Tame torsion and the tame inverse Galois problem

Fix a positive integer $g$ and a squarefree integer $m$. We prove the existence of a genus $g$ curve $C/\mathbb{Q}$ such that the mod $m$ representation of its Jacobian is tame. The method is to analyse the period matrices of hyperelliptic Mumford curves, which could be of independent interest. As an application, we study the tame version of the inverse Galois problem for symplectic matrix groups over finite fields.


Introduction
We say that a number field F is tame if F/Q is tamely ramified at every finite prime of F , and wild otherwise. The first result of this paper concerns the problem of finding, for fixed g and m, a (non-singular projective) curve C of genus g whose Jacobian J C has tame m-torsion field Q(J C [m]). If m is not squarefree, then Q(J C [m]) is wild, as it contains Q(ζ m ) (by the Weil pairing), which is wild above primes p for which p 2 |m. In that sense the result is the best possible.
Our strategy will be to reduce to the case where m = p is prime and show that it suffices to construct a curve whose p-torsion of the Jacobian is tamely ramified at p, which we then do with Mumford curves. To illustrate our Mumford curve approach to this problem, we explain the idea in the elliptic curve setting in the following example.
Example 1.2. Let E/Q p be an elliptic curve with split multiplicative reduction. Then E is isomorphic to a Tate curve and E(Q p ) ∼ = Q p × /q Z as Gal(Q p /Q p )-modules, for some q ∈ pZ p .
Moreover any such q gives rise to a Tate curve. In particular, Q p (E[p]) = Q p (ζ p , q 1/p ), and so, whenever q is a p-th power (say q = p p ), the extension Q p (E[p])/Q p is tamely ramified.
For our second result, recall that the classical inverse Galois problem asks, given a finite group G, if there is a Galois extension F/Q such that Gal(F/Q) ∼ = G? This is open in general, but known for certain class of groups including soluble groups and G = S n , A n , GSp 2g (F p ). Birch [Bir94,p.35] further asked whether F can also be taken to be tame? This is known as the tame inverse Galois problem.
Theorem 1.3 (=6.7). Fix a positive integer g and an odd prime p, such that there is a Goldbach triple for 2g + 2 not containing p. There is a curve C/Q of genus g such that Q(J C [p]) is tame, and Gal(Q(J C [p])/Q) ∼ = GSp 2g (F p ). See Conjecture 6.2 for the definition of a Goldbach triple. This is a (slightly) strengthened version of the Goldbach conjecture that predicts that such triples always exist. On the numerical side, we show that, consequently, GSp 2g (F p ) is tamely realisable as a Galois group over Q for g 10 7 and all p > 2; see Lemma 6.10 and the discussion afterwards.
Layout. In §2-5 we address the tame torsion question (Theorem 1.1). Specifically, in §2 we reduce the tame torsion question to odd primes p and show that it suffices to construct a curve whose p-torsion is tamely ramified at p. We then focus on Mumford curves, which are the rigid space generalisation of the Tate curve we used in Example 1.2. In §3, we review hyperelliptic Mumford curves and gather some basic results. We then compute an approximation to the period matrix in §4 and construct a suitable Mumford curve in §5. In §6 we give the application to the inverse Galois problem (Theorem 1.3).
Remark 1.4. Ensuring that the mod p representation is tamely ramified at p may also be done via imposing restrictions on the endomorphism algebra instead; for details, see [Bis]. The endomorphism approach does not require one to treat the (trivial) case p = 2 separately.
Notation. Throughout the paper, we denote G F = Gal(F /F ), the absolute Galois group of a field F C (hyperelliptic) non-singular projective curve C of genus g 1 J C Jacobian of C ζ m primitive m th root of unity In §3-5, we write K finite extension of Q p (p odd) O K , π, q K , e ring of integers of K, uniformiser, size of residue field, ramification degree | · | absolute value on K, normalised so that |π| = q −1 K K, C K an algebraic closure of K and its completion Γ Schottky group, see Definition 3.1 s i , a i , b i , c i , r i see Construction 3.5 2. Reduction to the prime case and ℓ = p First note that for squarefree m, the field Q(J C [m]) is the compositum of Q(J C [p j ]) for prime divisors p j |m, so it suffices to prove Theorem 1.1 when m = p is prime. In this section, we will reduce the question further to only needing to study the ramification of Q(J C [p])/Q at p via a result of Kisin of local constancy of Galois representations in ℓ-adic families, and deal with p = 2.
Lemma 2.1. Let m = p 1 p 2 · · · p n , with p j distinct primes. Let C/Q be a curve of genus g such that (i) C has semistable reduction at all primes ℓ 2g + 1 and ℓ|m; ) is the compositum of the fields Q(J C [p j ]) so it suffices to prove that these are all tame. Fix a prime p = p j ; we have to show that Q(J C [p])/Q is tamely ramified at ℓ for all primes ℓ; note that by condition (ii), we may assume that ℓ = p. If Proof. This is a special case of [Kis99, Theorem 5.1(1)]. Note that for N large enough, all f ≡ f mod p N have the same degree as f and are squarefree, and so define a p-adic family of hyperelliptic curves of the same genus.
With this theorem, we only need to construct genus g hyperelliptic curves C ℓ /Q ℓ at each prime ℓ 2g + 1 and ℓ|m and then glue them together with sufficient congruence conditions in order to realise the m-torsion as a tame extension.
To construct a curve C/Q p such that Q p (J C [p]) is semistable at p with Q p (J C [p j ]) ∼ = Q p (ζ p ) we will use the theory of Mumford curves. For simplicity in this approach however, we will assume that p is odd, so we briefly record below a curve that covers the case p = 2.
Proof. By assumption on the a i , the Newton polygon of h breaks completely, and [Dok18, Thm 1.2(6,7)] shows that J C is semistable and has totally toric reduction. Next, completing the square and replacing y by y/2 we see that C is isomorphic to As v 2 (N/2) > v 2 (h(0)), the polynomials h(x) − N/2 and h(x) + N/2 have the same Newton polygon as h, and so factor completely over Z 2 as well. It follows that J C [2] ⊂ J C (Q 2 ).

Mumford curves and Whittaker groups
In this paper, we will only need to concern ourselves with hyperelliptic Mumford curves in which case the Schottky group will be of a particular type called a Whittaker group. For more details on the background of Mumford curves in general, see [GvdP80].
From now on, we suppose that p 3 and let K/Q p be a finite extension. Let v be the normalised valuation on K, and O K , π, q K , e, | · | as in Notation 1.
(i) A point x ∈ P 1 (C K ) is a limit point of Γ if there exists y ∈ P 1 (C K ) and an infinite sequence (γ n ) ⊂ Γ with γ n distinct and lim γ n (y) = x. (ii) Γ is a Schottky group if it is discrete, free, and finitely generated.
Mumford curve of genus equal to the rank of Γ. (iv) Let Γ be a Schottky group. If the associated Mumford curve is hyperelliptic, then Γ is called a Whittaker group.
The construction of Mumford curves is analytic, so computing an algebraic model for them is in general difficult. The main approach is to construct a good fundamental domain, which we do via p-adic discs; we briefly set up some notation for them.
Notation. Let c, r ∈ C K . We define the open disc B(c, r) and closed disc B(c, r) with centre c and radius r as Definition 3.3. Let Γ be a Schottky group of rank g. Then a set F is called a good fundamental domain for Γ if: In this case, we say that the generators γ 1 , . . . , γ g are in good position. For hyperelliptic Mumford curves, one constructs a Whittaker group via a suitable choice of 2g + 2 points as follows: Definition 3.6. Let Z be a set of 2g + 2 distinct points. If the corresponding group Γ Z is a Whittaker group of rank g, then we say that Z is in good position.
Then it suffices to check the conditions (i) and (ii) of Definition 3.3 for these 2g discs to see if they define a good fundamental domain. (ii) We can always suppose that 0, 1, ∞ ∈ Z by applying a Möbius transformation.
This changes Γ Z to a conjugate subgroup and gives an isomorphic Mumford curve. (iii) Note that the construction requires a choice of pairing on Z. One can show that there is at most one pairing on Z such that it is in good position.
(iv) The equation of the hyperelliptic curve z−w(1) and W Z = s 1 , . . . , s g , s g+1 is the group generated by the associated involutions; the 2: (i) The points of Z are in good position; We will now implicitly assume these assumptions in the lemma whenever we deal with a Whittaker group. For two discs B, B ′ , we denote by d(B, B ′ ) the corresponding metric coming from the standard one on the Berkovich line P 1,an (see for example [MR15, p.7]).
|r i r j | . In particular, the minimum distance, m Γ 1 , between two distinct discs is min Proof. For the first part, note that the smallest disc containing B i and B j is B(c i , |c i − c j |) and the statement follows from the definition. For the second part, we get d( and note that the numerator is a unit as p = 2 and the centres c i are integral non-units. The minimum now follows.

Approximation of the period matrix
Let Γ be a Schottky group with generators γ 1 , . . . , γ g in good position. Let B 1 , . . . , B g , B ′ 1 , . . . , B ′ g be the associated disjoint discs defining the fundamental domain such that Notation. For a free group Γ = γ 1 , γ 2 , · · · , γ g , we let Γ n be the subset consisting of all elements of Γ of reduced word length at most n.
Proof. We prove the first part; the second part is analogous. First note that z−γa z−γγ i a − 1 = γγ i a−γa z−γγ i a . Since a ∈ ∂B ′ i , we have that γa, γγ i a ∈ B γ and hence |γγ i a − γa| diam(B γ ). Since γ = γ j , the discs B ′ j and B γ are disjoint, so let z ′ j , z γ be centres of We now return to the case where Γ is a Whittaker group and continue our notation from §3. The Jacobian J Ω Γ /Γ has a g × g period matrix Q = (Q ij ) whose entries can be computed in terms of Γ (see [GvdP80] VI.2) for any choice of non-conjugate ordinary points a, z.
Notation. For a subset S ⊂ Γ, let If S = Γ n , we write Q n ij for Q Γn ij ; clearly lim n→∞ Q n ij = Q ij .
Lemma 4.2. Let q ∈ K be such that max i =j Theorem 3.6] 2 , and max i =j .
Proof. Note first that such a q exists by Lemma 3.8. Using Lemma 4.2, we only need to consider the contributions from non-identity elements in Γ 1 . The result is then immediate from Lemma 4.1.
We will now compute Q α ij to get an explicit formula. By choosing the auxiliary parameters a, z carefully, we will not need to distinguish between the cases i = j and i = j, and we find that Q α ij = Q 0 ij with this choice. Lemma 4.4. Let a ∈ ∂B ′ i , z ∈ ∂B ′ j and assume a = z if i = j. Then a ≡ z mod Γ. Proof. We shall adapt the proof of [MR15, Lemma 2.4]. In fact, we shall prove that γa is contained in the of the open disc B h 1 (continuing notation from §2), where γ = h 1 · · · h m as a reduced word, unless γ ∈ {id, γ i }. Note that γ k (P 1 \B ′ k ) = B k and moreover γ k (∂B ′ k ) = B k for all k.
If h m = γ i then h m a ∈ B hm and hence iteratively we have γa ∈ B h 1 . If h m = γ i , then h m a ∈ ∂B i so if m 2, then h m−1 = γ −1 i so proceeding similarly we have γa ∈ B h 1 . Moreover, note γ i a = z since B ′ j = B i . Lastly, if γ = id, then a = z by assumption. Lemma 4.5. Let a = 2 − c i + r i and z = 2 − c j − r j . Then (i) a ∈ ∂B ′ i , z ∈ ∂B ′ j , and a and z are distinct mod Γ.
for all k. From this we compute explicitly that γ i a = c i − r i and similarly γ j z = c j + r j . Now the claim follows from (iii) This follows from (2), by setting i = j.
The claim now follows from

Tame torsion
Lemma 5.1. Let a ∈ 1 + π N O K for some positive integer N . If em N , then x m − a has a root in O K . In particular, every element of 1 + π em O K is an m th power for all m 1.
Proof. This is a simple application of Hensel's lemma, where we use the version that states there is a lift of a root a 0 (in the residue field) of a polynomial f if v(f (a 0 )) > 2v(f ′ (a 0 )), where we use f = x m − a and a 0 = 1.
Note that v(f (a 0 )) N by construction and f ′ (a 0 ) = m, so v(f ′ (a 0 )) = ev p (m) where v p is the standard p-adic valuation on Z. Now so v(f ′ (a 0 )) < em 2 and the result follows. Lemma 5.2. Let r i = π emα , c i = 2π emβ for some α, β > 0. Then Proof. We have (1 − c i ) 2 ∈ 1 + π em O K , so it is an m th power by Lemma 5.1.
Proof. Without loss of generality, suppose β i < β j . Then which is twice an m th power by Lemma 5.1. On the other hand, the denominator is which is also twice an m th power.
(v) Let Q = (Q ij ) denote the period matrix of the corresponding abelian variety. Then Q ij is an m th power for all 1 i, j, g.
(iii) We compute that for i = j, where we suppose i < j without loss of generality. Since α i > β j , we are done. (iv) First suppose i = j. If i = 1, then this follows directly from Lemma 5.2; the same proof also works for i = 1. Now suppose i = j. If i, j 2, then this is Lemma 5.3. If i = 1 or j = 1, then one can apply the same proof using the simplification c 1 = r 1 .
(v) By (iv), we have that Q 0 ij is an m th power. Now by Theorem 4.3, Lemma 4.5(4) and (iii), and 1 + π em b are m th powers (by Lemma 5.1), so is Q ij .
Lemma 5.5. Let J/K be an abelian variety with a Raynaud parameterisation J ∼ = K × g /Q.
Let m > 1 and suppose every entry the period matrix Q is an m th power in K. Then J[m] ∼ = µ g m × (Z/mZ) g as G K -modules. Here Z/mZ has a trivial action, and µ m = ζ m ⊂ K is the set of m th roots of unity, with natural action. In particular, K(J[m]) = K(ζ m ).
Theorem 5.6. Fix an integer g 1. Let • α i = 2g − i for 1 i g, and β i = g − i + 1 for 2 i g; • r 1 = c 1 = π pα 1 , and r i = π pα i , c i = 2π pβ i for 2 i g. Let C/K be the corresponding genus g hyperelliptic Mumford curve given by Construction 3.5. Then J C is semistable and K(J C [p]) = K(ζ p ).
Proof. Recall that all Mumford curves are semistable (see for example [GvdP80, Theorem 2.12.2]). By Theorem 5.4 with m = p, every entry of the period matrix of J C is a p th power; the statement now follows from Lemma 5.5 with m = p.
Theorem 5.7. Fix a positive integer g and squarefree integer m. Then there exists a non-singular projective curve C/Q of genus g such that Q(J C [m]) is tame.
Proof. By Kisin's result (Theorem 2.2) we need only choose a suitable genus g hyperelliptic curve C ℓ for the finite set of primes ℓ 2g + 1 and ℓ | m; if ℓ ∤ m we take C ℓ to be semistable at ℓ (e.g. good reduction at ℓ). For ℓ | m, ℓ = 2, Theorem 5.6 with K = Q ℓ , p = ℓ provides a construction of a genus g hyperelliptic curve C ℓ such that Q ℓ (J C ℓ [ℓ]) ∼ = Q ℓ (ζ ℓ ); similarly we can use Proposition 2.3 if ℓ = 2. We are now done by Lemma 2.1.
Remark 5.8. The same approach works to construct a curve C/Q p such that Q p (J C [p n ]) = Q p (ζ p n ) for any n 1 but note that this is wildly ramified if n = 1. However we can give global curves C/Q such that Q(J C [m])/Q(ζ m ) is a tame extension any odd integer m.

The tame inverse Galois problem
In this section, we investigate the tame version when G is of the form GSp 2g (F p ) via the mod p representation of abelian varieties.
Remark 6.1. An alternative approach to force surjectivity is to ensure End A = Z (to guarantee this, take Gal(f ) ∼ = S deg(f ) and apply [Zar00, Theorem 2.1]) and then apply Serre's open image theorem to obtain surjectivity for p sufficiently large 3 . There are two problems with this however: we do not know precisely what sufficiently large means and more importantly this says nothing for small p.
Theorem 6.4. Let p 5 be prime and let A/Q be a principally polarised abelian variety of dimension g. Suppose: (i) The G Q -action on A[p] is irreducible, primitive and contains a transvection; Before we state an explicit version of the above theorem, we need some quick definitions.
Definition 6.5. Let p be a prime and let f (x) = x m + a m−1 x m−1 + · · · + a 0 ∈ Z p [x] be a squarefree monic polynomial. Fix an integer t 1.
(i) We say that f is t-Eisenstein at p if v p (a i ) t for all i and v p (a 0 ) = t.
(ii) Let q 1 , · · · , q k be rational primes. We say that f is of type t−{q 1 , · · · , q k } if it can be factored over for some α i ∈ Z p such that α i ≡ α j mod p for i = j, g i (x) is t-Eisenstein of degree q i and the reduction mod p, h(x), of h(x) is separable with h(α i ) = 0 for all i.
Theorem 6.6. Let C/Q : y 2 = f (x) be a hyperelliptic curve of genus g and Jacobian J C . Assume 2g + 2 satisfies Conjecture 6.2 and let (q 1 , q 2 , q 3 ) be a Goldbach triple. Fix an odd prime p = q 1 , q 2 , q 3 . Choose primes p 1 , p 2 , p 3 > max(2g + 1, p) such that: • p 2 is a primitive root modulo q 1 and modulo q 2 ; • p 3 is a primitive root modulo q 3 ; • If p = 3, then moreover suppose that p 2 ≡ p 3 ≡ 1 mod 3. Suppose: ( Proof. This is a slight reformulation of [AD19, Theorem 6.2] where we can weaken some of the hypotheses since p is fixed. Suppose first that p 5. Then condition (i) implies the existence of a transvection [AD19, Lemma 2.9], whereas (ii) and (iii) imply that J C [p] is irreducible [AD19, Lemma 3.2]. Primitivity follows from iv) and v) (cf. [AD19, Remark 6.1]); the result now follows from Theorem 6.6. For the case p = 3, the same argument as [AD19, Theorem 6.5] holds. Corollary 6.7. Fix a positive integer g and assume 2g + 2 satisfies Conjecture 6.2. Fix an odd prime p. If there exists a Goldbach triple for 2g + 2 not containing p, then there exists a curve C/Q of genus g such that Gal(Q(J C [p])/Q) ∼ = GSp 2g (F p ) and Q(J C [p]) is tame.
Remark 6.8. If 2g + 2 satisfies Double Goldbach as well (Conjecture 6.3), then the conclusion holds for all odd primes p by applying the statement with the Goldbach triples (q 4 , q 5 , q 3 ) and (q 1 , q 2 , q 5 ). Double Goldbach has been numerically verified by Anni-Dokchitser (cf. [AD19, Remark 6.6]) to hold for all g 10 7 , excepting g = 1, 2, 3, 4, 5, 7, 13. Remark 6.9. Observe that if q = 2g + 1 is prime, then we do not need to use a Goldbach triple; imposing that f (x) has type 1 − {q} at some large prime ensures that J C [p] is an irreducible G Q -representation and we only then need to avoid p = q for the same result.
The reason for these exceptions is that the method of Anni-Dokchitser uses a Goldbach triple to ensure that J C [p] is an irreducible G Q -module when p is not in the Goldbach triple. Instead, we take a different approach to ensure that the mod p representation is surjective. (ii) For some prime ℓ = p of good reduction for C, the reduction mod p of the characteristic polynomial of a Frobenius element at ℓ is irreducible with nonzero trace.
Then ρ is surjective.
Proof. This is just a reformulation of [AdRK13, Corollary 2.2], where condition (i) forces the existence of a transvection (cf. proof of Theorem 6.6).
Condition (i) is easy to force at some large prime p 1 > max(2g + 1, p) so it just remains to exhibit curves which satisfy the second condition for each of our exceptional cases in order to give an affirmative answer to the tame inverse Galois problem in these cases as well. In the table below, we give polynomials f defining hyperelliptic curves, and a prime ℓ such that the image of Frob ℓ has the properties required for condition (ii).