Genus zero translates of three point ramified Galois extensions

Gunter Malle^{1}

Let N/k(t) be a regular Galois extension ramified at three points, with k/Q a number field, and L/k(t) another regular field extension, not necessarily Galois, ramified at the same points. If L is a rational function field k(u), then the translated field extension LN/L is again a Galois extension of a field for which the Hilbert irreducibility theorem holds, with Galois group H ≤ Gal(N/k(t)). In this paper for certain special cases of L/k(t) we derive sufficient conditions for the translated Galois extension to be defined over a proper subfield of k. These may be used to obtain regular Galois extensions over Q(t) for groups L_{2}(p) with primes p ≡ 2, 3 (mod 5). This includes an infinite set of primes p for which these groups were not previously known to occur as Galois groups over Q. Also, the Galois realizations obtained by Shih [2] in the setting of the modular group and using Shimuras theory of canonical systems of models are found quite naturally in our development.

1. Three point ramified genus zero translates

Let M̄/Q̄(t) be the maximal field extension of Q̄(t) unramified outside of three points p_{1}, p_{2}, p_{3} ∈ Q. Let further N̄/Q̄(t) be a finite Galois extension with group G = Gal(N̄/Q̄(t)). Denote by k the field of definition of that Galois extension; so there exists a regular Galois extension N/k(t) with Gal(N/k(t)) = G and NQ̄ = N̄. We will consider the following question: Let L̄/Q̄(t) be another subfield of M̄/Q̄(t), not necessarily Galois, and assume that this extension is also defined over k, namely there exists L/k(t) regular with LQ̄ = L̄. Then LN/L is also Galois, with the Galois group H = Gal(LN/L) ≤ G. More precisely, we have H = G iff L ∩ N = k(t). Using this construction, we want to understand a bit more about the field of definition of the translated Galois extension LN/L, and derive new rationality criteria.

A special case of this situation played an important role in Shih’s proof [2] that L_{2}(p) is a Galois group over Q(t) for primes p with one of (\frac{2}{p}) = −1,

1 The author gratefully acknowledges financial support by the Deutsche Forschungsgemeinschaft

(\frac{3}{p}) = −1 or (\frac{7}{p}) = −1. He only considered the case that L and N come from congruence subgroups of the modular group and he used deep results from Shimuras theory of canonical systems of models to derive his statement. It turns out that his results can be proved without using the context of modular groups.

Also, Matzat defined the notion of ‘verwandte Erzeugendensysteme’ in [1],III, §3, which is exactly the situation described above, and derived some results for the action of topological automorphisms, which also can be found as special cases of the development in this work.

As we are interested in constructing Galois extensions of number fields, we shall specialize the above situation as follows. The new Galois extension LN/L should still be an extension of a rational function field, hence L should have genus zero. The occurences of regular field extensions L/k(t) with L of genus zero were recently investigated by Guralnick and Thompson. But we will only be interested in cases where LN/L is ramified in few points (in three or four points, to be precise), because these seem to lead to interesting applications. So we will not have to appeal to the conclusions of Guralnick and Thompson.

Now if A denotes the Galois group of the Galois closure L̂ of L̄/Q̄(t), then we obtain a permutation representation of A on the cosets of Gal(L̂/L̄) of degree [L̄ : Q̄(t)] = (A : Gal(L̂/L̄)). By [1] it is possible to choose topological generators σ_{1}, σ_{2} and σ_{3} of the inertia subgroups of M̄/Q̄(t) at the ramified points p_{1}, p_{2}, p_{3}, such that σ_{1}σ_{2}σ_{3} = ι and σ_{1}, σ_{2}, σ_{3} generate Gal(M̄/Q̄(t)) as a profinite group. Hence G and also A are generated by the images of σ_{1}, σ_{2}, σ_{3} in the respective Galois groups.

Via the permutation representation of A on the cosets of Gal(L̂/L̄), the images σ̄_{i} in A act as permutations on d := [L : k(t)] = [L̄ : Q̄(t)] points. Then the cycles in a disjoint cycle decomposition of σ̄_{i} correspond bijectively to the points of L̄ lying over p̄_{i}, with ramification index equal to the length of the respective cycle (see [1], for example).

It is well known that L̄N/L̄ is ramified only in points lying over the ramified points p_{1}, p_{2}, p_{3} of N̄/Q̄(t). Moreover, if P_{i} lies over p_{i} with ramification index l_{i} in L/k(t), and e_{i} is the ramification index of p_{i} in N̄/Q̄(t), then the ramification index of P_{i} in L̄N/L̄ is equal to e_{i}/gcd(l_{i}, e_{i}). Hence it is possible to explicitely determine the ramification indices in LN/L from those of N/k(t) and L/k(t). We will see later how one can even find the class of elements of H which corresponds to LN/L by the Hurwitz classification [1], modulo conjugation and possibly the action of the braid group.

In particular, L̄N/L̄ ramifies precisely at those points P_{i} with e_{i} > gcd(l_{i}, e_{i}). First we are going to determine all possible L̄/Q̄(t) with the property that this happens at no more than three points. If the ramification indices e_{1}, e_{2}, e_{3} of N/k(t) satisfy 1/e_{1} + 1/e_{2} + 1/e_{3} ≥ 1 then we consider the group G not to be interesting, as it is then either isomorphic to one of Z_{n}, D_{n}, A_{4}, S_{4}, A_{5}, or it has a normal abelian subgroup of index at most six. Hence we will moreover assume ∑ 1/e_{i} < 1.

By the Hurwitz relative genus formula the genus g of L̄ satisfies

2g − 2 = d − #{cycles in σ̄_{1}, σ̄_{2}, σ̄_{3}}.

So if we want g = 0, this gives #cycles= d + 2. But we also require that L̄N/L̄ ramifies at exactly three points. So by the remark about the ramification indices, apart from exactly three cycles in σ̄_{1}, σ̄_{2}, σ̄_{3}, all cycle lengths in σ̄_{i} have to be multiples of e_{i}. Let us assume for the moment that all cycles apart from these three cycles indeed have length equal to e_{i}. Then the number of cycles in σ̄_{1}, σ̄_{2}, σ̄_{3} is equal to

\frac{d − n_{1}}{e_{1}} + \frac{d − n_{2}}{e_{2}} + \frac{d − n_{3}}{e_{3}} + 3,  (1)

where for each σ̄_{i} we have first subtracted the contribution n_{i} from the possible exceptional cycles (so n_{i} ≥ 0, ∑ n_{i} ≥ 3), and then added three to make up for these three cycles. So we have to find all solutions (d, n_{1}, n_{2}, n_{3}, e_{1}, e_{2}, e_{3}) in nonnegative integers of

d + 2 = \frac{d − n_{1}}{e_{1}} + \frac{d − n_{2}}{e_{2}} + \frac{d − n_{3}}{e_{3}} + 3,  (2)

where ∑ 1/e_{i} < 1.

Proposition 1. All possible solutions of the equation (2) under the restrictions d > 1, e_{1} ≤ e_{2} ≤ e_{3}, ∑ 1/e_{i} < 1, ∑ n_{i} ≥ 3 are given by the list of Table 1. Here, a sign ∗ for e_{i} means that e_{i} may take any positive integer value respecting the above restrictions.

Table 1. Solutions to (2)

\begin{matrix}d & n_{1} & n_{2} & n_{3} & e_{1} & e_{2} & e_{3} \\ 2 & 0 & 2 & 2 & 2 & ∗ & ∗ \\ 3 & 1 & 0 & 3 & 2 & 3 & ∗ \\ 3 & 0 & 0 & 3 & 3 & 3 & ∗ \\ 4 & 0 & 1 & 4 & 2 & 3 & ∗ \\ 4 & 0 & 0 & 4 & 2 & 4 & ∗ \\ 6 & 0 & 0 & 6 & 2 & 3 & ∗ \\ 6 & 0 & 2 & 1 & 2 & 4 & 5 \\ 8 & 0 & 2 & 1 & 2 & 3 & 7 \\ 8 & 0 & 0 & 3 & 2 & 4 & 5 \end{matrix}  \begin{matrix}d & n_{1} & n_{2} & n_{3} & e_{1} & e_{2} & e_{3} \\ 9 & 1 & 0 & 2 & 2 & 3 & 7 \\ 10 & 0 & 1 & 3 & 2 & 3 & 7 \\ 10 & 0 & 1 & 2 & 2 & 3 & 8 \\ 12 & 0 & 0 & 5 & 2 & 3 & 7 \\ 12 & 0 & 0 & 4 & 2 & 3 & 8 \\ 12 & 0 & 0 & 3 & 2 & 3 & 9 \\ 16 & 0 & 1 & 2 & 2 & 3 & 7 \\ 18 & 0 & 0 & 4 & 2 & 3 & 7 \\ 24 & 0 & 0 & 3 & 2 & 3 & 7 \end{matrix}

Proof. We can transform (2) to get

(1 − \frac{1}{e_{1}} − \frac{1}{e_{2}} − \frac{1}{e_{3}} )d = 1 − \frac{n_{1}}{e_{1}} − \frac{n_{2}}{e_{2}} − \frac{n_{3}}{e_{3}} .

It is well known that the minimum of 1 − ∑ 1/e_{i} for ∑ 1/e_{i} < 1 is attained for (e_{1}, e_{2}, e_{3}) = (2, 3, 7), so the above equality forces

\frac{1}{42}d ≤ 1 − \frac{n_{1}}{e_{1}} − \frac{n_{2}}{e_{2}} − \frac{n_{3}}{e_{3}} .

Visibly this bounds the degree d, and hence we get a finite number of solutions (d, n_{1}, n_{2}, n_{3}) (but, clearly, if n_{i} = d, there is no restriction to e_{i}). It is now a matter of calculation or of a computer program to completely determine the possible values. ⊓ ⊔

For the argument leading to Proposition 1 we had to assume that all cycles in each σ̄_{i}, apart from the three extra cycles, have length equal to e_{i}. Now if one of the lengths were a proper multiple of e_{i}, at least two cycles in formula (1) would fuse, the rest staying unchanged. Then formula (2) would read

d ≤ \frac{d − n_{1}}{e_{1}} + \frac{d − n_{2}}{e_{2}} + \frac{d − n_{3}}{e_{3}} ,

so 1/42d ≤ 0 by the estimate in the proof of the Proposition, which is absurd. Hence the triples in Table 1 in fact give all solutions for our original question. We next investigate to which field extensions L̄/Q̄(t) the solutions in Table 1 give rise by the Hurwitz classification.

Table 2. Cycle description of the Galois group for possible translation fields

d  cycle description of σ̄_{i}  A  e_{1} − e_{2} − e_{3}  remarks
2  (2) − (\underline{2}) − (\underline{1})^{2}  Z_{2}  2 − ∗ − ∗  regular
3  (3) − (3) − (\underline{1})^{3}  Z_{3}  3 − 3 − ∗  regular
3  (2)(\underline{1}) − (3) − (\underline{2})(\underline{1})  S_{3}  2 − 3 − ∗  [2]
4  (2)^{2} − (3)(\underline{1}) − (\underline{3})(\underline{1})  A_{4}  2 − 3 − ∗  [2]
4  (2)^{2} − (4) − (\underline{2})(\underline{1})^{2}  D_{4}  2 − 4 − ∗  imprimitive
6  (2)^{3} − (3)^{2} − (\underline{4})(\underline{1})^{2}  S_{4}  2 − 3 − ∗  imprimitive
6  (2)^{3} − (3)^{2} − (\underline{2})^{3}  S_{3}  2 − 3 − ∗  imprimitive
6  (2)^{3} − (4)(\underline{1})^{2} − (5)(\underline{1})  P GL_{2}(5)  2 − 4 − 5
8  (2)^{4} − (3)^{2}(\underline{1})^{2} − (7)(\underline{1})  L_{2}(7)  2 − 3 − 7
9  (2)^{4}(\underline{1}) − (3)^{3} − (7)(\underline{1})^{2}  L_{2}(8)  2 − 3 − 7
10  (2)^{5} − (3)^{3}(\underline{1}) − (7)(\underline{2})(\underline{1})  S_{10}  2 − 3 − 7
10  (2)^{5} − (3)^{3}(\underline{1}) − (8)(\underline{1})^{2}  P GL_{2}(9)  2 − 3 − 8
12  (2)^{6} − (3)^{4} − (8)(\underline{2})(\underline{1})^{2}  2^{2}.S_{4}  2 − 3 − 8  imprimitive
12  (2)^{6} − (3)^{4} − (9)(\underline{1})^{3}  3^{3}.A_{4}  2 − 3 − 9  imprimitive
18  (2)^{9} − (3)^{6} − (7)^{2}(\underline{2})(\underline{1})^{2}  2^{9}.L_{2}(8)  2 − 3 − 7  imprimitive
24  (2)^{12} − (3)^{8} − (7)^{3}(\underline{1})^{3}  L_{2}(7)  2 − 3 − 7  imprimitive

Proposition 2. The possible cycle shapes for generators σ̄_{1}, σ̄_{2}, σ̄_{3} of three point ramified genus zero extensions belonging to the parameters in Proposition 1 are as given in Table 2.

The underlined cycles are the three extra cycles corresponding to points at which ramification is going to occur in L̄N/L̄. The mention of ‘regular’ or ‘imprimitive’ in the last column indicates that the permutation representation for A is regular or imprimitive, so that the field L̄ is Galois over Q̄(t) in the first case, while in the second it can be obtained by composing two smaller extensions of the above type. Consequently, the latter case is not interesting for us.

Proof. By construction, the parameters in Table 1 already give the cycle shapes of σ̄_{1}, σ̄_{2} and σ̄_{3} up to the three extra cycles, and the latter are almost determined by the values of the n_{i}’s. Namely, σ̄_{i} has (d − n_{i})/e_{i} cycles of length e_{i} plus some cycles of lengths adding up to n_{i}. So from Table 1 we can easily write down a list of admissible cycle decompositions for the σ̄_{i}. Next, for each triple one has to determine whether there exists a transitive permutation group of the given degree d with three generators having that cycle description. This is easy to do for groups up to degree d = 10. I would like to thank Richard Parker for showing me a combinatorial method to complete the cases for 12 ≤ d ≤ 24. It turns out that whenever such a group exists (which is not always the case; for example there exists no transitive group for (2)^{4} − (4)^{2} − (5)(1)^{3}), generated by a triple of elements of S_{d} with required shapes, there is just one such triple up to conjugation in S_{d}. By [1] this means there exists a unique extension L̂/Q̄(t) with Galois group A and this branch cycle description, which is hence even defined over k = Q. So Table 2 is obtained. ⊓ ⊔

Now we will restrict our attention to those cases where at least one of the e_{i} is arbitrary. In particular, we want to understand the translated Galois extension LN/L in terms of the Hurwitz classification [1]. For this, canonical generators of the inertia subgroups over the ramification points of M̄/L̄ have to be determined. We will find these via the topological description of the Galois group of M̄/Q̄(t) as profinite completion of the fundamental group of a sphere with three points removed. In that language the Galois group of M̄/L̄ is the profinite completion of the fundamental group of a sphere with a finite number of points removed, the removed points corresponding to the cycles in the cycle decomposition of σ̄_{1}, σ̄_{2}, σ̄_{3}. This is so because L̄/Q̄(t) has genus zero, so is isomorphic to a rational function field Q̄(u) and corresponds topologically to a sphere. Canonical generators of that fundamental group can now be found as simple loops on the sphere belonging to C(u) around the ramification points. These may then be expressed in terms of the original loops for C(t), lifted to C(u), hence in terms of the σ_{i}. We get

Proposition 3. Let σ_{1}, σ_{2}, σ_{3} be the topological generators of the inertia subgroups of the Galois group Π = Gal(M̄/Q̄(t)) with σ_{1}σ_{2}σ_{3} = ι. Then topological generators for the inertia subgroups of Gal(M̄/L̄) with a genus zero extension L̄/Q̄(t) described by a triple of permutations as in Proposition 2 can be chosen as in Table 3.

Table 3. Canonical generators of the translated fundamental group

d  cycle dec.
2 (2) − (\underline{2}) − (\underline{1})^{2}  σ_{1}^{2}, σ_{2}^{2}, σ_{2}^{−1}σ_{3}σ_{2}, σ_{3}
3 (3) − (3) − (\underline{1})^{3}  σ_{1}^{3}, σ_{2}^{3}, σ_{2}^{−2}σ_{3}σ_{2}^{2}, σ_{2}^{−1}σ_{3}σ_{2}, σ_{3}
3 (2)(\underline{1}) − (3) − (\underline{2})(\underline{1})  σ_{1}^{2}, σ_{2}^{−1}σ_{1}σ_{2}, σ_{3}σ_{2}^{3}σ_{3}^{−1}, σ_{3}^{2}, σ_{1}σ_{3}σ_{1}^{−1}
4 (2)^{2} − (3)(\underline{1}) − (\underline{3})(\underline{1}) σ_{1}^{2}, σ_{1}^{−1}σ_{3}σ_{1}^{2}σ_{3}^{−1}σ_{1}, σ_{2}^{3}, σ_{2}^{−2}σ_{3}^{2}σ_{2}σ_{3}^{−2}σ_{2}^{2}, σ_{2}^{−2}σ_{3}^{3}σ_{2}^{2}, σ_{3}

Proof. As was already pointed out above, the new generators are easily determined by considering loops on the C(u)-sphere covering the C(t)-sphere. The pictures necessary to determine the right loops are omitted. It should be remarked that the generators are already chosen to have product equal to the identity, as is always possible. ⊓ ⊔

With the above Proposition we are able to identify the translated Galois extension LN/L in terms of the Hurwitz classification up to a possible action of the braid group. Namely, the generating triple of elements of H = Gal(LN/L) up to conjugation by inner automorphisms of H can be read of from Table 3 by taking into acount the restrictions on the ramification orders e_{i} in Gal(N/k(t)). Obviously the field extensions obtained in this way can not be algebraically conjugate to other Galois extensions with group H which do not occur as translates in the same way. So this gives some information on orbits under algebraic automorphisms of Galois extensions parametrized by generating triples of elements via the Hurwitz classification. Actually the first two cases of Table 3 were already investigated by Matzat in [1], and the resulting translated extensions correspond exactly to generating triples which are fixed points under the topological automorphisms (12) or (123) as defined in [1]. The other two cases were already used by Shih in the special context that both field extensions correspond to congruence subgroups of the modular group. Our results are more general, as they also apply to noncongruence subgroups.

Following [1] we define for a tuple C = (C_{1}, …, C_{r}) of conjugacy classes of a group G

Σ(G, C) := {(σ_{1}, …, σ_{r}) ∈ G^{r} | σ_{i} ∈ C_{i}, σ_{1} ⋯ σ_{r} = ι, ⟨σ_{1}, …, σ_{r}⟩ = G}

and

Σ^{i}(G, C) := Σ(G, C) mod Inn G.

Proposition 4. Let G be a finite group with trivial center. For each row j in Table 3 for which G does not have a proper subgroup of index at most d and each triple C_{j} = (C_{1}, C_{2}, C_{3}) of classes of G containing elements of orders e_{1}, e_{2}, e_{3} as in the specified row and Σ(G, C_{j}) nonempty, there is an injection

ϕ_{j} : Σ^{i}(G, C_{j}) → Σ^{i}(G, C′_{j}),

where C′_{j} is defined to consist of the classes of the images given explicitely below for each of the rows of Table 3:

(a) ϕ_{1}([σ_{1}, σ_{2}, σ_{3}]) = [σ_{2}^{2}, σ_{3}, σ_{1}^{−1}σ_{3}σ_{1}],

(b) ϕ_{2}([σ_{1}, σ_{2}, σ_{3}]) = [σ_{3}, σ_{2}σ_{3}σ_{2}^{−1}, σ_{2}^{2}σ_{3}σ_{2}^{−2}],

(c) ϕ_{3}([σ_{1}, σ_{2}, σ_{3}]) = [σ_{1}, σ_{2}^{−1}σ_{3}σ_{2}, σ_{3}^{2}],

(d) ϕ_{4}([σ_{1}, σ_{2}, σ_{3}]) = [σ_{2}, σ_{3}σ_{2}^{−1}σ_{3}σ_{2}σ_{3}^{−1}, σ_{3}^{3}].

Proof. It is clear that conjugate triples are sent to the same image by ϕ_{j}, so the maps ϕ_{j} are well defined if we can prove that the images also generate G. To see that, remark that G is isomorphic to Π/R, the factor group of the free profinite group Π on two generators by the relation subgroup R. Then H, the group generated by the images, is isomorphic to Π′/(R ∩ Π′), where Π′ is generated by the elements in the right hand column of Table 3 and (Π : Π′) ≤ d. By the isomorphism theorem, H ≅ Π′R/R ≤ G, with equality if G has no proper subgroups of index at most d, for example.

It remains to prove the injectivity. For this we construct a cross section for ϕ_{j}. In case (a), the image [σ_{2}^{2}, σ_{3}, σ_{1}^{−1}σ_{3}σ_{1}] is a fixed point under the topological automorphism η corresponding to (23) in the sense of [1], III, §2. Namely, the image of that representative under η is (σ_{2}^{2}, σ_{3}σ_{1}^{−1}σ_{3}σ_{1}σ_{3}^{−1}, σ_{3}); conjugate this representative by σ_{2} to get back (σ_{2}^{2}, σ_{3}, σ_{1}^{−1}σ_{3}σ_{1}), because σ_{1}^{2} = ι. Hence the topological automorphism acts as the inner automorphism σ_{2}, and because the center Z(G) is trivial, this uniquely determines σ_{2}. So from any representative of the image we can recover the preimage unambigiously.

The same applies in case (b). Here the image is a fixed point under the topological automorphism corresponding to a 3-cycle, and this again determines a unique element σ_{2} in G effecting the topological automorphism. The first two cases were already implicit in [1], III, §2.4, but injectivity of ϕ_{j} was not studied there.

In the remaining cases, the situation is a bit more complicated. Here the images are not in general fixed points under topological automorphisms. But we can compose ϕ_{j} with a second map ψ_{j} so that ψ_{j} ∘ ϕ_{j}([\underline{σ}]) is a fixed point under a group of topological automorphisms, and then apply the reasoning as above. If ψ_{j} ∘ ϕ_{j} is injective, then certainly ϕ_{j} is. In case (c), ψ_{3} := ϕ_{1}, i.e., ψ_{3}(ϕ_{3}([\underline{σ}])) = [σ_{2}^{−1}σ_{3}^{2}σ_{2}, σ_{3}^{2}, σ_{1}^{−1}σ_{3}^{2}σ_{1}] and this is fixed under the topological 3-cycle, which acts as the inner automorphism σ_{2}. This again gives us a cross section and proves injectivity. Also for (d), we translate by the second entry of Table 3, to get [σ_{2}σ_{3}^{3}σ_{2}^{−1}, σ_{3}^{3}, σ_{3}^{−2}σ_{2}σ_{3}^{3}σ_{2}^{−1}σ_{3}^{2}, σ_{3}^{−1}σ_{2}σ_{3}^{3}σ_{2}^{−1}σ_{3}]; this class is fixed under the involution (12)(34), acting like the inner automorphism σ_{1}. This establishes injectivity in the last case, proving the Proposition. ⊓ ⊔

The preceding Proposition can immediately be applied to fields of definition of Galois extensions. For this let G be a finite group with trivial center, C = (C_{1}, …, C_{r}) a class structure. Denote by k^{i} the abelian field extension of Q generated by the irrationalities of the irreducible characters of G on the classes C_{i} in C. Define the map N_{S} : [\underline{σ}] ↦ L\underline{σ} from Σ^{i}(G, C) to subfields L\underline{σ} of the maximal extension of C(t) unramified outside S = {p_{1}, …, p_{r}}, with Galois group G, via the Hurwitz classification as in [1], II, §1. Then Proposition 4 gives an injection from N_{S}(Σ^{i}(G, C)) to N_{S′}(Σ^{i}(G, C′)), where we choose a fixed ordering for the points S′ lying above those of S. Clearly the fields of definition of the images are contained in those of the preimages. More precisely we get :

Theorem 1. Let G be a finite group with trivial center, C = (C_{1}, C_{2}, C_{3}) a class structure. Then for each j = 1, …, 4 there is a disjoint decomposition

Σ^{i}(G, C) = ˙∪_{k} ϕ_{j}(Σ^{i}(G, C_{j,k})) ˙∪ R_{j},  (∗)

where the C_{j,k} run over all class structures of G with Σ^{i}(G, C_{j,k}) ∋ [\underline{τ}] ↦ ϕ_{j}([\underline{τ}]) ∈ Σ^{i}(G, C), and R_{j} contains those classes which can not be obtained as images under ϕ_{j}. This decomposition carries over to the fields L\underline{σ} ∈ N(Σ^{i}(G, C)) in the sense that the degree over k^{i} of a field of definition of any L\underline{σ} is at most the cardinality of the subset containing [\underline{σ}] in the decomposition (∗).

Proof. The first part is just Proposition 4, with the disjointness following from the explicit construction of preimages in the proof. The second part is an application of the Hurwitz classification. The Galois extensions L\underline{σ} belonging to [\underline{σ}] = ϕ_{j}([\underline{τ}]), [\underline{τ}] ∈ Σ^{i}(G, C_{j,k}) are just translates of the extensions classified by C_{j,k}, hence their fields of definition are contained in those for the original Galois extensions, which have degree at most ℓ^{i}(C_{j,k}) = |Σ^{i}(G, C_{j,k})| over k^{i} by [1], II, §4, Folgerung 2 . ⊓ ⊔

It should be clear that versions of Theorem 1 hold for all entries of Table 2. The remaining cases were not worked out because the values of the e_{i} are very restricted. But still they might be helpful in some cases.

As an application of the Theorem we give two examples for smaller fields of definition than could have been deduced by existing rationality criteria.

Example 1. Let G = S_{9}, C = ((2)^{3}, (7), (7)(2)) (the classes are denoted by their cycle shapes in S_{9}). Then on finds ℓ^{i}(C) = |Σ^{i}(G, C)| = 6. Current theorems give an estimate of (k : Q) ≤ 6 for the fields of definition k of corresponding S_{9}-extensions. But we have that ℓ^{i}((2)^{3}, (3)^{3}, (7)(2)) = 1, and ϕ_{3} maps this class into Σ^{i}(G, C), so by Theorem 1 one of the Galois extensions is even defined over Q. Indeed it turns out that the Galois group of the field of definition for all six extensions is isomorphic to S_{5} × 1.

Example 2. We take G = S_{8}, C = ((3)^{2}, (5)(2), (5)(2)), where ℓ^{i}(C) = 15. Here the topological automorphism (23) acts, and it turns out that it has nine fixed points. By III, §3, Folgerung 3 in [1], this yields a decomposition 9 + 6 for the degrees of fields of definition. But elements from C_{1} = ((2)^{3}, (6)(2), (5)(2)), C_{2} = ((2)^{4}, (6), (5)(2)) and C_{3} = ((2)^{4}, (3)^{2}(2), (5)(2)) all map into C under ϕ_{1}, with ℓ^{i}(C_{1}) = 7, ℓ^{i}(C_{2}) = 1, ℓ^{i}(C_{3}) = 1. Hence we even get a 1 + 1 + 7 + 6decomposition, and two of the Galois extensions are actually defined over Q. It is clear that the images under ϕ_{1} still generate S_{8}, although we have A_{8} as a subgroup of index two.

So Theorem 1 allows us to find better estimates for the degree of fields of definition of Galois extensions described by class tuples via Hurwitz classification. In the above two examples, we even obtained Galois realizations over Q. But both were just translates of other Galois extensions over Q with the same Galois group. So the Theorem mainly helps to construct further Galois realizations, if one has already been found. The only case in which really new Galois extensions over Q could be obtained would occur if it happened that |R_{j}| = 1 for some j. This would give rationality without having any class structure C_{j,k} of G with ℓ^{i}(C_{j,k}) = 1. But no example for this was found until now.

2. Twisting the genus zero translates

In this section, we want to understand the first part of result of Shih [2]. He also considered genus zero translates like the ones in the first section, but in addition applied some twisting to the branch points to reduce the degree of the field of definition. In the language of [1], he showed that under certain conditions the translated class structure had no fixed points under topological automorphisms. His approach works only for congruence subgroups of the modular group, but indeed a much more general result can be obtained.

For this, we go back to Proposition 4. In cases (a) and (b), the image class structure has a symmetry but all images are fixed under the corresponding topological automorphisms. So these do not allow a reduction of the degree of the field of definition. In the other two cases, no generic symmetries exist (see Example 1 for ϕ_{3}). But if in case (c) σ_{3} and σ_{3}^{2} (resp. σ_{3}, σ_{3}^{3} in case (d)) generate the same cyclic subgroup of G, then the second and third class are either the same or else algebraically conjugate. With n := |G| we define

C^{∗} := ⋃_{(ν,n)=1} C^{ν} := ⋃_{(ν,n)=1} (C_{1}^{ν}, …, C_{r}^{ν}), where C_{i}^{ν} := [σ_{i}^{ν}] for σ_{i} ∈ C_{i}.

Then [k^{i} : Q] = (C^{∗} : C) by [1], II, §4. A nonrational conjugacy class C is called semirational if all the primitive powers of an element σ ∈ C lie in the union C ∪ C̄ of C with a second class C̄. Because we are only interested in the last two cases of Proposition 4, we can assume C = (C_{1}, C_{2}, C_{3}) with e_{1} = 2 and e_{2} = 3.

Theorem 2. Let G be a group with trivial center, C = (C_{1}, C_{2}, C_{3}) with C_{i} containing elements of order e_{i} where e_{1} = 2 and e_{2} = 3, C_{2} is rational, C_{3} is semirational. Assume that C_{3}^{2} = C̄_{3} (C_{3}^{3} = C̄_{3} resp.). If ℓ^{i}(C) = 1 and the images under ϕ_{3} (ϕ_{4} resp.) of the two classes in Σ^{i}(G, C^{∗}) are interchanged by the topological automorphism (23) and still generate G, then G occurs as the Galois group of a regular field extension of Q(t).

Proof. By the assumption ℓ^{i}(C) = 1 there exists a Galois extension N/k(t) with group G and Hurwitz description C, where [k : Q] = 2 and k = k^{i} = Q(\sqrt{α}) is determined by the irrationality of the class C_{3} (remark that C_{1} is rational, as it contains involutions). We now translate N/k(t) by L according to the third or fourth entry in Table 3. Then from Proposition 4 we know the canonical generators of Gal(LN/L). In particular, if C_{3}^{2} = C̄_{3} in case (c) or C_{3}^{3} = C̄_{3} in case (d), they belong to the class structure (C_{1}, C_{3}, C̄_{3}), (C_{2}, C_{3}, C̄_{3}) respectively. These class structures have a symmetry, and if the ramification points corresponding to the second and third class are chosen to be algebraically conjugate in k over Q, the topological automorphism η interchanging these two points will act nontrivially. Now if the topological automorphism η interchanges the two images, then it acts like the algebraic automorphism of k/Q on the Galois extensions. Hence their product acts like an inner automorphism of G, and because Z(G) = 1, this implies that Q is a proper field of definition of LN/L by [1], II, §2, Satz 2. ⊓ ⊔

It should be remarked that the topological automorphism not always interchanges the two images. In that case we obtain four classes of tuples in the translated ramification structure, and η and the algebraic automorphism of k/Q generate the elementary abelian group of order four in their action on them. This is so because by Theorem 1 all Galois extensions are defined over k already.

By using the explicit formulae for η (in [1]) and ϕ we get:

Corollary 1. Let G, C be as in Theorem 2, with ℓ^{i}(C) = 1. Let [σ_{1}, σ_{2}, σ_{3}], [τ_{1}, τ_{2}, τ_{3}] be the two classes in Σ^{i}(G, C^{∗}). If one of

(a) C_{3}^{2} = C̄_{3} and [σ_{1}, σ_{3}^{2}, σ_{3}^{−2}σ_{2}^{−1}σ_{3}σ_{2}σ_{3}^{2}] ∼ [τ_{1}, τ_{2}^{−1}τ_{3}τ_{2}, τ_{3}^{2}] or

(b) C_{3}^{3} = C̄_{3} and [σ_{2}, σ_{3}^{3}, σ_{3}^{−2}σ_{2}^{−1}σ_{3}σ_{2}σ_{3}^{2}] ∼ [τ_{2}, τ_{3}τ_{2}^{−1}τ_{3}τ_{2}τ_{3}^{−1}, τ_{3}^{3}], then G occurs as a Galois group over Q(t).

From this we deduce the first part of Theorem 12 of Shih [2]:

Example 3. Let G = L_{2}(p), p > 2, and C = (C_{2}, C_{3}, C_{p}), where C_{i} contains elements of order i. Then C_{p} is semirational, and C_{p}^{2} = C̄_{p} iff (\frac{2}{p}) = −1,

C_{p}^{3} = C̄_{p} iff (\frac{3}{p}) = −1. It is easy to check ℓ^{i}(C) = 1 and then show that the Corollary applies. Hence if (\frac{2}{p}) = −1 or (\frac{3}{p}) = −1 then G is a Galois group over Q(t).

3. Twisting four point ramified genus zero translates

Clearly the development of the first two sections is not restricted to the special case that the translated Galois extension ramifies at only three points. But the more ramification points we allow, the more different cases will occur and the more complicated the formulae will become. We will therefore consider only the further case of four branch points for the translated Galois extension.

Following the course in the first section, we first have to find the admissible field extensions L̄/Q̄(t) of degree d and with L̄ of genus zero. If we assume again that all cycles apart from those which lead to ramification have length equal to e_{i}, we have to determine all solutions of

d + 2 = \frac{d − n_{1}}{e_{1}} + \frac{d − n_{2}}{e_{2}} + \frac{d − n_{3}}{e_{3}} + 4.  (3)

A computer search now shows that (3) has almost 200 solutions. We will not give them all here but consider them in a forthcoming paper. Here we concentrate on three particular cases to show that also the four point case yields interesting rationality criteria.

It is easily checked that the lines in the following table correspond to translation fields L/Q(t) of genus zero such that the translated Galois extension is ramified in precisely four points (again indicated by underlined cycles), if the ramification of the original extension N/k(t) is given by the e_{i}.

Table 4. Cycle description of the Galois group A for some translation fields L/Q(t)

d  cycle dec. of σ̄_{i}  A  e_{1} − e_{2} − e_{3}
6  (2)^{2}(\underline{1})^{2} − (3)^{2} − (\underline{5})(\underline{1})  L_{2}(5)  2 − 3 − ∗
6  (2)^{3} − (4)(\underline{1})^{2} − (\underline{5})(\underline{1})  P GL_{2}(5)  2 − 4 − ∗
8  (2)^{4} − (3)^{2}(\underline{1})^{2} − (\underline{7})(\underline{1})  L_{2}(7)  2 − 3 − ∗

The first and third field already implicitely occured in Shih’s work [2], as we will see in Example 4.

As in Section 2, we now want to start with a nonrational class structure C, translate the corresponding Galois extension by one of the possibilities of Table 4, and then twist the branch points of the resulting Galois extension so as to hopefully arrive at a Galois extension defined over Q. A necessary condition for the latter is that all primitive powers of any nonrational class in the translated class structure appear the same number of times. Assume that C is nonrational in C, and the primitive powers of its elements lie in C∪C^{e_{2}} ∪… C^{e_{s}}. If the translated class structure C′ is to satisfy the above symmetry condition, then certainly s ≤ 4. The most interesting case again seems to be s = 2. We will assume that the semirational class in C is C_{3}, because in most examples classes with elements of large order are more likely to be nonrational. The other possibilities will be investigated in a later paper.

In our cases, always two underlined cycles lie in σ̄_{3}, and the two other underlined cycles not in σ̄_{3} lie in the same σ̄_{i}. So we have two possibilities for the action of a topological automorphism η on the branch points of the translated Galois extension LN/L. Either η acts as the transposition (34), or as the double transposition (12)(34) on these points. It now has to be checked whether the branch point configuration of LN/L in our examples actually admits a topological automorphism with the right rationality properties. In the three branch point case of Section 2 this was trivial because P GL_{2}(Q) acts 3-transitive, and so all arrangements of points were equivalent. But for four branch points, it depends on the actual configuration.

As the degrees of L̄/Q̄(t) are at most eight, it is easy to determine a defining equation for L̄/Q̄(t) ramified at 1, 0, ∞ by the method described in [1]. It turns out that in no case the points corresponding to the underlined cycles form a configuration admitting a transposition (34) as topological automorphism. But all four point configurations admit a topological automorphism acting as a double transposition. By the method of Proposition 3, generators [τ_{1}, τ_{2}, τ_{3}, τ_{4}] for the inertia subgroups of the translated extension can be found. They are given in Table 5.

Table 5. Hurwitz description for the translated Galois group

d  cycle description  [τ_{1}, τ_{2}, τ_{3}, τ_{4}]  cond. on α
6 (2)^{2}(\underline{1})^{2} − (3)^{2} − (\underline{5})(\underline{1}) [σ_{1}, σ_{3}σ_{1}σ_{3}^{−1}, σ_{3}^{5}, σ_{3}^{−2}σ_{1}σ_{3}σ_{1}^{−1}σ_{3}^{2}] (α, 5) = 1
6 (2)^{3} − (4)(\underline{1})^{2} − (\underline{5})(\underline{1}) [σ_{2}, σ_{3}^{2}σ_{2}σ_{3}^{−2}, σ_{3}^{5}, σ_{3}^{−2}σ_{2}^{−1}σ_{3}σ_{2}σ_{3}^{2}]  −
8 (2)^{4} − (3)^{2}(\underline{1})^{2} − (\underline{7})(\underline{1}) [σ_{2}, σ_{3}^{2}σ_{2}σ_{3}^{−2}, σ_{3}^{7}, σ_{3}^{−3}σ_{2}^{−1}σ_{3}σ_{2}σ_{3}^{3}]  −

So again for each row j of the table we can define a map ϕ_{j} : Σ^{i}(G, C) → Σ^{i}(H, C′) by putting ϕ_{j}([\underline{σ}]) := [\underline{τ}], where [\underline{τ}] is given by the formulae in column 3 of Table 5.

As in the proof of Theorem 2 in the second section we now want to replace the transcendental u by v := \frac{au+b}{cu+d}, so that the topological automorphism (12)(34) acts like the field automorphism of k^{i}/Q on the set of branch points. Let k^{i} = Q(\sqrt{α}), α ∈ Q (remember that we are considering the case where C_{3} is semirational). Then without loss of generality we can send the two branch points above p_{3}, with inertia group generators in C_{3} ∪ C̄_{3}, to \{\sqrt{α}, −\sqrt{α}\}. As P GL_{2}(Q) acts 3-transitive, this leaves one degree of freedom for the twisting. But we also have to satisfy the second condition that the images of the first two branch points should be conjugate by Aut(k^{i}/Q). It turns out that this is not always possible. The resulting condition on α is given in the last column of Table 5. Here (a, b) denotes the Hilbert symbol (i.e., (a, b) = 1 iff ax^{2} + by^{2} = 1 has a solution in Q), and ”–” means that there is no condition on α.

We are now ready to state the rationality criterion for the derived Galois extensions:

Theorem 3. Let G be a group with trivial center, C = (C_{1}, C_{2}, C_{3}) with C_{1}, C_{2} rational, C_{3} semirational, and ℓ^{i}(C) = 1. Denote by e_{i} the order of elements in C_{i}, and let [σ_{1}, σ_{2}, σ_{3}], [τ_{1}, τ_{2}, τ_{3}] be the two classes in Σ^{i}(G, C^{∗}). Define

η([ρ_{1}, ρ_{2}, ρ_{3}, ρ_{4}]) := [ρ_{2}, ρ_{1}, ρ_{1}^{−1}ρ_{4}ρ_{1}, ρ_{2}ρ_{3}ρ_{2}^{−1}].

If one of (a) e_{1} = 2, e_{2} = 3, C_{3}^{5} = C̄_{3}, (α, 5) = 1 and η(ϕ_{1}([\underline{σ}])) ∼ ϕ_{1}([\underline{τ}]), or

(b) e_{1} = 2, e_{2} = 4, C_{3}^{5} = C̄_{3}, and η(ϕ_{2}([\underline{σ}])) ∼ ϕ_{2}([\underline{τ}]), or

(c) e_{1} = 2, e_{2} = 3, C_{3}^{7} = C̄_{3}, and η(ϕ_{3}([\underline{σ}])) ∼ ϕ_{3}([\underline{τ}]), and if ϕ_{i}([\underline{σ}]) also generates G, then G occurs as the Galois group of a regular field extension of Q(t).

Proof. Let N/k^{i}(t) be the Galois extension with group G corresponding to [σ], where k^{i} = Q(\sqrt{α}) is the abelian field of definition. Then because ℓ^{i}(C^{∗}) = 2, the automorphism of k^{i}/Q maps N/k^{i}(t) into the Galois extension corresponding to [\underline{τ}]. So clearly the two translates by any of the possibilities of Table 5 are also conjugate by Aut(k^{i}/Q). If the topological automorphism (12)(34) on the branch points also interchanges the two translates, then the product acts like an inner automorphism of G, and by the same argument as in the proof of Theorem 2 we are done.

Hence it remains to prove that the topological automorphism acts via the formula given for η. We will do this for case (b), the other cases being entirely similar. A defining equation for the stem field L̄ = Q̄(u) of the Galois extension L̂/Q̄(t) ramified at 1, 0, ∞ is given by

f(u, t) = (u − 1)^{4}(u^{2} − 6u + 25) + 256ut.

So the points lying above 0, ∞ corresponding to the underlined cycles in row 2 of Table 5 are 3 ± 4i, 0, ∞. If we replace u by the new variable v := \sqrt{α}\frac{u+5}{u−5}, then ramification occurs at ±2\sqrt{−α}, ±\sqrt{α}, and the topological automorphism (12)(34) is given by v ↦ −v. With the method of [1], III, §1, it is now very easy to deduce the action of the topological automorphism on classes of generators and to verify that the action is given by the formula for η in the Theorem. In the other cases, the same formula is obtained. ⊓ ⊔

The condition of generation is satisfied for example if G has no proper subgroups of index dividing 6, 8 respectively.

Example 4. Let G = L_{2}(p), p > 2, C = (C_{2}, C_{3}, C_{p}) as in Example 3. Apply part (c) of the Theorem to get L_{2}(p) as a Galois group over Q(t) iff (\frac{7}{p}) = −1. This is the third part of Shih’s result, which had previously defied all attempts of proof in the context of constructive Galois theory.

We can also see why the case n = 5 in Shih’s paper, which corresponds to row 1 of Table 5, did not work. Here α = (−1)^{\frac{p−1}{2}} p, and (α, 5) = 1 by our condition on the rationality of the branch locus. But to get C_{3}^{5} = C̄_{3}, one has to have (p, 5) = −1 which yields a contradiction.

But if we start from a different class structure, we obtain a new result for (\frac{5}{p}) = −1:

Theorem 4. Let p ≡ ±1 (mod 8) and (\frac{5}{p}) = −1. Then L_{2}(p) occurs as the Galois group of a regular field extension of Q(t) for the ramification structure (C_{4}, C_{4}, C_{p}, C̄_{p})^{∗}.

Proof. If p ≡ ±1 (mod 8), then G = L_{2}(p) contains elements of order four, which form a rational class C_{4}. So let us consider C = (C_{2}, C_{4}, C_{p}). It is easy to check that ℓ^{i}(C) = 1, and representatives of the two classes of generating systems in Σ^{i}(G, C^{∗}) are given by

[\underline{σ}] := [( \begin{matrix}0 & \sqrt{2} \\ −1/\sqrt{2} & 0 \end{matrix} ) , ( \begin{matrix}\sqrt{2} & −\sqrt{2} \\ 1/\sqrt{2} & 0 \end{matrix} ) , ( \begin{matrix}1 & 0 \\ 1 & 1 \end{matrix} )] ,

[\underline{τ}] := [( \begin{matrix}0 & \sqrt{2}/w \\ −w/\sqrt{2} & 0 \end{matrix} ) , ( \begin{matrix}\sqrt{2} & −\sqrt{2}/w \\ w/\sqrt{2} & 0 \end{matrix} ) , ( \begin{matrix}1 & 0 \\ w & 1 \end{matrix} )] ,

where \sqrt{2} denotes a fixed element β of IF_{p} with β^{2} = 2 and w is a fixed nonsquare in IF_{p}. Now

η(ϕ_{2}([\underline{σ}])) =

[( \begin{matrix}3\sqrt{2} & −\sqrt{2} \\ 13\sqrt{2}/2 & −2\sqrt{2} \end{matrix} ) , ( \begin{matrix}\sqrt{2} & −\sqrt{2} \\ 1/\sqrt{2} & 0 \end{matrix} ) , ( \begin{matrix}7 & −4 \\ 9 & −5 \end{matrix} ) , ( \begin{matrix}21 & −10 \\ 40 & −19 \end{matrix} )] ,

and

ϕ_{2}([\underline{τ}]) =

[( \begin{matrix}\sqrt{2} & −\sqrt{2}/w \\ w/\sqrt{2} & 0 \end{matrix} ) , ( \begin{matrix}3\sqrt{2} & −2\sqrt{2} \\ 13w\sqrt{2}/2 & −2\sqrt{2} \end{matrix} ) , ( \begin{matrix}1 & 0 \\ 5w & 1 \end{matrix} ) , ( \begin{matrix}−1 & −2/w \\ 2w & 3 \end{matrix} )] .

But these two are conjugate by

( \begin{matrix}−3/2 & 1 \\ −7w/2 & 3w/2 \end{matrix} ) ,

which has discriminant 5w/4, and this is a square in IF_{p} iff 5 is a nonsquare, i.e., iff (\frac{5}{p}) = −1. So in that case Theorem 3(b) applies and the statement is proved. ⊓ ⊔

With this result, the smallest prime p for which L_{2}(p) is not yet known to be a Galois group over Q(t) is p = 239.

References

1. Matzat, B. H.: Konstruktive Galoistheorie. (Lecture Notes 1284) Berlin - Heidelberg - New York: Springer 1987

2. Shih, K.-y.: On the construction of Galois extensions of function fields and number fields, Math. Ann. 207, 99-120 (1974)

Gunter Malle Mathematisches Institut und IWR Universität Heidelberg 6900 Heidelberg Federal Republic of Germany

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