I am looking for examples of CM fields whose Galois group is not abelian. By a CM field K I mean an imaginary quadratic extension of a totally real field $K_0$. If the extension is not Galois I take the Galois closure $L/K$.

Obviously the degree of such a field is even.

When $[K:Q]=2,$ K is an imaginary quadratic field, so its Galois group is $\mathbb{Z}/2\mathbb{Z}$.

My question is: what are the possible Galois groups for $[K:Q]=4$ and 6?

For instance, if I consider $K=K_0(\mu_3)$, with $K_0$ totally real of degree 4 and $\mu_3$ the roots of unity of order 3, what are the possible Galois groups?

Thanks for your help!

  • $\begingroup$ Why should a degree four CM field be Galois? $\endgroup$
    – cmfield
    Mar 19, 2013 at 9:43
  • $\begingroup$ Could you elaborate more on your last sentence "CM points will define CM fields with dihedral Galois group"? Thanks! $\endgroup$
    – cmfield
    Mar 19, 2013 at 9:46
  • $\begingroup$ When you take $K_0={\mathbb Q}({\sqrt 2}$ and $K=K_0[X]/(X^2+{\sqrt 2})$, it is clear that $[K:{\mathbb Q}]=4$ but it is not Galois. The Galois group is not abelian but is of order $8$. $\endgroup$ Mar 19, 2013 at 10:13
  • $\begingroup$ The Galois group is of the Galois closure of $K$. $\endgroup$ Mar 19, 2013 at 10:14
  • $\begingroup$ Sure! That's what I meant by "If the extension is not Galois..." So in degree 4 is it true that if the K is not abelian, then the Galois group is $D_4$? What about degree 6? $\endgroup$
    – cmfield
    Mar 19, 2013 at 10:17

1 Answer 1


Let $K_0$ be totally real of degree $2,3$ and $K/K_0$ a totally imaginary quadratic extension. Let $L_0$ and $L$ be the Galois closures of $K_0$ and $K$ over ${\mathbb Q}$. Then $Gal(L)$ maps onto $Gal (L_0)$ with kernel an abelian $2$ group (a vector space over ${\mathbb F}_2$) .

So the issue is: given a totally real number field $K_0$ over ${\mathbb Q}$ (of degree $2,3$), what can its Galois group be? It is easy to see that it can be cyclic of order $2$, cyclic of order $3$ or $S_3$.

  • $\begingroup$ Thanks for your answer. Do you mean $Gal(L_0)$ instead of $Gal(K)$ in line 2? $\endgroup$
    – cmfield
    Mar 19, 2013 at 14:23
  • $\begingroup$ yes, thanks. I do mean $Gal(L_0)$.Corrected now. $\endgroup$ Mar 19, 2013 at 14:48
  • $\begingroup$ Why does the kernel have order 2? If the kernel always has order $2$, then for Galois $K_0$, the CM field $K$ seems to be Galois too. Is this true? $\endgroup$
    – k.j.
    Feb 23, 2022 at 8:34

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