COUNTING MOD l SOLUTIONS VIA MODULAR FORMS

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1 COUNTING MOD l SOLUTIONS VIA MODULAR FORMS EDRAY GOINS AND L. J. P. KILFORD Abstract. [Something here] Contents 1. Introduction 1. Galois Representations as Generating Functions 1.1. Permutation Representation 1.. L-Series. Degree d =.1. Decomposition of Permutation Representation.. Example 4.. Relation with Theta Series 4 4. Degree d = Decomposition of Permutation Representation Cubics with Galois Group S Relation with Theta Series Examples Cubics with Galois Group A Degree d = Decomposition of Permutation Representation Quartics with Galois Group S Relation with Modular Forms Quartics with Galois Group A Quartics with Galois Group D 4 16 More Notes Introduction. Galois Representations as Generating Functions.1. Permutation Representation. Let P (z) be a monic polynomial of degree d with integral coefficients. If we denote the roots of P (z) by the set {a 1, a,..., a d }, then we assume that D = i<j (a i a j ) is nonzero; this is equivalent to saying the polynomial has no repeated factors. For each prime l, we wish to determine Mathematics Subject Classification. 11E5, 11E. Key words and phrases. Pythagorean Triples, Ternary Quadratic Forms, Sums of Squares. The authors would like to thank Norm Hurt for inspiration with this project. 1

2 EDRAY GOINS AND L. J. P. KILFORD the integer N P (l) = # { a F l P (a) mod l }. We will do so using Galois representations. Lemma.1. Let G be a group which acts continuously on a set {a 1, a,..., a d }, which we identify as the standard basis of R d : a 1 = [ 1 ] T, a = [ 1 ] T,..., ad = [ 1 ] T, Denote π : G GL d (C) as the permutation representation which sends σ to the matrix corresponding to a k σ(a k ). Then for each σ G we have tr π(σ) = # { a k Q σ(ak ) = a k }. Recall that the trace is invariant under conjugation and hence that this result is independent of the choice of basis for R d. Proof. Clear from the definitions. When {a 1, a,..., a d } is the collection of roots of P (z) of our monic, integral polynomial of degree d, the absolute Galois group G = Gal ( Q/Q ) acts continuously. Hence we have a permutation representation π P : G GL d (C) associated to this polynomial. The next result states how this representation is related to N P (l). Theorem.. Let P (z) be a monic, integral polynomial of degree d with nonzero discriminant. Denote π P : Gal ( Q/Q ) GL d (C) as the permutation representation acting on its roots. For all primes l, N P (l) = tr π P (Frob l ), where Frob l is (any lift of) the Frobenius automorphism. Proof. The absolute Galois group G = Gal ( Q/Q ) acts continuously on the module V = Q[a 1, a,..., a d ] consisting of formal sums of the roots of P (z), so the permutation representation is a continuous homomorphism π P : G GL(V ). Fix an embedding Q Q l, so that we have an embedding Gal ( Q l /Q l ) Gal ( Q/Q ). Recall that we have a short exact sequence {1} I l Gal ( Q l /Q l ) Gal ( Fl /F l ) {1}, where the right-most nontrivial object is a profinite group topologically generated by the Frobenius automorphism Frob l : z z l. Hence we may consider Frob l G to be a well-defined automorphism acting on the submodule V I l consisting of those v V which are fixed pointwise by I l. The restriction of the permutation representation to the decomposition subgroup at l may be identified as a continuous homomorphism π P : Gal ( ) F l /F l GL(V I l ). As P (z) is a monic, integral polynomial, each root a k Z. Denote a k as the image under the composition Z Z l F l. Then Gal ( ) F l /F l acts continuously on V = F l [a 1, a,..., a d ], so that { } { } { } ak F l Frobl (a k ) = a k = ak F l ak F l = a Fl P (a) mod l. By Lemma.1 we must have tr π P (Frob l ) = N P (l).

3 COUNTING MOD l SOLUTIONS VIA MODULAR FORMS.. L-Series. The relation in Theorem. allows for an obvious definition of generating function. We define the L-series associated to P (z) as the Euler product ( ) 1 V L(P, s) = det 1 π P (Frob l ) l s N P (n) = I l n s l mspec(z) for complex numbers s with real part greater than 1. We will see that in several cases we may associate this L-series with the L-series attached to a modular form.. Degree d =.1. Decomposition of Permutation Representation. Say that we have a monic quadratic polynomial P (z) = z + α z + β for some integers α and β such that the discriminant D = α 4 β is nonzero. We discuss the ideas above in more detail. Proposition.1. The permutation representation π P decomposes as the sum of the trivial representation 1 : Gal ( Q/Q ) {1} and a continuous 1-dimensional representation χ P : Gal ( Q/Q ) {±1}, so we have π p 1 χ P. In particular, for all primes l we have N P (l) = 1 + χ P (Frob l ), where Frob l is (any lift of) the Frobenius automorphism. Moreover, the L-series L(P, s) = ζ(s) L(χ P, s) factors as a product of the Riemann Zeta function and a Dirichlet L-series. Proof. The distinct roots of P (z) are a 1 = (α + D)/ and a = (α D)/. As the permutation group of {a 1, a } is generated by the -cycle (1 ), we may view these roots the standard basis vectors of R in order to compute the permutation representation π P : Gal ( Q/Q ) GL (C) associated to P (z): [ 1 a 1 = ] [ a = 1] [ = π P (σ) = [ We may simultaneously diagonalize these matrices: [ M = [ ] = M 1 π P (σ) M = [ ] ] n=1 if σ acts via the 1-cycle 1, if σ acts via the -cycle (1 ). 1 1 ] 1 1 ] if σ acts via 1, if σ acts via (1 ). Hence π P 1 χ P so that tr π P = 1 + χ P. Theorem. gives the desired result. It is easy to find an exact expression for this 1-dimensional representation. A Frobenius automorphism Frob l Gal ( Q/Q ) acts either nontrivially when P (z) has

4 4 EDRAY GOINS AND L. J. P. KILFORD no roots in F l or trivially otherwise. Hence we may identify χ P with the quadratic residue symbol: ( ) D N P (l) = 1 + χ P (Frob l ) = 1 + whenever l D. l.. Example. The quadratic polynomial P (z) = z +z+ has prime discriminant D = 11. For all primes l 11, we find the property { if l is a square modulo 11, and N P (l) = if l is not a square modulo 11. We can relate this to modular forms quite simply, because we can construct the theta series (which is also an Eisenstein series of weight 1 and level 11): Θ P (τ) = q x +xy+y = 1 + q + 4 q + q q q 9 + q 11 +, x,y Z where (as is standard) we define q := exp(πiz), where z is in the Poincaré upper half-plane. Note that the l th coefficient is N P (l)... Relation with Theta Series. More generally, we can associate a similar class of modular forms to a quadratic polynomial P (z) by considering a special class of examples. Proposition.. Fix a prime p mod 4. (1) Consider the monic, integral polynomial P (z) = z + z + p with discriminant D = p. The continuous 1-dimensional representation χ P : Gal ( Q/Q ) C can be expressed in terms of the quadratic residue symbol ( ) l χ P (Frob l ) = whenever l p. p () Let {Q k (x, y)} denote the collection of inequivalent reduced binary quadratic forms of discriminant D = p. In terms of q = e πiτ, we have the q-series expansion Θ P (τ) = ( h ) q Q k(x,y) = h + w N P (n) q n, x,y Z k=1 where h = h( p) is the class number of Q( p) and w = 6 or is the number of units in the ring of integers in Q( p). This is a modular form of weight 1 and level p with nebentype χ P. The L-series is related to this theta series via the Mellin transform Λ(P, s) = p s/ Θ P (i y) h y s dy ( ) p s w y = Γ(s) L(P, s) for Re(s) > 1. π Since L(P, s) = ζ(s) L(χ P, s), this function has a simple pole at s = 1. Hence there is meromorphic continuation to the half-plane consisting of those complex numbers s with positive real part, where we have a functional equation n=1 Λ(P, s) = i Λ(P, 1 s) for < Re(s) < 1.

5 COUNTING MOD l SOLUTIONS VIA MODULAR FORMS 5 For more details, see [?, pg. 58]. Proof. (1) First consider l p. Using Quadratic Reciprocity, ( ) ( ) ( ) D p l χ P (Frob l ) = = =. l l p Now consider l =. The mod reduction of P (z) has two rational roots if and only if the constant term (p + 1)/4 is congruent to modulo. Hence N P (l) = if and only if p 1 mod 8. According to [?, Ch. 6, ] we know that (/p) = 1 if and only if p ±1 mod 8. Since p mod 4, we see that 1 + χ P (Frob l ) = N P (l) = 1 + ( ) p = χ P (Frob l ) = ( ). p () As each Q k (x, y) = a k x + b k x y + c k y is a positive definite quadratic form, we may write Θ P (q) = h + a(n) q n in terms of a(n) = n=1 h k=1 { } # (x, y) Z Q k (x, y) = n. It is shown in [?, Theorem.] that this is a modular form of weight 1 and level p with nebentype χ P. It suffices to show a(l) = w N P (l) for all primes l p. We do so by finding a different expression for a(n). As p mod 4, the ring of integers in Q( p) is given by the set { O = x p y }. x, y Z We can express any quadratic form of discriminant D = p = b k 4 a k c k in the form ( a k Q k (x, y) = Q 1 a k x + b ) k 1 y, y where Q 1 (x, y) = x + x y + p + 1 y ; 4 so that a(n) is the number of α O such that N(α) = Q 1 (x, y) = a k n for some k. Note that the number of units u O is w = # { α O N(α) = 1 }. We know that N P (l) = 1,, or. We show that a(l) = w, w, or, respectively. Say that a(l). We can find α O such that N(α) = a k l for some k. Say that β O also satisfies N(β) = a k l. Since both p = α O and q = β O are prime ideals and O is a Dedekind domain, we have either α or its conjugate is u β for some unit u. Hence either a(n) = w or w depending on whether l ramifies in O (i.e., l = p) or l splits in O (i.e., l p), respectively. When l = p, the number of roots of P (z) modulo p is N P (p) = 1 because the only root is a (p 1)/ mod p. Hence a(p) = w N P (p). When l p, we can find integers x, y and k such that Q(x, y) = N(α) = a k l, so set a x/y mod l. Because P (a) Q(x, y)/y mod l we find that N P (l), and so N P (l) =. Again, a(l) = w N P (l). Finally, say that a(l) =. Then there are no integers x and y such that l = Q k (x, y) for any of the reduced binary quadratic forms. Hence l remains inert in O, so that N P (l) =. Again, a(l) = w N P (l).

6 6 EDRAY GOINS AND L. J. P. KILFORD 4. Degree d = 4.1. Decomposition of Permutation Representation. Say that we have a monic cubic polynomial P (z) = z + α z + β z + γ for some integers α, β, and γ such that the discriminant D = α β 4 β 4 α γ + 18 α β γ 7 γ is nonzero. The following result gives a decomposition of the permutation representation which is similar to the degree d = case above. Proposition 4.1. The permutation representation π P 1 Ind χ P decomposes as the sum of the trivial representation and a -dimensional representation which is induced from a continuous 1-dimensional representation χ P : Gal ( Q/Q( D) ) C. In particular, for all primes l we have N P (l) = 1 + tr Ind χ P (Frob l ) where Frob l is (any lift of) the Frobenius automorphism. Proof. The permutation group of a set {a 1, a, a } is generated by the cycles (1 ) and (1 ). Considering this set to be the roots of a polynomial P (z) with Galois group S = (1 ), (1 ), the reader may find the diagrams in Figure 1 of field extensions and their Galois groups to be a useful reference (we assume here that D is not a square). Figure 1. Subfields and Subgroups corresponding to S = (1 ), (1 ) Q(a 1, a, a ) {1} Q(a 1 ) Q(a ) Q(a ) ( ) (1 ) (1 ) Q( D) (1 ) Q S We may view these roots as the standard basis vectors of R in order to compute the permutation representation π P : Gal ( Q/Q ) GL (C) associated to P (z): 1 1 a 1 = 1 if σ acts via the 1-cycle 1; 1 a = 1 a = 1 1 = π P (σ) = 1 if σ acts via the -cycle (1 ); if σ acts via the -cycle (1 ). 1

7 COUNTING MOD l SOLUTIONS VIA MODULAR FORMS 7 Denoting ζ as a primitive cube root of unity, we will choose the matrix 1 ζ ζ 1 M = 1 ζ ζ = M 1 π P (σ) M = a 11 a a 1 a where the lower block matrix is defined by [ ] 1 if σ acts via the 1-cycle 1; 1 [ ] [ ] a11 a ρ P (σ) = 1 1 = a 1 a 1 [ ] ζ ζ if σ acts via the -cycle (1 ); if σ acts via the -cycle (1 ). Hence π P 1 ρ P for a -dimensional representation ρ P : Gal ( Q/Q ) GL (C). However, it is clear that this -dimensional representation is the induced representation associated with the 1-dimensional representation χ P defined by χ P (σ) = ζ k whenever σ Gal ( Q/Q( D ) acts via the cycle (1 ) k. By Theorem., we have N P (l) = 1 + tr Ind χ P (Frob l ). It is easy to see that the trace and determinant of this -dimensional representation are if σ acts via the 1-cycle 1; tr Ind χ P (σ) = if σ acts via the -cycle (1 ), (1 ) or ( ); 1 if σ acts via the -cycle (1 ) or (1 ). ( ) D det Ind χ P (Frob l ) = whenever l D. l 4.. Cubics with Galois Group S. The cubic polynomial P (z) = z z 1 has prime discriminant D =. Serre [?] showed that for all primes l, if l = x + x y + 6 y for some integers x and y, N P (l) = 1 if l is not a square modulo, and if l = x + x y + y for some integers x and y. He crucially used the fact that the imaginary quadratic field Q( ) has class number h =. We explain how to generalize this by studying all primes p such that the imaginary quadratic field Q( p) has an order with class number h =. We note here that there is an algorithm due to Goldfeld and Gross-Zagier ([?], [?]) which will give a complete list of imaginary quadratic fields of any given class number N, and that in [?] a list of all imaginary quadratic fields with class number less than 1 is given. The list of Q( p) with class number was given by Oesterlé [?].

8 8 EDRAY GOINS AND L. J. P. KILFORD Table 1. Imaginary Quadratic Fields Q( p) containing an Order O with Discriminant D and Class Number p P (z) D Q(x, y) R(x, y) z z 1 x + x y + 6 y x ± x y + y 1 z + z x + x y + 8 y x ± x y + 4 y 59 z + z x + x y + 15 y x ± x y + 5 y 8 z + z + z + 8 x + x y + 1 y x ± x y + 7 y 17 z + z + z + 17 x + x y + 7 y x ± x y + 9 y 19 z 8 z x + x y + 5 y 5 x ± x y + 7 y 11 z z + 11 x + x y + 5 y 5 x ± x y + 11 y 8 z + 4 z x + x y + 71 y 7 x ± 5 x y + 11 y 7 z + z + z 7 x + x y + 77 y 7 x ± x y + 11 y 1 z + z + z x + x y + 8 y 5 x ± x y + 17 y 79 z + z + z x + x y + 95 y 5 x ± x y + 19 y 499 z + 4 z 499 x + x y + 15 y 5 x ± x y + 5 y 547 z + z z x + x y + 17 y 11 x ± 5 x y + 1 y 64 z z x + x y y 7 x ± x y + y 88 z + 5 z 5 z + 88 x + x y + 1 y 1 x ± x y + 17 y 97 z + z 6 z x + x y + 7 y 1 x ± 9 x y + 19 y 11 z + z + z 1 44 x + 11 y x ± x y + 4 y 19 z z + 76 x + 19 y 4 x ± x y + 5 y 4 z + z z 17 x + 4 y 4 x ± x y + 11 y 67 z + z z 5 68 x + 67 y 4 x ± x y + 17 y 16 z 8 x x + 16 y 4 x ± x y + 41 y Proposition 4.. Let the cubic polynomial P (z), discriminant D, and binary quadratic forms Q(x, y) (the principal form) and R(x, y) be as in Table 1. For primes l or p, the number of roots of P (z) modulo l is if l = Q(x, y) for some integers x and y, N P (l) = 1 if l is not a square modulo p, and if l = R(x, y) for some integers x and y. Proof. The cubic polynomials P (z) in Table 1 have discriminant either D = p or D = 4 p. A quick check shows that each is irreducible, so the Galois group is the full symmetric group on letters. We consider two cases, following the argument in [?, p. 47]: Case #1: l is not a square modulo p. Using the construction in Proposition 4.1, ( ) ( ) ( ) [ ] D p l 1 det ρ P (Frob l ) = = = = 1 = ρ P (Frob l ) =. l l p 1 Hence N P (l) = 1 + tr ρ P (Frob l ) = 1. Case #: l is a square modulo p. Since det ρ P (Frob l ) = +1, we see that either [ ] [ ] 1 ζ ρ P (Frob l ) = or ρ 1 P (Frob l ) = ζ.

9 COUNTING MOD l SOLUTIONS VIA MODULAR FORMS 9 In particular, N P (l) = 1 + tr ρ P (Frob l ) equals either or. We differentiate these cases by considering ray class fields. Using Magma [?], it is easy to check the order [ ] { 1 + p 1 when D = p, O = Z + f Z having conductor f = when D = 4 p; has discriminant D and class number. Similarly, it is easy to check that inequivalent reduced binary quadratic forms of discriminant D are Q(x, y) and R(x, y) as in Table 1. By [?, Proposition.15], we can find integers x and y such that either l = Q(x, y) or l = R(x, y). Note that N(α) = Q(x, y) for any α O. The quadratic extension Q( p)/q is unramified away from p, whereas the cubic extension Q(a 1, a, a )/Q( p) associated with the splitting field of P (z) is unramified away from the conductor f. Following [?, 7.C], consider those subsets a Q( p) which are fractional ideals of O (i.e., nonzero finitely generated O-modules), proper (i.e., β a a if and only if β O), and relatively prime to the conductor (i.e., a + f O = O). Let Pic(O) denote the quotient group of these ideals modulo those which are principal (i.e., a = α O for some α Q( p). Then [?, Prop. 7.19, Prop 7., Thm. 8.] together imply that Pic(O) Gal (Q(a 1, a, a )/Q( p)). As det ρ P (Frob l ) = +1, we see that l splits completely in the ring of integers of Q( p). Then N P (l) = if and only if l splits completely in the ray class field Q(a 1, a, a ), and by by the isomorphism above, this happens if and only if l O = p 1 p for some principal primes p 1 = α 1 O and p = α O. Since N(α) = Q(x, y), we conclude that N P (l) = when l = Q(x, y), and N P (l) = when l = R(x, y). 4.. Relation with Theta Series. Given a cubic, monic polynomial P (z), we have seen in Proposition 4.1 that the permutation representation π P decomposes as the sum of the trivial representation and a -dimensional representation: π P 1 Ind χ P. The latter component is induced from a continuous 1-dimensional representation χ P : Gal ( Q/Q( D) ) C. In particular, the L-series factors as L(P, s) = ζ(s) L(χ P, s), where the first factor is the Riemann zeta function and the second is a Hecke L- function. Hecke [?] showed that such functions have meromorphic continuation to the entire complex plane via a functional equation. It is well-known that the Hecke L-function is associated to a modular form through a theta correspondence ; see [?, Prediction 1.8.4]. We give an explicit correspondence for a special class of examples. Corollary 4.. Let the cubic polynomial P (z), discriminant D, and binary quadratic forms Q(x, y) and R(x, y) be as in Table 1. In terms of q = e πiτ, we have the q-series expansion f P (τ) = 1 q Q(x,y) q R(x,y) = a(n) q n ; x,y Z n=1

10 1 EDRAY GOINS AND L. J. P. KILFORD where a(l) = tr Ind χ P (Frob l ) for all primes l. Moreover, this is a normalized cusp form of weight 1 and level D which satisfies L(P, s) = ζ(s) L(f P, s). Proof. Following the argument in the proof of Proposition., we have the following two theta series: { q Q(x,y) = 1 + A(n) q n N P (l) if l = Q(x, y) for some x, y; ; A(l) = otherwise. x,y Z x,y Z q R(x,y) = 1 + n=1 B(n) q n ; B(l) = n=1 { if l = R(x, y) for some x, y; otherwise. (Here l p, and N P (l) = 4 if D is divisible by 4 and otherwise.) Hence the difference is the series N P (l) if l = Q(x, y) for some x, y; q Q(x,y) q R(x,y) = C(n) q n ; C(l) = if (l/p) = 1; and x,y Z n=1 if l = R(x, y) for some x, y. In particular, C(1) = and C(l) = (tr Ind χ P (Frob l )). Since L(χ P, s) = L(f P, s), the result follows Examples. Consider the cubic polynomial P (z) = z z 1 with discriminant D = as first studied by Serre. There is a unique normalized cusp form of weight 1 and level, so we find the theta correspondence f P (τ) = 1 q x +xy+6y q x +xy+y = η(τ) η( τ) x,y Z in terms of the Dedekind eta function η(τ) = q 1/4 n 1 (1 qn ). Note that the l th coefficient of this modular form is a(l) = N P (l) 1. Some explicit data can be found in Table. Table. Counting Roots of P (z) = z z 1 modulo l and Coefficients of η(τ) η( τ) = n 1 a(n) qn l a l N P (l),,1,9,1,41,47,71,7, ,7,11,17,19,7,4,5,61, ,... Similarly, consider the cubic polynomial P (z) = z +z +z 1 with discriminant D = 44. There is a unique normalized cusp form of weight 1 and level 44, so we find the theta correspondence f P (τ) = 1 q x +11y q x +xy+4y = η( τ) η( τ). x,y Z Again, note that the l th coefficient of this modular form is also a(l) = N P (l) 1. Explicit data can be found in Table.

11 COUNTING MOD l SOLUTIONS VIA MODULAR FORMS 11 Table. Counting Roots of P (z) = z + z + z 1 modulo l and Coefficients of η( τ) η( τ) = n 1 a(n) qn l a l (f) N P (l),5,,1,7,59,67,71,89,97, ,7,1,17,19,9,41,61,7,79,8, ,5,..., Cubics with Galois Group A. The cubic polynomials P (z) = z +z z 1 and P (z) = z +z z 1 have discriminants D = 7 and D = 19, respectively. Since these discriminants are square, the Galois groups of the splitting fields are abelian. Proposition 4.4. Fix a prime p 1 mod. Let ζ p be a primitive pth root of unity, and let P (z) be that cubic which is the minimal polynomial of the resolvent 1 p 1 a ζ k p, k=1 where a is a generator of the cyclic group (Z/p Z). Then for primes l p, { if l p 1 1 mod p; N P (l) = otherwise. Proof. The splitting field of P (z) is contained in the cyclotomic field Q(ζ p ). This splitting field must be the unique cubic subfield, so that P (z) has Galois group corresponding to the -cycle (1 ). (Compare with Figure 1, where D = p is not a square.) The -dimensional representation ρ P constructed in Proposition 4.1 is reducible; in fact, the permutation representation decomposes as π P 1 χ P χ P for some cubic character χ P : Gal ( Q/Q ) ζ. This character must be the cubic residue symbol, i.e., χ P (Frob l ) l (p 1)/ (mod p) for all primes l p. The result follows from Theorem.. 5. Degree d = Decomposition of Permutation Representation. Say that we have a monic quartic polynomial P (z) = z 4 + α z + β z + γ z + δ for some integers α, β, γ, and δ such that the discriminant D = α β γ 4 β γ 4 α γ + 18 α β γ 7 γ 4 4 α β δ + 16 β 4 δ + 18 α β γ δ 8 α β γ δ 6 α γ δ β γ δ 7 α 4 δ α β δ 18 β δ 19 α γ δ + 56 δ is nonzero. The following result is a generalization of Propositions.1 and 4.1. Proposition 5.1. The permutation representation π P 1 Sym ρ P decomposes as the sum of the trivial representation and a -dimensional representation which the symmetric square of a -dimensional representation ρ P : Gal ( Q/Q ) GL (C). In particular, for all primes l we have [ N P (l) = 1 + tr ρ P (Frob l )] det ρ P (Frob l ),

12 1 EDRAY GOINS AND L. J. P. KILFORD where Frob l is (any lift of) the Frobenius automorphism. Figure. Subfields and Subgroups corresponding to S 4 = (1 ), (1 4) Q(a 1, a, a, a 4 ) {1} Q(θ 1, θ, θ ) Q( D) 4 6 Q(θ 1 ) Q Q(a ) 4 6 V 4 D A 4 S 4 S Proof. The permutation group of a set {a 1, a, a, a 4 } is generated by the cycles (1 ), (1 ) ( 4), (1 ) and (1 4). Considering this set to be the roots of a polynomial P (z) with Galois group S 4 = (1 ), (1 4), the reader may find the diagrams in Figure of field extensions and their Galois groups to be a useful reference (note that here D is not a rational square). Recall that the following are subgroups of the symmetric group: S 4 = (1 ), (1 4) A 4 = (1 ) ( 4), (1 ) V 4 = (1 ) ( 4), (1 4) D 4 = (1 ) ( 4), (1 4) S = (1 ), (1 ) Note that θ 1 = a 1 a + a a 4 is a root of the resolvent cubic P (z) = z β z + ( α γ 4 δ ) z + ( γ α δ + 4 β δ ) which also has discriminant D, and θ and θ are the other roots of this cubic.

13 COUNTING MOD l SOLUTIONS VIA MODULAR FORMS 1 We may view these roots as the standard basis vectors of R 4 in order to compute the permutation representation π P : Gal ( Q/Q ) GL 4 (C) associated to P (z): 1 a 1 = a = 1 a = 1 a 4 = = π P (σ) = if σ acts via (1 ); if σ acts via (1 ) ( 4); if σ acts via (1 ); and if σ acts via (1 4). Denoting ζ 8 as a primitive eighth root of unity, we will choose the 4 4 matrix M as below to conjugate the matrix of π P (σ) with; this reveals the following structure: ζ8 ζ8 1 M = 1 ζ8 1 ζ ζ8 ζ8 = M 1 π P (σ) M = a a b b a c a d + b c b d 1 ζ8 1 ζ8 1 c c d d where the lower block matrix is the image of the symmetric square representation GL (C) GL (C) defined by [ ] a Sym a11 a 11 a 11 a 1 a 1 1 = a a 1 a 11 a 1 a 11 a + a 1 a 1 a 1 a a 1 a 1 a a

14 14 EDRAY GOINS AND L. J. P. KILFORD and each matrix is defined by ρ P (σ) = [ ] a11 a 1 = a 1 a ± ± 1 ζ 8 ζ 8 [ 1 1 [ ] ] [ ] ± 1 ζ 8 ζ 8 ζ 8 ζ8 ζ8 ζ8 if σ acts via (1 ); if σ acts via (1 ) ( 4); if σ acts via (1 ); and ± [ ] ζ 8 ζ8 if σ acts via (1 4). This shows that the 4-dimensional representation π : S 4 GL 4 (C) may be decomposed into a trivial 1-dimensional representation 1 : S 4 GL 1 (C) and the symmetric square of a -dimensional representation ρ : S 4 GL (C). Note that regardless of the sign ±, for any σ S 4 we have [ ] a b ρ(σ) = ± = [tr ρ(σ)] = a + a d + d and det ρ(σ) = a d b c; c d so that tr Sym ρ(σ) = a + (a d + b c) + d = [tr ρ(σ)] det ρ(σ). We remark here that by Tunnell s theorem [?] it is known that the octahedral representation ρ : S 4 GL (C) is modular.

15 COUNTING MOD l SOLUTIONS VIA MODULAR FORMS 15 [This hasn t been checked!] 5.. Quartics with Galois Group S Relation with Modular Forms. As in the degree case, we can, using work of Langlands and Tunnell, associate a modular form to the continuous odd irreducible -dimensional representation that we obtain in the degree 4 case, and as before this modular form has weight 1. Let us consider two examples; ( if we take p = 8 and P (x) = x 4 x 1, then the modular form f S 1 Γ (8); ( 8)) of weight 1 that corresponds to the representation ρ has Fourier expansion beginning f(q) = q + q q q 4 q 5 + O(q 6 ). This example was first found by Serre. Putting this together, this means that we have that, if a l is the l th Fourier coefficient of f, then ( ) l N P (l) = 1 + a l, if l 8. 8 Similarly, if we ( take p = 491 and P (x) = x 4 x x + x 1, then the modular form g S 1 Γ (8); ( 8)) of weight 1 that corresponds to the representation ρ has Fourier expansion which begins g(q) = q + q q q 4 q 6 + q 7 + q 11 + O(q 1 ), and we have, if b l is the l th Fourier coefficient of g, then ( ) l N P (l) = 1 + b l, if l In general, the field of definition of the modular form is going to be the imaginary quadratic field Q( ). This will be precisely the field of definition because the modular form is a cusp form of weight 1: indeed, the norms of its Fourier coefficients a p are bounded by by the Ramanujan-Petersson conjecture Quartics with Galois Group A 4. The case where the Galois group is A 4 is known as the tetrahedral case; it is known by work of Hecke that this is modular. There are some references to tetrahedral forms in the literature; Tate [?] and Chinburg [?] both considered a particular tetrahedral modular form of level 1 = 7 19 corresponding to the polynomial P (x) = x 4 + x 7x + 4; this form has Fourier expansion which begins q + (ω 1 1)q + ω 1 q ω 1 q 5 + O(q 6 ), where ω 1 has minimal polynomial x 4 x + 1. Another form was considered by Crespo in [?]; this corresponds to the polynomial P (x) = x 4 x + x x + and has Fourier expansion beginning q iq q 5 iq 11 + iq 15 q 17 iq 19 iq iq 7 + iq 1 + O(q ) where i is the normal square root of 1. By specializing Proposition 5.1, we have the result that N( l (f) ) = (a l ) because the discriminant is always square in the alternating case, so = 1. D p

16 16 EDRAY GOINS AND L. J. P. KILFORD Table 4. Hilbert Class Fields of Imaginary Quadratic Fields Q( p) with an Order O with Discriminant D and Class Number 4 p P (z) D Q(x, y) R(x, y) 17 z 4 + 8z x + xy + 9y x ± xy + 6y 7 z z x + xy + 7y 7x ± 4xy + 11y 97 z z x + xy + 49y 7x ± xy + 14y 19 z z x + xy + 97y 11x ± 8xy + 19y 5.5. Quartics with Galois Group D 4. We used Magma ([?]) to find the Hilbert class fields of the imaginary quadratic fields in this table. 4 if l ( is ) represented by the principal quadratic form x + py, l N P (l) = if p = 1 and ( ) 1 l = 1, and if l is represented by either Q or R. More Notes You mentioned in your that there are explicit formulas for N p (f) given certain conditions on the primes p. Let me use the description above in terms of Galois representations to prove the following: Proposition Let f(x) = x x 1. Then N p (f) = or whenever p = a + a b + 6 b for some integers a and b.. Let f(x) = x 4 x + x x +. Then N p (f) =, 1, or 4 whenever p or 7.. Let f(x) = x 4 x 1. Then N p (f) =, 1, or 4 whenever p = a + a b + 71 b for some integers a and b. Proof. For f(x) = x x 1, we use the ideas in the proof of Proposition to construct the following table: σ p (1) (1 ) (1 ) tr ρ f (σ p ) 1 det ρ f (σ p ) N p (f) 1 Hence N p (f) = or whenever det ρ f (σ p ) = 1. Similarly, for either f(x) = x 4 x + x x + or f(x) = x 4 x 1, we use the ideas in the proof of Proposition to construct the following table: σ p (1) (1 ) (1 ) ( 4) (1 ) (1 4) [tr ρ f (σ p )] 4 1 det ρ f (σ p ) N p (f) 4 1 Hence N p (f) =, 1, or 4 whenever det ρ f (σ p ) = 1. We focus on the determinant of the associated -dimensional representation. Denote a 1, a,..., a n as the roots

17 COUNTING MOD l SOLUTIONS VIA MODULAR FORMS 17 of f(x). We know that the local Galois group G p = Gal ( F p /F p ) acts on these roots, so in particular it acts on the quantity for Wilton s (199) example; Disc(f) = (a i a j ) = i<j 7 for Crespo s (1997) example; and 8 for Serre s () example. ( Disc(f) ) Since det ρ f = ±1 and σ = ± Disc(f) for any σ G Q, we see that det ρ f (σ p ) = 1 Disc(f) σ p ( Disc(f) ) = ( Disc(f) p ). When f(x) = x 4 x + x x +, we see that det ρ f (σ p ) = +1 whenever p, 7. Otherwise, in the other two cases, Disc(f) = l for some prime l (mod 4). It suffices to show that det ρ f (σ p ) = +1 whenever p = a + a b + l + 1 b 4 for some integers a and b. Indeed, since p does not divide b, we have ( 1 + a ) ( ) = l+ p = det ρ f (σ p ) = b b ( l p ) ( ) 1 + a/b = = +1. p This completes the proof! Proposition. Let ρ : G GL(V ) be a representation acting on a -dimensional vector space V defined over a field k with characteristic not dividing n. The representation ρ n = ρ ρ ρ acting on the n -dimensional vector space V n = V V V decomposes as [ (det ρ) Sym ρ ] Sym ρ for n = ; ρ n [ (det ρ) Sym 1 ρ ] [ (det ρ) Sym 1 ρ ] Sym ρ for n = ; [ (det ρ) Sym ρ ] [ (det ρ) Sym ρ ] Sym 4 ρ for n = 4. Proof. We give an elementary proof of this well-known fact using explicit matrices. First, we define ρ n and Sym n 1 ρ. Upon choosing a basis {e 1, e,..., e m } for V, we identify V n as that m n -dimensional k-vector space spanned by the tensor products e i1 e i e in. Hence, when m =, we find the following matrices

18 18 EDRAY GOINS AND L. J. P. KILFORD (ρ n ) (σ) by considering how each σ G acts: [ ] a b ρ(σ) = c d a a b a b b (ρ ρ) (σ) = a c a d b c b d a c b c a d b d c c d c d d a a b a b a b a b a b a b b a c a d a b c a b d a b c a b d b c b d a c a b c a d a b d a b c b c a b d b d (ρ ρ ρ) (σ) = a c a c d a c d a d b c b c d b c d b d a c a b c a b c b c a d a b d a b d b d a c a c d b c b c d a c d a d b c d b d a c b c a c d b c d a c d b c d a d b d c c d c d c d c d c d c d d a a b a b a b a b a b a b b a c a d a b c a b d a b c a b d b c b d a c a b c a d a b d a b c b c a b d b d (ρ ρ ρ ρ) (σ) = a c a c d a c d a d b c b c d b c d b d a c a b c a b c b c a d a b d a b d b d. a c a c d b c b c d a c d a d b c d b d a c b c a c d b c d a c d b c d a d b d c c d c d c d c d c d c d d We identify Sym n 1 V as that n-dimensional k-vector subspace which is the image of the map V n V n that sends e i1 e i e in σ S n e σ(i1) e σ(i) e σ(in). Hence we have the following matrices: [ ] Sym 1 a b ρ(σ) = c d Sym ρ(σ) = a a b b a c a d + b c b d c c d d a a b a b b Sym ρ(σ) = a c a d + a b c a b d + b c b d a c b c + a c d b c d + a d b d c c d c d d a 4 4 a b 6 a b 4 a b b 4 a c a d + a b c a b d + a b c a b d + b c b d Sym 4 ρ(σ) = a c a c d + a b c a d + 4 a b c d + b c a b d + b c d b d a c a c d + b c a c d + b c d a d + b c d b d c 4 4 c d 6 c d 4 c d d 4 In general, a Mathematica command which generates the matrix is Sym[n_Integer, rho_matrix] := Table[ CoefficientList[ (rho[[1,1]] + rho[[1,]]*x)^(n-k) * (rho[[,1]] + rho[[,]]*x)^k,

19 COUNTING MOD l SOLUTIONS VIA MODULAR FORMS 19 x], {k,, n} ] In what follows, let D = a d b c to aid notation. It is easy to check that ρ G ρ GL (k) Sym q ρ n GL n(k) (det ρ) p det k det G GL (k) GL q+1 (k) GL q+1 (k) k ] [ ] det [ρ n (σ) = D n n 1 and det (det ρ) p Sym q ρ (σ) = D (p+q)(q+1)/. We wish to decompose ρ n = p+q=n ] m q [(det ρ) p Sym q ρ for some multiplcities m q. Clearly necessary conditions and q < n and n = p+q=n m q (q + 1). Consider n =. Denote the 4 4 matrix M = = M 1 (ρ ρ) (σ) M = D a a b b a c a d + b c b d. c c d d Since det M =, we see that the decomposition M 1 (ρ ρ) M = (det ρ) Sym ρ holds for fields k of characteristic different from. Now consider n =. We define the 8 8 matrix M to be M = By explicit computation, we see that the conjugate matrix M 1 (ρ ρ ρ) (σ) M is given by Da Db Dc Dd Da Db Dc Dd a a b a b b a c a d + a b c a b d + b c b d a c b c + a c d b c d + a d b d c c d c d d Since det M = 9 we see that the decomposition. M 1 (ρ ρ ρ) M = [(det ρ) ρ] [(det ρ) ρ] Sym ρ holds for fields k of characteristic different from. Finally, consider n = 4. We define the matrix M to be

20 EDRAY GOINS AND L. J. P. KILFORD Purdue University, 15 North University Street, West Lafayette, IN 4797, United States address: University of Bristol, Department of Mathematics, University Walk, Bristol, BS8 1TW, United Kingdom address: