Abelian extensions with squarefree discriminant

Question

Is it true that for all $2 \leq n \in \mathbb N$ we can find an abelian extension of the rationals with squarefree discriminant and degree n? Are there even infinitely many?

Some even degrees may be handled by cyclotomic fields but certainly not all. For quadratic numberfields there are infinitely many: $\mathbb Q(\sqrt{d})$, for squarefree $d>0$ and $d \equiv 1 \mod 4$.

I found this paper implying the existence of infinitely many extensions of squarefree discriminant and fixed degree n, but they all seem to have $S_n$ as galois group of their normal closure.

Answer 1

The discriminant of an abelian extension is computed by the conductor-discriminant formula, which writes the discriminant as the product over all the characters of the Galois group of the conductor of the splitting field of each such character.

In order for the discriminant to be square free, we need in particular each splitting field to appear just once in this product. Since a character $\chi$ and its inverse $\chi^{-1}$ have the same splitting field, we thus need each non-trivial character to be of order two, and hence the Galois group should be an elementary abelian two group.

A further analysis shows that in fact the Galois group should be of order one or two. E.g. already a biquadratic field such as $\mathbb Q(\sqrt{2},\sqrt{5})$ has discriminant $8 \cdot 5 \cdot 40$ (because the non-trivial characters of its Galois group have fixed fields $\mathbb Q(\sqrt{2})$, $\mathbb Q(\sqrt{5})$, and $\mathbb Q(\sqrt{10}),$ whose conductors are $8,$ $5$, and $40$ respectively), which is not square free.

In conclusion, for an abelian extension to have square free discriminant, it should be of degree $1$ or $2$.