I am looking for an extension $F/\mathbb{Q}$ with the following properties:

  1. $F/\mathbb{Q}$ is Galois with $\mathrm{Gal}(F/\mathbb{Q}) \simeq A_4$.
  2. $F$ is totally real.
  3. The prime $2$ has full decomposition group.
  4. For every odd prime $p$ that ramifies in $F/\mathbb{Q}$, the primes above $p$ all have decomposition group of odd order. (In other words, every such prime is totally split in the biquadratic sub-extension cut out by the subgroup $V_4$ of $A_4$).

Using John Jones' number fields database http://hobbes.la.asu.edu/NFDB/ I found 7670 totally real quartic extensions of $\mathbb{Q}$, each of which with normal closure a totally real $A_4$-extension of $\mathbb{Q}$. Using Magma, I found that 792 of these also have full decomposition group at 2. But then none of these also satisfy the last condition.

Is there some reason why no such extension exists, or is it just the case that I haven't looked hard enough?

Note that if no such extensions exist, the reason cannot be purely local in nature. For example, let $F$ be the Galois closure of $x^4 - 108x^2 - 136x - 8$. Then $F/\mathbb{Q}$ is a totally real $A_4$ extension with full decomposition group at $2$. The other primes dividing the discriminant are 17 and 241. The primes above 241 behave in the way that I want, but those above 17 do not. It appears that something similar always happens, i.e., there is always at least one odd prime dividing the discriminant that doesn't behave in the way that I want.

Also note that is there is only one $A_4$-extension of a $p$-adic field (for any $p$), and this is an $A_4$-extension of $\mathbb{Q}_{2}$. That means that no odd prime can have full decomposition group and that there is no local obstacle to the condition on $2$. Moreover, the unique $A_4$ extension of $\mathbb{Q}_2$ has a cubic unramified subextension, so any $A_4$ extension of $\mathbb{Q}$ with full decomposition group at 2 must also be ramified at (at least) one odd prime.

  • $\begingroup$ Hi Henri, one thing I do not understand: by $V_4$ do you mean the Klein $4$-group? Then the field fixed by it is a cubic extension of $\mathbb{Q}$, or may be I do not understand what you mean by "cut out"? $\endgroup$ Feb 28, 2017 at 16:46
  • $\begingroup$ I should have been clearer - thanks for pointing this out. Yes, by $V_4$ I mean the Klein 4-subgroup of $A_4$. Let $K$ be the subfield of $F$ fixed by $V_4$. Then the biquadratic extension I am referring to is $F/K$. $\endgroup$ Feb 28, 2017 at 16:49
  • $\begingroup$ Just out of curiosity: have you computed the class number (or, better, the class group) of the cubic field in your example where $241$ and $17$ ramify? I cannot access a Pari calculator now, and I am pretty curious... $\endgroup$ Feb 28, 2017 at 17:07
  • $\begingroup$ @Filippo: In the example above, the cubic subfield is only ramified at 241 and has class number 1. Both the quartic subfield and $K$ itself have class number 2. (Calculated using magma under assumption of GRH.) $\endgroup$ Feb 28, 2017 at 22:41

1 Answer 1


There is no such field. In fact, let $F$ be an $A_4$-extension of the rationals, and let $K$ denote the cyclic cubic subfield of $F$. Then $F = K(\sqrt{\alpha},\sqrt{\alpha'}\,)$ for some $\alpha \in K$ such that $\alpha\alpha'\alpha''$ is a square in ${\mathbb Q}$. Here $\alpha'$ is the conjugate of $\alpha$ etc.

Every odd prime $p$ that ramifies is assumed to split in $F/K$, hence it must ramify in $K/{\mathbb Q}$. Thus $F/K$ is unramified outside $(2)$.

The prime $2$ must be inert in $K/{\mathbb Q}$ by assumption 3. But then we can choose $\alpha$ to have odd norm, since $\alpha\alpha'\alpha''$ cannot be a rational square if $2$ divides $\alpha$ exactly. Since no odd prime ramifies in $F/K$, the ideal $(\alpha)$ must be a square of an ideal, and $\alpha$ must be totally positive by assumption 2.

This means that $\alpha \in Sel^+(K)$ (I am using the notation and a few results from https://arxiv.org/pdf/1108.5674.pdf from now on; the Selmer group $Sel^+(K)$ consists of all $\alpha K^{\times 2}$ for which $\alpha$ is totally positive and $(\alpha)$ is the square of an ideal). The $2$-rank of this group is $\rho^+ + s = \rho^+$ in our case (Thm. 2.2) since $K$ is totally real and thus $s = 0$. Here $\rho^+$ is the rank of the $2$-class group in the strict sense. By Thm. 7.2., we have $\rho^+ = \rho$ in our case $(n = 3)$, hence the $2$-rank of $Sel^+(K)$ is equal to the $2$-rank of the class group.

Since every quadratic unramified extension is generated by the square root of an element in $Sel^+(K)$, and since these groups have the same $2$-rank, every extension $K(\sqrt{\alpha})$ for $\alpha \in Sel^+(K)$ must be a subfield of the Hilbert class field of $K$. But since $2$ is inert in $K$, it is principal, and thus splits in every subfield of the Hilbert class field. This contradicts assumption 3.


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