In this question I asked whether, given a Galois field $K$ with Galois group $H$, it was possible to find a Galois extension $L/K/\mathbb{Q}$ of a desired degree with Galois group $G$ and $H \leq G$ in a desired way - taking into consideration the obvious degree considerations.

The answer was in the negative with counterexample a cyclotomic field. I know in most cases the answer can be yes, by taking another Galois field disjoint from the given field $K$ and then taking a compositum, but this results in a field like $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$ rather than $\mathbb{Z}/nm\mathbb{Z}$. This led me to this question:

Given a cyclotomic field $K$ (or subfield there of) with Galois group $H$, is it possible to find a Galois extension $L/K/\mathbb{Q}$ of desired degree with cyclic Galois group $G$ and $H \leq G$? Again the answer must be "no" with a counterexample given in my previous question, but I am wondering is it sometimes possible, i.e. what are the obstructions to doing this?

The case of $\mathbb{Q}(i)$ was done in the previous question with obstruction essentially being complex conjugation (though this should probably correspond to some cohomological obstruction?). The next smallest degree examples would be $\mathbb{Q}(\zeta_7)^+$ or $\mathbb{Q}(\zeta_9)^+$ in degree 3 or $\mathbb{Q}(\zeta_5)$ in degree 4. Are there such obstructions or more generally, are necessary conditions to being able to do this?

  • 2
    $\begingroup$ For $L/K/Q$ with $K/Q,L/Q$ Galois then $Gal(K/Q)$ is always a quotient of $Gal(L/Q)$. The quotients of a cyclic group are cyclic. If $K/Q$ is cyclotomic then $Gal(K/Q) $ is $\Bbb{Z/nZ}^\times$ quotiented by $\pm c$ if $n$ is even, it is cyclic iff $n = p^r$ and $L = Q(\zeta_{p^{r+s}})$ is a larger cyclotomic cyclic field. And $Gal(K/Q)$ is isomorphic to a subgroup of $Gal(L/Q)$, different from the canonical relation which is $Gal(K/Q) = Gal(L/Q)/Gal(L/K)$. $\endgroup$
    – reuns
    Jul 26, 2019 at 22:15
  • $\begingroup$ To get answers from people who really understand this it may help to use either algebraic-number-theory or class-field-theory as a tag. $\endgroup$ Jul 29, 2019 at 8:02
  • $\begingroup$ @JyrkiLahtonen Thanks for the great suggestion! $\endgroup$
    – TinyTim
    Jul 29, 2019 at 23:15

3 Answers 3


First of all, the formulation doesn't make much sense; surely you want to ask for an extension $L/\mathbf{Q}$ such that the inclusion $K \subset L$ induces a surjection

$$G = \mathrm{Gal}(L/\mathbf{Q}) \rightarrow \mathrm{Gal}(K/\mathbf{Q}) = H.$$

There is then a second ambiguity as to whether you want to fix the choice of this map (note that the last isomorphism is canonical so this makes a difference: if $K = \mathbf{Q}(\sqrt{-1},\sqrt{5})$ so $\mathrm{Gal}(K/\mathbf{Q}) = (\mathbf{Z}/2 \mathbf{Z})^2$ and if $$G = (\mathbf{Z}/2) \oplus (\mathbf{Z}/4)$$ then two of the three surjective maps will not correspond to fields $L$ but the third does. Of course the question with the map unspecified is just the question ranging over the finitely many choices of map.

The typical way to study (central) extensions is via class field theory and the Brauer Group, but since everything is in the abelian context that is not really necessary. Instead, make some reductions:

  1. Because abelian groups are canonically direct sums of their $p$-Sylow subgroups, you can immediately reduce to the case where $H$ and $G$ have $p$-power order.

  2. If $G = G' \oplus \Gamma$ where the map $G \rightarrow H$ factors through $G'$, you can replace $G$ by $G'$. That is because if $L'/K$ is an extension with Galois group $G'$, you can always find a disjoint abelian extension with Galois group $\Gamma$ and then take the compositum. From basic facts about $p$-groups (or even more basic facts about abelian groups), this means that we can assume that $G$ and $H$ have the same number of generators, and that $G/p=H/p$.

Explicitly, one should think about the case

$$G = \prod \mathbf{Z}/p^{a_i + b_i} \mathbf{Z}, H = \prod \mathbf{Z}/p^{a_i} \mathbf{Z}.$$

In the version of the problem where the map is specified, we can even reduce to the case where $G$ and $H$ are cyclic. (This requires a small lemma: if $L_i/K_i$ are the extensions, then one wants to make sure that the $L_i$ are all disjoint, since otherwise they only give rise to a group $G'$ surjecting onto $H$ with $G' \subset G$. But any non-trivial intersection of the $L_i$ would contain a field of degree $p$, but the degree $p$ subfields of their compositum correspond to quotients of $G'/p = H/p$ (since $G/p = H/p$), and thus one is OK since the $K_i$ are disjoint.

This paragraph now contains the key claim: If an $L$ exists (with $G/p = H/p$) then an $L$ exists which has the following property:

$L/\mathbf{Q}$ is only ramified at the (finite) primes where $K/\mathbf{Q}$ is ramified.

We now prove this. It suffices to consider the cyclic case $L = L_i$ and $K = K_i$, since if all the $L_i$ are unramified at $q$ then so is their compositum. So we have a surjection:

$$\mathbf{Z}/p^{a+b} \mathbf{Z} = \mathrm{Gal}(L/\mathbf{Q}) \rightarrow \mathrm{Gal}(K/\mathbf{Q}) = \mathbf{Z}/p^b \mathbf{Z}.$$

Suppose that the image of inertia at the prime $q$ has order $p^r$. That means that, for a prime $v$ above $q$, the extension $K_v/\mathbf{Q}_q$ is ramified of degree $p^r$. But local class field theory implies that every unramified extension of $\mathbf{Q}_q$ is contained in extension of $\mathbf{Q}_q$ containing all $q$-power roots of unity. Moreover, this extension can be "globalized," that is, there exists an extension $E/\mathbf{Q}$ of the same degree $p^r$ which is contained in the $q^{\infty}$ roots of unity and such that $\mathrm{Gal}(E/\mathbf{Q})$ is cyclic of order $p^r$. So now consider the compositum of $L$ and $E$. Note that $L$ and $E$ are disjoint, because any intersection would contain a degree $p$ extension, but (since $L$ is cyclic) it has a unique such extension which is contained in $K$ and $K$ is unramified at $q$. So now

$$\mathrm{Gal}(L.E/\mathbf{Q}) = \Gamma = \mathbf{Z}/p^{a+b} \mathbf{Z} \oplus \mathbf{Z}/p^{r} \mathbf{Z},$$

and the inertia group $I$ at $q$ is still cyclic of order $p^r$, because $L$ and $E$ locally give the same type of ramified extension. Moreover, the map to $\mathrm{Gal}(K/\mathbf{Q})$ is projection onto the second factor. But this map factors through the quotient by $I$ because $K$ is unramified. And, by construction, the quotient by $I$ is cyclic of order $p^{a+b}$ (generated by $[(1,0)]$). Thus we can replace $L$ by an extension $L'$ which is ramified at exactly the same primes except for $q$. Repeat with all primes where $L$ is ramified by $K$ is not. This is the crossing with a cyclic extension which gets used in many places, in particular in the proof of the Kronecker-Weber theorem.

Now using the Kronecker-Weber theorem, if $K$ is ramified only at primes dividing $N$, then $K \subset \mathbf{Q}(\zeta_N)$ if $(N,p) = 1$ or $\mathbf{Q}(\zeta_M,\zeta_{p^{\infty}})$ if $N=Mp$. Hence $H$ is canonically a quotient of $$\mathbf{Z}^{\times}_p \times \prod_{q|M} (\mathbf{Z}/q \mathbf{Z})^{\times}$$ if $N = pM$ or $$\prod_{q|M} (\mathbf{Z}/q \mathbf{Z})^{\times}$$ otherwise.

So now for $G$ (with the same number of generators) to arise, it is sufficient and necessary that $G$ is also a quotient of this group. (Clearly you can specify the map to $H$ or not if you like.) Of course, you need to understand cyclotomic fields enough to understand what the specific map from the group above to $H$ is (slightly annoying in the case when $p = q = 2$.

Examples 1: $H$ cyclic of order $p$, and $G$ cyclic of order $p^{2}$.

Then such an $L$ exists if and only if all of the following are satisfied.

  1. If $K$ is ramified at a prime $q \ne p$, then $q \equiv 1 \mod p^2$.
  2. If $K$ is ramified at $p$, then $K \otimes \mathbf{Q}_2$ is a ramified quadratic extension of $\mathbf{Q}_2$. The complete list of quadratic extensions is $\mathbf{Q}_2(\sqrt{d})$ with

$$d = -1, -5, 2, 10, -2, -10, 5.$$

The last one is unramified and so doesn't occur in this context, and otherwise the condition is that the localization of $K$ at the prime $2$ does not contain (equal) the cases $d = -1, -5, -2, -10$, i.e. it is $d = 2$ or $d = 10$.

Example 2: If $H$ is cyclic of order $p^a$ and $G$ is cyclic of order $p^{a+b}$ with $b > 0$ (pretty much the general case) then you win if and only if:

  1. If $K$ is ramified at a primes $q \ne p$ with $e_q = p^{r}$, then $q \equiv 1 \mod p^{r+b}$.
  2. If $K$ is ramified at $p$, and $p = 2$, then the localization of $K$ at the prime $2$ does not contain the quadratic fields corresponding to $d = -1,-5, -2, -10$.

Anything but a full answer, but posting my way of handling the two cubic extension fields that you listed. I also think it sheds a bit more light on the more general question.

Any finite abelian group is a direct product of its $p$-subgroups. I think (?) that the components with $p\neq3$ won't pose a problem for the cubic extension as we are forced to have those compositums anyway.

Claim 1. There does not exist a field $L$ containing $K=\Bbb{Q}(\zeta_7+\zeta_7^{-1})$ such that $L/\Bbb{Q}$ is cyclic Galois of degree nine.

Proof. Assume contrariwise that such an $L$ exists. By Kronecker-Weber $L$ is contained in a cyclotomic field $\Omega=\Bbb{Q}(\zeta_M)$. Clearly $7\mid M$, and we can write $M=7^am$ with $a\ge1$ and $\gcd(7,m)=1$. The group $\Bbb{Z}_{7^m}^*$ is cyclic of order $2\cdot3\cdot7^{a-1}$ and thus has a unique cyclic subgroup of order three. An application of the Chinese Remainder Theorem tells that there exists an integer $u$ (unique modulo $M$) such that $$ \begin{aligned} u&\equiv2\pmod 7,\\ u^3&\equiv1\pmod {7^a},\\ u&\equiv1\pmod m. \end{aligned} $$ Consider the automorphism $\sigma$ of $\Omega$ determined by $\zeta_M\mapsto\zeta_M^u$. We can use the automorphism $\sigma$ the same way Lord Shark the Unknown used complex conjugation in his answer to your earlier question. By basic Galois theory $\sigma(L)=L$, and $\sigma$ has order three. Because the restriction of $\sigma$ to $K$ is non-trivial, its restriction to $L$ has order three as well. As $Gal(L/\Bbb{Q})$ is cyclic of order nine, it follows that $\sigma\vert_L=\tau^3$ for some automorphism $\tau$ of $L$. But $Gal(K/\Bbb{Q})$ is cyclic of order three, implying that the restriction of $\tau^3$ to $K$ must be the identity. A contradiction.

On the other hand

Claim 2. For any power of three $3^a$, $a>0$, the field $K=\Bbb{Q}(\zeta_9+\zeta_9^{-1})$ is contained in a cyclic extension of degree $3^a$.

Proof. The real subfield of $\Bbb{Q}(\zeta_{3^{a+1}})$ works.

Further remarks:

  • The method of my proof of Claim 1 fails in the situation of Claim 2 because all the lifts of automorphisms of $\Bbb{Q}(\zeta_9)$ of order three to automorphisms of $\Bbb{Q}(\zeta_{27})$ have orders divisible by nine. Considering their orders modulo the cyclic subgroup generated by complex conjugation does not change this.
  • I think that compositions will allow us to confirm your hunch (in the case of Claim 2) also when the $3$-part of $G$ is non-cyclic.
  • Similar obstructions seem to occur for a subfield of $\Bbb{Q}(\zeta_p)$, $p$ a prime, when a prime factor of $p-1$ is involved in an essential way. Leaving as an exercise the proof of a non-existence of a field $L$ containing $\Bbb{Q}(\zeta_{11}+\zeta_{11}^{-1})$ such that $L/\Bbb{Q}$ is cyclic of degree $25$ :-)
  • I don't know of a cohomological interpretation of this. Probably those are buried in the application of Kronecker-Weber.

As already noticed in previous answers, the latest version of your question can be reduced to the following embedding problem (EP) : given a Galois extension of number fields $K/k$ with cyclic group H, given a cyclic group G of which H is a quotient, produce a criterion for the existence of a Galois extension L/K/k with group $\cong G$ s.t. the surjection $G\to H$ coincides with the natural surjection $Gal(L/k)\to Gal(K/k)$. As explained below, this can be solved using the classical methods of EP, at least theoretically.

Recall that over arbitrary fields, given a group extension $1\to A\to G\to H\to 1$ and an isomorphism $\gamma: H\cong Gal(K/k)$, the strong EP consists in constructing (if possible) a tower of Galois extensions $L/K/k$ with an isomorphism $\beta: G\cong Gal(L/k)$ such that, in the commutative diagram obtained by linking the two horizontal exact sequences $1\to Gal(L/K)\to Gal(L/k) \to Gal(K/k) \to 1$ and $1\to A\to G\to H\to 1$ using $\beta$ and $\gamma$, the induced homomorphism $\alpha: A \to Gal(K/k)$ is an isomorphism. The EP is called weak if $L/k$ is only required to be a Galois algebra in the sense of Hasse. In the case - to which we'll stick from here - of an abelian kernel A, the previous group extension is described by a 2-cocycle class $\epsilon\in H^2(H, A)$ and the corresponding EP is denoted $(K/k, A, \epsilon)$. A criterion of Hoechsmann (see [H]) states that $(K/k, A, \epsilon)$ admits a weak solution iff inf $(\epsilon)=0$, where inf is the inflation map $H^2(H, A)\to H^2(G_k, A)$, where $G_k$ is the absolute Galois group of $k$. If $k$ is a number field, a weak solution exists iff a strong solution exists (Ikeda's theorem, see [I]). But this property is not available for local fields $k_v$. In compensation, using local cohomological duality, Neukirch has expressed Hoechsmann's criterion in terms of characters. For a question of notations, it will be more convenient to explain the local approach via characters in [N] after explaining Poitou's generalization to global fields. The starting point in [P] is the introduction of the idèle class group $C_K$ of a number field K , which has the same cohomological properties as the multiplicative group ${K_v}^*$ of a local field. In particular, $H^2(H, C_K)$ injects canonically into $H^2(G_k, C)$, where $C$ is the direct limit of the $C_K$'s. For any character $\chi:A\to C$ defined over $k$ (i.e invariant under the action of the absolute Galois group of $k$), the values of $\chi$ can in particular be viewed in $C_K$. This allows to define the image $\chi^*(\epsilon)$ of $\epsilon$ in $H^2(H, C_K)$. Using global cohomological duality, Poitou has shown that Hoechsmann's criterion, inf$(\epsilon)=0$, is equivalent to the nullity of the $\chi^*(\epsilon)$ for all the characters $\chi:A\to C$ defined over $k$. This is the exact global analogue of Neukirch's local conditions inf$(\epsilon_v)=0$.

Summarizing : Moreover, using the definition of the idèles and the idèle classes, it can be shown that the natural map $\pi^*:Hom_k(A,I)\to Hom_k(A,C)$, where $I$ is the direct limit of the idèle groups, has a finite cokernel. Hence the EP $(K/k, A, \epsilon)$ over number fields admits a strong solution iff the local Neukirch conditions are fulfilled and the $\chi^*(\epsilon)$ vanish for a (finite) set of $\chi$ representing the cokernel of $\pi^*$.

It remains to apply the above theoretical considerations to your question. As already noticed, one can consider only the case where $G$ and $A$ are cyclic $p$-groups. This has been done in detail in [P], at the end of §§2 and 3. Two major simplifications occur : first $H^2(H,A)\cong A$ if A is central; second, there is at most one global condition. The complete answer is quite accessible, except in the case when $A$ is a cyclic 2-group. This "special case" corresponds to the difficulties classically encountered in the so called "Grunwald-Wang theorem", see Artin-Tate, CFT, chap.10 ./.

[H] K. Hoechsmann, Zum Eibettungsproblem, J. reine angew. Math., 229 (1968), 81-106

[I] M. Ikeda, Zur Existenz eigentlicher galoisscher Körper beim Einbettungsproblem, Hamb. Abh., 24 (1960),126-121

[N] J. Neukirch, Über das Einbettungsproblem der algebraischen Zahlentheorie, Inventiones Math., 21 (1973), 59-116

[P] G. Poitou, Conditions globales pour les problèmes de plongement à noyau abélien, Ann. Inst. Fourier, 29,1 (1979), 1-14

  • $\begingroup$ There is one thing which might be obscured by this answer: as a easy consequence of the (global and local) Kronecker-Weber theorem (as explained in my answer), all obstructions to lifting one cyclic abelian extension of $\mathbf{Q}$ to another are local (that is, there is no Grunewald-Wang phenomena). This is an important point which makes it possible to write down a fairly simple general answer. This is something that one could miss by passing immediately to the general theory. $\endgroup$
    – user687721
    Aug 5, 2019 at 22:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .