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Mathematical Research Letters 10, 205–217 (2003) ICOSAHEDRAL Q-CURVE EXTENSIONS Edray Herber Goins 1. Introduction Let Ρ Be A

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Mathematical Research Letters 10, 205–217 (2003) ICOSAHEDRAL Q-CURVE EXTENSIONS Edray Herber Goins 1. Introduction Let Ρ Be A Mathematical Research Letters 10, 205–217 (2003) ICOSAHEDRAL Q-CURVE EXTENSIONS Edray Herber Goins √ Abstract. We consider elliptic curves defined over Q ( 5) which are either 2- or 3-isogenous to their Galois conjugate and which have an absolutely irreducible mod 5 representation. Using Klein’s classical formulas which associate an icosahedral Galois extension K/Q with the 5-torsion of an elliptic curve, we prove that there is an association of such extensions generated by quintics x5 + Ax2 + Bx+ C satisfying AB = 0 with the aforementioned elliptic curves. 1. Introduction Let ρ be a continuous two-dimensional complex Galois representation, and consider the composition ρ (1) ρ : GQ −−−−→ GL2(C) −−−−→ PGL2(C) As ρ is continuous, there exists a finite extension K which is the field fixed by the kernel of ρ; we consider only the icosahedral Galois extensions i.e. Gal(K/Q) A5. Klein [8], motivated by the isomorphism A5 PSL2(F5), showed in 1884 that any icosahedral Galois extension K/Q is contained in the extension Q(E[5]x) generated by the x-coordinates ofthe 5-torsion√ froma suitable elliptic curve E defined over some quadratic extension of Q( 5). Conversely,√ Shih [13] observed in 1978 that for any elliptic curve E defined over Q( 5) which possesses a p- isogeny λ : E → Eσ with its Galois conjugate√ for a prime p ≡ 2, 3 mod 5 such σ that λ (E[p]) = (ker λ) is rational over Q( 5) and Gal (Q(E[5]x)/Q(ζ5)) A5, there is indeed such an icosahedral Galois extension K ⊂ Q(E[5]x). (Shih’s result is stronger than stated, but this will suffice for the exposition at hand.) In this exposition, I present an explicit characterization ofShih’s result when p =2, 3. My main result is the following. √ Theorem 1.1. Let E be an elliptic curve over Q( 5) which is either 2- or 3- isogenous to its Galois conjugate, and which has an absolutely irreducible mod 5 representation.Then there exists an icosahedral Galois extension K/Q contained in the field Q(E[5]x) generated by the x-coordinates of the 5-torsion of E.Explic- 5 2 itly, K(ζ5)=Q(E[5]x), and K is the splitting field of a quintic x +Ax +Bx+C with AB =0. Received November 13, 2002. Key words and phrases. Galois representations, icosahedral representations, Q -curves. This research was sponsored in part by the National Physical Science Consortium (NPSC), and by the Max Planck Institute for Mathematics in Bonn, Germany. 205 206 EDRAY HERBER GOINS Conversely, if K/Q is an icosahedral Galois extension that is the splitting 5 2 Q field of a quintic x + Ax + Bx√ + C with AB =0, then K(ζ5)= (E[5]x) for some elliptic curve E over Q( 5) that is isogenous to its Galois conjugate. Following the terminology in Ribet [10], an elliptic curve defined over a number field which is isogenous to its Galois conjugates is called a Q-curve; we call an elliptic curve as in the proposition an icosahedral Q-curve, and in turn call such extensions K icosahedral Q-curve extensions with E its associated curve. In particular, any quintic in Bring-Jerrard form x5 − x + C which generates an icosahedral Galois extension actually generates an icosahedral Q-curve extension associated 2-isogenous Q curve. As the ultimate goal is to understand icosahedral Galois representations ρ,I also prove the following. Proposition 1.2. Let K/Q be an icosahedral Q-curve extension.There exist → C → Q Galois representations ρ0 : GQ GL2( ) and ρ5 : GQ GL2( 5) such that 1. ρ0 and ρ5 are odd and continuous. 2. K is the field fixed by the kernel of ρ0. · −1 − ∗ − ∗ 3. ρ5 has nebentype det ρ0 ω5 , where det ρ0 is either ( 1/ ) or ( 3/ ) and ω5 is a primitive Dirichlet character mod 5. 4. ρ0 ≡ ρ5 mod λ5 for some prime above 5. 5. The residual representation ρ = ρ0 mod λ5 has image Z(F5) · SL2(F5) consisting of those matrices in GL2(F5) with square determinant. Conversely, let ρ be a icosahedral Galois representation, and denote K as the field fixed by the kernel of its projectivization.If K/Q is an icosahedral Q-curve extension, then ρ is a twist (of a conjugate) of ρ0. The main results above are general enough for me to prove the following. Corollary 1.3. Let ρ be an icosahedral Galois representation which is unrami- fied outside of {2, 5, ∞}.Then ρ is residually modular. For example, the icosahedral Galois representation considered by Buhler [1] in 1978 may be realized on the 5-torsion ofa modular Q-curve because it can be associated to the splitting field 5 x5 +20x+16, a fact which seems to have never been exploited in the literature. Explicitly, an associated Q-curve is √ √ (2) y2 = x3 + 5 − 5 x2 + 5 x. My interest in stating such a result is to give a partial converse to Buzzard, Dickinson, Shepherd-Barron, and Taylor [3] where they work with complex Ga- lois representations which are unramified at {2, 5}. This result says that ifan odd representation is ramified precisely at {2, 5} then it is residually modular. ICOSAHEDRAL Q -CURVE EXTENSIONS 207 2. Lectures on the Icosahedron We begin by showing the explicit relationship between a certain class ofquin- tics and elliptic curves. Most ofthe exposition that follows in this section is motivated by both Klein [8] and Klute [9]. The group A5 may be realized as the group ofrotations ofthe icosahedron. This Platonic Solid has 12 vertices which may be inscribed on the unit sphere, so that after projecting to the extended complex plane we have a natural group action on the complex numbers 4 ν 2 3 ν (3) ζ5 + ζ5 ζ5 , ζ5 + ζ5 ζ5 for ν ∈ F5; ∞; and 0; generated by the three fractional linear transformations εz+1 1 (4) Sz= ζ z, T z = , and Uz= − ; 5 z − ε z 4 where ζ5 is a primitive fifth root of unity and ε = ζ5 + ζ5 is a fundamental unit. By considering a homogeneous polynomial which vanishes at the vertices, we find three polynomials which are “invariant” under A5 S, T, U: c (z)= z20 +1 − 228 z15 − z5 + 494 z10 4 30 25 5 20 10 (5) c6(z)= z +1 + 522 z − z − 10005 z + z 5 ∆(z)= −z z10 +11z5 − 1 3 − 2 related by c4 c6 = 1728 ∆. (∆ vanishes at the vertices, c4 at the midpoints of the faces, and c6 at the midpoints ofthe edges. Consult Klein [8] and Klute [9] for more details.) Since polynomials are not actually “invariant” under action by fractional linear transformations we consider instead homogeneous rational functions. Define 2 2 2 z2 +1 z2 − 2 εz− 1 z2 +2ε−1 z − 1 λ(z)= − 10 5 − (6) z (z +11z 1) −125 z5 µ(z)= z10 +11z5 − 1 (The numerator of λ(z) vanishes at the fixed points of T, UV4 – the Klein Viergruppe – while the numerator of µ(z) vanishes at one ofthe vertices.) λ(z) has nontrivial action by S so that we may associate a polynomial ofdegree 5, while µ(z) has trivial action by S so that we may associate a polynomial of degree 6. Lemma 2.1. Define c4(z), c6(z), ∆(z), λ(z) and µ(z) as in (5) and (6). 1. We have the identities c3 c3 (µ2 +10µ +5)3 (7) j := 4 = 1728 4 =(λ +3)3 (λ2 +11λ + 64) = 3 − 2 ∆ c4 c6 µ 208 EDRAY HERBER GOINS 2. For each m, n ∈ Q, the five resolvents m n (8) xν = ν + ν ν 2 ν λ(ζ5 z)+3 [λ(ζ5 z)+3][λ(ζ5 z) +10λ(ζ5 z) + 45] 5 2 are roots of the quintic x + Am,n,j x + Bm,n,j x + Cm,n,j where 20 6 mn2 + n3 A = − 2 m3 +3m2 n + 432 m,n,j j 1728 − j 5 3 m2 n2 +2mn3 n4 (9) B = − m4 − 864 + 559872 m,n,j j 1728 − j (1728 − j)2 1 m3 n2 15 mn4 +4n5 C = − m5 − 1440 + 62208 m,n,j j 1728 − j (1728 − j)2 3. Conversely, given a quintic q(x)=x5 + Ax2 + Bx+ C over Q such that A4 −5B3 +25ABC is nonzero, there exist m, n, j ∈ Q( 5 · Disc(q)) such that A = Am,n,j, B = Bm,n,j and C = Cm,n,j. Remark. The discriminant of q(x)=x5 + Ax2 + Bx+ C is Disc(q)=−27 A4 B2 + 256 B5 (10) + 108 A5 C − 1600 AB3 C + 2250 A2 BC2 + 3125 C4 Proof. The identities stated are easily verified using a symbolic calculator. Con- versely, by eliminating m and n in (9), we find that j is a root ofthe equation 5 2 − 3 − 2 5 2 3 (11) δ j 1728 γ4 γ6 + δ j + 1728 γ4 =0 where 54 · δ = A4 − 5B3 +25ABC 2 5 4 2 5 5 3 12 5 · γ4 = 128 A B − 144 B − 192 A C − 600 AB C + 1000 A2 BC2 + 3125 C4 3 10 10 6 3 2 6 (12) 12 5 · γ6 = 1728 A + 10400 A B + 405000 A B − 180000 A7 BC− 1170000 A3 B4 C + 1725000 A4 B2 C2 − 2025000 B5 C2 − 1800000 A5 C3 + 2812500 AB3 C3 − 4687500 A2 BC4 − 9765625 C6 The lemma follows since (11) has discriminant 5 · Disc(q) and m and n may be expressed in terms of j. Lemma 2.2. Let q(x)=x5 + Ax2 + Bx+ C be a quintic over Q such that A4 − 5B3 +25ABC is nonzero, and denote K as its splitting field.There exists an elliptic curve E over Q( 5 · Disc(q)) such that K( 5 · Disc(q)) is the field generated by sum xP + x2P of x-coordinates of the 5-torsion of E.We say that such an E is associated to q(x).
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