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).



Upper bound for discriminant of Galois closure
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 $|H|= \tilde{N}/N$.



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> 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 $G/H$.
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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.


