I am currently reading this paper counting $S_3$-sextic fields

  • Manjul Bhargava and Melanie Matchett Wood, The density of discriminants of $S_3$-sextic number fields, Proc. Amer. Math. Soc. 136 (2008), 1581-1587. doi:10.1090/S0002-9939-07-09171-X

I'm trying to verify how they obtained the asymptotic constant for $p = 2, 3$.

My attempt:

$$c_2 =\sum_A \mu_2(A) \frac{D(A)}{D(\bar{A})^{1/3}}$$ where the sum $A$ is over all etale cubic $\mathbb{Q}_2$ algebra $A$, $\mu_2(A)$ is the relative density as defined on the paper on page 3 (or 1583), and $\bar{A}$ is the $S_3$ closure of $A$. To compute $\mu_2(A)$, I referred to Davenport-Heilbronn paper

in particular Lemma 18 and Lemma 19 in the arXiv version. To look up all the possible extensions, I went over to LMFDB at https://www.lmfdb.org/padicField/.

For example, if $A$ is the cubic extension of $\mathbb{Q}_2$ with ramification index 3 and inertia degree 1, the relative density would be $\frac{\mu(\mathcal{U}_p(1^3))}{\mu(\mathcal{U}_p)}$. The discriminant of $A$, $D(A)$, would be $2^2$ since the discriminant exponent $c=2$, and the discriminant of the $S_3$-closure of $A$ would be $2^4$. I computed this by looking at the Galois splitting model provided in LMFDB, and finding the discriminant of the $p$-part of splitting field.

Doing this for all cubic etale algebra of $\mathbb{Q}_2$, I have that $$c_2 = \frac{\mu(\mathcal{U}_p(111))}{\mu(\mathcal{U}_p)}\frac{1}{1} + \frac{\mu(\mathcal{U}_p(12))}{\mu(\mathcal{U}_p)}\frac{1}{1} + \frac{\mu(\mathcal{U}_p(3))}{\mu(\mathcal{U}_p)}\frac{1}{1} + \frac{\mu(\mathcal{U}_p(1^21))}{\mu(\mathcal{U}_p)}\left(2\frac{p^2}{p^{2/3}} + 4\frac{p^3}{p^{3/3}}\right) + \frac{\mu(\mathcal{U}_p(1^3))}{\mu(\mathcal{U}_p)} \frac{p^2}{p^{4/3}} =\frac{1+2p^{1/3}+4p+p^{-4/3}}{1+p^{-1}+p^{-2}}$$

The 2 and 4 are because there are 2 quadratic extensions of $\mathbb{Q}_2$ with $e=2$ and $c=2$, and 4 quadratic extensions of $\mathbb{Q}_2$ with $e=2$ and $c=3$. I thought maybe I should replace 2 with 2/6, and 4 with 4/6 since there are 6 quadratic extensions with $e=2, f=1$, but it still does not work. Doing the same for $p=3$, I have

$$c_3 = \frac{\mu(\mathcal{U}_p(111))}{\mu(\mathcal{U}_p)}\frac{1}{1} + \frac{\mu(\mathcal{U}_p(12))}{\mu(\mathcal{U}_p)}\frac{1}{1} + \frac{\mu(\mathcal{U}_p(3))}{\mu(\mathcal{U}_p)}\frac{1}{1} + \frac{\mu(\mathcal{U}_p(1^21))}{\mu(\mathcal{U}_p)}\left(2\frac{p}{p^{1/3}}\right) + \frac{\mu(\mathcal{U}_p(1^3))}{\mu(\mathcal{U}_p)}\left(2\frac{p^3}{p^{7/3}} + 3\frac{p^4}{p^{4/3}} + \frac{p^4}{p^{8/3}} + 3\frac{p^5}{p^{11/3}}\right)$$

Additional questions:

  1. Is my computation of the relative density correct? Or have I misunderstood the definition from the paper?

  2. To compute the discriminant of the $S_3$ closure of a cubic extension $A$ of $\mathbb{Q}_p$, is it right to compute the $p$-part of the discriminant of the splitting field of the Galois splitting model of the particular cubic extension $A$?

These are vastly different from what they had on their paper. Obviously, I have a severe misunderstanding/error in my calculation and I would appreciate if someone can point out where my mistakes are in particular.

Thanks in advance!

  • $\begingroup$ Please use a high-level tag like "nt.number-theory". I added this tag now. $\endgroup$
    – GH from MO
    Feb 11 at 20:46
  • 1
    $\begingroup$ @GHfromMO Sorry, first time. And thanks! $\endgroup$ Feb 11 at 20:56
  • 1
    $\begingroup$ Why was this downvoted? At first glance this seems like a pretty appropriate question. $\endgroup$ Feb 18 at 23:21
  • $\begingroup$ I"m not an expert on this, I just tidied up the citations to be more informative and stable. I can't judge the number theory content so I'm not sure what precisely is wrong with it. $\endgroup$
    – David Roberts
    Feb 19 at 0:38
  • $\begingroup$ @DavidRoberts Thanks! Sorry you had to do that. I'll be very clear and explicit next time. $\endgroup$ Feb 19 at 3:12

1 Answer 1


I will do a worked example for the tame case and show that it agrees with the formula from their paper. You should then be able to adapt this to the case $p=2,3$.

Let $p > 3$. I use formula (7) from the paper. We need to show that $$\sum_{[L:\mathbb{Q}_p] = 3} \frac{1}{|\mathrm{Aut}(L)|D(\bar{L})^{1/3}} = 1 + 1/p + 1/p^{4/3}$$ where $\bar{L}$ denotes the $S_3$-closure of $L$ and $D$ its discriminant. Here $L$ runs over cubic etale $\mathbb{Q}_p$-algebras. The local factor $c_p$ in Theorem 2 is obtained by multiplying this by $(1-1/p)$.

We consider each case in turn.

  1. $L= \mathbb{Q}_p^3. |\mathrm{Aut}(L)| = 6, D(\bar{L}) = 1.$
  2. $L = \mathbb{Q}_p \times K$, $K$ quadratic. $|\mathrm{Aut}(L)| = 2, D(\bar{L}) = 1$ if $K$ is unramified and $D(\bar{L}) = p^3$ if $K$ is ramified.
  3. $L$ cyclic cubic. $|\mathrm{Aut}(L)| = 3, D(\bar{L}) = 1$ if $L$ is unramified and $D(\bar{L}) = p^4$ if $L$ is ramified.
  4. $L$ non-Galois cubic. $|\mathrm{Aut}(L)| = 1, D(\bar{L}) = 1$ if $L$ is unramified and $D(\bar{L}) = p^4$ if $L$ is ramified.

We work this all out for the case $p = 5$. There is a unique unramified quadratic extension and two ramified quadratic extensions. There is a unique cyclic cubic extension, which is unramified, and a unique non-Galois cubic extension, which is ramified. (You can get all this from the LMFDB.)

All together we obtain: $$\frac{1}{6} + \left(\frac{1}{2} + \frac{2}{2p}\right) + \frac{1}{3} + \frac{1}{p^{4/3}} = 1 + \frac{1}{p} + \frac{1}{p^{4/3}}$$ as required.

(The $S_3$-closure is a bit different from the Galois closure, see the paper for more details. This explains some of the strange exponents in $p$ which appear above, as the index of the subgroup in $S_3$ plays a role in the exponent. This may be where your mistake lies. These exponents should be checked again carefully!)

  • $\begingroup$ Hi, thank you very much for pointing out where I made the mistake. It did not occur to me to use the formula from (7). I went over the construction for the $S_K$-closure of an etale algebra but frankly, the structure of is a bit over my head. To compute discriminant of a number field, I know we generally find the $\text{Tr}(x_ix_j)$ for a set of integral basis, but I'm confused how one does that when the basis is a set of tensors. Could you please refer me to a source where I can learn how to compute discriminants such as the above? Thank you in advance! $\endgroup$ Feb 21 at 16:25
  • $\begingroup$ I'm not entirely sure I'm afraid. I came up with the discriminant exponents by looking at Section 2.1 of Bhargava's "Higher composition laws III: The parametrization of quartic rings". This explains to the definition of the $S_3$-closure and some examples. You shouldn't try to work directly with the definition of the discriminant, but rather relate the discriminant of the $S_3$-closure to the cubic algebra, in a similar way to Bhargava and Wood do at the start of Section 2. $\endgroup$ Feb 22 at 11:21

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