S_{4} AND S̃_{4} EXTENSIONS OF ℚ

RAMIFIED AT ONLY ONE PRIME.

Darrin Doud

University of Illinois Department of Mathematics Urbana, IL 61801

August 4, 1998

Abstract. For each prime p in a certain family of odd primes, we construct an S_{4} extension of ℚ unramified outside p. We show that for all p ≡ 3 (mod 8) in our family, this S_{4} extension embeds in an S̃_{4} extension, which is also unramified outside p. Invoking Serre’s conjecture (in a proven case) allows us to relate the splitting of primes in these extensions to certain modular forms of level 1.

1. Introduction

In this paper we will be interested in analyzing the structure of certain S_{4} extensions which are ramified at only one prime. We give a conjecturally infinite family of primes, and for each such prime p we define an explicit polynomial which has an S_{4} extension unramified outside p as its splitting field. We show that when p ≡ 3 (mod 8), the given S_{4} extension can be enlarged to an S̃_{4} extension which is unramified outside p (where S̃_{4} is the central extension of S_{4} by ±1 isomorphic to GL_{2}(𝔽_{3})). Finally, a proven case of Serre’s conjecture allows us to produce a modular form of level 1 which encodes the splitting of the primes in this S̃_{4} extension.

2. S_{4} extensions

Theorem 1. Let p be an odd prime such that

(−1)^{(p−1)/2}p = p^{∗} = (256n^{3} − k^{2})/27,

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 11R21,11F11.

2 S_{4} AND S̃_{4} EXTENSIONS OF ℚ

with k divisible by 27 if n is divisible by 3. Suppose that p^{∗} ≠ 1 + 4n. Then the splitting field of f(x) = (x + n)^{4} − p^{∗}x is an S_{4} extension of ℚ which is ramified only at p.

Remarks. Given the conditions on n and k, we see that k must be odd. Also, either n is exactly divisible by 3, or n ≡ 1 (mod 3). Any p of this form must be congruent to either 3 or 5 mod 8, since p^{∗} ≡ −k^{2}/27 ≡ 5 (mod 8). There are conjecturally infinitely many such primes, even if we fix n or k. In fact, such primes are not rare: the ones below 1000 are 59, 107, 139, 229, 283, 307, 331, 419, 491, 499, 733, 883.

This list can be shown to be complete. Lemmas 1 and 2 show that 3 cannot be on the list, since if p were 3, we would have a quartic field with discriminant dividing −27, and such a field cannot exist[6]. For any other prime p to be on the list, we must have p ≡ 3 or 5 (mod 8), and the class number of ℚ(\sqrt{p^{∗}}) must be divisible by three. Other than the primes above, only eight primes less than 1000 satisfy these conditions. Seven of these (83, 211, 379, 547, 563, 643, and 971) can be eliminated by observing that a certain modular form of weight 1+(p^{2}−1)/8 does not exist. The last (907) can be eliminated by showing the nonexistence of certain integral points on an elliptic curve, which we do not describe in this paper.

Fix p^{∗} = (256n^{3} − k^{2})/27, as in the theorem, and define f(x) = (x + n)^{4} − p^{∗}x. We see easily that (n, p) = (k, p) = 1, so that f is irreducible (since f(x − n) is Eisenstein). The discriminant of f is k^{2}p^{∗}^{3}. Let K_{4} be the field ℚ(α) where α is a root of f. We define t(k) to be the unique element in {−1, 0, 1} such that t(k) ≡ k (mod 3).

Lemma 1. If (n, 3) = 1, the elements

\{1, α, α^{2}, \frac{1}{k} (α^{3} + \frac{13n − t(k)k}{3} α^{2} + \frac{67n^{2} + (n + 1)t(k)k}{9} α − 3n^{3})\}

are all algebraic integers.

Proof. The only question in this lemma is whether the fourth element (which we will denote by β) is an algebraic integer. Since (256n^{3} − k^{2})/27 is an integer, (k, 3) = 1, and in fact, n ≡ 1 (mod 3). Thus, 13n − t(k)k ≡ 0 (mod 3). Using the fact that n^{3} ≡ 1 mod 9, we see that k^{2} ≡ 4 (mod 9), so that k ≡ ±2 (mod 9) and t(k)k ≡ −2 (mod 9). Hence,

n(67n^{2} + (n + 1)t(k)k) ≡ 67n^{3} + (n^{2} + n)t(k)k (mod 9)
≡ 4 + 2t(k)k (mod 9)
≡ 0 (mod 9).

Since n is invertible mod 9, 67n^{2} + (n + 1)t(k)k ≡ 0 (mod 9). This shows that at least kβ is an algebraic integer. We may easily compute the four conjugates of β over ℚ (by replacing α by each of the four roots of f in turn); denoting these conjugates by β = β_{0}, β_{1}, β_{2}, β_{3}, the polynomial

∏_{i=0}^{3}(x − β_{i})

can be computed by using a computer algebra system (such as Mathematica) and the relations between symmetric polynomials in roots of f and coefficients of f. This polynomial is x^{4} + a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0}, where

a_{3} = \frac{k + t(k)(16n^{2} + 4n)}{9}
a_{2} = \frac{nk^{2} − t(k)(4n^{2} + n)k − (288n^{4} + 48n^{3} + 2n^{2})}{3^{3}}
a_{1} = \frac{1}{3^{9}} (t(k)k^{4} − (30n + 2)k^{3} + t(k)(229n^{3} − 144n^{2} − 30n − 1)k^{2}
+(8976n^{4} + 1592n^{3} + 81n^{2})k − t(k)(10944n^{5} + 1200n^{4} + 148n^{3}))
a_{0} = \frac{1}{3^{8}} (−t(k)n^{3}k^{3} + (−n^{5} − 29n^{4} + 2n^{3})k^{2}

−t(k)(192n^{6} + 112n^{5} − 28n^{4} + n^{3})k − (9216n^{7} − 1632n^{6} + 112n^{5} − n^{4}))

We now note that kβ has minimal polynomial x^{4} + ka_{3}x^{3} + k^{2}a_{2}x^{2} + k^{3}a_{1}x + k^{4}a_{0}, which has integer coefficients. Thus, the denominators of the a_{i} must divide k^{4}. However, (k, 3) = 1, so the a_{i} must all be integers. □

Lemma 2. If 3|n, then the elements

\{1, α, \frac{1}{3}(α^{2} − t(n/3)α), \frac{3}{k} (α^{3} + \frac{13n}{3}α^{2} + \frac{67n^{2}}{9} α − 3n^{3})\}

are all algebraic integers.

Proof. Let γ be the third element of this set, and δ be the fourth. Using Mathematica, as before, we find that the minimal polynomial of δ is

x^{4} + \frac{k}{3} x^{3} + \frac{32n^{3} + k^{2}}{27} x^{2} + \frac{176kn^{3} + k^{3}}{729} x + \frac{k^{2}n^{3}}{81} ,

which is easily seen to have integral coefficients (since 3|n and 27|k).

The minimal polynomial of γ is

x^{4} − \frac{4}{3}(n^{2} + t(n/3)n)x^{3} + \frac{2210n^{4} + 1092t(n/3)n^{3} + 162n^{2} − 8k^{2}n − 3t(n/3)k^{2}}{243} x^{2}

+ \frac{1}{19683}( − 13156n^{6} + 32724t(n/3)n^{5} + 18900n^{4} + (296k^{2} + 3996t(n/3))n^{3}

− 162t(n/3)k^{2}n^{2} − 108k^{2}n + (−k^{4} − 27t(n/3)k^{2}))x

+ \frac{27n^{8} − 148t(n/3)n^{7} + 162n^{6} + 108t(n/3)n^{5} + (t(n/3)k^{2} + 27)n^{4}}{2187} .

We note that the coefficient of x^{3} and the constant coefficient are trivially integers, since 3|n and 27|k. The coefficient of x^{2} is an integer, since 2210n^{4} +1092t (\frac{n}{3}) n^{3} =

3^{4} (2210 (\frac{n}{3})^{4} + 364t (\frac{n}{3}) (\frac{n}{3})^{3}) = 3^{4} (\frac{n}{3}) (2210(n/3) − 364t(n/3)) is a multiple of

3^{5}. To show that the coefficient of x is an integer, we need only show that 3^{9} divides −13156n^{6}+18900n^{4}+3996t(n/3)n^{3}. Letting m = n/3, this is equivalent to showing that −13156m^{3} + 2100m + 148t(m) ≡ 0 (mod 27). Note that m is relatively prime to 3. By checking each possible value of m mod 27, we see that this is true. □

Lemma 3. K_{4} has discriminant p^{∗}^{3}.

Proof. Lemmas 1 and 2 show that the discriminant of K_{4} divides p^{∗}^{3} (since ℤ[α] has index at least k in the ring of integers). Using that (n, p) = 1, a simple application of Dedekind’s Criterion ([2], pp. 299-300) shows that the order ℤ[α] is p-maximal, and thus the discriminant of K_{4} is p^{∗}^{3}. □

Lemma 4. If p^{∗} ≠ 1 + 4n, then the Galois group of f is S_{4}.

Proof. We note that the resolvent cubic of f(x − n) is x^{3} − 4np^{∗}x − p^{∗}^{2}. This is reducible exactly when it has a rational root. It is easy to see that ±1, ±p^{∗}^{2}, are not roots. Now, p^{∗} is a root exactly when p^{∗} = 1 + 4n, and −p^{∗} is a root exactly when p^{∗} = −1 + 4n. Since p^{∗} ≡ 1 (mod 4), the second case can never happen, and we have excluded the first in our hypothesis. Then, since the resolvent cubic defines a field contained in the splitting field of f, we see that the Galois group has cardinality divisible by 3. In fact, since the discriminant of the resolvent cubic is −k^{2}p^{∗}^{3}, it has Galois group S_{3}, so the Galois group of f must have S_{3} as a quotient. Thus, the Galois group of f is S_{4}. □

This completes the proof of Theorem 1, since by Lemma 4, the splitting field of f is an S_{4} extension of ℚ, and by Lemma 3 it is unramified outside p.

We note that Theorem 1 does not give all S_{4} extensions unramified outside p. In particular it misses all those whose quartic subfield has discriminant p^{∗}. This is not a great loss, since such quartic fields have small discriminant, and are easy to find in tables of number fields. Theorem 1 also misses some S_{4} extensions whose quartic subfield has discriminant p^{∗}^{3}, even for p ≡ 3 or 5 (mod 8).

Remark. If we relax the condition on n and k somewhat, and allow p^{∗} = 1 + 4n, then we have that 1 + 4n = (256n^{3} − k^{2})/27. Solving this for k, we find that k^{2} = (4n − 3)(8n + 3)^{2}. Thus, 4n − 3 = m^{2}, so 4n + 1 = m^{2} + 4 = p^{∗}, and we see that p^{∗} = m^{2} + 4.

In this case, the splitting field of f has no cubic subfield, and is hence either D_{4}, the Klein four group, or cyclic of order 4. The first two possibilities are eliminated because in both cases the splitting field would contain more than one quadratic subfield, and hence would be ramified outside p. In fact, for any prime p = m^{2} + 4, if we let n = (p − 1)/4, then f(x) = (x + n)^{4} − p^{∗}x defines a cyclic quartic field ramified only at p.

[figure: K; K′_{12}; K_{12}; K_{8}; K′_{6}; L; K_{6}; K_{4}; K_{3}; ℚ(\sqrt{p^{∗}}); ℚ] Diagram 1

3. The splitting of p

We now proceed to determine exactly how p splits in K, the splitting field of f. We have seen that K is a degree 24 extension of ℚ, with Galois group isomorphic to S_{4}. Diagram 1 gives an inclusion diagram for the subfields of K (the degree of a field over ℚ is indicated by its subscript, and L is Galois over ℚ with Galois group S_{3}). This diagram shows only one representative of each conjugacy class of subfields of K, and K_{6}, K_{12}, and K′_{12} are fixed fields of a four cycle, a double transposition, and a transposition, respectively.

Let \sqrt{α} be one of the square roots of α. Then \sqrt{α} is a root of f(x^{2}) = (x^{2} + n)^{4} − p^{∗}x^{2}. This polynomial factors over ℚ(\sqrt{p^{∗}}) as

f(x^{2}) = ((x^{2} + n)^{2} − \sqrt{p^{∗}}x)((x^{2} + n)^{2} + \sqrt{p^{∗}}x).

We now know that ℚ(\sqrt{α}) is of degree at most 8 over ℚ, and contains α and \sqrt{p^{∗}} = ±(α + n)^{2}/\sqrt{α}. Thus, ℚ(\sqrt{α}) = K_{8}. Since the discriminant of f(x^{2}) is 2^{8}n^{4}k^{2}p^{∗}^{6}, and the discriminant of K_{8} is divisible by at least p^{∗}^{6}, we may find the factorization of p in K_{8} by factoring f(x^{2}) (mod p). To calculate this factorization, we use the fact that n is a quadratic residue mod p (since n^{3} ≡ k^{2}/256 (mod p)). We find that for p ≡ 3 (mod 8), f(x^{2}) factors as (x^{2} + n)^{4}, but for p ≡ 5 (mod 8), f(x^{2}) factors as (x − m)^{4}(x + m)^{4}, where m is a square root of −n mod p. In terms of primes, this says that the unique prime above p in K_{4} is inert in K_{8} when p ≡ 3 (mod 8), but it splits when p ≡ 5 (mod 8). We note that p must ramify in ℚ(\sqrt{p^{∗}}), and it must split into three primes in L. Thus, the number g of primes lying over p in K is divisible by 3, and each has ramification index e divisible by 4. If p ≡ 5 (mod 8), we have that g is even, so that g = 6 and e = 4, and the inertial degree f = 1. On the other hand, if p ≡ 3 (mod 8), we have that g = 3, e = 4, and f = 2. This information, together with the fact that the inertia group is cyclic tells us that up to conjugation the inertia field of K at p is K_{6} and the decomposition field is K_{3} (resp. K_{6}) if p ≡ 3 (mod 8) (resp. p ≡ 5 (mod 8)).

4. S̃_{4} extensions

We will now investigate the problem of extending our family of S_{4} extensions to S̃_{4} extensions. Recall [8, p. 97] that S̃_{4} is the unique central extension of S_{4} in which each transposition lifts to an element of order 2, and each double transposition lifts to an element of order 4. We first show that if p ≡ 5 (mod 8), there is no S̃_{4} extension of ℚ containing K. For such a prime the maximal real subfield of K is K_{12} (since f has no real roots, the maximal real subfield cannot contain K_{4}), so that complex conjugation in K would correspond to a double transposition. Then if K were contained in some S̃_{4} extension, complex conjugation in this extension would be a lift of a double transposition, and have order 4, which is impossible.

In the case p ≡ 3 (mod 8), ℚ(\sqrt{p^{∗}}) is imaginary quadratic, so the following theorem of Bayer and Frey applies:

Theorem 2. [1] If K is a Galois extension of ℚ with Galois group S_{4}, with K_{4} one of its quartic subfields, and the quadratic subfield of K is imaginary, then K can be extended to an S̃_{4} extension if and only if the local invariant e_{K,p} is trivial in the Brauer group of ℚ_{p} for each odd prime p which ramifies in K. The invariants e_{K,p} may be calculated from the table below:

Factorization of p𝔒_{K_{4}} e_{K,p}
𝔭_{1}^{4}  (−1)^{(p^{2}−1)/8}(−1)^{(p−1)/2}
𝔭_{1}^{3}𝔭′_{1}  1
𝔭_{1}^{2}𝔭′_{1}𝔭′′_{1}  1
𝔭_{1}^{2}𝔭′_{2}  −1
𝔭_{1}^{2}𝔭′_{1}^{2}  (−1)^{(p−1)/2} · (d, p)_{p}
𝔭_{2}^{2}  (−1)^{(p+1)/2}

Applying this to our extensions, in which p is the only ramified prime, ramifies totally in the quartic extension, and is congruent to 3 mod 8, we see that an S̃_{4} extension does exist. Finally, Serre [8, Corollary 2.1.8] shows that the existence of any S̃_{4} extension of ℚ containing K guarantees the existence of one which is unramified outside p, and this extension is unique. Thus, for any p ≡ 3 (mod 8) of the specified form, we have an example of an S̃_{4} extension unramified outside p.

Table I

Conjugacy Class  1 2 3 4 5 6  7  8
Order  1 2 2 3 4 6  8  8
Size  1 1 12 8 6 8  6  6
χ_{0}  1 1 1 1 1 1  1  1
χ_{1}  1 1 −1 1 1 1  −1  −1
χ_{2}  2 2 0 −1 2 −1  0  0
χ_{3}  2 −2 0 −1 0 1 \sqrt{−2} −\sqrt{−2}
χ_{4}  2 −2 0 −1 0 1 −\sqrt{−2} \sqrt{−2}
χ_{5}  3 3 1 0 −1 0  −1  −1
χ_{6}  3 3 −1 0 −1 0  1  1
χ_{7}  4 −4 0 1 0 −1  0  0

We see easily that this new extension (which we will denote by K̃) must be ramified at p over K, since if it is unramified the inertia group must be cyclic of order 4. Letting s̃ generate the inertia group, we note that the restriction s of s̃ to K must also have order 4 (and would thus be four cycle). However, then s^{2} would be a double transposition, and would lift to s̃^{2}, which would then have to have order 4, giving us a contradiction.

5. Galois Representations

Let p be a prime congruent to 3 (mod 8), of the form indicated above, and let K̃ be the corresponding S̃_{4} extension. Examining the character table of S̃_{4} (Table I) shows that there are two two-dimensional irreducible faithful complex representations of S̃_{4}, which are actually defined over ℤ(\sqrt{−2}). Choosing one and composing the projection G_{ℚ} → Gal(K̃/ℚ) with it, we get a two-dimensional irreducible complex Galois representation ρ_{0} : G_{ℚ} → GL_{2}(ℤ(\sqrt{−2})). We may reduce this representation mod 𝔭 for one of the primes 𝔭 of ℚ(\sqrt{−2}) lying over p. Since p splits in ℚ(\sqrt{−2}), this gives rise to a representation ρ : G_{ℚ} → GL_{2}(𝔽_{p}), with the fixed field of the kernel of ρ equal to K̃. This representation is unramified outside p, by construction; hence its Artin conductor is a power of p. Since we know the ramification groups at p in K̃, we may compute the Artin conductor; it is p^{2}. We note that the central element of S̃_{4} has fixed field K; this central element cannot be complex conjugation, since K is not real. Hence complex conjugation maps to a non-scalar matrix, so the representation ρ is odd.

Given a two-dimensional odd irreducible representation ρ : G_{ℚ} → GL_{2}(\overline{𝔽}_{p}), we have the following conjecture:

Conjecture. (Serre) There is a cusp form f of some integer weight k greater than 1, some level N and some character ϵ, which is a simultaneous eigenform for the Hecke operators, such that if f has Fourier expansion at infinity ∑ a_{n}q^{n}, and ℓ is an unramified prime for ρ which does not divide Np,

Tr(ρ(Fr_{ℓ})) = ã_{ℓ} and det ρ(Fr_{ℓ}) = { \atop ϵ(ℓ)ℓ^{k−1}},

where the map x ↦ x̃ from the algebraic integers to \overline{𝔽}_{p} denotes reduction modulo some place lying above p.

In fact, in [9], Serre gives a recipe which predicts the exact weight, level, and character of the eigenform which corresponds to f. So far, little progress has been made on the main part of Serre’s conjecture, but the following, (often known as the “epsilon-conjecture”) has been proven (for p > 3):

Theorem 3. Given ρ as above, if there is an eigenform f which corresponds to ρ, then there is an eigenform which corresponds to ρ and has the weight, level and character predicted in [9].

For a discussion of this theorem, see [3] and [7].

The level predicted by Serre is just the prime to p part of the Artin conductor, which in this case is a power of p, so the desired level is 1. The predicted character is a character of ℤ/Nℤ, where N is the predicted level. Clearly, it is the trivial character.

In order to calculate the weight predicted by Serre, we need to distinguish between the two Galois representations corresponding to the two irreducible faithful two dimensional mod p representations of S_{4}. Let I be the inertia group (in G_{ℚ}) of a place above p, and choose a generator x of its tame quotient I_{t}. We have two surjective fundamental characters Ψ, Ψ′ : I_{t} → 𝔽_{p^{2}}^{∗}, and we choose a square root of

−2 in 𝔽_{p} by \sqrt{−2} = Ψ^{(p^{2}−1)/8}(x) + Ψ′^{(p^{2}−1)/8}(x). We then choose the representation ρ such that ρ(x) = \sqrt{−2}. (The other representation will take x to −\sqrt{−2}.) Our choice of ρ guarantees that ρ|_{I_{t}} decomposes as the direct sum of Ψ^{(p^{2}−1)/8} and Ψ′^{(p^{2}−1)/8}. Now

\frac{p^{2} − 1}{8} = \frac{3p − 1}{8} + p\frac{p − 3}{8} = b + pa,

with a = (p−3)/8 and b = (3p−1)/8. Serre’s prediction for the weight is 1+pa+b = 1 + (p^{2} − 1)/8. If we had chosen the other representation (with ρ(x) = −\sqrt{−2}), a similar calculation would have predicted a weight of 1 + 5(p^{2} − 1)/8.

Let ρ_{0} be a lift of ρ to GL_{2}(ℤ(\sqrt{−2})), and let 𝔭 be a prime above p in ℚ(\sqrt{−2}) such that the reduction of ρ_{0} mod 𝔭 is ρ. Then ρ_{0} is an odd, absolutely irreducible, complex two dimensional representation with image isomorphic to S̃_{4}. By the theorem of Langlands-Tunnell[11], ρ_{0} satisfies the conditions of the theorem of Weil-Langlands[10]. Hence, there is a weight 1 eigenform f = ∑ a_{n}q^{n} (of level equal to the Artin conductor of ρ_{0} and character equal to the determinant of ρ_{0}), such that for ℓ ≠ p, a_{ℓ} = Tr(ρ_{0}(Fr_{ℓ})).

If we multiply f by the Eisenstein series E_{p−1} of weight p−1 (which is congruent to 1 mod p), we obtain a modular form fE_{p−1} of weight p. This form might not be an eigenform for all the Hecke operators, but we do have that fE_{p−1}|_{T_{ℓ}} ≡ a_{ℓ}fE_{p−1} (mod 𝔭) for all Hecke operators T_{ℓ} with ℓ ≠ p. A lemma of Deligne-Serre [4, 6.11] then gives the existence of a cuspidal eigenform g = ∑ b_{n}q^{n} (possibly defined over a larger field) which has eigenvalues b_{ℓ} ≡ a_{ℓ} (mod 𝔭′) for some prime 𝔭′ lying over 𝔭, and for ℓ ≠ p. If \overline{b}_{n} is the reduction of b_{n} mod 𝔭′, we see that \overline{b}_{ℓ} = Tr(ρ(Fr_{ℓ})) for ℓ ≠ p.

We have now proven the existence of a characteristic p eigenform of weight p corresponding to ρ. By the epsilon conjecture, there is thus an eigenform of the weight, level, and character predicted by Serre, which corresponds to ρ.

6. Splitting of Primes

For a prime ℓ, not equal to p, we can describe the splitting of ℓ in K̃ by determining the Frobenius of ℓ (up to conjugacy). We note from the character table of S̃_{4}, that the order of the Frobenius element is uniquely determined by its trace, and can be read directly off of the character table, except when the trace is 0, in which case the Frobenius has order 2 or 4. If the Frobenius is of order 2, and has trace 0, it cannot be the central element, so it must be a lift of a transposition in S_{4}, and its fixed field contains K_{12}, but not K, so its fixed field cannot contain ℚ(\sqrt{p^{∗}}). If the Frobenius is of order 4, then it is a lift of a double transposition in S_{4}, and its fixed field contains ℚ(\sqrt{p^{∗}}). In the first case, l cannot split completely in ℚ(\sqrt{p^{∗}}), while in the second case it must. Since the splitting of ℓ in ℚ(\sqrt{p^{∗}}) is completely determined by the Legendre symbol (\frac{ℓ}{p}), we see that if the trace of the Frobenius

is 0, the Frobenius has order 2 if (\frac{ℓ}{p}) = −1, and order 4 if (\frac{ℓ}{p}) = 1. Thus, we can easily determine the conjugacy class of the Frobenius of ℓ from its trace, which is the ℓth coefficient of the Fourier series of the modular form corresponding to ρ.

7. Calculation of Eigenforms

Direct calculation of the eigenform of weight 1 + (p^{2} − 1)/8 is possible, but difficult, since the smallest prime in which we are interested is 59, which gives a weight 436 eigenform. However, Edixhoven has shown [5] that every system of eigenvalues of a characteristic p cusp form comes from the eigenvalues of a cusp form of weight at most p + 1 via the θ operator. More precisely, if f = ∑ a_{n}q^{n} is a modular form (mod p) of weight k, θf = ∑ na_{n}q^{n} is a modular form (mod p) of weight k + p + 1, and we have:

Theorem 4. (Edixhoven) Let f be a cusp form of type (N, k, ϵ). Then there is a cusp form g of type (N, k′, ϵ) and an integer m such that θ^{m}g and f have the same Fourier expansion (mod p), and k′ ≤ p + 1.

This shows that the modular form f which corresponds to our representation (and has weight 1 + (p^{2} − 1)/8) is obtained from a modular form g of weight (p + 5)/4 by applying the (p − 3)/8th power of θ.

We note that upon reduction, the modular form f has coefficients in 𝔽_{p} (in fact, its Fourier coefficients a_{ℓ} for ℓ ≠ p a prime are in the set \{0, ±1, ±2, ±\sqrt{−2}\} of values of the trace of ρ), so that in fact g has coefficients in 𝔽_{p}.

To calculate g explicitly, we begin with a basis for the space of modular forms of weight k and level 1: a suitable basis is the set {G_{4}^{a}G_{6}^{b} : 4a + 6b = k}, where G_{4} and G_{6} are the Eisenstein series of weights 4 and 6, respectively. These forms all have integer coefficients–we reduce them mod p for ease of calculation. It is a simple matter to compute the action of a Hecke operator on this space (in terms of the basis); we find the eigenvectors of the resulting matrix which correspond to eigenvalues in 𝔽_{p}. The space spanned by these eigenvectors contains g, since g has eigenvalues in 𝔽_{p}, and is stable under the action of all the Hecke operators. Thus, we may choose another Hecke operator, and repeat the process on this smaller space. We repeat this until the basis we are left with consists of simultaneous eigenvectors of the Hecke operators. One of these will be the Eisenstein series of weight k, the rest will be cusp forms. We apply the θ operator to the cusp forms the appropriate number of times, eliminate any of the resulting forms which have Fourier coefficients outside the allowable set (described above). This will reduce the choices for f considerably–if more than one form remains, we calculate the splitting for well chosen primes in order to distinguish which corresponds to our Galois representation. We give several examples.

For p = 59, there is a single cuspidal eigenform g of weight (p + 5)/4 = 16. Then θ^{7}g must be the eigenform we want. As a mod p eigenform, g = 7(\overline{G}_{4}^{4} − \overline{G}_{4}\overline{G}_{6}^{2}), where \overline{G}_{4} and \overline{G}_{6} are the mod p reductions of G_{4} and G_{6}. If we calculate θ^{7}g, the coefficients in the Fourier expansion are as follows:

\begin{matrix}ℓ & 2 & 3 & 5 & 7 & 11 & 13 & 17 & 19 & 23 & 29 & 31 & 37 & 41 & 43 & 47 & 53 & 59 & 61 & 67 & 71 \\ a_{ℓ} & 36 & 1 & 58 & 58 & 23 & 0 & 0 & 58 & 0 & 58 & 0 & 0 & 1 & 0 & 0 & 58 & 0 & 23 & 23 & 0 \end{matrix}

For p = 107, there are two cuspidal eigenforms of weight (p+5)/4 = 28. As mod p eigenforms, they are

80\overline{G}_{4}^{7} + 34\overline{G}_{4}^{4}\overline{G}_{6}^{2} + 100\overline{G}_{4}\overline{G}_{6}^{4},

and

51\overline{G}_{4}^{7} + 92\overline{G}_{4}^{4}\overline{G}_{6}^{2} + 71\overline{G}_{4}\overline{G}_{6}^{4}.

If we apply θ^{13} to both of them, we get the following coefficients:

\begin{matrix}ℓ & 2 & 3 & 5 & 7 & 11 & 13 & 17 & 19 & 23 & 29 & 31 & 37 & 41 & 43 & 47 & 53 & 59 & 61 \\ a_{ℓ} & 0 & 1 & 76 & 0 & 106 & 1 & 76 & 1 & 1 & 0 & 0 & 106 & 1 & 31 & 0 & 106 & 0 & 106 \\ b_{ℓ} & 1 & 96 & 52 & 63 & 13 & 104 & 59 & 22 & 35 & 93 & 104 & 32 & 101 & 42 & 80 & 27 & 41 & 46 \end{matrix}

It is clear that the first of these is the form we want.

If we set p = 283, compute the cuspidal eigenforms of weight 72, and compute θ^{35} of each such form, we immediately eliminate all but the two forms with the following Fourier coefficients:

\begin{matrix}ℓ & 2 & 3 & 5 & 7 & 11 & 13 & 17 & 19 & 23 & 29 & 31 & 37 & 41 & 43 & 47 \\ a_{ℓ} & 0 & 156 & 0 & 282 & 282 & 1 & 127 & 0 & 282 & 282 & 0 & 156 & 282 & 127 & 127 \\ b_{ℓ} & 127 & 0 & 127 & 1 & 282 & 282 & 127 & 127 & 282 & 1 & 156 & 127 & 282 & 0 & 0 \end{matrix}

As mod p eigenforms, these two forms correspond to θ^{35} of

212\overline{G}_{4}^{18} + 166\overline{G}_{4}^{15}\overline{G}_{6}^{2} + 83\overline{G}_{4}^{12}\overline{G}_{6}^{4} + 91\overline{G}_{4}^{9}\overline{G}_{6}^{6} + 162\overline{G}_{4}^{6}\overline{G}_{6}^{8} + 148\overline{G}_{4}^{3}\overline{G}_{6}^{10} + 270\overline{G}_{6}^{12},

and

−3\overline{G}_{4}^{18} + 11\overline{G}_{4}^{15}\overline{G}_{6}^{2} + 189\overline{G}_{4}^{12}\overline{G}_{6}^{4} + 37\overline{G}_{4}^{9}\overline{G}_{6}^{6} + 245\overline{G}_{4}^{6}\overline{G}_{6}^{8} + 87\overline{G}_{4}^{3}\overline{G}_{6}^{10}.

These two forms are associated to two non-isomorphic quartic fields of discriminant −283^{3} which correspond to the two representations

−283 = \frac{256(−3)^{3} − 27^{2}}{27} and − 283 = \frac{256(4)^{3} − 155^{2}}{27} .

To determine which form corresponds to which field, it will suffice to determine the splitting of one prime. If we let K̃ be the S̃_{4} extensions corresponding to the first formula, then its quartic subfields are defined by the polynomial f(x) = (x − 3)^{4} + 283x. Now disc(f) is odd, and f is irreducible (mod 2), so 2 is totally inert in the quartic subfield of K̃. This implies that the order of the Frobenius at 2 is divisible by 4. However, a_{2} = 0, and (\frac{2}{283}) = −1, so the first form above corresponds to a field with Frobenius at 2 of order 2. Hence, K̃ corresponds to the second form, and the first form must correspond to the S̃_{4} extension derived from the polynomial (x + 4)^{4} + 283x.

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