A table of A_{5}-fields

Jacques Basmaji, Ian Kiming

For an extension K of Q with Galois group isomorphic to A_{5} , we shall by a root field of K mean any extension of Q of degree 5 contained in K . Table 1 is a table containing all A_{5}-extensions of Q which are non-real and for which the discriminant of a root field is at the most 2083^{2}. This table was used in section 4 of I, and our present purpose is to describe how the table was obtained.

Our starting point is the following theorem of Hunter.

Theorem. (cf. [3]) Suppose that F/Q is an extension of degree 5 and discriminant D . Then there is an algebraic integer θ in F such that

F = Q(θ) ,

and such that, denoting by θ_{1} = θ, θ_{2}, θ_{3}, θ_{4}, θ_{5} the conjugates of θ ,

∑_{i} θ_{i} ∈ {0, ±1, ±2} ,

and

∑_{i} |θ_{i}|^{2} ≤ (\frac{8}{5}|D|)^{\frac{1}{4}} .

Let D be a positive real number. If K/Q is a non-real Galois extension with Galois group isomorphic to A_{5} such that the discriminant of a root field of K is bounded by D , then we conclude that K is the splitting field of a polynomial

f(X) = X^{5} + a_{1}X^{4} + a_{2}X^{3} + a_{3}X^{2} + a_{4}X + a_{5} ∈ Z[X] ,

such that

a_{1} ∈ {0, ±1, ±2} ,

and, denoting by θ_{1}, θ_{2}, θ_{3}, θ_{4}, θ_{5} the roots of f ,

∑_{i} |θ_{i}|^{2} ≤ (\frac{8}{5}D)^{\frac{1}{4}} .

We want to bound the coefficients of f .

Suppose first that a_{1} = 0 . Now, f(X) has 1 real root and 2 pairs of complex conjugate roots, so we may write:

θ_{1} = r , θ_{2} = σ_{1} + it_{1} , θ_{3} = σ_{1} − it_{1} , θ_{4} = σ_{2} + it_{2} , θ_{5} = σ_{2} − it_{2} ,

with r, σ_{1}, σ_{2}, t_{1}, t_{2} ∈ R . We have:

r + 2σ_{1} + 2σ_{2} = ∑_{i} θ_{i} = 0 ,

i.e.:

r = −2(σ_{1} + σ_{2}) ,

We then compute:

∑ |θ_{i}|^{2} = 6σ_{1}^{2} + 8σ_{1}σ_{2} + 6σ_{2}^{2} + 2t_{1}^{2} + 2t_{2}^{2} ,
a_{2} = − 3σ_{1}^{2} − 4σ_{1}σ_{2} − 3σ_{2}^{2} + t_{1}^{2} + t_{2}^{2} ,
a_{3} = 2σ_{1}^{3} + 8σ_{1}^{2}σ_{2} + 8σ_{1}σ_{2}^{2} + 2σ_{2}^{3} + 2σ_{1}t_{1}^{2} + 2σ_{2}t_{2}^{2} ,

and

a_{4} = − 4σ_{1}^{3}σ_{2} − 7σ_{1}^{2}σ_{2}^{2} − 4σ_{1}σ_{2}^{3} − 3σ_{1}^{2}t_{2}^{2} − 3σ_{2}^{2}t_{1}^{2} − 4σ_{1}σ_{2}t_{1}^{2}
− 4σ_{1}σ_{2}t_{2}^{2} + t_{1}^{2}t_{2}^{2} .

From this it follows immediately that

|a_{2}| ≤ \frac{1}{2} ∑ |θ_{i}|^{2} .

Put α_{1} = σ_{1}^{2} + t_{1}^{2} and α_{2} = σ_{2}^{2} + t_{2}^{2} . Then:

∑ |θ_{i}|^{2} = 4σ_{1}^{2} + 8σ_{1}σ_{2} + 4σ_{2}^{2} + 2α_{1} + 2α_{2} ,
a_{3} = 8σ_{1}^{2}σ_{2} + 8σ_{1}σ_{2}^{2} + 2σ_{1}α_{1} + 2σ_{2}α_{2} ,

and

a_{4} = − 4σ_{1}^{2}α_{2} − 4σ_{1}σ_{2}α_{1} − 4σ_{1}σ_{2}α_{2} − 4σ_{2}^{2}α_{1} + α_{1}α_{2} .

If we fix the value of ∑ |θ_{i}|^{2} to be β , we may then seek the extremal points of a_{3} and a_{4} considered as functions of σ_{1}, σ_{2}, α_{1}, α_{2} under the restricitons

α_{1} ≥ σ_{1}^{2} and α_{2} ≥ σ_{2}^{2} .

It turns out to be unproblematic to find these extremal points by applying the Langrangian method. We shall not repeat the computations here, but merely note that one finds the following possible extremal values for a_{3} and a_{4} :

±\frac{1}{3\sqrt{30}}β^{\frac{3}{2}} , ±\frac{1}{3\sqrt{3}}β^{\frac{3}{2}} , ±\frac{1}{5}β^{\frac{3}{2}} , ±\frac{1}{2\sqrt{5}}β^{\frac{3}{2}} , for a_{3} ,

and

−\frac{4}{61}β^{2} , −\frac{1}{20}β^{2} , −\frac{3}{80}β^{2} , 0 , \frac{1}{16}β^{2} , \frac{1}{15}β^{2} , \frac{3}{40}β^{2} , for a_{4} .

For a_{5} we may use the inequality between arithmetic and geometric means:

(|a_{5}|^{2})^{\frac{1}{5}} = (Π|θ_{i}|^{2})^{\frac{1}{5}} ≤ \frac{1}{5} ∑ |θ_{i}|^{2} ,

i.e.:

|a_{5}| ≤ (\frac{1}{5} ∑ |θ_{i}|^{2})^{\frac{5}{2}} .

Noting that we may well assume that a_{5} is positive, we conclude that (in the case a_{1} = 0):

−\frac{1}{2}β(D) ≤ a_{2} ≤ \frac{1}{2}β(D) ,
−\frac{1}{2\sqrt{5}}β(D)^{\frac{3}{2}} ≤ a_{3} ≤ \frac{1}{2\sqrt{5}}β(D)^{\frac{3}{2}} ,
−\frac{4}{61}β(D)^{2} ≤ a_{4} ≤ \frac{3}{40}β(D)^{2} ,
1 ≤ a_{5} ≤ \frac{1}{5}β(D)^{\frac{5}{2}} ,

with

β(D) = (\frac{8}{5}D)^{\frac{1}{4}} .

If a_{1} ≠ 0 , we first note that the bounds for a_{5} obviously still hold. Secondly, we consider the polynomial

f̃(X) = f(X − \frac{1}{5}a_{1}) = X^{5} + ã_{2}X^{3} + ã_{3}X^{2} + ã_{4}X + ã_{5} ,

where:

\begin{matrix}a_{2} = & ã_{2} + \frac{2}{5}a_{1}^{2} , \\ a_{3} = & ã_{3} + \frac{3}{5}a_{1}ã_{2} + \frac{2}{25}a_{1}^{3} , \\ a_{4} = & ã_{4} + \frac{2}{5}a_{1}ã_{3} + \frac{3}{25}a_{1}^{2}ã_{2} + \frac{1}{125}a_{1}^{4} . \end{matrix}

Now, f̃(X) has the roots θ_{i} + \frac{1}{5}a_{1} , i = 1, 2, 3, 4, 5 , and we find (using ∑ θ_{i} = −a_{1}) :

∑_{i} |θ_{i} + \frac{1}{5}a_{1}|^{2} = −\frac{1}{5}a_{1}^{2} + ∑_{i} |θ_{i}|^{2} .

The above considerations then give us bounds for ã_{2}, ã_{3}, ã_{4} in terms of β̃(D) = −\frac{1}{5}a_{1}^{2} + β(D) , and this of course gives us immediately bounds for ã_{2}, ã_{3}, ã_{4} .

For the case:

D = 2083^{2} ,

we obtain the following bounds:

\begin{aligned}For a_{1} = 0 : −25 & ≤ a_{2} ≤ 25 , \\ −82 & ≤ a_{3} ≤ 82 , \\ −172 & ≤ a_{4} ≤ 197 , \\ 1 & ≤ a_{5} ≤ 337 . \end{aligned}

\begin{aligned}For a_{1} = ±1 : −25 & ≤ a_{2} ≤ 25 , \\ −97 & ≤ a_{3} ≤ 97 , \\ −207 & ≤ a_{4} ≤ 231 , \\ 1 & ≤ a_{5} ≤ 337 . \end{aligned}

\begin{aligned}For a_{1} = ±2 : −23 & ≤ a_{2} ≤ 27 , \\ −108 & ≤ a_{3} ≤ 108 , \\ −236 & ≤ a_{4} ≤ 260 , \\ 1 & ≤ a_{5} ≤ 337 . \end{aligned}

This gives us about 7, 7·10^{9} polynomials, and these polynomials are now investigated computationally in the following steps:

(A) The polynomial is eliminated, if its discriminant is not a square modulo some prime in {3, 5, 7, 11, 13, 17, 19, 23, 29} .

(B) For the remaining polynomials the polynomial discriminant is caculated, and the polynomial is eliminated, if this discriminant is not a square in Z .

(C) The reducible survivors from (B) are eliminated.

(D) For the remaining polynomials it is tested whether the splitting field has Galois group isomorphic to A_{5} or not. If not, the polynomial is eliminated.

(E) Among the remaining polynomials, those are selected for which the field discriminant of a field obtained by adjoining 1 root of the polynomial to Q is ≤ 2083^{2} .

(F) These selected polynomials are put in classes such that 2 polynomials are in the same class, if and only if their splitting fields are identical. From each class, 1 polynomial is chosen.

We shall briefly describe the methods used to perform these steps.

Ad (A), (B) : The discriminant (mod p or over Z) of a polynomial

f(X) = X^{5} + a_{1}X^{4} + a_{2}X^{3} + a_{3}X^{2} + a_{4}X + a_{5} ∈ Z[X] ,

is the resultant R(f, f′) , where

f′(X) = 5X^{4} + 4a_{1}X^{3} + 3a_{2}X^{2} + 2a_{3}X + a_{4} ,

i.e., it is the determinant of the matrix:

( \begin{matrix}1 & a_{1} & a_{2} & a_{3} & a_{4} & a_{5} & 0 & 0 & 0 \\ 0 & 1 & a_{1} & a_{2} & a_{3} & a_{4} & a_{5} & 0 & 0 \\ 0 & 0 & 1 & a_{1} & a_{2} & a_{3} & a_{4} & a_{5} & 0 \\ 0 & 0 & 0 & 1 & a_{1} & a_{2} & a_{3} & a_{4} & a_{5} \\ 5 & 4a_{1} & 3a_{2} & 2a_{3} & a_{4} & 0 & 0 & 0 & 0 \\ 0 & 5 & 4a_{1} & 3a_{2} & 2a_{3} & a_{4} & 0 & 0 & 0 \\ 0 & 0 & 5 & 4a_{1} & 3a_{2} & 2a_{3} & a_{4} & 0 & 0 \\ 0 & 0 & 0 & 5 & 4a_{1} & 3a_{2} & 2a_{3} & a_{4} & 0 \\ 0 & 0 & 0 & 0 & 5 & 4a_{1} & 3a_{2} & 2a_{3} & a_{4} \end{matrix} ) .

Introducing

c_{i} = −2a_{i+1} + a_{1}a_{i} , for i = 1, 2, 3, 4 ,

and

c_{5} = a_{1}a_{5} ,

we see that the discriminant of f(X) is the determinant of:

( \begin{matrix}1 & a_{1} & a_{2} & a_{3} & a_{4} & a_{5} \\ c_{4} − a_{1}c_{1} & c_{3} − a_{2}c_{1} & c_{2} − a_{3}c_{1} & c_{1} − a_{4}c_{1} & −a_{5}c_{1} & 0 \\ 0 & c_{4} − a_{1}c_{1} & c_{3} − a_{2}c_{1} & c_{2} − a_{3}c_{1} & c_{1} − a_{4}c_{1} & −a_{5}c_{1} \\ 0 & c_{1} & c_{2} & c_{3} & c_{4} & c_{5} \\ 0 & −a_{1} & −2a_{2} & −3a_{3} & −4a_{4} & −5a_{5} \\ 0 & 5 & 4a_{1} & 3a_{2} & 2a_{3} & a_{4} \end{matrix} )

This determinant is computed using Gauß elimination to triangulize this matrix.

Of course one could also use explicit formulas expressing the polynomial discriminant as a polynomial in the a_{i}’s , but the above method is considerably faster.

In step (A) it is convenient first to generate tables of all polynomials over Z/Zp for which the discriminant is a square in Z/Zp , for p = 3, 5, 7, 11, 13, 17, 19, 23, 29 . A polynomial f(X) ∈ Z[X] is then tested by reducing its coefficients modulo p and checking whether this reduction occurs in the table belonging to p . If this is not the case for one of the above primes, the polynomial is eliminated.

Ad (C), (F) :

Here the problem is to factorize a polynomial f(X) ∈ Z[X] either over Q (for step (C)) or over an extension of degree 5 over Q (for step (F)), where of course the latter case is the most complicated. The algorithm given in [6] was used.

Ad (D) :

The Galois group of the splitting field of any polynomial which is tested in step (D) is isomorphic to a transitive subgroup of A_{5} . The question is whether this Galois group is solvable or not. Let us recall how this question can be resolved by use of the ‘Cayley resolvent’.

Consider the symmetric group S_{5} on the symbols 1, 2, 3, 4, 5 , and in it the permutations:

σ = (1 2 3 4 5) , τ = (1 2 4 3) .

Then σ and τ^{2} generate a subgroup of A_{5} isomorphic to the dihedral group of order 10 . The elements

1, (1 2 5), (1 5 2), (1 4 5), (1 5 4), (2 3 5)

form a complete set of (right) coset representatives of ⟨σ, τ^{2}⟩ in A_{5} . Of course, S_{5} = A_{5} ∪ τA_{5} . Consider the field Q(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) with the action of S_{5} through permutation of the x_{i}’s . Let a_{1}, a_{2}, a_{3}, a_{4}, a_{5} be the elementary symmetric polynomials in the x_{i} , i.e.:

∏_{i}(t − x_{i}) = t^{5} + a_{1}t^{4} + a_{2}t^{3} + a_{3}t^{2} + a_{4}t + a_{5} .

Let v = x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{4} + x_{4}x_{5} + x_{1}x_{5} . Then both v and τv are invariant under ⟨σ, τ^{2}⟩ . Hence the element

u_{1} = v − τv = x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{4} + x_{4}x_{5} + x_{1}x_{5} − x_{2}x_{4} − x_{1}x_{4} − x_{1}x_{3} − x_{3}x_{5} − x_{2}x_{5} is also invariant under ⟨σ, τ^{2}⟩ . It has 6 translates under A_{5} :

u_{2} = (1 2 5)u_{1} = x_{2}x_{5} + x_{3}x_{5} + x_{3}x_{4} + x_{1}x_{4} + x_{1}x_{2}
− x_{4}x_{5} − x_{2}x_{4} − x_{2}x_{3} − x_{1}x_{3} − x_{1}x_{5} ,
u_{3} = (1 5 2)u_{1} = x_{1}x_{5} + x_{1}x_{3} + x_{3}x_{4} + x_{2}x_{4} + x_{2}x_{5}
− x_{1}x_{4} − x_{4}x_{5} − x_{3}x_{5} − x_{2}x_{3} − x_{1}x_{2} ,
u_{4} = (1 4 5)u_{1} = x_{2}x_{4} + x_{2}x_{3} + x_{3}x_{5} + x_{1}x_{5} + x_{1}x_{4}
− x_{2}x_{5} − x_{4}x_{5} − x_{3}x_{4} − x_{1}x_{3} − x_{1}x_{2} ,
u_{5} = (1 5 4)u_{1} = x_{2}x_{5} + x_{2}x_{3} + x_{1}x_{3} + x_{1}x_{4} + x_{4}x_{5}
− x_{1}x_{2} − x_{1}x_{5} − x_{3}x_{5} − x_{3}x_{4} − x_{2}x_{4} ,
u_{6} = (2 3 5)u_{1} = x_{1}x_{3} + x_{3}x_{5} + x_{4}x_{5} + x_{2}x_{4} + x_{1}x_{2}
− x_{3}x_{4} − x_{1}x_{4} − x_{1}x_{5} − x_{2}x_{5} − x_{2}x_{3} .

Furthermore,

τu_{1} = −u_{1} , τu_{2} = −u_{2} , τu_{3} = −u_{4} ,
τu_{4} = −u_{6} , τu_{5} = −u_{3} , τu_{6} = −u_{5} .

Define b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6} by:

∏_{i}(y − u_{i}) = y^{6} + b_{1}y^{5} + b_{2}y^{4} + b_{3}y^{3} + b_{4}y^{2} + b_{5}y + b_{6} .

Now, b_{1}, b_{3}, b_{5} are skew-symmetric in the x_{i}’s , and are therefore multiples of

Δ = ∏_{i<j}(x_{j} − x_{i})

in Q(x_{1}, …, x_{5}) . Since b_{1}, b_{3}, b_{5}, Δ are homogenous in the x_{i}’s of degrees 2, 6, 10 and 10 respectively, we conclude that:

b_{1} = b_{3} = 0 , \frac{b_{5}}{Δ} ∈ Q .

Substituting (for example) 0, 1, −1, 2, −2 for x_{1}, …, x_{5} , one easily finds:

b_{5} = −32Δ .

The elements b_{2}, b_{4}, b_{6} are symmetric in the x_{i}’s , and they are therefore polynomials in the a_{i}’s . Using the computer-algebra system MAPLE, we found:

b_{2} = − 3a_{2}^{2} + 8a_{1}a_{3} − 20a_{4} ,

b_{4} = 3a_{2}^{4} − 16a_{1}a_{2}^{2}a_{3} + 16a_{1}^{2}a_{3}^{2} + 16a_{1}^{2}a_{2}a_{4} − 64a_{1}^{3}a_{5} + 16a_{2}a_{3}^{2} − 8a_{2}^{2}a_{4}
− 112a_{1}a_{3}a_{4} + 240a_{1}a_{2}a_{5} + 240a_{4}^{2} − 400a_{3}a_{5} ,

b_{6} = − a_{2}^{6} + 8a_{1}a_{2}^{4}a_{3} − 16a_{1}^{2}a_{2}^{3}a_{4} − 16a_{1}^{2}a_{2}^{2}a_{3}^{2} + 64a_{1}^{3}a_{2}a_{3}a_{4} − 64a_{1}^{4}a_{4}^{2} − 16a_{2}^{3}a_{3}^{2}
+ 28a_{2}^{4}a_{4} + 64a_{1}a_{2}a_{3}^{3} − 112a_{1}a_{2}^{2}a_{3}a_{4} + 48a_{1}a_{2}^{3}a_{5} − 128a_{1}^{2}a_{3}^{2}a_{4}
+ 224a_{1}^{2}a_{2}a_{4}^{2} − 192a_{1}^{2}a_{2}a_{3}a_{5} + 384a_{1}^{3}a_{4}a_{5} + 224a_{2}a_{3}^{2}a_{4} − 64a_{3}^{4}
− 176a_{2}^{2}a_{4}^{2} − 80a_{2}^{2}a_{3}a_{5} − 64a_{1}a_{3}a_{4}^{2} + 640a_{1}a_{3}^{2}a_{5} − 640a_{1}a_{2}a_{4}a_{5}
− 1600a_{1}^{2}a_{5}^{2} + 320a_{4}^{3} − 1600a_{3}a_{4}a_{5} + 4000a_{2}a_{5}^{2} .

The conclusion is, that if a polynomial

f(X) = X^{5} + a_{1}X^{4} + a_{2}X^{3} + a_{3}X^{2} + a_{4}X + a_{5} ∈ Q[X]

is irreducible and has discriminant d^{2} with d ∈ Q , then its Galois group is isomorphic to A_{5} , if and only if the polynomial:

y^{6} + b_{2}y^{4} + b_{4}y^{2} − 32dy + b_{6} ,

where b_{2}, b_{4}, b_{6} are defined in terms of the a_{i}’s by the above formulas, does not have a rational root. Remark: The above formula for b_{4} does not quite coincide with the formula given on p.99 in [2].

Ad (E) : To compute this field discriminant we used the computer-algebra system SIMATH. The algorithm involved is described in [4], chap. 4. See also [1].

In all, 238 fields emerged from this process. Only one of them, namely the field no. 176 in table 1, is real; it is marked with a star for this reason. To test whether a given field was real, we used the following well-known method (cf. [5], p. 259 and p. 284):

The irreducible polynomial:

f(X) = X^{5} + a_{1}X^{4} + a_{2}X^{3} + a_{3}X^{2} + a_{4}X + a_{5} ∈ Z[X]

has 5 real roots if and only if the matrix

( \begin{matrix}4a_{4}^{2}−10a_{3}a_{5} & 3a_{3}a_{4}−15a_{2}a_{5} & 2a_{2}a_{4}−20a_{1}a_{5} & a_{1}a_{4}−25a_{5} \\ 3a_{3}a_{4}−15a_{2}a_{5} & 6a_{3}^{2}−10a_{2}a_{4}−20a_{1}a_{5} & 4a_{2}a_{3}−15a_{1}a_{4}−25a_{5} & 2a_{1}a_{3}−20a_{4} \\ 2a_{2}a_{4}−20a_{1}a_{5} & 4a_{2}a_{3}−15a_{1}a_{4}−25a_{5} & 6a_{2}^{2}−10a_{1}a_{3}−20a_{4} & 3a_{1}a_{2}−15a_{3} \\ a_{1}a_{4}−25a_{5} & 2a_{1}a_{3}−20a_{4} & 3a_{1}a_{2}−15a_{3} & 4a_{1}^{2}−10a_{2} \end{matrix} )

and its principal minors all have positive determinant.

The 238 fields are listed in table 1. For each field the smallest possible conductor of a lifting to a 2-dimensional Galois representation over Q of (one, hence any of) the 2-dimensional projective representation(s) associated with the field, is displayed. This minimal conductor can be computed from the knowledge of the structure of the associated local projective representations of Gal(\overline{Q}_{p}/Q_{p}), where p runs over the ramified prime numbers, by using the theory given in I. But, as has been shown by Buhler (cf. [2]), it may also in most cases (and in fact in all cases of table 1) be determined alone from the knowledge of the type of factorization of the ramified prime numbers in a root field of the field in question.

The notation of table 1 is as follows: \sqrt{d} denotes the squareroot of the discriminant of a root field of the given A_{5}-extension. “polynomial” denotes a generating equation of the field given by its coefficients (a_{1}, a_{2}, a_{3}, a_{4}, a_{5}). “conductor” denotes the abovementioned minimal conductor of liftings of the associated projective representations. Under “ramified primes” the type of factorization of each of the ramified prime numbers in the root field is given. The notation:

p = f_{1}^{e_{1}} … f_{s}^{e_{2}}

means that (p) factorizes as:

(p) = p_{1}^{e_{1}} … p_{s}^{e_{s}} ,

where the residue degree of the prime p_{i} is f_{i} .

Table 2 lists the 238 fields ordered by the size of the minimal conductor mentioned above. The notation in table 2 is the same as for table 1.

It should be noted that the book [2] of J. P. Buhler contains also a table of A_{5}- fields with small discriminants of the corresponding root fields. In fact, one of the examples occuring in his table served as a motivation for fixing the bound D ≤ 2083^{2} above. The main difference between his table and ours is that the above analysis of the bounds of the polynomial coefficients is lacking in [2]; this means that one does not know for what bound on D his table is complete (though it is claimed in [2] that the table in [2] is complete for the bound D ≤ 200^{2} , cf. pp. 42-43 in [2]). But for use in section 4 of I (and in VI) we must have completeness for the bound D ≤ 2083^{2} , and for this bound the table in [2] is certainly not complete as a comparison with our table 1 shows. Thus, for the purposes in section 4 of I ( and in VI) the table of Buhler, though it served as a starting point, would have been insufficient.

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