is found :

1948 : S. Wang, A counter-example to Grunwald's theorem, Ann. Math. 49 (1948), 1008–1009.

and the theorem is corrected :

1950 : S. Wang, On Grunwald's theorem, Ann. Math. 51 (1950), 471–484.

twice in the same year :

---- : H. Hasse, Zum Existenzsatz von Grunwald in der Klassenkörpertheorie, J. reine angew. Math. 188 (1950), 40–64.

A quarter of a century later, a simpler proof is given :

1974: J. Neukirch, Eine Bemerkung zum Existenzsatz von Grunwald-Hasse-Wang, J. Reine Angew. Math. 268/269 (1974), 315–317.

but more than half a century later, corrections to the corrections are required :

2007 : W-D. Geyer & C. Jensen, Embeddability of quadratic extensions in cyclic extensions. Forum Math. 19 (2007), no. 4, 707–725.

2011 : P. Morton, A correction to Hasse's version of the Grunwald-Hasse-Wang theorem. J. Reine Angew. Math. 659 (2011), 169–174.

Addendum (2013/05/18)

I'm afraid the above list of errors and corrections might look a bit negative, so let me add a positive note (which will also save you 30,00 € or $42.00 by not having to read it here) :

In 1933, van der Waerden asked in the Jahresbericht : Which quadratic fields can be embedded in cyclic quartic fields ? Solutions were provided by four people, among them Hasse, who generalised the problem to : Under which conditions can a degree-$l$ ($l$ prime) cyclic extension $K_1$ of a number field $K$ be embedded into a degree-$l^n$ cyclic extension $K_n$ of $K$ ?

A. Scholz sent in a "solution" to this problem in 1935 which essentially claimed that the obstructions are purely local in nature. But Hans Richter, a doctoral student of van der Waerden, knew already that there is an exception when $l=2$, so a Scholtz-Richter correction to Scholz's paper was required. In a sense, Richter anticipated not only Wang's counterexample to Grunwald's theorem but also its solution, without mentioning it explicitly as such.

Answer by Chandan Singh Dalawat for Proof of Bloch-Kato conjecture of K-theory?

2013-05-14T04:21:19Z

On 22 June 2013, Joël Riou is going to give a Bourbaki talk on

La conjecture de Bloch-Kato [d'après M. Rost et V. Voevodsky].

La conjecture de Bloch-Kato énonce que pour tout corps $k$ et tout nombre premier $l$ différent de la caractéristique de $k$, l'algèbre de K-théorie de Milnor de $k$ modulo $l$ (qui est définie par générateurs et relations) s'identifie à une algèbre de cohomologie galoisienne associée à $k$. La démonstration de cet énoncé, qui admet de nombreuses applications, utilise de façon essentielle d'une part les théories motiviques (cohomologie, homotopie, opérations de Steenrod) et d'autre part des constructions géometriques de variétés algébriques ayant des propriétés remarquables par rapport à des symboles en K-théorie de Milnor.

Usually the notes are put up on http://www.bourbaki.ens.fr/ some time after the actual talk. You can also write to Joël Riou directly at http://www.math.u-psud.fr/~riou/.

Chapters 1--4 of the Artin-Tate notes on Class Field Theory

2010-10-14T07:35:20Z

Emil Artin and John Tate held a seminar on class field theory at Princeton University in 1951--1952. Their notes were published in 1967 by Benjamin (New York), but the first four chapters covering (among other things) "the fundamentals of algebraic number theory" and "local class field theory" were omitted from the printed version.

Question. Are the notes of Chapters 1--4 available to you ?

Request. Can you make them electronically accessible to the mathematical community ?

Postscript. Parts of Hasse's Klassenkörperbericht (Bericht über neuere Untersuchungen und Probleme aus der Theorie der algebraischen Zahlkörper) are available online at the Göttingen library :

Jber. deutsch. Math.-Verein. Teil I : 35 (1926), 1--55, Teil Ia : 36 (1927), 233--311.

Ulf Rehmann (Bielefeld) and Keith Dennis (Cornell) have promised to put Teil II (Jber. deutsch. Math.-Verein., Ergänzungsband 6 (1930), 1--204) online soon.

Addendum. In an interview which has appeared recently in the Notices of the AMS, Tate makes some remarks about the genesis of the Artin-Tate notes.

Which hard mathematical problems do you have to solve to earn bitcoins ?

2013-04-20T03:17:41Z

A virtual currency called bitcoins has been in the news recently. It is said that in order to "mine" bitcoins, you have to solve hard mathematical problems.

Now, there are two kinds of mathematical problems. The difference is best explained by the following beautiful quotation from Langlands :

[T]here is an appealing fable that I learned from the mathematician Harish-Chandra, and that he claimed to have heard from the French mathematician Claude Chevalley. When God created the world, and therefore mathematics, he called upon the Devil for help. He instructed the Devil that there were certain principles, presumably simple, by which the Devil must abide in carrying out his task but that apart from them, he had free rein. Both Chevalley and Harish-Chandra were, I believe, persuaded that their vocation as mathematicians was to reveal those principles that God had declared inviolable, at least those of mathematics for they were the source of its beauty and its truths. They certainly strived to achieve this. If I had the courage to broach in this paper genuine aesthetic questions, I would try to address the implications of their standpoint. It is implicit in their conviction that the Devil, being both mischievous and extremely clever, was able, in spite of the constraining principles, to create a very great deal that was meant only to obscure God’s truths, but that was frequently taken for the truths themselves. Certainly the work of Harish-Chandra, whom I knew well, was informed almost to the end by the eﬀort to seize divine truths.

Question Which kind of mathematical problems do you have to solve in order to mine bitcoins ?

Let me clarify that I'm not interested in mining, only in knowing whether the problems are divine or devilish.

Answer by Chandan Singh Dalawat for Galois group of constructible numbers

2013-05-06T12:35:03Z

It is always a good idea to look at the local question first. What is the maximal pro-$p$ extension of $\mathbf{Q}_p$ ? There is a vast literature on the subject, starting with Demushkin whose results are exposed in an old Bourbaki talk by Serre on the Structure de certains pro-p-groupes (d'après Demuškin), (http://www.numdam.org/numdam-bin/fitem?id=SB_1962-1964__8__145_0). They turn out to be extremely interesting groups which satisfy a kind "Poincaré duality" and have a striking presentation mysteriously similar to the presentation of the fundamental group of a compact orientable surface. Now I have to go for my yoga class; I hope you can find more on the web by yourself.

Addendum. Another good idea which I record here although I'm sure you don't need to be reminded of it is that in the local situation one should first ask what happens at primes $l\neq p$. What is the maximal pro-$l$ extension of $\mathbf{Q}_p$ ? This is much easier to answer because the only possible ramification is tame. You have the maximal unramified $l$-extension, which is a $\mathbf{Z}_l$-extension, and then a totally ramified extension of group $\mathbf{Z}_l(1)$, the projective limit of the roots of $1$ of $l$-power order. As a pro-$l$ group it admits the presentation

$\langle\tau,\sigma\mid\sigma\tau\sigma^{-1}=\tau^p\rangle$.

All this can be found in Chapter 16 of Hasse's Zahlentherie and in a paper by Albert in the Annals in the late 30s.

Addendum 2. A third good idea is to look at the function field analogues, wherein you replace $\mathbf{Q}$ by a function field $k(X)$ over a finite field $k$ of characterisitc $p$, where $X$ is a smooth projective absolutely irreducible curve over $k$. There would again be two cases, according as you are looking at the maximal pro-$l$ extension for a prime $l\neq p$ or for the prime $l=p$; the former should be much easier. You can simplify the problem by replacing $k$ by an algebraic closure. We are then in the setting of Abhyankar's conjecture (1957) which was very useful to Grothendieck as a first test for his theory of schemes. In any case, the conjecture was settled by Raynaud and Harbatter in the early 90s.

Answer by Chandan Singh Dalawat for Are finite index subgroups of inertia closed?

2013-04-24T16:14:12Z

Here is another way to see that the wild inertia group $P_K=\mathrm{Gal}(\bar K|T)$ (where $T$ is the maximal tamely ramified extension of $K$ (a finite extension of $\mathbf{Q}_p$) in $\bar K$, an algebraic closure of $K$) is not finitely generated (as a profinite group). One doesn't need class field theory, only Kummer theory and the fact that a pro-$p$ group $G$ is finitely generated if and only if the $\mathbf{F}_p$-space $G/D$ is finite, where $D$ is the closed subgroup generated by all commutators and all $p$-th powers in $D$.

In the case at hand, $P_K/D=\mathrm{Gal}(T(\root p\of{T^\times})|T)$, and it can be easily seen that the $\mathbf{F}_p$-space $T^\times/T^{\times p}$ is infinite.

D K Faddeev's construction of quaternionic fields

2013-04-16T07:11:10Z

I have come across the following reference to D K Faddeev's construction of quaternionic fields in the book The Embedding Problem in Galois Theory by Ishkhanov, Lur'e and Faddeev :

[45] D. K. Faddeev, Construction of algebraic domains whose Galois group is the quaternionic group, Leningrad. Gos. Univ. Uchen. Zap. 3 (1937), no. 17, 17--23,

which is presumably the same as an item in the bibliography of Faddeev's survey article ТЕОРИЯ ГАЛУА (В МИАНе) :

[16] Фаддеев Д. К. Построение алгебраических областей, группой Галуа которых является группа кватернионов. — Учен. зап. ЛГУ, 1936, т. 17, с. 17—25.

Edit KConrad has kindly provided a link to the English translation of this survey (Спасибо, Кит). Faddeev says

In a paper of mine in 1937 [16], the problem of constructing fields with quaternionic groups over $\mathbf{Q}$ was solved. The algebraic part of the construction can be extended to any field of characteristic different from $2$ which has a sufficient number of quadratic extensions. The arithmetic part allows one to give an algorithm for construction of fields over $\mathbf{Q}$ in the order of growth of their discriminants.

I couldn't find this paper at mathnet.ru, nor is it listed in the Zentralblatt. There is a later paper with a similar title

Construction of fields of algebraic numbers whose Galois group is a group of quaternion units, C. R. (Dokl.) Acad. Sci. URSS, n. Ser. 47, 390-392 (1945)

which is listed but not reviewed in the Zentralblatt; it is not be found at mathnet.ru either.

Question. What does Faddeev prove in these papers ?

Answer by Chandan Singh Dalawat for Algebraic $p$-adic integers mod $p$

2013-04-06T05:25:49Z

This is an extended comment.

Let $\Gamma=\mathrm{Gal}(\overline{\mathbf{Q}}_p|\mathbf{Q}_p)$. Your question pertains to the filtration $V\subset T\subset\Gamma$, where $T$ is the inertia subgroup and $V$ the ramification subgroup, so that $\overline{\mathbf{Q}}_p^T$ is the maximal unramified extension of $\mathbf{Q}_p$ and $\overline{\mathbf{Q}}_p^V$ is the maximal tamely ramified extension of $\mathbf{Q}_p$. The quotient $\Gamma/T$ is $\hat{\mathbf{Z}}$, and the quotient $T/V$ can be identified with the group of roots of $1$ of order prime to $p$.

But the filtration (in the upper numbering) on $\Gamma$ is much finer, and it is indexed by the reals $\geqslant-1$; we have $\Gamma^0=T$ and $\Gamma^{0+}=V$. This filtration is separated in the sense that the intersection of all $\Gamma^u$ (for $u\geqslant-1$) is trivial, so $\overline{\mathbf{Q}}_p$ is the inductive limit of the various $K^u=\overline{\mathbf{Q}}_p^{\Gamma^u}$. As a result the $\overline{\mathbf{F}}_p$-algebra $\overline{\mathbf{Z}}_p/p\overline{\mathbf{Z}}_p$ is the inductive limit (increasing union) of $\mathfrak{o}^u/p\mathfrak{o}^u$, where $\mathfrak{o}^u$ is the ring of integers of $K^u$, and $u>0$

It might be interesting to be able to say something sensible about each individual quotient $\mathfrak{o}^u/p\mathfrak{o}^u$ (and also about the quotients corresponding to $\Gamma^{u+}$, the union of all $\Gamma^v$ such that $v>u$).

New Geometric Methods in Number Theory and Automorphic Forms

2013-04-04T05:05:14Z

The MSRI is organising a programme with the above title from Aug 11, 2014 to Dec 12, 2014. Here is a short description from their website :

The branches of number theory most directly related to the arithmetic of automorphic forms have seen much recent progress, with the resolution of many longstanding conjectures. These breakthroughs have largely been achieved by the discovery of new geometric techniques and insights. The goal of this program is to highlight new geometric structures and new questions of a geometric nature which seem most crucial for further development. In particular, the program will emphasize geometric questions arising in the study of Shimura varieties, the $p$-adic Langlands program, and periods of automorphic forms.

Question Which new geometric structures, techniques and insights have been crucial for this recent progress ?

Answer by Chandan Singh Dalawat for Elements of the history of mathematics

2013-01-22T11:07:42Z

Different people wrote different parts. Weil indicates somewhere that he wrote the Note historique about infinitesimal calculus. A perusal of Archives de l'Association des Collaborateurs de Nicolas Bourbaki might provide further hints.

Other possible sources of information are articles and interviews by Henri Cartan, Jean Dieudonné, Armand Borel, and Laurent Schwartz about their involvement with Bourbaki.

Among the living, Pierre Cartier, Roger Godement and Jean-Pierre Serre should be able to provide you ample information. Bourbaki's biographer Liliane Beaulieu might be another source of information.

Addendum (2013/03/31) Bourbaki's editor Hermann has donated all his papers to the Bibliothèque Nationale de France. See their announcement (http://www.datapressepremium.com/RMDIFF/1805134//archivesbourbaki2.pdf) and an article in Libération's science blog (http://sciences.blogs.liberation.fr/home/2012/11/les-archives-bourbaki-%C3%A0-la-bnf.html).

Answer by Chandan Singh Dalawat for Have we ever proved any non-solvable case of reciprocity without the Langlands program ?

2013-03-28T09:26:44Z

This is a very interesting question and would make an excellent topic for a doctoral thesis in the history of mathematics. I will interpret the question as

Which pre-Langlands results, problems, and theories --- apart from what is easily deducible from the theory of $\;\mathrm{GL}_1$ (from Gauß to Tate) --- can now be considered a part of the Langlands programme ?

There is nothing original in my answer : everything is gleaned from the writings of Langlands, Serre and Weil. I may have misrepresented some of their words, and in any case our future doctoral candidate will have to delve deeper into the original sources.

Fricke & Klein (1912) observe that the modular curve $X_0(11)$ of level $\Gamma_0(11)$ is defined by the equation $\sigma^2=1-20\tau+56\tau^2-44\tau^3$.

Hasse (193?) asks a doctoral student (Pierre Humbert) to prove that the $L$-function of an elliptic curve $E$ over $\mathbf{Q}$ (defined as the product over various primes $p$ of the $\zeta$-function of $E$ modulo $p$) is entire and satisfies a functional equation. Humbert sagely decides to work on quadratic forms with Siegel instead.

Weil (1951) asks in his report Sur la théorie du corps de classes for a galoisian interpretation of the whole idèle class group of a number field (as opposed to the quotient of the said group by the connected component of the identity), analogous to the galoisian interpretation in the function field case. See http://mathoverflow.net/questions/41318 in this regard.

Weil (1952) shows that certain elliptic curves with complex multiplications (such as $y^2=x^4+1$) are modular.

Deuring (1953--1957) proves (following a suggestion by Weil) that all elliptic curves with complex multiplications are modular.

Eichler (1954) proves that the $L$-function of $X_0(N)$ is essentially the product of Hecke $L$-functions attached to cuspidal eigenforms of weight $2$ and level $N$. This was generalised by Shimura (1958) and completed by Igusa (1959).

Taniyama (1955) asks at the Tokyo-Nikko conference a somewhat imprecise question which some interpret as implying that one can prove Hasse's conjecture for $E$ by showing that $E$ is modular.

Shimura (1966) explicitly determines the reciprocity law for the splitting of rational primes in the number field obtained by adjoining the $l$-torsion ($l$ prime) of the Fricke curve $X_0(11)$ in terms of the coefficient $c_l$ of $q^l$ in the modular form $$ q\prod_{n>0}(1-q^n)^2(1-q^{11n})^2 $$ (but only for $l<100$ for which he could check that the mod-$l$ representation is surjective).

Weil (1967) proves that if an elliptic curve over $\mathbf{Q}$ is modular, then it has to be modular of level equal to its conductor, and assigns the Übungsaufgabe to the interested reader to show that every elliptic curve over $\mathbf{Q}$ is indeed modular.

Around this time Langlands wrote a letter to Weil and changed the world.

Answer by Chandan Singh Dalawat for local field and number field

2013-03-28T04:33:04Z

First of all, it is not correct to say that for a local field $K$, the group $K^\times/K^{\times 2}$ has order $1,2,4$, or $8$. To get a counter-example, think of a sufficiently ramified extension of $\mathbf{Q}_2$, or the local field $\mathbf{F}_2((T))$.

To compute this group for any local field $K$ with (finite) residue field $k$ of cardinality $q=p^f$ ($p$ prime), you have to use the structure of the multiplicative group $K^\times$, which turns out to be isomorphic --- after you choose a uniformiser of $K$ --- to the product $\mathbf{Z}\times k^\times \times U_1$, where $U_1$ is the group of $1$-units (Einseinheiten, the kernel of the map $\mathfrak{o}^\times\to k^\times$, where $\mathfrak{o}$ is the ring of integers of $K$). Now remark that $k^\times$ is cyclic of order $q-1$ and $U_1$ is a $\mathbf{Z}_p$-module (which is finitely generated of rank $[K:\mathbf{Q}_p]$ in the characteristic-$0$ case).

Answer by Chandan Singh Dalawat for The unification of Mathematics via Topos Theory

2013-03-18T14:42:00Z

It would be presumptuous on my part to attempt to answer this question, but I want to share with other MOers this recent paper

http://www.ihes.fr/~lafforgue/math/TheorieCaramello.pdf

of Laurent Lafforgue and this video

https://sites.google.com/site/logiquecategorique/Contenus/20130227_Lafforgue

of one of his recent lectures, [dont] le but [] est de poser cette question (inspirée par la théorie de Caramello) : l'indépendance de $l$ de la cohomologie $l$-adique et la correspondance de Langlands sont-elles des équivalences de Morita entre topos classifiants ?

Here is a quote from the paper : La théorie de Caramello... offre déjà un très grand nombre d'exemples d'équivalences de Morita et de leurs applications. Ces exemples sont étonnament divers et ils apparaissent presque toujours comme surprenants. Beaucoup d'énoncés auraient été très difficiles à démontrer, et plus encore à imaginer, sans les topos et sans les méthodes de calcul que la théorie des topos classifiants et des équivalences de Morita rend possibles et naturelles. Quand on songe que la correspondance de Langlands ressemble beaucoup à une equivalence de Morita et qu'elle en est peut-être une, on se dit que le champ ouvert à cette théorie est immense.

Décomposition des nombres premiers dans des extensions non abéliennes

2013-01-19T04:49:08Z

Gauß famously determined the cubic character of $2$ in his Disquisitiones : $2$ is a cube modulo a prime number $p\equiv1\mod3$ if and only if $p=x^2+27y^2$ for some $x,y\in\mathbf{Z}$. This implies that the prime numbers which split completely in the $\mathfrak{S}_3$-extension $\mathbf{Q}(\root3\of1,\root3\of2)$ of $\mathbf{Q}$ are precisely the ones which are $\equiv1\pmod3$ and represented by the quadratic form $X^2+27Y^2$.

This was generalised by Philippe Satgé in 1977 in a paper whose title I have borrowed. He shows for example that the prime numbers which split completely in the $\mathfrak{S}_3$-extension $\mathbf{Q}(\root3\of1,\root3\of5)$ of $\mathbf{Q}$ are precisely the ones which are $\equiv1\pmod3$ and represented by one of the quadratic forms $$ X^2+XY+169Y^2,\qquad 343X^2-131XY+13Y^2. $$ I believe that similar results about all $\mathfrak{S}_3$-extensions (of $\mathbf{Q}$) can now be recovered using the known cases of Langlands reciprocity, as illustrated around the same time by Serre for the splitting fields of $T^3-T-1$ and $T^3+T-1$, which are the maximal unramified abelian extensions of $\mathbf{Q}(\sqrt{-23})$ and $\mathbf{Q}(\sqrt{-31})$ respectively, in his Modular forms of weight one and Galois representations, pp. 193–268 of Algebraic number fields: L-functions and Galois properties (Proc. Sympos., Univ. Durham, Durham, 1975), Academic Press, London, 1977.

But Satgé's theorem is applicable to more general extensions : it is applicable to a $G$-extension $K$ of $\mathbf{Q}$ whenever the finite group $G$ contains a commutative normal subgroup $H\subset G$ such that

(*) the transfer (Verlagerung) map $G/G'\to H$ is trivial, and

(**) the order of $H$ is odd if the field $K^H$ is totally real of degree $>2$ (over $\mathbf{Q}$).

Question. Can these more general results of Satgé be recovered from known cases of Langlands reciprocity ?

Answer by Chandan Singh Dalawat for Algebraic number theory: building and simplifying

2013-02-06T05:18:10Z

A major simplification in algebraic number theory occurred in the beginning of the 20th century when Hensel explicitly introduced his $\mathfrak{p}$-adic numbers. Compare the original cumbersome definition of the Hilbert symbol with the modern definition using local fields.

Another major simplification occurred in the 40s when Chevalley put all these local fields together into the ring of adèles. Compare the original cumbersome formulation of class field theory in terms of ray class groups with the modern formulation in terms of idèle class characters.

A third major simplification occurred in the 50s when Tate rederived results of Hecke and others about the analytic properties (analytic continuation and functional equations) of certain $\zeta$- and $L$-functions using Fourier analysis on the said adèles and idèles.

As far as Wiles's proof of Fermat's Last Theorem is concerned, I believe that many of the ingredients (such as modularity lifting theorems) have since been vastly simplified and generalised. You should ask the experts, or a more specific question here.

So the trend towards greater simplicity and generality will continue, but there is no reason to believe that the proof will one day fit into a small volume accessible to undergraduates. Is there such a proof of the Kronecker-Weber theorem, which is now more than a hundred years old ?

Addendum I didn't want to create the impression that simplifications have not been made in recent times. Consider for example the local-to-global principle for the existence of rational points. Various people (Lind, Reichardt, Selmer, Cassels, Swinnerton-Dyer, ...) found examples of the failure of this principle, but these examples could be understood from a unified and simplified perspective only after Manin (1970) introduced his obstruction based on the Brauer group.

Since then, Skorobogatov (1999) has found examples of the failure of the local-to-global principle which cannot be accounted for by the Manin obstruction. I get the impression that such examples are beginning to be understood from a general point of view only now, and that the theory of étale homotopy is being used to find a whole hierarchy of obstructions; see for example Homotopy Obstructions to Rational Points by Yonatan Harpaz and Tomer M. Schlank (http://arxiv.org/abs/1110.0164).

It might be objected that the introduction of such a high-level theory into what was initially an elementary question cannot be called a simplification. However, most mathematicians would concur that it is indeed a simplification when a general theory serves to illustrate various disparate phenomena. This simplification has not yet been fully worked out.

Answer by Chandan Singh Dalawat for How does "modern" number theory contribute to further understanding of $\mathbb{N}$?

2012-12-02T04:48:54Z

Let me illustrate the point with a specific and well-known example: the congruent number problem, perhaps the oldest open problem in mathematics. An integers $n>0$ is said to be a congruent number if it can be written as the area of right-angled triangle with rational sides. Which integers are congruent in this sense ?

Using his method of infinite descent, Fermat proved that $1$ is not congruent, thereby settling a conjecture of Fibonacci.

The best current answer to the problem, which depends heavily on the mathematics of the 20th century, is Tunnell's conjectural characterisation of congruent numbers. You can find many expository articles on the web (for example by Karl Rubin, Pierre Colmez, Franz Lemmermeyer's translation of an article by Guy Henniart, and John Coates). My own attempt can be found on the arXiv. Koblitz has written a whole book about it.

More 20th century mathematics (and perhaps even some 21st century mathematics) will be needed for the proof that this conjectural characterisation is indeed correct.

Very substantial progress has been made recently by Ye Tian. See for example his preprint or the video of his Moscow talk.

Try doing this with just $\mathbf{N}$ (and $\zeta$) !

Addendum. You might also be interested in Lang's article Mordell's review, Siegel's letter to Mordell, Diophantine Geometry, and 20th Century Mathematics in the Gazette (1995) 63.

Addendum 2. To get a broad overview of Number Theory today, you may want to consult Introduction to Modern Number Theory: Fundamental Problems, Ideas and Theories by Yu. I. Manin and Alexei A. Panchishkin.

Addendum 3 (2013/01/19). John Coates has written a short account of Tian's result in the PNAS 109 (52), 21182–21183.

Answer by Chandan Singh Dalawat for How to solve n! mod p?(p is a prime)

2013-01-18T05:17:14Z

There are two aspects to understanding ``$n!\mod p$''. The first question is : What is the exact power of $p$ dividing $n!$ ? Calling this power $p^{v_p(n!)}$, the second question is : What is $n!/p^{v_p(n!)}$ modulo $p$ ?

Answers to both questions can be found in Chapter 17 of Hasse's Number Theory. Let $$ n=a_0+a_1p+\cdots+a_rp^r\qquad(a_i\in[0,p[) $$ be the expansion of $n$ in base $p$, and define $$ s_n=a_0+a_1+\cdots+a_r,\quad t_n=a_0!a_1!\cdots a_r!. $$ With this notation, it is easy to prove that $$ v_p(n!)={n-s_n\over p-1},\quad {n!\over (-p)^{v_p(n!)}}\equiv t_n\mod p. $$

Answer by Chandan Singh Dalawat for Old books still used

2012-12-29T07:12:53Z

I'm surprised that nobody has mentioned Serre's Corps locaux (Local Fields), his Cohomologie galoisienne (Galois cohomology) and his Représentations linéaires des groupes finis (Linear representations of finite groups).

Other eternal texts in Number Theory include Artin's Algebraic numbers and algebraic functions and the Artin-Tate notes on Class field theory, Hasse's Zahlentheorie and his Klassenkörperbericht, Hecke's Vorlesungen über die Theorie der Algebraischen Zahlen, Weyl's Algebraic Theory of Numbers, and Hilbert's Zahlbericht.

Answer by Chandan Singh Dalawat for Video lectures of mathematics courses available online for free

2011-10-28T02:36:42Z

The Eilenberg Lectures at Columbia. So far, the topics have been:

Benedict Gross, on number theory and representation theory
Edward Frenkel, on Langlands program and quantum field theory
Sergiu Klainerman, on the mathematical theory of general relativity

Answer by Chandan Singh Dalawat for Given a number field $K$, when is its Hilbert class field an abelian extension of $\mathbb{Q}$?

2012-12-28T07:11:24Z

I happened to come across this question again today. In some cases at least, the Hilbert class field $H$ of an abelian extension $K$ of $\mathbf{Q}$ will have to be abelian over $\mathbf{Q}$ for purely algebraic reasons.

Let $F$ be any field, $K|F$ an abelian extension of group $G=\mathrm{Gal}(K|F)$ and containing a primitive $n$-th root of unity for some $n>1$, $\omega:G\to(\mathbf{Z}/n\mathbf{Z})^\times$ the cyclotomic character giving the action of $G$ on $\mu_n$, and $H|K$ an abelian extension of exponent dividing $n$. Then $H=K(\root n\of D)$ for some subgroup $D\subset K^\times/K^{\times n}$, by Kummer theory. It can be checked that $H|F$ is galoisian if and only if $D$ is $G$-stable. When such is the case, the conjugation action of $G$ on $\mathrm{Gal}(H|K)$ coming from the short exact sequence $$ 1\to\mathrm{Gal}(H|K)\to\mathrm{Gal}(H|F)\to G\to1 $$ is trivial if and only if $G$ acts on $D$ via $\omega$. In this situation ($H=K(\root n\of D)$ for some subgroup $D\subset(K^\times/K^{\times n})(\omega)$), a sufficient condition for $H$ to be abelian over $F$, is that the order of $G$ be prime to $n$, because then $\mathrm{Gal}(H|F)=\mathrm{Gal}(H|K)\times\mathrm{Gal}(K|F)$.

I'm sure this situation can be realised when $F=\mathbf{Q}$, for example when the finite abelian extension $K$ has odd degree $[K:\mathbf{Q}]$, $n=2$, the class group of $K$ has order ($1$ or) $2$, and $H$ is the Hilbert class field of $K$. In this case the extension $H|\mathbf{Q}$ will be necessarily abelian.

Answer by Chandan Singh Dalawat for Elementary proof of Mordell's theorem

2012-12-17T14:11:16Z

In the meantime I have been able to consult Cassels's Lectures on Elliptic Curves. He proves Mordell's theorem in Chapter 15, which has these two interesting footnotes :

${}^{16}$ This is the only place where the use of algebraic number theory is unavoidable. If she does not know the theory, the reader should take it on trust that it is very like the rational case. But see next footnote.

${}^{17}$ This line of argument proves the finiteness of $\mathfrak{G}/2\mathfrak{G}$ without algebraic number theory at the expense of a fairly substantial study of binary quartic forms.

The main text says at this point:

In fact this is what Birch and Swinnerton-Dyer did in their historic computations [Notes on elliptic curves. I, II. J. reine angew. Math. 212 (1963), 7--25, 218 (1965), 79--108].

It must be added that it is this approach via binary quartic forms which has made the recent spectacular advances by Arul Shankar and Manjul Bhargava possible; see arXiv:1006.1002.

What happened to Emmy Noether's Zukunftsphantasie ?

2012-12-10T04:27:00Z

Recenly I came across Peter Roquette's article On the history of Artin's $L$-functions and conductors (23 July 2003) in which he talks about some letters from Emil Artin and Emmy Noether to Helmut Hasse in the early 1930s.

Artin is trying to give the definitive form to the definition of his $L$-functions (to include ramified and archimedean places), and has proved what Hasse calls the Führerdiskriminantenproduktformel : for a finite galoisian extension $L|K$ of number fields with group $G=\mathrm{Gal}(L|K)$, the discriminant $\mathfrak{d}$ of $L|K$ can be decomposed as the product

$$ \prod_{\chi}\mathfrak{f}(\chi,L|K)^{\chi(1)} $$

extending over all characters $\chi$ of $G$, where $\mathfrak{f}(\chi,L|K)$ denotes the conductor of $\chi$ (as defined by Artin).

Emmy Noether writes to Hasse that she is looking for a decomposition formula for the diﬀerent $\mathfrak{D}$ of $L|K$ which would yield Artin’s product formula for the discriminant $\mathfrak{d}$ after applying the norm map $N_{L|K}$. Perhaps this is what she calls her Zukunftsphantasie (a fantasy for the future).

Question. Is there such a decomposition of the different $\mathfrak{D}$ ?

Answer by Chandan Singh Dalawat for The origin of the root number $w(C/ℚ)=±1$ (the sign of the functional equation)

2012-12-12T06:07:24Z

In the specific context of elliptic curves $C$ over (any finite extension of) $\mathbf{Q}$, the root number has been analysed by

Rohrlich (David), Galois theory, elliptic curves, and root numbers, Compositio Math. 100 (1996), no. 3, 311–349. Numdam.

Kobayashi (Shin-ichi), The local root number of elliptic curves with wild ramification, Math. Ann. 323 (2002), no. 3, 609–623. Springer.

and

Dokchitser (Tim) & Dokchitser (Vladimir), Root numbers of elliptic curves in residue characteristic 2, Bull. Lond. Math. Soc. 40 (2008), no. 3, 516–524. Oxford Journals.

Answer by Chandan Singh Dalawat for Algebraic maximal extension and algebraic closure

2012-12-04T12:43:18Z

If you take the compositum $K=TP$ of the maximal tamely ramified extension $T$ of $\mathbf{Q}_p$ with the cyclotomic $\mathbf{Z}_p$-extension $P$ of $\mathbf{Q}_p$, then $K$ is not algebraically closed, its residue field is $\bar{\mathbf{F}}_p$, and the value group is $\mathbf{Q}$.

Answer by Chandan Singh Dalawat for expository papers related to quantum groups

2012-12-03T03:37:19Z

Verdier, Jean-Louis

Groupes quantiques

Séminaire Bourbaki, 29 (1986-1987), Exposé No. 685, 15 p. numdam.org

Answer by Chandan Singh Dalawat for local class field theory (Norm map)

2011-03-22T04:01:39Z

Here is another interesting case where the image of the norm map can be written down explicitly. Let $F$ be a finite extension of $\mathbf{Q}_p$ containing a primitive $p$-th root $\zeta$ of $1$, and denote the filtration on the multiplicative group $F^\times$ by $$ \ldots\subset U_2\subset U_1\subset\mathfrak{o}^\times\subset F^\times. $$ We thus get the extensions $L_n=F(\root p\of{U_n})$. It can be checked that $L_{pe_1+1}=F$, where $e_1$ is the absolute ramification index of $F$ divided by $p-1$, and that $L_{pe_1}$ is the unramified degree-$p$ extension of $F$, so the image of the norm map $L_{pe_1}^\times\to F^\times$ is $\mathfrak{o}^\times F^{\times p}$.

What is the image of the norm map $L_{n}^\times\to F^\times$ for other $n$ ? Local class field theory and a certain orthogonality relation for the Kummer pairing (see for example the last section of arXiv:0711.3878) can be used to answer this question. Basically, for $n\in[1,pe_1]$, $N(L_n^\times)=U_mF^{\times p}$, where $m=pe_1+1-n$.

There are similar results for elementary abelian $p$-extensions of finite extensions of $\mathbf{F}_p((t))$. See for example the last section of arXiv:0909.2541.

These two papers have appeared in J. Ramanujan Math. Soc. 25 (2010), no. 1, 25–80, and 25 (2010), no. 4, 393–417.

There are other instances where the image of the norm map can be computed explicitly. This happens for the cyclotomic extension $K_m$ of $\mathbf{Q}_p$ obtained by adjoining $\root{p^m}\of1$. It can be shown that $p\in N(K_m^\times)$, and that the image of the units of $K_m$ under the norm map down to $\mathbf{Q}_p$ is $1+p^m\mathbf{Z}_p$. See for example Artin, Algebraic numbers and algebraic functions, p. 208, or Neukirch, Class Field Theory, p. 45.

Answer by Chandan Singh Dalawat for Not especially famous, long-open problems which anyone can understand

2012-11-27T08:20:59Z

The Kurepa conjecture : For every odd prime $p$, one has $$ 0!+1!+\cdots+(p-1)!\not\equiv0\pmod p $$ A proof was claimed and published in 2004 but the claim was withdrawn in 2011. See also my comment on the accepted answer to MO24265.

Answer by Chandan Singh Dalawat for Existence of (smooth) models

2009-12-31T12:38:38Z

I'm happy to present my example of a smooth projective surface $X$ over $K=\mathbb{Q}_p$ ($p$ prime) such that $X(K)\neq\emptyset$, whose $l$-adic cohomology groups are unramified (for all primes $l$) and which still has bad reduction : there is no smooth $\mathbb{Z}_p$-scheme whose generic fibre is $X$. (The method works for any finite extension of $\mathbb{Q}_p$ and was worked out a few years ago.)

The surface $X$ is going to be a conic bundle over $\mathbb{P}_1$ with four degenerate fibres, so it is a rational surface in the sense of being $\bar K$-birational to $\mathbb{P}_2$. It will be clear that the example is not isolated.

If $p$ is odd, let $d\in\mathbb{Z}_p^\times$ be a unit which is not a square, and take $d=5$ if $p=2$, so that $K(\sqrt{d})|K$ is the unramified quadratic extension.

Let $e_1, e_2$ be two distinct units of $K$. We take $X$ to be the surface in $\mathbb{P}({\cal O}(2)\oplus{\cal O}(2)\oplus{\cal O})$ (coordinates $y:z:t$) over $\mathbb{P}_1$ (coordinates $x:x'$) defined by the equation $$ y^2-dz^2=xx'(x-e_1x')(x-e_2x')t^2. $$ I claim that this $X$ has all the properties stated above, if $v_p(e_1-e_2)>0$.

First, $X(K)\neq\emptyset$ because each degenerare fibre is a pair of intersecting lines conjugated by $\mathrm{Gal}(\bar K|K)$.

Secondly, the $l$-adic cohomology is unramified because the action of $\mathrm{Gal}(\bar K|K)$ on the Picard group $\mathrm{Pic}(\bar{X})$ of $\bar X=X\times_K\bar K$ factors via the quotient $\mathrm{Gal}(K(\sqrt{d})|K)$.

Finally, $X$ has bad reduction because its Chow group $A_0(X)_0$ of $0$-cycles of degree $0$ is $\mathbb{Z}/2\mathbb{Z}$ (cf. prop. 1 of arXiv:math/0302156), and a theorem of Bloch (th. 0.4, On the Chow groups of certain rational surfaces, Annales scientifiques de l'École Normale Supérieure, Sér. 4, 14 no. 1 (1981), p. 41-59, available at Numdam) asserts that if a conic bundle has good reduction, then its Chow group of $0$-cycles of degree $0$ is $0$.

Addendum (in response to a question in an email I received). One can show moreover that no smooth projective surface $Y$ over $\mathbf{Q}_p$ which is $\mathbf{Q}_p$-birational to $X$ can have good reduction. This follows from the facts recalled above and the theorem of Colliot-Thélène and Coray (which can be found in Fulton's Intersection theory) : $A_0(Y)_0$ is isomorphic to $A_0(X)_0$.

Answer by Chandan Singh Dalawat for Simple Tamagawa number calculations

2012-11-03T06:54:43Z

It is a pity you don't read French. For those who do (or after you've learnt to read it), I would recommend Annexe B in the undergraduate text Éléments d’analyse et d’algèbre (et de théorie des nombres) by Pierre Colmez, and also his popular article Un autre monde est possible.

Addendum. The basic idea for computing the volume of $\mathrm{SL}_2(\mathbf{A})/\mathrm{SL}_2(\mathbf{Q})$ is that this space can be identified with $$ (\mathrm{SL}_2(\mathbf{R})/\mathrm{SL}_2(\mathbf{Z})) \times \mathrm{SL}_2(\mathbf{Z}_2) \times \mathrm{SL}_2(\mathbf{Z}_3) \times \mathrm{SL}_2(\mathbf{Z}_5) \times\cdots $$ so the volume in question is the product of the volumes of the various factors. The computation of a difficult integral shows that the volume of $\mathrm{SL}_2(\mathbf{R})/\mathrm{SL}_2(\mathbf{Z})$ is $\zeta(2)$, and an easy computation shows that the volume of $\mathrm{SL}_2(\mathbf{Z}_p)$ is $1-{1\over p^2}$, for every prime $p$. Finally, the eulerian product $\zeta(2)=\prod_p(1-{1\over p^2})^{-1}$ allows you to conclude that the volume of $\mathrm{SL}_2(\mathbf{A})/\mathrm{SL}_2(\mathbf{Q})$ is $1$.

Answer by Chandan Singh Dalawat for Algebraic extensions of p-adic closed fields

2011-06-15T11:16:04Z

Here is a completely elementary proof of the fact that a finite extension $F$ of $\mathbf{Q}_p$ has only finitely many extensions of any given degree $n>0$. As $F$ has a unique unramified extension of any given degree, we are reduced to the case of totally ramified extensions. Next, it is easily seen that there are only finitely many (totally ramified) extensions of degree prime to $p$ (essentially because $F^\times/F^{\times m}$ is finite for any $m$ prime to $p$), so we are reduced to the case $n=p^r$. Passing to the galoisian closure and using the fact that any $p$-group admits a filtration whose successive quotients are of order $p$, we are reduced to the case $n=p$. Now, degree-$p$ extensions of $F$ become cyclic when translated to $K$, the extension of $F$ obtained by adjoining to $F$ the $(p-1)$-th roots of everything in $F$; this follows from "Galois's last theorem" (MO24081), see for example arXiv:1005.2016v3. Two degree-$p$ extensions of $F$ give rise to the same cyclic extension of $K$ if and only if they are conjugate over $F$, and the number of conjugates of a degree-$p$ extension $E$ is $1$ (if $E$ is cyclic over $F$) or exactly $p$ (otherwise). As $K$ automatically contains a primitive $p$-th root of $1$, cyclic extensions of $K$ correspond to $\mathbf{F}_p$-lines in $K^\times/K^{\times p}$, which is finite. In fact, the compositum of all degree-$p$ extensions of $F$ is the extension of $K$ obtained by adjoining the $p$-th roots of everything in $K$. Done. (It is at this point that the proof breaks down if $F$ had been a finite extension of $\mathbf{F}_p((t))$; such $F$ have infinitely many separable (indeed cyclic) extensions of degree $p$.)

Comment by Chandan Singh Dalawat

2013-05-17T12:00:38Z

This answer deserves to be included in "The Best of MatheOverflow".

Comment by Chandan Singh Dalawat

2013-05-16T04:13:32Z

When I say direct sum, I mean the direct sum of B(R) and B(Q_p) for all primes p. The sum is indexed by the places of Q, not by the natural numbers.

Comment by Chandan Singh Dalawat

2013-05-16T04:09:52Z

B(Q) might be abstractly isomorphic to this direct sum (and that might suffice for an algebraist), but the point is that B(Q_p) is naturally isomorphic to Q/Z and B(R) to Z/2Z, and that there are natural maps from B(Q) to B(Q_p) and to B(R) which embed B(Q) into the direct sum; this natural embedding is not an isomorphism.

Comment by Chandan Singh Dalawat

2013-05-16T03:05:28Z

only those which are coherent in the sense that their invariants add up to 0. This is a manifestation of reciprocity laws in arithmetic.

Comment by Chandan Singh Dalawat

2013-05-16T03:05:05Z

B(Q) is not quite equal to the direct sum you write in the edit in response to Emerton's comment. It injects into the direct sum, and the image is equal to the kernel of the "sum of the components" from this direct sum onto Q/Z. This is a way of expressing the fact that a central simple algebra over Q gives rise to a family of central simple algebras over Q_p (for every prime p, including p=\infty), but you don't get all such local families from a global object,

Comment by Chandan Singh Dalawat

2013-05-10T02:36:51Z

Joël, it must be Dirichlet's Vorlesungen (with many Supplements by Dedekind), partially translated into English by John Stillwell.

Comment by Chandan Singh Dalawat

2013-05-06T18:02:31Z

... or in Cohomology of number fields by Jürgen Neukirch, Alexander Schmidt and Kay Wingberg. Thanks, Timo.

Comment by Chandan Singh Dalawat

2013-05-06T12:20:07Z

I would look at Kraus-Oesterlé (ams.org/mathscinet-getitem?mr=1166121) and and the papers which refer to it.

Comment by Chandan Singh Dalawat

2013-05-06T09:20:56Z

To get the exact analogue of the question for primes $p\neq2$, perhaps it would be a good idea to start with $\mathbf{Q}(\zeta_p)$ instead of $\mathbf{Q}$.

Comment by Chandan Singh Dalawat

2013-05-06T07:14:36Z

Start with $K_0={\bf Q}$ and let $K_1$ be the maximal abelian extension of $K_0$ of exponent $2$, so $K_1=K_0(\sqrt{K_0^\times})$. Repeating the process $K_{n+1}=K_n(\sqrt{K_n^\times})$ indefinitely gives your field $\mathscr{C}$. Clearly the group $\mathrm{Gal}(K_{1}|K_0)$ can be explicitly described. It would be nice to be able to describe $\mathrm{Gal}(K_{n}|K_0)$ for all $n$.

Comment by Chandan Singh Dalawat

2013-05-06T03:20:06Z

Something that Persiflage wrote only a few days ago is related to your question : galoisrepresentations.wordpress.com/2013/05/04/…

Comment by Chandan Singh Dalawat

2013-04-23T03:11:46Z

Pourquoi François, Joël et Rabelais sont-ils en train de discuter des maths en anglais ?

Comment by Chandan Singh Dalawat

2013-04-20T16:10:46Z

Read the last chapter of Simon Singh's Code Book.

Comment by Chandan Singh Dalawat

2013-04-20T16:08:07Z

The same goes for BSD, Joël, in spite of people having proved it for some 10% of the cases.

Comment by Chandan Singh Dalawat

2013-04-20T10:41:32Z

Isn't that truly amazing !

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What happened to Emmy Noether's *Zukunftsphantasie* ?

Answer by Chandan Singh Dalawat for The origin of the root number $w(C/ℚ)=±1$ (the sign of the functional equation)

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What happened to Emmy Noether's Zukunftsphantasie ?