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Since the discriminant of a number field $K \neq \mathbb{Q}$ is bounded from below by an exponential of the degree $[K:\mathbb{Q}]$, for instance by Minkowski's Geometry of Numbers bound, there are finitely many number fields $K$ below a given discriminant bound $|D_{K/\mathbb{Q}}| \leq X$. Let $\mathcal{S}(X)$ be that finite set.

Question. What is a good guess for the order of growth of $\# \mathcal{S}(X)$ as a function of $X$? Any reasons to expect (or to doubt) that this count should be asymptotic to a regular (say, smooth) function of $X$?

If so, what about finer statistics of number fields at large (unconstrained degrees with $|D_{K/\mathbb{Q}}|$ as the sole parameter), such as the function of typical degree $d(X) := \frac{1}{\#\mathcal{S}(X)} \sum_{K \in \mathcal{S}(X)} [K:\mathbb{Q}]$, the mean Euler-Kronecker invariant $\gamma_K$, or the mean minimum height of a generator of a random number field, $$ h := \lim_{X \to \infty} \frac{1}{\#\mathcal{S}(X)}\sum_{K \in \mathcal{S}(X)} \min_{\alpha: \, K = \mathbb{Q}(\alpha)} h(\alpha) \in [0,\infty]. $$ Assuming this extended number $h$ exists, should it be $0$, $\infty$, or a positive finite number? The former alternative would mean that when number fields are ordered by discriminant, a random one can be generated by a root of an integer polynomial whose coefficients are asymptotically sub-exponential in the degree. The second alternative would mean that a typical number field is complicated to describe, in the sense of requiring super-exponentially large coefficients to write down a generating element. The third alternative says simply that typical number fields have Bogomolov's height gap property.

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    $\begingroup$ Do you know about the known results where the degree is fixed, such as the Davenport–Heilbronn results and the more recent results for quintics? $\endgroup$
    – Greg Martin
    Commented Nov 15, 2018 at 1:50
  • $\begingroup$ @GregMartin: Yes, and so we have a certain linear (conjectured) asymptotic in $X$ in every fixed degree. The unconstrained degrees case seems like generally a wholly different world, e.g. all of the further questions I highlighted are meaningless in the bounded degree regime. That is unless it somehow turns out that the large degree fields make a negligible contribution to the count. For a modest start, perhaps one could limit to answering even this: does $\#\mathcal{S}(X)/X$ tend to infinity with $X$? Certainly the simple radicals $\sqrt[n]{b}$ generate only an $O(X)$ number of fields. $\endgroup$
    – Vesselin Dimitrov
    Commented Nov 15, 2018 at 2:13
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    $\begingroup$ If I understand correctly, Conjecture 1.2 of Manjul's paper "Mass Formulae for Extensions of Local Fields, and Conjectures on the Density of Number Field Discriminants" suggests that the limit as $X$ goes to infinity of the number of degree $n$ $S_n$ fields with discriminant $\leq X$ divided by $X$ decreases rapidly with $n$. Indeed, it goes like $r_2(S_n)/ (2 \cdot n!) \prod_{d=2}^{\infty} \zeta(d)$ where $r_2(S_n)$ is the number of $2$-torsion elements of $S_n$. $\endgroup$
    – Will Sawin
    Commented Nov 15, 2018 at 3:10
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    $\begingroup$ This suggests an answer, but first one has to (1) see if this holds for small $X$ relative to $n$ and (2) generalize to arbitrary groups. $\endgroup$
    – Will Sawin
    Commented Nov 15, 2018 at 3:11
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    $\begingroup$ Given all of the uncertainty, I'd be inclined to define $\overline{h}=\limsup_{X\to\infty}$ and $\underline{h}=\liminf_{X\to\infty}$. At least then you have two well-defined quantities in $[0,\infty]$, and if it turns out that they are not equal, that would be quite interesting, too. $\endgroup$
    – Joe Silverman
    Commented Nov 15, 2018 at 3:26

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