Uniform boundedness of the number of number fields having fixed discriminant

Question

Let $d$ be an integer. It is a well-known theorem, attributed to Hermite and Minkowski, which asserts that the number of number fields $K$, allowed to have any degree over $\mathbb{Q}$, having discriminant $\Delta_K = d$ is finite.

Let $S(d)$ be the number of isomorphism classes of number fields of discriminant $d$. The Hermite-Minkowski theorem is the assertion that $S(d) < \infty$ for all $d \in \mathbb{Z}$. Is $S(d)$ uniformly bounded? That is, does there exist a positive integer $N$ such that $S(d) \leq N$ for all $d \in \mathbb{Z}$?

We can refine the question and instead only count (isomorphism classes of) number fields of fixed degree over $\mathbb{Q}$. Let $S_n(d)$ be the number of isomorphism classes of number fields of discriminant equal to $d$ and degree equal to $n$. Does there exist a number $N(n)$ such that $S_n(d) \leq N(n)$ for all $d \in \mathbb{Z}$?

Observe that the answer is yes for $n = 2$. Indeed we find that $S_2(d) \leq 1$ for all $d \in \mathbb{Z}$, with equality if and only if $d$ is a fundamental discriminant.

Answer 1

The answers to both questions is no. For instance for $n=3$, if $p_1$,... $p_k$ are primes congruent to $1$ modulo $3$ there exist $(3^k-1)/2$ cyclic cubic fields of discriminant equal to $(p_1...p_k)^2$ by elementary class field theory.