# How do I calculate the discriminant of a galois closure and its other subfields?

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?

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## 3 Answers

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.)

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Thanks for the concise answer about the power of p dividing the subfields. As to the mentioned fields for which this fails, as I said, I am not interested in those - p is totally ramified in all these failures. Can you give a hint for as to how I might prove the result? –  Dror Speiser Nov 25 '09 at 0:51
I don't remember what special ramification means, but I've edited my answer to give some sketch of how you should attack a problem like this. –  David Speyer Nov 28 '09 at 14:01

According to my hand computation, the claim is not true for $K = \mathbb{Q}(\sqrt[3]{3})$. Have you checked this case?

It is true that a prime ramifies in $K$ if and only if it ramifies in the Galois closure of $K$. There might be a formula like the one you are describing that is true for primes greater than $d$.

Warning: this part of the answer was written without access to my usual number theory references, so there may be errors. It isn't meant to be read by itself anyway; it is meant to be a guide to a good book on algebraic number theory, like Neukirch or Janusz.

# How you should approach a question like this

Your fundamental strategy to be to relate the order of ramification of $p$ in $L$ and in the Galois closure of $L$. This does not precisely determine the power of $p$ dividing the discriminant, but we have the following

Fact: Let $L/K$ be an extension of number fields and $p$ a prime of $K$ with ramification indices $e_1$, $e_2$, ... $e_r$ and residue field extensions of degree $f_1$, $f_2$, ..., $f_r$. Then the power of $p$ dividing $D_{L/K}$ is at least $\sum (e_i-1) f_i$, with equality if and only if all the $e_i$ are less than relatively prime to the characteristic of $p$.

From now on, I will address the question of how to relate the orders of ramification in $L$ and in the Galois closure. If you need to deal with the case that some of the $e_i$ are greater than or equal to the characteristic of $p$, you should read about higher ramification groups.

Let $M$ be the Galois closure of $L$, let $G$ be $\mathrm{Gal}(M/K)$, and let $H$ be the fixed field of $L$. Fix a prime $p$ of $K$, and1 $\mathfrak{p}$ a prime over $p$. Let $D \subseteq G$ be the decomposition group of $\mathfrak{p}$ and $I \subset D$ the inertia group. $I$ is normal in $D$; the quotient $D/I$ is cyclic and has a canonical generator $F$ called the Frobenius.

I believe that FC's computation was for the case $G=S_n$, $D=I=\{ e, (12) \}$.

In $M$, the primes lying over $p$ are in bijection with $G/D$. Each of them has $f=|D/I|$ and $e=|I|$.

Let $X=G/H$ (a set with $G$-action). In $L$, the primes above $p$ are in bijection with the $D$-orbits in $X$. Let $O$ be such a $D$-orbit, corresponding to a prime $q_O$. Because $I$ is normal, $O$ breaks up as a union of $I$-orbits all of the same cardinality. Then $q_O$ has $e$ equal to the cardinality of these $I$-orbits, and $f$ equal to the number of them.

Using these ideas, you should be able to relate ramification in $L$ and $M$ for any $G$ and $H$ which interest you.

1 I fix $\mathfrak{p}$ only for expositional purposes. Changing $\mathfrak{p}$ will conjugate $(D,I,F)$. Probably the right way to think of all of this stuff is working up to conjugacy.

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Thanks! You're right about the coprime to p issue; I'll fix that. I remember that there is something surprising which goes wrong for $e>p$, even for tame ramification. Apparently that wasn't it though. –  David Speyer Nov 28 '09 at 21:54
Very nice, thanks! By the way, I think there's a typo in the definition of $H$, which is the subgroup associated to $L$ and not its fixed field (obviously). –  Niccolo' Feb 3 at 23:27

The problem of computing the relative discriminant of an extension of number fields can be reduced to the computation of the relative discriminant of a local kummerian extension of degree equal to the residual characteristic, which is easy to solve. See for example arXiv:0711.3878v1 [math.NT], p. 51.

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