[Math] Galois groups of CM fields

nt.number-theory

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!

Best Answer

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 $S_3$.

Related Solutions

[Math] What are the roots of unity in abelian extensions of imaginary quadratic fields

Just as a minor warning: even if the conductor is $1$, there might be nontrivial roots of unity in the class field: take $K = {\mathbb q}(\sqrt{-5}\,)$ and ${\mathfrak c} = (1)$; then the ray class field is the Hilbert class field $K(\sqrt{-1})$, which contains the 4th roots of unity. The roots of unity in the Hilbert class field (i.e. for conductor $1$) lie in the genus class field and can be computed easily.

Any additional roots of unity must come from ramified extensions; a necessary condition for the $p$-th roots of unity to lie in the ray class field must be that the ry class number, which is easily computed, be divisible by $p-1$ (or $(p-1)/2$ if the genus class field contains the quadratic subfield of the $p$-th roots of unity).

Degree 17 Number Fields Ramified Only at 2

Thank you for calling this problem to my attention. I computed $K$ en route to AWS (though this year's topics are a rather different flavor of number theory...). After some simplification (gp's $\rm polredabs$), it turns out that the field $K$ is generated by a root of $$ f(x) = x^{17} - 2x^{16} + 8x^{13} + 16x^{12} - 16x^{11} + 64x^9 - 32x^8 - 80x^7 $$ $$ \qquad\qquad\qquad {} + 32x^6 + 40x^5 + 80x^4 + 16x^3 - 128x^2 - 2x + 68. $$ gp quickly confirms:

p = Pol([1,-2,0,0,8,16,-16,0,64,-32,-80,32,40,80,16,-128,-2,68])
poldegree(p)
F = nfinit(p);
factor(F.disc)

returns

[2 79]

So the field has discriminant $2^{79}$.

Let $F = {\bf Q}(\zeta_{64}^{\phantom{0}} - \zeta_{64}^{-1})$, with $F/\bf Q$ cyclic of degree $16$ of $\bf Q$. It was known that $K$ is contained in an unramified cyclic extension $L/F$ of degree $17$. Hence $L(\zeta_{17})$ is a Kummer extension of $F_{17} := F(\zeta_{17})$. Now $F_{17}$ is a $({\bf Z} / 16 {\bf Z})^2$ extension of $\bf Q$. Fortunately it was not necessary to work in the unit group of this degree $256$ field: the Kummer extension is $F_{17}(u^{1/17})$ with $u \in F_{17}^* / (F_{17}^*)^{17}$ in an eigenspace for the action of ${\rm Gal}(F_{17}/{\bf Q})$, which makes it fixed by some index-$16$ subgroup of this Galois group. Thus $u$ could be found in one of the cyclic degree-$16$ extension of $\bf Q$ intermediate between $\bf Q$ and $F_{17}$, and eventually a generator of $K$ turned up whose minimal polynomial, though large, was tractable enough for gp's number-field routines to do the rest.

Best Answer

Related Solutions

[Math] What are the roots of unity in abelian extensions of imaginary quadratic fields

Degree 17 Number Fields Ramified Only at 2

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