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Algebraic number fields, Algebraic integers, Arithmetic Geometry, Elliptic Curves, Function fields, Local fields, Arithmetic groups, Automorphic forms, zeta functions, $L$-functions, Quadratic forms, Quaternion algebras, Homogenous forms, Class groups, Units, Galois theory, Group cohomology, Étale cohomology, Motives, Class field theory, Iwasawa theory, Modular curves, Shimura varieties, Jacobian varieties, Moduli spaces

4 votes

Old question of Serre on discriminants of a sequence of polynomials

To reiterate Serre's question (Open Problem 2.5 in Odlyzko's survey): Must there be only finitely many polynomials having root discriminant below a given bound? With this answer I just want to note …
Vesselin Dimitrov's user avatar
7 votes
Accepted

Finite Nontrivial Unramified Towers of Number Fields

It is certainly possible that $1 < [L:F] < \infty$, i.e. that the extension $F^{\mathrm{un}}/F$ be finite and non-trivial. The simplest example of this is $F = \mathbb{Q}(\sqrt{-5})$. Its Hilbert clas …
Vesselin Dimitrov's user avatar
16 votes

Special topics to include in course in algebraic number theory

Here are a few suggestions, though within number theory. Chebotarev's original proof of the density theorem (in the logarithmic form). Or at least the infinitude of totally split primes as well as t …
Vesselin Dimitrov's user avatar
9 votes
Accepted

Is the infimum of Salem numbers > 1?

I believe it is the general opinion, at least among those working in diophantine approximations, that the extreme case of Salem numbers (the question of the title) would be just as difficult as the fu …
Vesselin Dimitrov's user avatar
14 votes
0 answers
641 views

Can there be arbitrarily many cubic fields unramified outside $\{p,\infty\}$?

Observe, trivially, that since quadratic fields correspond to rational integers modulo squares (viz. discriminants), there are (roughly about, but certainly at most) $2^{|S|+1}$ quadratic fields unram …
Vesselin Dimitrov's user avatar
13 votes

Estimate number of solutions in the Roth's theorem

For a fixed $\alpha$, the number $N_{\alpha}(\epsilon)$ is bounded by a polynomial function of $1/\epsilon$. The proof of this requires either Faltings's product theorem, or Esnault and Viehweg's mult …
Vesselin Dimitrov's user avatar
3 votes
0 answers
100 views

Independence of number fields generated by roots of Littlewood polynomials

Let $\mathcal{R}_d \subset \bar{\mathbb{Q}}$ be the set of all roots of degree $d$ polynomials with $\{-1,0,1\}$ coefficients and $$ c(d) := \min_{\substack{ \alpha, \beta \in \mathcal{R}_d \\ \alpha^ …
Vesselin Dimitrov's user avatar
4 votes

Asymptotics for algebraic numbers of height less than one

As I indicated above, $\log{N(d)} \asymp d^2$ reduces to a problem about irreducible polynomials, the very likely affirmative answer to which would prove the lower bound $n(h,d) \geq (h-o(1))d^2$, as …
Vesselin Dimitrov's user avatar
14 votes
3 answers
925 views

Asymptotics for algebraic numbers of height less than one

The question. Is an asymptotic equivalent known or conjectured for the number $N(d)$ of $\alpha \in \bar{\mathbb{Q}}$ with $h(\alpha) < 1$ and $[\mathbb{Q}(\alpha):\mathbb{Q}] \leq d$? The rather cr …
Vesselin Dimitrov's user avatar
11 votes
2 answers
1k views

Can there be a power basis for a totally real field of high degree?

A number field $K$ is said to have a power basis if there is an $\alpha \in K$ such that the full ring of integers $O_K$ is the $\mathbb{Z}$-linear span of $1,\alpha,\alpha^2,\ldots,\alpha^{\deg{K}-1} …
Vesselin Dimitrov's user avatar
9 votes
0 answers
263 views

How small may the discriminant of an $S_d$-field be?

In every degree $d$, the Galois closure of the typical number field has the maximal possible Galois group $S_d$. Denote by $f(d)$ the least absolute value of a discriminant of an $S_d$-field of degree …
Vesselin Dimitrov's user avatar
19 votes
Accepted

Multizeta function values

The elements of $S$ are conjectured to be $\mathbb{Q}$-linearly independent, and so a basis for the $\mathbb{Q}$-linear span of the multiple zeta values. This is what Francis Brown accomplished at t …
Vesselin Dimitrov's user avatar