How can one show that an $A_4$-extension of $\mathbb{Q}$ ramified at only one prime must be totally real?

  • 3
    $\begingroup$ To answer this, we need to know how much you already know, what you’ve tried, and also whether it’s a homework problem. $\endgroup$
    – Lubin
    Commented Jul 12, 2015 at 18:34
  • $\begingroup$ @Lubin Not a homework problem (at least not mine). I've tried ideas from Algebraic Number Theory and Class Field Theory with no luck but thought I may have missed something easy. $\endgroup$
    – sharding4
    Commented Jul 13, 2015 at 1:13
  • $\begingroup$ Did you ever find an answer to this? $\endgroup$
    – peter a g
    Commented Jul 27, 2015 at 0:02
  • $\begingroup$ @peterag I haven't found a simple answer. I think it should follow fairly trivially from basic facts on two dimensional odd Galois Representations, but I'm beginning to think there isn't a solution using less machinery. $\endgroup$
    – sharding4
    Commented Jul 27, 2015 at 0:31
  • 1
    $\begingroup$ I believe I have computed correctly that every cyclotomic unit en.wikipedia.org/wiki/Cyclotomic_unit in the cubic subfield of $\mathbb{Q}(\zeta_{163})$ has the same sign under all real embeddings. Unfortunately, I believe the group of cyclotomic units has index $4$ in the group of all units in this case. $\endgroup$ Commented Apr 16, 2018 at 13:24

2 Answers 2


$\def\Z{\mathbf{Z}}$ $\def\Q{\mathbf{Q}}$ $\def\PSL{\mathrm{PSL}}$ $\def\SL{\mathrm{SL}}$ $\def\inv{\mathrm{inv}}$ $\def\Br{\mathrm{Br}}$ $\def\Gal{\mathrm{Gal}}$ $\def\F{\mathbf{F}}$ $\def\R{\mathbf{R}}$ $\def\C{\mathbf{C}}$

This is a consequence of class field theory.

After some obvious reductions, one can assume that $p \equiv 1 \pmod 3$, and that the inertia and decomposition groups at $p$ are given by $I_p = D_p = \Z/3\Z \subset A_4$.

Note that $A_4 = \Gal(K/\Q) = \PSL_2(\F_3)$. Consider the problem of lifting such a representation to $\SL_2(\F_3)$, which is a non-split central extension by $\Z/2\Z$. Group theory says that the obstruction to such a lifting lies in

$$H^2(G_{\Q},\Z/2\Z) = H^2(G_{\Q},\mu_2) = \Br(\Q)[2],$$

the $2$-torsion in the Brauer group of $\Q$. By the Albert-Brauer-Hasse-Noether theorem (class field theory) the Brauer group injects into the sum of local Brauer groups, so the only obstructions are local. Similarly (again by class field theory), the image of $\Br(\Q)$ in the sum of the local Brauer groups is the kernel of the invariant map $\inv$ to $\Q/\Z$, so it follows that any non-trivial class must be non-trivial in $\Br(\Q_v)$ for at least two places $v$ of $\Q$. (For $2$-torsion classes, one even sees it has to be ramified at an even number of $v$, but that is not relevant here).

There is no obstruction to lifting away from the two places $p$ and $\infty$ because unramified representations trivially lift. There is also no obstruction at the prime $p$, because $I_p = D_p = \Z/3\Z$, and $G_{\Q_p} \rightarrow \Z/3\Z \rightarrow \PSL_2(\F_3)$ trivially lifts to $\SL_2(\F_3)$. Thus the only possible obstruction can occur at $\infty$. But, by the observation above, the image of a non-trivial element of $\Br(\Q)$ must be ramified at at least two primes. Thus there is no local obstruction at $\infty$ either.

The result is immediate: if complex conjugation was non-trivial in $\PSL_2(\F_3)$, there would be a local obstruction at $\infty$ since the lift of any order $2$ element in $\PSL_2(\F_3)$ to $\SL_2(\F_3)$ has order $4$, so a non-trivial map $\Z/2\Z = \Gal(\C/\R) \rightarrow \PSL_2(\F_3)$ cannot lift to $\SL_2(\F_3)$. Hence the $A_4$-extension is totally real, as desired.

  • $\begingroup$ Very clear. Thank you. One question though. How do we know that the $A_4$ extension lifts to $\SL_2(\F_3)$? $\endgroup$
    – sharding4
    Commented Apr 18, 2018 at 11:26
  • $\begingroup$ I'm confused. First, I don't directly use the fact that the $A_4$ extension lifts to $\mathrm{SL}_2(\mathbf{F}_3)$. Second, the argument I give proves that it does lift. So what was your question again? $\endgroup$
    – Infinity
    Commented Apr 19, 2018 at 2:55
  • $\begingroup$ Sorry. I hadn't properly appreciated your point that any non-trivial class must be non-trivial for at least two places. Now I see how the result must follow from that, but I will need to give that a little more thought. Thanks again for your answer. $\endgroup$
    – sharding4
    Commented Apr 19, 2018 at 3:08
  • $\begingroup$ Not unexpectedly whatever CFT I picked up from wading thru Milne's lecture notes 12 years ago turned out to be insufficient for me to be able to follow all the details. I come annoyingly close at several points :-). Well, I take that as an incentive to try and learn more (if I still can). $\endgroup$ Commented Apr 19, 2018 at 19:38

Here is a partial answer as discussed in the comments. If a correct solution will follow from this I will update the answer accordingly. I hope this will happen!

Let $L/\mathbb{Q}$ be an $A_4$-extension ramified only at $p$. As $A_4$ contains $V_4$ as a normal subgroup of index $3$, we have a corresponding subfield $L/K/\mathbb{Q}$, with $K/\mathbb{Q}$ of degree $3$ and Galois. Then $K/\mathbb{Q}$ is also ramified only at $p$, since it has to ramify at some prime (Minkowski) and is unramified at all primes $q\neq p$ as $L$ is. The Kronecker-Weber theorem now implies that $K\subset \mathbb{Q}(\zeta_{p^k})$ for some $k\in\mathbb{Z}_{>0}$.

If $p\neq 3$ then $p$ is tamely ramified in $K$ and hence $K\subset\mathbb{Q}(\zeta_p)$, which forces $p\equiv 1\bmod 3$.

If $p=3$ then $p$ is wildly ramified in $K$, which implies $k\geq 2$. But $\mathbb{Q}(\zeta_{3^k})/\mathbb{Q}$ is a cyclic extension of degree $\varphi(3^k)=2\cdot 3^{k-1}$. As the degree of this cyclic extension is divisible by $3$ it contains a unique cubic subfield, hence $K=\mathbb{Q}(\zeta_9+\zeta_9^{-1})$. Thus in this case we have $k=2$, i.e. $K\subset\mathbb{Q}(\zeta_9)$.

The idea is now as follows: $K$ is totally real and $L/K$ an Abelian extension, to which we can apply class field theory to study the ramification at the infinite primes of $K$. Proving that $L/K$ is unramified at the infinite primes will prove that $L$ is totally real.

I first prove that $L/K$ is unramified at the finite primes of $K$. Note that $p$ is totally ramified in $K$, say as $p\mathcal{O}_K=\mathfrak{p}^3$, since $K/\mathbb{Q}$ is Galois of prime degree. Thus we need to show that $L/K$ is unramified over $\mathfrak{p}$. Note that $\mathrm{Gal}(L/K)\cong V_4$, so there are three intermediate quadratic subfields $M_1,M_2,M_3$ of the extension $L/K$. These three fields are all conjugate to each other in the big extension $L/\mathbb{Q}$, hence $p$ has the same factorisation type in each $M_i$: if $\phi\in\mathrm{Gal}(L/\mathbb{Q})$ sends $M_i$ to $M_j$, then applying $\phi$ to the factorisation of $p$ in $M_i$ yields the factorisation of $p$ in $M_j$. Moreover, as $K/\mathbb{Q}$ is normal and $\mathfrak{p}$ is the unique prime of $K$ lying over $p$, we can also conclude that $\mathfrak{p}$ has the same factorisation type in all the $M_i$. This implies that $\mathfrak{p}$ is either unramified in all the $M_i$, hence also in their compositum $L$, or that $\mathfrak{p}$ is ramified in all the $M_i$. In this second case we must have that $\mathfrak{p}$ is totally ramified in $L$, since $e_{\mathfrak{p}}=2$ would imply that some $M_i$ equals the inertia field $T_{\mathfrak{p}}$ in the extension $L/K$, which cannot happen as $\mathfrak{p}$ ramifies in $M_i$.

Now if $\mathfrak{p}$ would be totally ramified in $L$, then $p$ is also totally ramified in $L$. If now $E$ is a subfield of $L/\mathbb{Q}$ corresponding with a subgroup of $A_4$ of order $3$, hence index $4$. The prime $p$ is also totally ramified in $E$, say as $p\mathcal{O}_E=\mathfrak{q}^4$. As $E/\mathbb{Q}$ has normal closure $L$ we have that $\Delta_E=\Delta(\mathcal{O}_E)$ has to be a square: the discriminant of the minimal polynomial of an integral primitive element for $E/\mathbb{Q}$ is a square. However, as $p\equiv 0,1\bmod 3$ we have $p\neq 2$, so $p$ is tamely ramified in $E$. Hence $\mathfrak{D}_E=\mathfrak{q}^3$, which yields $|\Delta_E|=p^3$ upon taking norms. This contradicts that $\Delta_E$ is a square, so we conclude that $\mathfrak{p}$ is unramified in $L$.

As $\Delta_K$ is a square, we have $1=\mathrm{sign}(\Delta_K)=(-1)^s$, where $s$ is the number of pairs of complex embeddings of $K$. As $s\leq 1$ we have $s=0$, i.e. $K$ is totally real. Now either $L$ is totally real or totally complex. For as $\Delta_E$ is a square we also have $\mathrm{sign}(\Delta_E)=1$, hence $E$ is totally real or totally complex. If it is totally real then $L$ is totally real as the compositum of the totally real fields $K$ and $E$, while if $E$ is totally complex then also $L$ is totally complex as any real embedding $\tau:L\to\mathbb{C}$ restricts to a real embedding of $E$.

Now assume for contradiction that $L$ is not totally real, so that it is totally complex. If we let $\infty_i:K\to\mathbb{C}$ be the infinite primes of $K$ for $i\in\{1,2,3\}$, then $L$ being totally complex, each $\infty_i$ being real and $L/K$ being unramified at the finite primes implies that the abelian extension $L/K$ has conductor $\mathfrak{f}=\mathfrak{f}_{L/K}=\infty_1\infty_2\infty_3$.

Now let $H$ be the Hilbert class field, and $H^{+}$ the narrow Hilbert class field of $K$. Class field theory tells us that $H^{+}/H$ is a Galois extension with group an elementary abelian $2$-group of order $2^r[\mathcal{O}^{*}:\mathcal{O}^{*}_{+}]^{-1}$, where $\mathcal{O}=\mathcal{O}_K$ and $\mathcal{O}^{*}_{+}$ is the subgroup of $\mathcal{O}^{*}$ consisting of those units $u\in\mathcal{O}^{*}$ with $\infty_i(u)>0$ for all $i$, and $r=3$ is the number of real embeddings of $K$.

Now $L\cap H/K$ is unramified, which implies $L\cap H=K$ as $L$ is totally complex: clearly $L\cap H\neq L$ as $L/K$ is not unramified, and if $L\cap H=M_i$ for some $i$ then $M_i$ is totally real, but as $M_i$ is conjugate to the other $M_j$ this would yield that $L$ is totally real as the compositum of the totally real $M_i$ and $M_j$ for $j\neq i$, which contradicts our assumption that $L$ is not totally real. Thus we have $L\cap H=K$. As $[L:K]=4$ we also have $[LH:H]=4$. However we also have $H\subset LH\subset H^{+}$, so $$ [LH:H]\leq[H^{+}:H]=8[\mathcal{O}^{*}:\mathcal{O}^{*}_{+}]^{-1}. $$ Thus if we can show that $[\mathcal{O}^{*}:\mathcal{O}^{*}_{+}]>2$ then we have the desired contradiction to conclude that $L$ must be totally real. In any case we have $-1\in\mathcal{O}^{*}\setminus \mathcal{O}^{*}_{+}$, so to obtain the contradiction we need to show that the quotient $\mathcal{O}^{*}/\mathcal{O}^{*}_{+}$ is not generated by the image of $-1$, i.e. we need to find a unit $u\in\mathcal{O}^{*}$ not equal to $1$ or $-1$, for which $\mathrm{sign}(\infty_i(u))\neq\mathrm{sign}(\infty_j(u))$ for certain distinct $i,j\in\{1,2,3\}$.

Now in the special case where $p=3$ and $K=\mathbb{Q}(\zeta_9+\zeta_9^{-1})$ one can show this directly: the minimal polynomial of $x=\zeta_9+\zeta_9^{-1}$ is $X^3-3X+1$, which shows that $x\in\mathcal{O}_K^{*}$, and one can see that $X^3-3X+1$ has both a positive and a negative real root.

In general however, the structure of $\mathcal{O}_K^{*}$ when $p\equiv 1\bmod 3$ seems too mysterious to always make sure one can find a unit with the desired properties discussed above, so this is where I'm stuck.

I hope that someone else can find a way to do this!


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