# On a Galois group arising from an iterated map.

1. Introduction. The aim of this paper is to study the Galois group of a certain factor of a 4-th dynatomic polynomial. In general, the 4-th dynatomic polynomial for the polynomial map [sigma] is defined by

[[PHI].sub.4,[sigma]](x) = [[sigma].sup.4](x) - x/[[sigma].sup.2](x) - x,

where [[sigma].sup.i] is the i-fold iteration of [sigma] with itself (see  for details).

Dynatomic polynomials have been intensively studied by Morton. For example, he computed the Galois group of [[PHI].sub.3,[sigma]] (x) with [sigma](x) = [x.sup.2] + a , and in particular, he was led to an analogue of Kummer theory for cyclic cubic extensions by using the map [sigma](x) = [x.sup.2] - 1/4 ([s.sup.2] + 7) over the base field without cube roots of unity . He also proved that the dynatomic curve [[PHI].sub.4,[sigma]](x) = 0 with [sigma](x) = [x.sup.2] + a has no rational points, i.e., [[PHI].sub.4,[sigma]](x) has no rational roots for rational values of a .

In this paper, we consider the 4-th dynatomic polynomial [[PHI].sub.4,[sigma]] with [sigma](x) = [x.sup.3] + ax. The polynomial [[PHI].sub.4,[sigma]](x) has degree 72 and it has a factor:

(1.1) f(a, x) = [x.sup.8] + 3a[x.sup.6] + 3[a.sup.2][x.sup.4] + ([a.sup.2] + 1)a[x.sup.2] + [a.sup.2] + 1.

We shall investigate the Galois groups of the polynomial f a, x) over Q(a) and its specializations over Q.

In general, the Galois group of a dynatomic polynomial is isomorphic to a subgroup of a wreath product . We show that the polynomial f(a, x) has a Galois group which is isomorphic to the whole wreath product [C.sub.4] [??] [C.sub.2] over the function field Q(a) (see Theorem 2.1).

The group [C.sub.4] [??] [C.sub.2] has order 32 and has the following presentation:

<[[sigma].sub.1],[[sigma].sub.2],[tau]|[[sigma].sup.4.sub.1] = [[sigma].sup.4.sub.2] = [[tau].sup.2] = 1, [[sigma].sub.1][[sigma].sub.2] = [[sigma].sub.2][[sigma].sub.1], [tau][[sigma].sub.1][tau] = [[sigma].sub.2]>.

Every Galois extension L/Q with this Galois group can be obtained as a class field of a certain quadratic field. By choosing the signature of L carefully, we can find such an extension that is a class field of a real quadratic field and that has an odd Artin representations of degree 2 induced from a character corresponding to the real quadratic field. This group [C.sub.4] [??] [C.sub.2] is known to be a minimal group with this property (see ). This is a strong motivation to construct Galois extensions with this Galois group systematically.

The outline of this paper is as follows: In Section 2, we show that the splitting field of the polynomial f (a, x) is a [C.sub.4] [??] [C.sub.2]-extension over the function field Q(a). In the rest of this paper, we are concerned with the Galois groups of the specializations f(a, x) with various a [member of] Q. In Section 3, we determine a condition for the irreducibility of f(a, x) for specific values of a in Q. For a [member of] Q, let [[summation].sup.a.sub.f] be the splitting field of f(a; x) over Q. In Section 4, we give a condition for the Galois group Gal([[summation].sup.a.sub.f]/Q) to be isomorphic to [C.sub.4] [??] [C.sub.2], and compute the signature of [[summation].sup.a.sub.f]. In Section 5, we classify the Galois group Gal([[summation].sup.a.sub.f]/Q) when it is smaller than [C.sub.4] [??] [C.sub.2].

2. The Galois group over a function field. In this section, we prove the following main theorem.

Theorem 2.1. The Galois group of f (a,x) over Q(a) is isomorphic to [C.sub.4] [??] [C.sub.2].

Proof. By a straightforward computation, we can check f (a,x)|f (a, [sigma](x)). Hence if [alpha] is a root of f(a,x), then so is [sigma]([alpha]).

The roots of f(a, x) fall into two distinct orbits under [sigma]. To be more specific, if we define

[[alpha].sub.1] = [1/2] [square root of -3a - [square root of [a.sup.2] - 8] + [square root of 8 + 2[a.sup.2] - 2a[square root of [a.sup.2] - 8]],

[[alpha].sub.2] = [1/2] [square root of -3a + [square root of [a.sup.2] - 8] + [square root of 8 + 2[a.sup.2] + 2a[square root of [a.sup.2] - 8]],

then the two orbits are {[[sigma].sup.j]([[alpha].sub.1])} and {[[sigma].sup.j]([[alpha].sub.2])} for 0 [less than or equal to] j [less than or equal to] 3. If we set [[lambda].sub.i](x) = [[PI].sup.3.sub.j=0](x - [[sigma].sup.j]([[alpha].sub.i])) for i = 1, 2, then [[lambda].sub.i](x) are polynomials in Q([square root of [a.sup.2] - 8])[x] of degree 4. Let Li be the splitting field of [[lambda].sub.i](x) over Q([square root of [a.sup.2] - 8]) . Since [sigma] has order 4, the extensions [L.sub.i]/Q([square root of [a.sup.2] - 8]) are cyclic of degree 4. Let [K.sub.i] be the intermediate field of [L.sub.i]/Q([square root of [a.sup.2] - 8]) such that [[K.sub.i] : Q([square root of [a.sup.2] - 8])] = 2. The fields [K.sub.1] and [K.sub.2] are explicitly given by

(2.1) [K.sub.1] = Q([[alpha].sup.2.sub.1]) = Q([square root of 8 + 2[a.sup.2] - 2a[square root of [a.sup.2] - 8]]),

(2.2) [K.sub.2] = Q([[alpha].sup.2.sub.2]) = Q([square root of 8 + 2[a.sup.2] + 2a[square root of [a.sup.2] - 8]]).

Since

[square root of 8 + 2[a.sup.2] - 2a[square root of [a.sup.2] - 8]][square root of 8 + 2[a.sup.2] + 2a[square root of [a.sup.2] - 8]] = 8[square root of [a.sup.2] + 1] [??] Q([square root of [a.sup.2] - 8]),

we have [K.sub.1] [not equal to] [K.sub.2]. Let [[SIGMA].sub.f] be the splitting field of f(a,x) over Q([square root of [a.sup.2] - 8]). Since the field [[SIGMA].sub.f] is the composit um of L1 and L2, the Galois group G' of [[SIGMA].sub.f]/Q([square root of [a.sup.2] - 8]) is isomorphic to [C.sub.4] x [C.sub.4].

The group G' is generated by the following automorphisms:

(2.3) [mathematical expression not reproducible]

(2.4) [mathematical expression not reproducible]

If we set

(2.5) [mathematical expression not reproducible]

then this map t is an extension of the generator of Gal(Q([square root of [a.sup.2] - 8])/Q(a)) to Gal([[SIGMA].sub.f]/Q(a)).

If we set [G.sub.0] = <[[sigma].sub.1], [[sigma].sub.2], [tau]>, then the generators of [G.sub.0] satisfy [[sigma].sup.4.sub.1] = [[sigma].sup.4.sub.2] = [[tau].sup.2] = 1, [[sigma].sub.1][[sigma].sub.2] = [[sigma].sub.2][[sigma].sub.1] and [tau][[sigma].sub.1][tau] = [[sigma].sub.2]. Thus [G.sub.0] is isomorphic to [C.sub.4] [??] [C.sub.2]. Since the field [[summation].sub.f] is an extension over Q(a) of degree 32, the group Gal([[summation].sub.f]/Q(a)) is isomorphic to [C.sub.4] [??] [C.sub.2].

Next, we describe some intermediate fields of [[summation].sub.f]/Q(a) for our later use. The subgroups of index 2 in [C.sub.4] [??] [C.sub.2] = <[[sigma].sub.1], [[sigma].sub.2], [tau]> are

<[[sigma].sup.2.sub.1], [[sigma].sub.1][tau]), <[[sigma].sub.1], [[sigma].sub.2]>, <[[sigma].sup.2.sub.1], [[sigma].sub.1][[sigma].sub.2], [tau]>.

The quadratic fields over Q(a) corresponding to these subgroups are

(2.6) [mathematical expression not reproducible]

(2.7) [mathematical expression not reproducible]

(2.8) [mathematical expression not reproducible]

with v = [square root of [a.sup.2] + 1].

Proposition 2.2. The quadratic extensions of [k.sub.2] inside [[SIGMA].sub.f] are given by the following

[M.sub.1] = Q([square root of (v - 1)(v - 3)]),

[M.sub.2] = Q([square root of (v + 1)(v + 3)]),

[M.sub.3] = Q([square root of v(v - 1)]),

[M.sub.4] = Q([square root of v(v - 3)]),

[M.sub.5] = Q([square root of v(v+3)]),

[M.sub.6] = Q([square root of v (v - 1)(v -3)(v + 3)]).

The Galois groups of the extensions [[SIGMA].sub.f]/[M.sub.i] (i = 3, 4, 5, 6) are

[mathematical expression not reproducible]

Proof. We can show our assertions by calculating the fixed subgroups in <[[sigma].sub.1], [[sigma].sub.2], [tau]> corresponding to these fields. We omit the detail.

3. Irreducibility under specializations. The Hilbert irreducibility theorem guarantees that there are infinitely many a [member of] Q such that f (a, x) is irreducible and that the Galois group of f(a, x) over Q is isomorphic to [C.sub.4] [??] [C.sub.2]. In the next section, we shall give an explicit description of such rational a's. In this section, we give a criterion for the irreducibility of the specialization f (a, x) with a in Q. Recall that [[SIGMA].sup.a.sub.f] is the splitting field of the specialization f(a, x) with a in Q.

Theorem 3.1. The specialization of the polynomial f (a, x) with a 2 Q is irreducible if and only if a is not one of the following forms with a rational solution (A, B) of the Diophantine equation [A.sup.2] - 2[B.sup.2] = 1:

(31) [2A/B];

(3.2) [+ or -] 2(A + B)(A + 2B)/ B(2A + 3B).

Proof. We recall that f (a, x)|f (a, [sigma](x)). Let [[alpha].sub.1] and [[alpha].sub.2] be the roots of f (a, x) given in the proof of Theorem 2.1. By [[sigma].sup.4]([[alpha].sub.i]) = [[alpha].sub.i] (i = 1, 2), we have [[sigma].sup.2]([[alpha].sub.i]) = -[[alpha].sub.i].

Now we consider the following six polynomials:

[mathematical expression not reproducible]

with 1 [less than or equal to] i, j [less than or equal to] 2 and i [not equal to] j.

We shall show that f(a; x) is reducible if and only if one of the fields [k.sub.1], [M.sub.1] and [M.sub.2] coincides with Q.

At first, if [k.sub.1] = Q, [M.sub.1] = Q or [M.sub.2] = Q, then f(a; x) is obviously reducible over Q.

Conversely, we assume that f(a; x) is reducible over Q. Let [beta] be a root of an irreducible factor of f(a, x). Since f(a, x)|f(a, [sigma](x)), we see that -[beta] and [+ or -][sigma]([beta]) are also roots of f(a, x). Similarly, if [gamma] is a root of f(a, x) which is different from [+ or -][beta] and [+ or - ][sigma]([beta]), then so are -[gamma], [+ or -][sigma](7). Now we set g(x) = (x - [beta]) (x - [sigma]([beta]))(x + [beta])(x + [sigma]([beta])) and h(x) = (x - [gamma])(x - [sigma]([gamma]))(x + [gamma])(x + [sigma]([gamma])), and we obviously have f (a, x) = g(x)h(x). Hence the pair (g(x),h(x)) coincides with one of ([[lambda].sub.1](x), [[lambda].sub.2](x)), ([[mu].sub.1](x), [[mu].sub.2](x)) or ([v.sub.1](x), [v.sub.2](x)). Thus we get [k.sub.1] = Q, [M.sub.1] = Q or [M.sub.2] = Q.

Next we consider the conditions for [k.sub.1], [M.sub.1] or [M.sub.2] to coincide with Q.

We first consider the case [k.sub.1] = Q, equivalently [square root of [a.sup.2] - 8] [member of] Q. We can show that this condition is equivalent to a = 2A/B with a rational solution (A, B) of the Diophantine equation [A.sup.2] - 2[B.sup.2] = 1.

Next, if [M.sub.1] = Q, then we get v [member of] Q because [M.sub.1] [contains] [k.sub.2]. Noting that [v.sup.2] = [a.sup.2] + 1, we can write a in the form a = ([n.sup.2] - 1)/(2n) with n [member of] Q. This equation yields v = ([n.sup.2] + 1)/(2n). Therefore [M.sub.1] = Q is equivalent to the condition that

(v - 1)(v - 3) = [((n - 1)/(2n)).sup.2] ([(n - 3).sup.2] - 8)

is a square. If there exists q in [Q.sup.x] such that [(n - 3).sup.2] - 8 = [q.sup.2], then we have

[(n - 3/q).sup.2] - 2 [(2/q).sup.2] = 1.

If we set n - 3 = 2A/B with (A, B) satisfying [A.sup.2] - 2[B.sup.2] = 1, then the element (v - 1)(v - 3) is a square. Hence we can get the following equality:

a = 2(A + B)(A + 2B)/B(2A + 3B).

The converse is clear.

We can treat the case [M.sub.2] = Q similarly. Indeed, if n + 3 = 2A/B where (A, B) satisfies [A.sup.2] - 2[B.sup.2] = 1, then the element (v + 1)(v + 3) is a square. Thus, in this case, a has the form

a = 2(A - B)(A - 2B)/B(2A - 3B)

Replacing the sign of B implies (3.2). The converse is clear again.

Remark 3.2. We can obtain infinitely many non-isomorphic fields if we specialize a [member of] Q. To prove this, it is enough to show that there are infinitely many quadratic fields [k.sub.1] when a runs through the rational integers. This follows from the result of Estermann .

4. Non-degenerate case. In this section, we see exactly when the Galois group of a specialization f (a, x) with a [member of] Q is isomorphic to [C.sub.4] [??] [C.sub.2].

Theorem 4.1. We assume that the specialization f(a, x) with a [member of] Q is irreducible. The Galois group of f (a, x) is isomorphic to [C.sub.4] [??] [C.sub.2] if and only if a [not equal to] [n.sup.2] - 1/2n with a rational number n.

Proof. Since f(a,x) is irreducible, it follows from Theorem 2.1 that the extensions [L.sub.i]/[k.sub.1] are cyclic extensions of degree 4 and we have [k.sub.1] [not equal to] Q.

If Gal([a.summation over (f)]/Q) is isomorphic to [C.sub.4] [??] [C.sub.2], then [a.summation over (f)]/[k.sub.1] is an extension of degree 16, hence we get [K.sub.1] [not equal to] [K.sub.2]. By (2.1) and (2.2), the fields are [K.sub.1] [not equal to] [K.sub.2] if and only if [square root of [a.sup.2] + 1] [??] Q, equivalently a does not have the form ([n.sup.2] - 1)/(2n) with n [member of] Q.

Conversely, if a [not equal to] ([n.sup.2] - 1)/(2n) for any n [member of] Q, then the extensions [L.sub.1]/[k.sub.1] and [L.sub.2]/[k.sub.1] are distinct cyclic extensions of degree 4 because [K.sub.1] [not equal to] [K.sub.2]. Moreover [k.sub.1]/Q is a quadratic extension because the polynomial f(a, x) is irreducible; hence we get [[summation].sup.a.sub.f] : Q] = 32.

The complex conjugation lies in one of the conjugacy classes of order less than or equal to 2. The following conjugacy classes of G are of order less than or equal to 2:

Cl(1), Cl([[sigma].sup.2.sub.1][[sigma].sup.2.sub.2]) of length 1;

Cl([[sigma].sup.2.sub.1]) of length 2;

C1([tau]) of length 4.

The following theorem describes the signature of [[summation].sup.a.sub.f] whose Galois group is isomorphic to [C.sub.4] [??] [C.sub.2].

Proposition 4.2. We assume that the specialization f(a, x) with a 2 Q has the Galois group isomorphic to [C.sub.4] [??] [C.sub.2].

(i) If a < -2[square root of 2], then Sa is a real field.

(ii) If -2[square root of 2] < a < 2[square root of 2], then [[summation].sup.a.sub.f] is an imaginary field and the complex conjugation lies in C1([tau]).

(iii) If 2[square root of 2] < a, then [[summation].sup.a.sub.f] is a CM-field (i.e., the complex conjugation lies in Cl([[alpha].sup.2.sub.1][[sigma].sup.2.sub.2]) contained in the center of the group).

Proof. By the proof of Theorem 2.1, the group [C.sub.4] [??] [C.sub.2] is generated by [[sigma].sub.1], [[sigma].sub.2] and [tau] defined by (2.3), (2.4) and (2.5), respectively. Let [[alpha].sub.1] and [[alpha].sub.2] be the roots of f (a, x) defined in the proof of Theorem 2.1. The quadratic fields contained in [[summation].sup.a.sub.f] are [k.sub.0], [k.sub.1] and [k.sub.2] (see (2.6), (2.7) and (2.8)). In particular, [k.sub.2] is a real quadratic field for any a [member of] Q.

(i) If a < -2[square root of 2], then it is easy to see that the four elements [[alpha].sup.2.sub.1], [[alpha].sup.2.sub.2], [[sigma].sub.1] [([[alpha].sub.1]).sup.2] and [[sigma].sub.2][([[alpha].sup.2]).sup.2] are positive. This gives the result.

(ii) If -2[square root of 2] < a < 2[square root of 2], then [k.sub.1] and [k.sub.0] are imaginary quadratic fields. The field [k.sub.2] is contained in the totally imaginary quartic field Q([square root of [a.sup.2] - 8], [square root of [a.sup.2] + 1]) and the fixed group of this quartic field is <[[sigma].sup.2.sub.1], [[sigma].sub.1][[sigma].sub.2]>. On the other hand, the fixed subgroup of [k.sub.2] is <[[alpha].sup.2.sub.1],[[sigma].sub.1][[sigma].sub.2],[tau]>. This implies that the complex conjugation lies in the conjugacy class of [tau].

(iii) If 2[square root of 2] < a, then both [[alpha].sup.2.sub.1] and [[alpha].sup.2.sub.2] are negative. The field [[summation].sup.a.sub.f] contains subfields N = Q([[alpha].sub.1][[alpha].sub.2]), [N.sub.1] = Q([[alpha].sub.1],[[alpha].sup.2.sub.2]) and [N.sub.2] = Q([[alpha].sup.2.sub.1],[[alpha].sub.2]]) of degree 16. Since both [[alpha].sup.2.sub.1] and [[alpha].sup.2.sub.2] are negative, the fields [N.sub.1] and [N.sub.2] are totally imaginary. Thus the field [[summation].sup.a.sub.f] = Q([[alpha].sub.1],[[alpha].sub.2]) is also totally imaginary. On the other hand, the field N is the composite field of all [M.sub.i]'s in Proposition 2.2. We can show that N is totally real by examining the generators. Since the fixed subgroup of N is <[[sigma].sup.2.sub.1][[sigma].sup.2.sub.2]>, the complex conjugation acts as [[sigma].sup.2.sub.1][[sigma].sup.2.sub.2].

Remark 4.3. By Proposition 4.2, if -2[square root of 2] < a < 2[square root of 2], then Gal([[summation].sup.a.sub.f]/Q) has an odd faithful irreducible 2-dimensional complex representation induced from a character corresponding to the real quadratic field [k.sub.1].

In the paper , they constructed [C.sub.4] [??] [C.sub.2]-extensions with the complex conjugation lying in Cl([[sigma].sup.2.sub.1]).

5. Degenerate cases. In this section, we classify the Galois groups Gal([[summation].sup.a.sub.f]/Q) when it is smaller than [C.sub.4] [??] [C.sub.2] and the polynomial f (a, x) is irreducible over Q.

By Theorem 4.1, Gal([[summation].sup.a.sub.f]/Q) = [C.sub.4] [??] [C.sub.2] if and only if a = ([n.sup.2] - 1)/(2n) with n [member of] Q. Then we have v = ([n.sup.2] + 1)/(2n) [member of] Q and this implies [k.sub.2] = Q.

Since the Galois group of f(([n.sup.2] - 1)/(2n),x) over the function field Q(n) is Gal([[summation].sub.f]/Q(n)) = Gal([[summation].sub.f]/[k.sub.2]) [congruent to] [Q.sub.8] [??] [C.sub.2], the Galois group of a specialization f(([n.sup.2] - 1)/(2n),x) with n [member of] Q is isomorphic to a subgroup of [Q.sub.8] [??] [C.sub.2]. If f (a, x) is irreducible with a specific a [member of] Q and the Galois group of f (a,x) is smaller than [Q.sub.8] [??] [C.sub.2], then we have [[[summation].sup.a.sub.f] : Q] = 8. Hence, in this case, f (a, x) is an irreducible Galois polynomial. The fields [M.sub.1] and [M.sub.2] in Proposition 2.2 cannot coincide with Q by the proof of Theorem 3.1. Hence from Proposition 2.2, it follows that one of [M.sub.3], [M.sub.4], [M.sub.5] or [M.sub.6] has to coincide with the base field Q. Therefore, we conclude that the Galois group of f(a, x) is isomorphic to one of the groups [D.sub.4], [C.sub.2] x [C.sub.4], [Q.sub.8] by the same proposition.

Proposition 5.1. We assume that a = [n.sup.2] - 1/2n for some n [member of] Q.

(i) If there exists Y 2 Q which satisfies [Y.sup.2] = [n.sup.2] + 1, then Gal([[summation].sup.a.sub.f]/Q) [congruent to] [D.sub.4].

(ii) If there exists Y [member of] Q which satisfies [Y.sup.2] = [n.sup.4] - 6[n.sup.3] + 2[n.sup.2] - 6n + 1 or [Y.sup.2] = [n.sup.4] + 6[n.sup.3] + 2[n.sup.2] + 6n + 1, then Gal([[summation].sup.a.sub.f]/Q) [congruent to] [C.sub.4] x [C.sub.2].

(iii) If there exists Y 2 Q which satisfies [Y.sup.2] = ([n.sup.2] + 1)([n.sup.2] - 6n + 1)([n.sup.2] + 6n + 1), then Gal([[summation].sup.a.sub.f]/Q) [congruent to] [Q.sub.8].

(iv) If none of the conditions above holds, then Gal([[summation].sup.a.sub.f]/Q) [congruent to] Q [??] [C.sub.2].

Proof. (i) If there exists a rational number Y satisfying [Y.sup.2] = [n.sup.2] + 1, then we have [square root of v(v [+ or -] 1)] = (n [+ or -] 1)/(2n)Y [member of] Q; and hence, [M.sub.3] = Q. Thus we get Gal([[summation].sup.a.sub.f]/Q) [congruent to] [D.sub.4].

(ii) If there exists a rational number Y satisfying [Y.sup.2] = [n.sup.4] - 6[n.sup.3] + 2[n.sup.2] - 6n + 1, then we have [square root of v(v - 3)] = (n - 1)/(2n)Y [member of] Q; and hence, [M.sub.4] = Q. If there exists a rational number Y which satisfies [Y.sup.2] = [n.sup.4] + 6[n.sup.3] + 2[n.sup.2] + 6n + 1, then we get [M.sub.5] = Q similarly. Thus in the cases where [M.sub.4] = Q or [M.sub.5] = Q, we have Gal([a.summation over (f)]/Q) [congruent to] [C.sub.4] x [C.sub.2].

(iii) If there exists Y [member of] Q such that [Y.sup.2] = ([n.sup.2] + 1)([n.sup.2] - 6n + 1)([n.sup.2] + 6n + 1), then we have [square root of v(v - 1)(v - 3)(v + 3)] = (n - 1)/(4[n.sup.2])Y [member of] Q. This implies [M.sub.6] = Q. Therefore, we get Gal([[summation].sup.a.sub.f]/Q) [congruent to] [Q.sub.8].

(iv) If none of the conditions in (i) and (ii) and (iii) is satisfied, then none of the fields [M.sub.i] (i = 3, 4, 5, 6) coincides with Q. Hence, we get Gal([[summation].sup.a.sub.f]/Q) = ([[sigma].sup.2.sub.1],[[sigma].sub.1][[sigma].sub.2],[tau]> [congruent to] [Q.sub.8] [??] [C.sub.2].

Remark 5.2. (i) The curve [Y.sup.2] = [n.sup.4] - 6[n.sup.3] + [n.sup.2] - 6n + 1 in Proposition 5.1 (ii) is a non-singular plane curve of genus 1 and has a rational point (0:1:0) in the projective coordinates. Therefore it has a Weierstrass model E : [Y.sup.2]Z - 6XYZ - 54Y[Z.sup.2] = [X.sup.3] + 14[X.sup.2]Z + 45X[Z.sup.2] with (y : n : z) [??] (2[n.sup.2]z - 6n[z.sup.2] + 2y[z.sup.2] - 7[z.sup.3] : 4[n.sup.3] - 12[n.sup.2]z + 4nyz - 14n[z.sup.2] : [z.sup.3]). The Mordell-Weil group of E is E(Q) = <(-9 : 0 : 1), (9 : 126 : 1)> [congruent to] Z/2Z [direct sum] Z.

Since the inverse map gives n = 4[X.sup.3] - 12[X.sup.2]Z + 4XYZ - 14X[Z.sup.2], the point (9 : 126 : 1) on E gives a = 24/7, for example. In general, these corresponding a's have huge heights. All these elliptic curve computation were done with Magma . (ii) The genus 2 curve

C : [Y.sup.2] = ([n.sup.2] + 1)([n.sup.2] - 6n + 1)([n.sup.2] + 6n + 1)

appeared in Proposition 5.1 (iii) has rational points (1 : [+ or -]1 : 0) and (0 : [+ or -]1 : 1) in the projective coordinates. These points are irrelevant for our purpose. It is very probable that these are all the rational points on C. The anonymous referee suggested us to use the elliptic Chabauty method by Bruin and Stoll  to prove this assertion. We describe the method here.

We decompose the right-hand side of the defining equation of C as a product of

A(n) = (n + i)([n.sup.2] - 6n + 1)

and

B(n) = (n - i)([n.sup.2] + 6n + 1) [member of] Q(i)[n].

The resultant computation shows [delta] = gcd(A(n),B(n)) | [i.sup.2][(1 + i).sup.14][3.sup.4]. We consider an elliptic curve [E.sub.[delta]] : 6[z.sup.2] = A(n) defined over Q(i). We shall compute the rational points on E6 over Q(i) whose n-coordinates are rational and substitute the value of n to C to find the corresponding Y. Since the point (n, z) on [delta][z.sup.2] = A(n) corresponds to the point (n, dz) on [d.sup.2][delta][z.sup.2] = A(n), it suffices to consider square-free [delta]'s. Thus we may assume 6 [member of] {1, i, 3, 3i, 1 + i, 3(1 + i)}. If 6 2 {1, 3i, 3(1 + i)}, then we find rank [E.sub.[delta]] = 0 and the n-coordinates of the torsion points are 1, which gives (1 : [+ or -]1:0) on C. For the other 6, we have rank [E.sub.[delta]] = 1. Using Magma, we can compute the subgroup E' of [E.sub.[delta]](Q(i)) of an odd finite index. We apply the elliptic Chabauty method with the map u : E' [right arrow] [P.sup.1], (X : Y : Z) [right arrow] (X : Z) to find the subset of E' whose image under u is contained in [P.sup.1](Q). The program successfully finds some points on [E.sub.[delta]] with rational n-coordinates but, at this moment, we cannot guarantee that they are all. For example, when [delta] = 3, the program finds three possible points on [E.sub.3]

{(0:1:0),(-5/4: [+ or -] 46 + 69i/8 :1)},

but the bound of the number of the possible points is greater than 3.

doi: 10.3792/pjaa.94.43

Acknowledgments. The author would like to thank the anonymous referee for his/her helpful and constructive comments on Proposition 5.1. The author also would like to express his gratitude to Prof. Masanari Kida and Dr. Genki Koda for their helpful discussion during the preparation of this paper.

References

 W. Bosma, J. Cannon and C. Playoust, The Magma algebra system. I. The user language, J. Symbolic Comput. 24 (1997), no. 3 4, 235 265.  N. Bruin and M. Stoll, The Mordell-Weil sieve: proving non-existence of rational points on curves, LMS J. Comput. Math. 13 (2010), 272 306.

 T. Estermann, Einige Satze uber quadratfreie Zahlen, Math. Ann. 105 (1931), no. 1, 653 662.

 M. Kida and G. Koda, Isoclinism classes of Galois groups of number fields. (Preprint).

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Masamitsu SHIMAKURA

Takushoku University Daiichi High School, 4-64-5 Ominami, Musashimurayama, Tokyo 208-0013, Japan

(Communicated by Kenji FUKAYA, M.J.A., April 12, 2018)

2010 Mathematics Subject Classification. Primary 11R32; Secondary 12F10, 12F20.
Author: Printer friendly Cite/link Email Feedback SHIMAKURA, Masamitsu Japan Academy Proceedings Series A: Mathematical Sciences Report 9JAPA May 1, 2018 5000 Inequalities for free multi-braid arrangements. Non-left-orderable surgeries on negatively twisted torus knots. Group theory Groups (Mathematics) Mappings (Mathematics) Maps (Mathematics)