5
$\begingroup$

For an introduction to Leopoldt's conjecture, see part 3, Chapter X of "Cohomology of Number Fields", which is freely available here.

Write Leo($K$,$p$) for Leopoldt's conjecture for the number field $K$ at the prime $p$. If $K$ is an abelian extension of either $\mathbb{Q}$ or an imaginary quadratic field, then we know Leo($K$,$p$) for all primes $p$. (This is due to Ax and Brumer - see reference above for more details.)

Now let $K$ be a cubic extension of $\mathbb{Q}$. I am interested in Leo($K$,3). If $K/\mathbb{Q}$ is Galois then we are done. If $K$ has mixed signature, then its Galois closure $L$ is a totally complex $S_3$ extension of $\mathbb{Q}$ and thus is a cyclic extension of an imaginary quadratic field. Thus we know Leo($L$,3) and this implies Leo($K$,3) (Leopoldt's conjecture is inherited by subfields).

My question is: are there any cases for which Leo($K$,3) is known when $K$ is a totally real non-Galois cubic extension of $\mathbb{Q}$? By the above reasoning, it would suffice to find a totally real $S_3$ extension $F/\mathbb{Q}$ for which Leo($F$,3) is known.

Improve this question
$\endgroup$
3
  • 4
    $\begingroup$ For an explicit field $K$ one could just check experimentally that the Leopoldt conjecture is true at 3, basically by computing the regulator and checking it's non-zero mod $3^m$ for some large $m$. $\endgroup$
    – znt
    Commented Nov 20, 2016 at 16:27
  • $\begingroup$ @znt Can you email me at the address on my website? $\endgroup$
    – Henri Johnston
    Commented Feb 8, 2019 at 17:22
  • $\begingroup$ Fabio Ferri and I have now proven the following result: given any finite set of prime numbers $\mathcal{P}$,there exists an infinite family $\mathcal{F}$ of totally real $S_{3}$-extensions of $\mathbb{Q}$ such that Leopoldt's conjecture for $F$ at $p$ holds for every $F \in \mathcal{F}$ and $p \in \mathcal{P}$. See arxiv.org/abs/2301.05700 $\endgroup$
    – Henri Johnston
    Commented Jan 17, 2023 at 9:43

1 Answer 1

Reset to default
7
$\begingroup$

Out of curiosity, I thought I'd follow up on znt's suggestion.

Let $K = \mathbf{Q}(\alpha)$ where $\alpha^3 - 13\alpha + 7=0$. Then $K$ is totally real and non-Galois, and the prime 3 is totally inert in $K$. The unit group of $K$ is generated by $(u_1, u_2) = (\alpha^2 + 3\alpha - 2, \alpha^2 - 4\alpha + 2)$.

There are 3 embeddings of $K$ into the unramified cubic extension $L$ of $\mathbf{Q}_3$. Sage happens to like the polynomial $y^3 + 2y + 1$ as a defining polynomial for $L$; if $c$ is a root of this, then for $i = 0, 1, 2$ there's an embedding $\sigma_i$ for which $\sigma_i(\alpha) = c + i \bmod 3$.

Now one finds that the matrix with $i,j$ entry $\log \sigma_i(u_j)$ for $1 \le i, j \le 2$ has determinant

3^3 + 3^7 + 2*3^8 + 3^9 + 3^10 + 2*3^12 + 2*3^15 + 3^17 + 3^18 + 2*3^19 + 3^20 + O(3^21)

which is not zero. Hence Leopoldt's conjecture holds for this extension at 3.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thanks, David and znt! So in principle we can check Leo($K$,$p$) for any particular (totally real) number field $K$ and prime $p$ computationally. I'll hold off accepting your answer for the moment to see if there are any other answers that take a different approach. $\endgroup$
    – Henri Johnston
    Commented Nov 21, 2016 at 11:58
  • $\begingroup$ This article: Buchmann, Johannes; Sands, Jonathan, An algorithm for testing Leopoldt's conjecture, J. Number Theory 27 (1987), no. 1, 92–105 gives an algorithm that uses a reformulation of Leopoldt's conjecture that is more amenable to computation. However, as far as I know, there isn't an implementation of this algorithm in Magma or Sage. $\endgroup$
    – Henri Johnston
    Commented Nov 22, 2016 at 12:32
  • $\begingroup$ Tommy Hofmann, Andreas Nickel and I have come up with a new algorithm described in the appendix of this article: dx.doi.org/10.1112/jlms.12310 The magma code is available on Tommy's website: thofma.com/software $\endgroup$
    – Henri Johnston
    Commented Jan 17, 2023 at 9:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .