Given a number field K of dimension d over Q, and galois closure of dimension d! over Q (i.e galois group Sd), can we relate the discriminant of the galois closure to that of the discriminant of K? Assume no special ramification happens in the closure or the other subfields, for example if the discriminant of K is a prime.


Tests in sage indicate that the discriminant of the galois closure is $\Delta\_K^{\frac{d!}{2}}$, and that the discriminants of the other subfields are also powers of $\Delta\_K$, but the power has something to do with the corresponding subgroup of Sd. (Not just the size of the subgroup)


Is there a way to prove the first indication, and thoughts about the second?



How do I calculate the discriminant of a galois closure and its other subfields?
It's true for $d = 2$. Even for $d = 3$, it fails miserably. If $K = \mathbf{Q}(p^{1/3})$ and $p \ne 3$ then $p^2$ exactly divides $\Delta\_{K}$, whereas $p^4$ exactly divides the discriminant of the Galois closure. (As David points out, things are even worse for $p = 3$.) Not to mention the fact that $p$ doesn't divide to any power the discriminant of $\mathbf{Q}(\sqrt{-3})$ which is contained inside the Galois closure of $K$.


About the only time the power of $p$ dividing the discriminant of the Galois closure is $n!/2$ times the power of $p$ dividing the discriminant is when $p$ *exactly* divides $\Delta\_K$. In this case, the power of $p$ dividing the fixed field of $H$ in $S\_n$ is equal to


$$(n-2)! \cdot \frac{\text{the number of $2$-cycles which do not lie in $H$}}{|H|}$$


This is always an integer, and it answers your question when $\Delta\_K$ is squarefree. (The general case having already seen to be false.) 


