Field of minimum discriminant of a given degree

Question

Hermite and Minkowski proved, separately, that the number of (isomorphism classes) of number fields of bounded discriminant is finite. The way this is usually stated today is that one has the bound

$$\displaystyle |D_K| \geq \frac{n^{2n}}{(n!)^2} \left(\frac{\pi}{4}\right)^n,$$

if $K$ is a number field with $[K : \mathbb{Q}] = n$. The above inequality gives an explicit lower bound for the smallest possible discriminant of a field of degree $n$ over the rationals.

Let $f(n)$ denote this minimum; i.e.,

$$f(n) = \inf_{K : [K : \mathbb{Q}] = n} |D_K|$$

and $K^{(n)}$ to be a field which achieves this minimum (that is, $|D_{K^{(n)}}|=f(n)$).

Is it expected that the Galois group of the Galois closure of $K^{(n)}$ to be $S_n$?

This is trivially true for $n = 2$, and also true for $n = 3$.

Another way to ask the question is the following. For any number field $K$, denote by $G(K)$ the Galois group of the Galois closure of $K$. Define the function $g(n)$ by

$$\displaystyle g(n) = \begin{cases} 1 & \text{ if } G(K^{(n)}) \cong S_n \\ 0 & \text{ otherwise.} \end{cases}$$

Is it expected that $\lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^N g(n)$ exists and equal to one?

Answer 1

That's already false for $n=4$, when the field of smallest $|D|$ has dihedral Galois group, and indeed of the ten smallest $|D|$ only two come from fields with Galois group $S_4$. [Explicitly, $f(4) = 117$, achieved by ${\bf Q}(x)$ with $x^4-x^3-x^2+x+1 = 0$.] I found this quickly on the LMFDB: http://www.lmfdb.org/NumberField/?degree=4

Computing $f(n)$ gets hard quickly as $n$ increases. For $n \leq 11$ the smallest $|D|$ in the LMFDB belongs to a field of Galois group $S_n$ for $n=2,3,5,7,9,11$ but not $4,6,8,10$. I don't know what is the largest $n$ for which the first LMFDB field is proved to attain the minimal $|D|$.