An octahedral-elliptic type equality on Br_{2}(k)

Joan-C. Lario and Anna Rio

Departament de Matemàtica Aplicada II Universitat Politècnica de Catalunya February 2, 1995

Abstract. Let k be a field of characteristic ≠ 2, 3 which does not contain the cubic roots of unity. Given a quartic irreducible polynomial f ∈ k[x] with discriminant −3 in k^{∗}/k^{∗}^{2} and Galois group S_{4}, we describe in a explicit way the equivalence between the obstruction of a Galois embedding problem of octahedral type attached to f and the obstruction to the existence of an elliptic curve having the splitting field of f as field of x-coordinates of its 3-torsion points.

1. Introduction. Let k be a field of characteristic ≠ 2, 3 which does not contain the cubic roots of unity. Let f = x^{4} + a x^{2} + b x + c ∈ k[x] be an irreducible polynomial with discriminant d_{f} ≡ −3 (mod k^{∗ 2}) and splitting field K such that Gal(K/k) ≃ S_{4}. Since the group PGL(2, 3) is isomorphic to the symmetric group S_{4}, this is equivalent to giving a Galois represetation

ϱ: Gal( \overline{k}/k) ⟶ PGL(2, 3)

whose image is the whole group, and such that the fixed field of det ϱ is the quadratic field k(\sqrt{−3} ).

We consider the embedding problem given by the diagram

Gal( \overline{k}/k) u u ff ff ff ff ff fl

Gal(K/k) y u

1  wF_{3}^{∗}  wGL(2, 3)  wPGL(2, 3)  w1,

and we claim that its solvability is equivalent to the existence of an elliptic curve E, defined over k, such that the linear liftings of ϱ are the representations giving the action of Gal( \overline{k}/k) on the 3-torsion points of the twists of E by quadratic characters of k, and we provide a method to obtain it explicitly. In fact, if the embedding problem is solvable, there are infinitely many such elliptic curves and the method gives a parametrization of this family.

Research of the authors was supported by DGICYT PB93-0815 and PB93-0034, respectively.

Email addresses: lario@ma2.upc.es and rio@ma2.upc.es

2. A Galois embedding problem of octahedral type. Since GL(2, 3) is the double cover of S_{4} in which transpositions lift to elements of order two, the Galois embedding problem described above is the one studied by Serre in [1] (this group extension is denoted S̃_{4} there). The problem is solvable if and only if the element in the Brauer group of k

η = w(Q_{K_{1}}) ⊗ (2, d_{f})

is trivial. Here w denotes the Witt invariant, K_{1} = k(x_{1}), where x_{1} is a root of f, and Q_{K_{1}} is the quadratic form Tr_{K_{1}/k}(X^{2}).

Proposition 1. Let f = x^{4} + a x^{2} + b x + c ∈ k[x] be an irreducible polynomial with discriminant d_{f} ≡ −3 (mod k^{∗ 2}). Let δ = 2 a^{3} + 9 b^{2} − 8 a c. The embedding problem is solvable if and only if

(−6a, δ) = 1,

provided that a δ ≠ 0. The embedding problem is always solvable when a δ = 0.

If a and δ are both non-zero, the matrix

P = ( \begin{matrix}1/2 & 0 & a^{2} & b(4 a^{3} + 27 b^{2} − 48 a c)/2 \\ 0 & 1 & −3 b & 4 a^{4} + 21 a b^{2} − 24 a^{2} c + 32 c^{2} \\ 0 & 0 & 2 a & −2 b (a^{2} + 12 c) \\ 0 & 0 & 0 & δ \end{matrix} ) ∈ GL(4, k)

gives

P^{t} · Q_{K_{1}} · P = ⟨1, −2 a, −2a δ, −δ d_{f}⟩,

which under our assumptions is equivalent to ⟨1, −2a, −2 a δ, 3 δ⟩. Therefore the Witt invariant of the trace form is (−6 a, δ)(−1, 3) and the obstruction given by Serre’s formula is (−6a, δ)(−1, 3)(2, −3). Since (−1, 3)(2, −3) = (2, −1)(3, −2) = 1 the result follows. In the case a δ = 0, the obstruction can be written as (−1, 3)(2, −3) and is therefore trivial.

3. Tschirnhausen relation and 3-torsion of elliptic curves. Let E be an elliptic curve defined over k. When we consider a Weierstraß model y^{2} = x^{3} − 27 c_{4} x − 54 c_{6} over k for E, the roots of the polynomial

g(x) = x^{4} − 54 c_{4} x^{2} − 216 c_{6} x − 243 c_{4}^{2}

are the four x-coordinates of the 3-torsion points on this affine model. A change of model for E yields a new polynomial, but the splitting field of g(x) in a fixed algebraic closure of k remains the same. Denote by K this splitting field and observe that k(\sqrt{−3}) ⊂ K. We call such a field of 3-torsion elliptic type. Our goal is to describe the quartic polynomials having splitting field of 3-torsion elliptic type. For the sake of simplicity we have restricted ourselves to the case of irreducible polynomials with S_{4} Galois closure.

Let f(x) = x^{4} + a x^{2} + b x + c be an irreducible polynomial in k[x]. Observe that if a^{2} + 12 c = 0, then the roots of f(x) are the x-coordinates of the 3-torsion points of the elliptic curve y^{2} = x^{3} + a/2 x + b/4. Note that its discriminant is Δ = −(8 a^{3} + 27 b^{2}) ≠ 0. Otherwise there is an obstruction for the splitting field of f to be of 3-torsion elliptic type.

Proposition 2. Let f(x) = x^{4} + a x^{2} + b x + c be an irreducible polynomial in k[x] with discriminant d_{f} = −3 d^{2} and a^{2} + 12 c ≠ 0. Let K be the splitting field of f(x) and assume Gal(K/k) ≃ S_{4}. Let α = 2 a^{3} + 9 b^{2} − 8 a c − 3 d and β = −2 a^{3} − 27 b^{2} + 72 a c + 9 d. Then K is of 3-torsion elliptic type if and only if

(−6 a, −3 α β) = 1,

provided that a α ≠ 0. The field K is always of 3-torsion elliptic type when a α = 0.

To see this, first observe that any other polynomial having splitting field K is obtained from f(x) by a Tschirnhausen transformation:

f_{∗}(x) = Irr(y_{1}, k, x),

where y_{1} = m x_{1}^{3} + n x_{1}^{2} + p x_{1} + q with (m, n, p) ∈ k^{3} − {0}. Since the Galois group of f is S_{4}, any such f_{∗}(x) is of degree 4. We will take q = (3 b m + 2 a n)/3 in order to consider only reduced polynomials (with no cubic term) and hence the question turns out to be whether there exist m, n, and p such that f_{∗}(x) = x^{4}+a_{∗}x^{2}+b_{∗}x+c_{∗} satisfies a_{∗}^{2} + 12 c_{∗} = 0.

It is easy to see that the above Tschirnhausen transformation yields:

(a^{2} + 12 c)(a_{∗}^{2} + 12 c_{∗}) = Q(m, n, p) \overline{Q}(m, n, p) ,

where

Q(m, n, p) =(2 a^{4} + 9 a b^{2} − 6 a^{2} c + 24 c^{2} − 3 a d)/2 m^{2} + b (a^{2} + 12 c) m n+
(−4 a^{3} − 9 b^{2} − 16 a c + 3 d)/2 m p + (2 a^{3} + 9 b^{2} − 8 a c − 3 d)/2 n^{2}+
(a^{2} + 12 c) p^{2}

and \overline{Q}(m, n, p) is obtained from Q by the conjugation d ↦ −d, denoted in what follows by a bar. Observe here that β\overline{β} = 4 (a^{2} + 12 c)^{3} ≠ 0.

If a α ≠ 0, the quadratic form Q is equivalent to ⟨a^{2}+12 c, 2 α, −α β d_{f} (a^{2}+12 c)⟩ and the condition for this to represent 0 is

(−1, −6 β)(2 α, 3 α β)(a^{2} + 12 c, −3 α β) = 1,

which is equivalent to the one given in the statement. On the other hand, if a α = 0 the quadratic form always represents 0.

At this point it is worth noting that both quadratic forms Q and \overline{Q} represent zero or they do not at the same time, since

(−6 a, −3 α β)(−6 a, −3 \overline{α} \overline{β}) = (−6 a, (a^{2} − 4 c)^{2} + 6 a b^{2}) = 1.

To end this section, we describe explicitly the family of elliptic curves over k with common 3-torsion elliptic type field. We begin with an irreducible polynomial f(x) = x^{4}+a x^{2}+b x+c having a^{2}+12 c = 0. For any (m, n, p) ∈ k^{3}−{0} a reduced Tschirnhausen transformation of f has the form f_{∗}(x) = x^{4} + a_{∗}x^{2} + b_{∗}x + c_{∗}, with

a_{∗}^{2} + 12 c_{∗} = (8 a^{3} + 27 b^{2}) Q_{1}(m, n, p) Q_{2}(m, n, p)

where

Q_{1}(m, n, p) = a m^{2} + n^{2} − m p

and

Q_{2}(m, n, p) = (7 a^{2} m^{2} + 12 b m n + 8 a n^{2} − 20 a m p + 12 p^{2})/36.

The rational points of these two irreducible components, namely the pair of conics C_{1} : Q_{1}(X, Y, Z) = 0 and C_{2} : Q_{2}(X, Y, Z) = 0, parametrize all the elliptic curves over k having the splitting field of f(x) as field of x-coordinates of its 3-torsion points. It is easy to see that to each rational point corresponds an elliptic curve along with its quadratic twists. One also checks that the four intersection points [X_{i} : Y_{i} : Z_{i}] of these two conics give the roots Y_{i}/X_{i} of the polynomial f.

The rational points of the conics can be parametrized by the following birational maps

q_{1} : P^{1} → C_{1}
t ↦ [1 : t : a + t^{2}]

q_{2} : P^{1} → C_{2}
t ↦ [2 a(3 a + 8 t^{2}) : −8 t (a^{2} + 3 b t) : a (7 a^{2} + 12 b t + 8 a t^{2})]

After some algebraic symbolic computations, we can give the elliptic curves over k in Weierstraß form

y^{2} = x^{3} + A_{i}(t) x + B_{i}(t),

where i = 1, 2, corresponding to the points q_{i}(t) ∈ C_{i}, for any t ∈ P^{1}(k). We also give their discriminants Δ_{1}(t) and Δ_{2}(t). In the following formulae we have set Δ = −(8 a^{3} + 27 b^{2}) = Δ_{1}(∞):

A_{1}(t) =\frac{−2 a^{3} − 9 b^{2} − 12 a b t − 24 a^{2} t^{2} + 72 b t^{3} + 24 a t^{4}}{48} ;

A_{2}(t) =\frac{Δ a (3 a^{3} + 48 a^{2} t^{2} + 192 b t^{3} − 64 a t^{4})}{12} ;

B_{1}(t) =(−3 a^{3} b − 9 b^{3} − 8 a^{4} t − 18 a b^{2} t + 30 a^{2} b t^{2} − 180 b^{2} t^{3}−
180 a b t^{4} − 96 a^{2} t^{5} + 72 b t^{6})/288 ;

B_{2}(t) =Δ (9 a^{6} b − 48 a^{7} t − 360 a^{5} b t^{2} − 1440 a^{3} b^{2} t^{3} + 960 a^{4} b t^{4}−
256 a^{2} (4 a^{3} + 9 b^{2}) t^{5} − 1536 b (a^{3} + 3 b^{2}) t^{6})/36 ;

Δ_{1}(t) = − Δ f(t)^{3} = −27 Δ Q_{2}(q_{1}(t))^{3} ;

Δ_{2}(t) = − \frac{Δ^{2}}{27} Q_{1}(q_{2}(t))^{3} .

4. An equality in the Brauer group. Now we are able to show the equivalence between the problems considered in the two previous sections: on one side the solvability of an embedding problem of octahedral type and on the other the existence of an elliptic curve with a given mod 3 projective Galois representation.

Theorem 3. Let k be a field of characteristic ≠ 2, 3 which does not contain the cubic roots of unity. Let f(x) ∈ k[x] be an irreducible quartic polynomial with discriminant d_{f} ≡ −3 (mod k^{∗ 2}) and splitting field K, such that Gal(K/k) ≃ PGL(2, 3). Then, the following three conditions are equivalent:

1) the Galois embedding problem GL(2, 3) ↠ PGL(2, 3) ≃ Gal(K/k) is solvable;

2) the field K is of 3-torsion elliptic type;

3) there exists f̃(x) = x^{4} +ã x^{2} +b̃ x+c̃ ∈ k[x] with splitting field K and ã = −6.

It is clear that 2) implies 1), since the field obtained adjoining all the coordinates of the 3-torsion points provides a solution to the embedding problem. Note that here the class η is (c_{4}, −3Δ) = 1. Besides, using the previous results we have that 3) implies both 1) and 2). Therefore the only thing we have to check is that 1) implies 3). In order to do that we proceed as before, considering a reduced Tschirnhausen transformation of f(x) = x^{4} + a x^{2} + b x + c. Then −6ã is the quadratic form

((−24 a^{3} + 9 b^{2}+72 a c) m^{2} + 12 (a^{2} − 4 c) n^{2} − 24 a p^{2}+
84 a b m n + 48 (a^{2} − 2 c) m p − 72 b n p)/4.

Using the same notation as in the second section, set δ = 2 a^{3}+9 b^{2}−8 a c. If a δ ≠ 0 this form is equivalent to ⟨−6 a, 6 a δ, −3 δ d_{f}⟩, and therefore to ⟨−6 a, 6 a δ, δ⟩. The obstruction for it to represent 1 is then (−6a, δ), the obstruction for the solvability of the embedding problem. If a = 0 the quadratic form always represents 1 and if δ = 0 the condition d_{f} ≡ −3 (mod k^{∗ 2}) implies that a should be congruent to −6 mod k^{∗ 2}.

Remark. Serre suggested that we also point out the fact that the implication 1) ⟹ 2) of the theorem can be proved directly by introducing the twisted modular curve X(3, ρ) attached to ρ : Gal(\overline{k}/k) → GL(2, 3), a linear lifting of ϱ. Since the modular curve X(5) has also genus zero, the analogous result mod 5 can be obtained.

5. Examples. As first example take the polynomial f(x) = x^{4} + 2 x^{2} + 88 x + 101 which has discriminant −3 2^{8} 19^{2} and Galois group isomorphic to S_{4}. The class η in Br_{2}(Q) is (−3, 266), which ramifies at 2 and 3. Hence, the Galois closure of f(x) is not of 3-torsion elliptic type.

Set up this second example by taking the polynomial f(x) = x^{4} −18x^{2} −48x+81 which is irreducible over Q and has discriminant d_{f} = −2^{18} 3^{7}. Now we have η = (3, 1) = 1. Note that neither a^{2} + 12c = 0 nor a ≡ −6 (mod Q^{∗}^{2}). The first step in our algorithm requires to find a Tschirnhausen transformation such that

f_{∗}(x) = Resultant[f(t), x − (mt^{3} + nt^{2} + pt + q), t] = x^{4} + a_{∗}x^{2} + b_{∗}x + c_{∗}

is irreducible and a_{∗} becomes −6. Once we fix q = (3 b m + 2 a n)/4 = 7/32, each rational point in 3X^{2} − 3Y^{2} + Z^{2} = 1 provides a concrete instance for m, n, and p. Choosing X = Y = 0 and Z = 1, one gets m = −1/288, n = −1/96, and p = 7/96. Thus, we can take

f_{∗}(x) = x^{4} − 6x^{2} − 16x − 12 .

The new discriminant is −2^{10} 3^{5} and its invariant a^{2} + 12 c is non-zero. Let f(x) = f_{∗}(x) and remove f_{∗}(x).

In this case, the second step of the algorithm ensures us that rational points on X^{2} − 2Y^{2} + Z^{2} = 0 or X^{2} − 10 Y^{2} − 15Z^{2} = 0 yield Tschirnhausen transformations to f_{∗}(x) = x^{4} + a_{∗}x^{2} + b_{∗}x + c_{∗} having a_{∗}^{2} + 12 c_{∗} = 0. If we start with the point on the first conic given by X = Y = Z = 1, we get f_{∗}(x) = x^{4} − 33 x^{2} + 772 x − 363/4 which corresponds to the 3-torsion elliptic type field attached to E : y^{2} = x^{3} − 33/2 x + 193 with conductor 2^{8} 3^{4} 29. Among the elliptic curves parametrized by the two conics, four have minimal conductor; namely, using the terminology of [2], one conic contains the curves 768 B and 768 F, while 768 D and 768 H can be found in the other conic.

The authors would like to thank J. Quer for his valuable comments during the elaboration of this work.

Bibliography

[1]. J-P. Serre, L’invariant de Witt de la forme Tr(x^{2}), Comm. Math. Helv. 59 (1984), 651–676. [2]. J.E. Cremona, Algorithms for modular elliptic curves, Cambridge University Press, 1992.