Twists of Hessian Elliptic Curves and Cubic Fields

In this paper we investigate Hesse’s elliptic curves Hμ : U + V 3 + W 3 = 3μUV W, μ ∈ Q − {1}, and construct their twists, Hμ,t over quadratic fields, and H̃(μ, t), μ, t ∈ Q over the Galois closures of cubic fields. We also show that Hμ is a twist of H̃(μ, t) over the related cubic field when the quadratic field is contained in the Galois closure of the cubic field. We utilize a cubic polynomial, R(t; X) := X + tX + t, t ∈ Q − {0,−27/4}, to parametrize all of quadratic fields and cubic ones. It should be noted that H̃(μ, t) is a twist of Hμ as algebraic curves because it may not always have any rational points over Q. We also describe the set of Q-rational points of H̃(μ, t) by a certain subset of the cubic field. In the case of μ = 0, we give a criterion for H̃(0, t) to have a rational point over Q.


Introduction
In 1840's L. O. Hesse (1811-74) investigated plane curves in a series of papers, and found an interesting family of elliptic curves of form on the projective plane P 2 (U : V : W ) .If µ = 1, it is non-singular, and is an elliptic curve with points of order 3 defined over Q(µ) where Q is the field of rational numbers.
Our concerns in this paper are to introduce systematically twists of the curves with µ in Q − {1} over quadratic fields and the Galois closures of cubic fields, and, in the latter case, to describe the set of the rational points of the curve over Q by a certain subset of the related cubic field.Here we mean by twists of the elliptic curves those as algebraic curves.This is because the twists we construct may not have any rational points over Q even though the curves themselves are defined over the field.
It parametrizes not only all cubic fields but also all quadratic ones; we exclude the values 0 and −27/4 of the parameter t to save R(t; X) from multiple roots; indeed, the discriminant of the polynomial in X is −t 2 (4t+ 27).Let ξ be a root of R(t; X) = 0 in the complex number field C and put K t := Q(ξ); if R(t; X) = 0 has a root r in Q, then we pick it up as ξ = r and have K t = Q; in this case, obviously, we have t = −r 3 /(r + 1).
Let Kt be the splitting field of the cubic polynomial R(t; X) over Q.If t runs over all such rational values as K t = Q, then Kt covers all quadratic fields as Q( −(4t + 27)) besides Q.
First in Section 2 we list some known facts on the Hessian curves.Then in Section 3 we define a family of curves H(µ, t), µ, t ∈ Q, µ = 1, t = 0, −27/4.Each of the curves is of genus 1 and a twist of H µ over the splitting field Kt as algebraic curves.In Section 4 we describe the set of Q-rational points of H(µ, t) for such t as R(t : X) is irreducible over Q by introducing a certain subset of the cubic field K t .
In general, the curve H(µ, t), though defined over Q, may not have any rational points over Q.If K t = Q, however, it has at least one rational point over Q, and hence is an elliptic curve defined over Q; it is a quadratic twist of H µ if Kt is not Q but a quadratic field.
In Section 5, we show that the curve on the projective plane P 2 (u : v : w) is a twist of H(µ, t) over K t , and also a quadratic twist of H µ (Theorem 5.1).
In Section 6, we give simple affine forms A µ,t and A µ of 3-isogenies of both H µ,t and H µ , respectively, and show that the former is the quadratic twist of the latter over Q( √ d t ) (Proposition 6.1).We have infinitely many one parameter subfamilies of H(µ, t) which have rational points over Q.In the final Section 7 we show a few simple ones; namely, Proposition 1.1.Suppose that the cubic polynomial R(t; X) is irreducible over Q for t ∈ Q − {0, −27/4}.Then H(µ, t) has a rational point over Q in each of the following three cases: (1) in case of µ = t/3+1 = 1, the point Here we denote the set of Q-rational points of H(µ, t) by H(µ, t)[Q].In the special case of µ = 0, we give a necessary and sufficient condition on t for H(0, t) to have a rational point over Q; Proposition 1.2.In case of µ = 0, the curve H(0, t) has a rational point over It should be noted that we have infinitely many those rational functions µ = f (t) of t for each of which the one-parameter family H(f (t), t) has a rational point over Q(t).In deed, for example, take those rational functions of t, x = x(t), y = y(t), z = z(t), with coefficients in Q which satisfy Det(M (x, y, z)) = 0. (The 3×3 matrix M (x, y, z) will be given in Section 3 to define H(µ, t).) Then we are able to determine µ = µ(t) as a rational function of t in Q(t) so as to have the point (x(t) : It may be of some help for interested readers to see the recent works [3], [4] and [5] of the author to understand the background of the present article.

Some facts on the Hessian curves H µ
In this section we pick up some facts on the Hessian curves H µ defined in the introduction from the textbooks of D. Husemöller [2, Chapter 4], and L. J. Mordell [6,Chapter 3].
(2) If µ = 1, then H µ is singular and consists of a line and a quadratic curve: Proposition 2.2.The affine elliptic curve is a 3-isogeny of an affine model The isogeny is given by x = −uv, y = −v 3 .
Theorem 2.3.Suppose that 3µ belongs to the ring of rational integers Z.
Remark 2.4.Mordell uses the tangential method to show the contents of this theorem.The supposition that 3µ is an integer is essential for the proof of the theorem.

The Twist H(µ, t)
As we introduced them in Section 1, let R(t; X) be the cubic polynomial and, define the fields K t and Kt as follows: let ξ be a root of R(t; X) = 0 in the complex number field C and put K t = Q(ξ); if the equation R(t; X) = 0 has a root r in Q, then we pick it up as ξ = r and have K t = Q.Let Kt be the splitting field of the cubic polynomial R(t; X) over Q.Then the degree [ Kt : Q] varies over all the divisors of 6, and the degree [K t : Q] is either 3 or 1.Now let us introduce an algebraic curve H(µ, t) for µ, t ∈ Q, t = 0, −27/4 which is a twist of H µ as algebraic curves over the splitting field Kt .We define it by making use of 3 × 3 matrices.Put The following proposition is easily verified; the proof is omitted: Proposition 3.1.The characteristic polynomial of the matrix Ξ is equal to the cubic polynomial R(t; X).
To define our curve, take independent variables x, y, z and put where 1 3 is the unit matrix of size 3.The curve H(µ, t) is defined on the projective plane P 2 (x : y : z) by where Tr and Det denote the trace and the determinant of matrices, respectively.
It should be noted again that, in general, the curve H(µ, t) may not have any rational points over Q, even though it is defined over Q.
Proof.Let α, β and γ be the distinct three roots of R(t : X) = 0 in C. Since R(t : X) is the characteristic polynomial of Ξ, these roots are the eigen values of Ξ; let x α , x β and x γ be eigen vectors in the 3-dimensional column vector space over Kt for the corresponding eigen values, respectively, and put Then this is a 3 × 3 matrix with entries in Kt .Since the eigen values are different from each other, A is invertible.By definition, furthermore, we have Therefore, we see Since α, β and γ are different from each other, the matrix , and then we see a ij ∈ Kt , 1 ≤ i, j ≤ 3, and Now we take to have and, hence, These E j , j = 1, 2, 3, belong to Kt -subspace of dimension 3 spanned by 1 3 , Ξ, Ξ 2 in the space of 3 × 3 full matrix algebra over Kt .It is clear that E j , j = 1, 2, 3, are linearly independent over Kt .Therefore, we can find such b ij ∈ Kt , 1 ≤ i, j ≤ 3, as we have Then we see the equality determine a linear transformation over Kt which send (x, y, z) to (u, v, w).Since we have it is clear that we have the equations This shows that the curve H(µ, t) is isomorphic to H µ over Kt .
Remark 3.3.Actually, ( b ij ) is the inverse matrix of ( a ij ); hence, we have and x = a 11 u + a 12 v + a 13 w, y = a 21 u + a 22 v + a 23 w, z = a 31 u + a 32 v + a 33 w.

The Set of Q-rational Points of H(µ, t)
In this section, we assume that the cubic polynomial R(t; X) with t ∈ Q is irreducible over Q.As in the preceding section, let α, β and γ be the distinct three roots of R(t : X) = 0 in C.They are conjugates to each other by our present assumption.Pick up ξ := γ without loss of generality.Proof.The proposition immediately follows from the following lemma.
Lemma 4.2.The notation and the assumptions being as above, we have Kt = Q(ξ, α, β), and The discriminant D of the quadratic factor is given by The proof of the lemma is easily obtained and hence omitted.
Let H(µ, t)[Q] be the set of all Q-rational points on the curve H(µ, t).To describe this set we define a subset S µ,t of the cubic field K t by Here Tr Kt/Q and N Kt/Q are the trace map and the norm map of K t over Q, respectively.It is clear that the multiplicative group Q × naturally acts on the set S µ,t .We denote the set of orbits by S µ,t /∼ Q × .Theorem 4.3.Let the notation and the assumptions be as above.Then there exists a canonical one-to-one and onto correspondence between the two sets, S µ,t /∼ Q × and H(µ, t)[Q].
Proof.An element η of K t is uniquely expressed as Then the conjugates of η over Q are given by Therefore, if we use the notation in the proof of Theorem 3.2 in the preceding Section 3, we have It is, hence, clear that η = a + b ξ + c ξ Proof.If Kt = K t , then the lemma is obvious.Suppose Kt K t .Let σ and τ be the generators of the Galois groups Gal( Kt /K t ) and Gal( Kt /Q( √ d t ), respectively.Then we have σ 2 = τ 3 = id.Since Kt = K t ( √ d t ), we have , and ξ = α and ξ = β by our notation, we have Kt = K t (ξ ) = K t (ξ ), and hence ξ σ = ξ and ξ σ = ξ .Therefore, η σ = η and η σ = η .We also have η σ = η.We now see the lemma is now clear.
Lemma 4.6.We have: Proof.The proof is omitted because it is obtained by strait-forward calculation.
Proof.Just replace the x, y, z in the identity of the lemma by x 3 , y 3 , z 3 , respectively.

5.
The Twist H µ,t of H(µ, t) over K t In this section we give a twist H µ,t of H(µ, t) over K t , and also show that H µ,t is a quadratic twist of the Hessian curve H µ .Let H µ,t be the curve on the projective plane P 2 (u : v : w) defined by where d t = −(4t + 27) as above.
(1) The curve H(µ, t) is isomorphic to H µ,t over K t .
(2) The curve H µ,t is a twist of the Hessian curve , and an elliptic curve defined over Q with a Q-rational point (u : v : w) = (0 : 1 : 0).The isomorphism is given by Proof.This time we only use the root ξ of R(t; X) = 0.By Lemma 4.2 we have The discriminant D of the quadratic factor was also given by Put ρ := ξ 2(2ξ + 3) and Then we easily see Since R(t; X) does not have any multiple roots, we have 3ξ 2 + t = 0. Therefore, Define 3 × 3 matrices E 1 and E 2 this time by 'inserting' X := Ξ as (The constant terms of Q and P should be replaced by scalar matrices.)These belong to the linear span by 1 3 , Ξ, Ξ 2 over K t in the space of full matrix algebra of size 3 over K t .By definition and from R(t; Ξ) = 0, we easily see Since E 2 i = E i , the eigen values of E i , i = 1, 2, should be either 0 or 1.Let us show rank(E 2 ) = 1; then we also have rank(E 1 ) = 2 because As in the proof of Theorem 3.2, we diagonalize Ξ over Kt ; namely, Then we see Since R(t; X) does not have any multiple roots, we see P (t; ξ) = 0.This shows rank(E 2 ) = rank((3ξ 2 + t) −1 P (t; Ξ)) = 1.
Let V = K 3 t be the 3-dimensional column vector space over K t , and put Take a non-zero vector x from V 2 .Then V 2 = K t x, E 2 x = x, and Ξ x = ξx.
As for V 1 , there is such a vector a ∈ V 1 as a and (Ξ + (ξ/2)1 3 )a are linearly independent over K t .Indeed, suppose that we have (Ξ+(ξ/2)1 3 )a = a a with a ∈ K t .Then we have Ξ a = (a − ξ/2)a; hence, the number (a − ξ/2) belongs to K t and is an eigen value of Ξ.This cannot be ξ because the latter has an eigen vector x in V 2 .Therefore it is one of the roots α and β of P (t; X).This shows that both α and β belong to K t .We may assume a − ξ/2 = α without losing generality.Take an eigen vector b ∈ K 3 t for β; that is, Ξ b = β b, and so (Ξ − β1 3 )b = 0. Then we have Since a and b are eigen vectors belonging to different eigen values, they are linearly independent over K t .This shows that the vector (α + ξ/2)a + (β + ξ/2)b is not equal to any scalar multiple of a = a + b.Hence we obtained the desired vector a in place of a which we picked up at the beginning.Now for our choice of x ∈ V 2 and a ∈ V 1 , put x 1 := a, x 2 := ρ −1 (Ξ + (ξ/2)1 3 )a, where ρ = ξ/(2(2ξ + 3)) as we picked it up at the beginning of the proof, and x 3 := x.Then the 3 × 3 matrix B := (x 1 , x 2 , x 3 ) over K t is regular.Moreover, we easily see On the vector space V 1 , we have Hence we see Therefore, the linear subspace spanned by the three matrices E 1 , ρ −1 (Ξ + (ξ/2)1 3 )E 1 , E 2 over K t in the full matrix algebra of size 3 coincides with the one spanned by 1 3 , Ξ, Ξ 2 over K t .Expressing now as we have a linear transformation over K t which relates (x, y, z) to (u, v, w), and also

Thus we obtained
Tr M (x, y, z)3 = 2 u 3 + 3d t uv 2 + w 3 , From this we immediately see the statement (1) of the theorem.As for (2) of the theorem, it is easily confirmed by a straightforward way; the calculation is omitted.

The 3-Isogeny of H µ,t
By Proposition 2.2 in Section 2, we gave a simple affine model A µ of the 3-isogeny of the Hessian elliptic curve H µ for µ = 1; namely, A µ : y 2 + 3µxy + y = x 3 .
(1) The elliptic curve A µ,t is an affine model of a 3-isogeny of the elliptic curve H µ,t defined by Proof.For simplicity, put f := u/(−w) and g := v/(−w).Then we have Since we have x = (2X − 1)/(3µ), we easily see that the defining equation of A µ is equivalent to the equation Hence A µ,t is nothing but the standard quadratic twist of A µ over Q( √ d t ) combined with the above transformation (x, y) → (X, Y ).

Some cases of non-empty H(µ, t)[Q]
As we stated in Introduction, we now show the two propositions to demonstrate some cases of non-empty H(µ, t)[Q].
In the special case of µ = 0 we give a necessary and sufficient condition on t for H(0, t) to have a rational point over Q. Ω = a −1 Θ because it generates A t which is isomorphic to a cubic field.)Therefore, we have and hence, f = g + 1/3 from a −3 Tr(Θ 3 ) = 0. Thus we obtain Here we have g = −Det(Ω) = 0 and Then by multiplying the both sides of the equation ( * ) by g 3 (2g/3 − 1/27) −3 , we have Ξ 3 + t Ξ + t 1 3 = 0. Finally put h := 9g.Then we obtain Since Ξ also generates A t over Q, we have It is now clear that the equation M (x, y, z) = M (x , y , z ) with determines a linear transformation (x, y, z) → (x , y , z ) over Q which induces an isomorphism of H(0, t) to H(0, t ) defined over Q.Conversely, suppose that the above ( * * ) is satisfied by Ξ and t in place of Ξ and t , respectively, for simplicity.Put Θ := (2h − 1)h −1 Ξ − 1.Then we have We also have Tr(Ξ) = 0.It follows from (Ξ 3 + t Ξ) 2 = t 2 1 3 that Thus we found such Θ ∈ A t as we have Tr Θ 3 = 0.The proof of Proposition 7.1 is completed.
Remark 7.2.In the proof of the final part of Proposition 7.1, we saw that the isomorphism of H(0, t) to H(0, t ) was obtained from a linear transformation of K t = K t defined by the base change {1, ξ, ξ 2 } and {1, ξ , ξ 2 } of K t as a Q-space.In the forthcoming paper [1] by Akinari Hoshi and the author, a necessary and sufficient condition for K t = K t is obtained.Namely, Proposition 7.3.Let the notation and the assumptions be as above; in particular, suppose that R(t; X) with t ∈ Q is irreducible over Q.Then K t = K t for t ∈ Q if and only if there exists such an element u ∈ Q as Since the condition K t = K t is reciprocal in t and t , so should the latter condition of the proposition be.Indeed, let T and U be two independent variables, and put Then we have a rational endomorphism of the rational function field k(T, U ) of two variables, Φ : k(T, U ) → k(T, U ), by assigning (T , U ) to (T, U ).The endomorphism Φ is involutive; that is, Φ • Φ is equal to the identity map.Hence Φ is an automorphism of k(T, U ) and an involutive Cremona transformation of dimension 2.
We have [ Kt : Q] = 3 if and only if d t is a square in Q; in the case, we have Kt = K t .(2) We have [ Kt : Q] = 6 if and only if d t is not a square in Q; in the case, Kt contains the quadratic field Q( √ d is a cubic field.Proposition 4.1.The splitting field Kt of the cubic polynomial R(t; X) over Q contains √ d t , d t = −(4t + 27), and Kt = K t ( √ d t ).(1) t ), and Kt = K t ( √ d t ).
2∈ K t belongs to S µ,t if and only if the point (a : b : c) on the projective plane P 2 (x : y : z) belongs to H(µ, t)[Q].The theorem is now clear.We transfer the well-known tangential method on elliptic curves into our S µ,t .t η η 3 − η 3 in Kt also belongs to K t .The conjugates of ρ : and hence, on one hand,X 2 − d t Y 2 = (f 2 − d t g 2 )3by a straightforward way.On the other hand, by dividing the both sides of the defining equation of H µ,t , we obtain2X −1 = −3µ(f 2 − d t g 2 ).Therefore, we have X 2 − d t Y 2 = ((2X − 1)/(−3µ))3.This is equivalent to the defining equation of A µ,t .The first statement of the proposition is proved.To see the second one, set