Upper bound for discriminant of Galois closure In Lang's Algebraic Number Theory book he uses a certain bound on the discriminant of the Galois closure of a number field $K$ without proof stating that it is an easy exercise. Let $\tilde{K}$ be the Galois closure of $K$. Set $[K:\mathbb{Q}] = N$ and $[\tilde{K}:\mathbb{Q}] = \tilde{N}$. Let us also denote the absolute values of the discriminants as $d_K$ and $d_{\tilde{K}}$. Then $d_{\tilde{K}} \leq d_{K}^{\frac{\tilde{N}}{2}}$.
What I tried was the following. Let $L$ be the subfields of $\tilde{K}$ such that $K$ and $L$ are linearly disjoint, and $\tilde{K} = KL$, then $[L:\mathbb{Q}] = \frac{\tilde{N}}{N}$. Now we have that $d_{\tilde{K}} = d_{K}^{\frac{\tilde{N}}{N}}d_{L}^{N}$.
But I do not see how I could proceed further. Any ideas on this would be extremely helpful. Thank you.
Edit:
This fact is found on page number 327 of Lang's Algebraic Number Theory book. (It is stated in the last few line of the page).
 A: One approach is to prove the corresponding fact on the Artin conductor side. Let $G$ be the Galois group ok $\tilde{K}/\mathbb Q$ and $H$ the Galois group of $\tilde{K}/K$, so that $|G|= \tilde{N}$ and $|H|= \tilde{N}/N$.

At a prime $p$, the Artin conductor of the regular representation of $G$ is at most $|G|/2$ times the Artin conductor of the permutation representation of $G$ acting on $G/H$.

To prove this, note that the Artin conductor is a linear combination over ramification groups $G_i$ of the codimension of the $G_i$-invariants of a given representation. It suffices to prove that the codimension of the $G_i$-invariants acting on the regular representation is at most $|G|/2$ times the codimension of the $G_i$-invariants acting on $|G|/H$.
Since the permutation representation of $G$ on $G/H$ is faithful, if the codimension of the $G_i$ invariants is $0$ then the codimension of the $G_i$-invariants of the regular representation is $0$.
If the codimension of the $G_i$-invariants on $G/H$ are at least $2$, then certainly the codimension of the $G_i$-invariants on the regular representation are at most the dimension $|G|$ of the regular representation.
In the remaining case, when the coinvariants are one-dimensional, because $G_i$ is finite, $G_i$ must act on the orthogonal complement of the invariants as a finite cyclic group, and because the permutation representation is integral, that group must have order $2$. So its invariants on the regular representation are exactly $|G|/2$, dimensional, and the codimension of the invariants is also $|G|/2$.
So the inequality is satisfied in all three cases, as desired.
With the Artin conductor inequality checked, we just need to convert it to an equality of discriminants with the conductor-discriminant formula - i.e. that the discriminant of $K$ over $\mathbb Q$ is the product over $p$ of $p$ raised to the Artin conductor of the permutation representation $G/H$, and similarly the discriminant of $\tilde{K}$ over $\mathbb Q$ is the product over $p$ of $p$ raised to the Artin conductor of the regular representation. The more general, non-Galois version of this is stated, e.g., as Corollary 11.8 in Neukirch's algebraic number theory book. I couldn't follow Lang's notation well enough to see if it's explained there.
If you want to complain that this isn't an "easy exercise", I understand, and I am only unsure whether Lang had a different notion of "easy exercise" than I do, or whether he was thinking of a different proof.
