aa r X i v : . [ m a t h . N T ] M a y M¨obius transforms and cyclic equations
Kurt GirstmairInstitut f¨ur MathematikUniversit¨at InnsbruckTechnikerstr. 13/7A-6020 Innsbruck, [email protected]
Abstract
We present a simple method for the construction of polynomials with cyclic Ga-lois groups, hoping to encourage a reader with some background in algebra to makecomputations of his/her own.
Let K be a subfield of the field C of complex numbers. A randomly chosen polynomial f ∈ K [ X ] of degree 5 is likely to be irreducible with the symmetric group S as Galoisgroup. So, if x , . . . , x are the zeros of f in C , an arbitrary permutation of the zeros like (cid:18) x x x x x x x x x x (cid:19) defines an automorphism of the splitting field K [ x , . . . , x ] of f over K . This means that arelation P ( x , . . . , x ) = 0 , where P is a polynomial in five variables with coefficients in K , remains valid if the abovepermutation is applied to the zeros x , . . . , x .At first glance this property looks nicely. But a somewhat deeper knowledge of algebraicequations shows that • the equation f = 0 is not solvable by radicals; • the splitting field K [ x , . . . , x ] of f is a vector space over K of dimension 5 ! = 120; • the zero x , say, cannot be expressed in terms of x alone — which would mean that x ∈ K [ x ]; not even in terms of x , x , x alone but only in terms of x , x , x and x .For example, if K = Q , f = X + 2 X − X + X − X + 1 ∈ Q [ X ] is such a polynomial, asthe software package MAPLE easily shows.At the other end of the scale are polynomials of degree 5 with cyclic Galois group, suchas f = X − X − X + 12 X + 5 X −
1. Indeed, the splitting field of f over Q equals Q [ x ],which means that it is a vector space of dimension 5 over Q . In particular, all of the other1eros of f can be expressed as polynomials in x of degree (at most) 4, these polynomialsbeing P = ( − X + 98 X − X − X + 6) / ,P = (5 X − X − X + 60 X + 98) / ,P = ( X − X + 58 X + 38) / ,P = (12 X − X + 69 X + 75 X − / P , for instance, in f , f ( P ) = P − P − P + 12 P + 5 P − , and computing the residue of f ( P ) modulo f . This residue should be zero.As to the Galois group of f , i. e., the group of automorphisms of the field Q [ x ], one canshow that it equals h ( x x x x x ) i , the group generated by the cycle ( x x x x x ) ∈ S .This cycle maps x to x , x to x , . . . , x to x . Here x j = P j ( x ), j = 2 , . . . ,
5. Note,however, that this only works since we have numbered the polynomials P j appropriately.Changing this numbering would mean that we had to choose a different cycle as the generatorof the Galois group.Methods for the construction of polynomials f with cyclic Galois group over K are known.A fundamental paper with explicit examples depending on one parameter is [2]. The methodwe present here has several advantages for a nonprofessional, namely, • it is easy to understand and simple to handle if one has a software package like MAPLEat hand; • the interrelation of the zeros of f is a priori clear; in particular, analogues of the abovepolynomials P , . . . , P are easily found; • it is also easy to number the zeros in such a way that a certain cycle (like the aboveone) generates the Galois group.A disadvantage of this method consists in the fact that one has to work with an appropri-ate extension field K of Q in certain cases (instead of Q ). However, this disadvantage iscompensated by the fact that the method yields polynomials over Q with other interestingGalois groups (like dihedral groups, see Section 4).Possibly this method is not new, but we do not know where it could be found in theliterature.What kind of knowledge does this article presume? We hope that the basic conceptsof Galois theory suffice for the understanding of the method (Section 2). Sections 3 and 4require a bit more. But numerous examples in the text should be helpful for feeling one’sway. Of the many books on Galois theory we list only [1, 3, 6], each of which has its ownmerits. As above, let K be a subfield of C and A ∈ GL (2 , K ), the group of invertible 2 × K . The matrix A defines the M¨obius transform C r K → C r K : z A ◦ z ;indeed, if A = (cid:18) a bc d (cid:19) , then A ◦ z = az + bcz + d . A ◦ ( B ◦ z ) = ( AB ) ◦ z , which means that we may write A ◦ B ◦ z forthis item. In particular, A ◦ A ◦ . . . ◦ A ◦ z = A n ◦ z , if the matrix A occurs n times on theleft hand side.The main idea of the method is as follows. Suppose we are given a zero x of an irreduciblepolynomial f ∈ K [ X ] of degree n ≥
3. Then one may hope that the other zeros are x = A ◦ x , x = A ◦ x , . . . , x n = A n − ◦ x . In order that the Galois group of f equals h ( x , . . . , x n ) i , one must have A n ◦ x = x . Let A n = (cid:18) a ′ b ′ c ′ d ′ (cid:19) . (1)Then c ′ x + ( d ′ − a ′ ) x − b ′ = 0 holds. In the case c ′ = 0, x is a zero of a quadraticpolynomial in K [ X ] and of the irreducible polynomial f of degree n ≥
3. This is impossible.Hence c ′ = 0, but then d ′ − a ′ = 0 and b ′ = 0. Accordingly, A n is the diagonal matrix a ′ · I ,where I is the 2 × A ? For this purpose we put N = (cid:26) n, if n is odd;2 n, if n is even.Let ζ ∈ C be a primitive N th root of unity. In most cases we choose ζ = ζ N = e πi/N . Weneed a 2 × A = (cid:18) a bc d (cid:19) with entries in K that is similar to (cid:18) rζ rζ − (cid:19) for some r ∈ K. (2)Similarity here means similarity over C . Since the matrix in (2) has two different eigenvalues,it suffices that A has the same characteristic polynomial, i. e., the same trace and the samedeterminant. So we have the conditions a + d = r ( ζ + ζ − ) and A = ad − bc = ( rζ )( rζ − ) = r (3)for the entries a , b , c , d . Because the matrix A n is similar to ± r n · I , it must coincide withthis multiple of the unit matrix. Hence it has the desired property. Example.
Let n = 6, so N = 12. We obtain, for ζ = ζ , ζ + ζ − = 2 cos(2 π/
12) = √
3, soone possibly thinks that one has to choose K = Q [ √ r plays a fruitful role,since one can choose r = √
3, so A may be a rational 2 × a + d = √ · √ ad − bc = √ = 3. Accordingly, we may choose A = (cid:18) − (cid:19) , say.How to find the polynomial f whose roots are x , . . . , x n ? For this purpose we considerthe rational function field K ( X ) = (cid:26) f f : f , f ∈ K [ X ] , f = 0 (cid:27) . For A as above and g ∈ K ( X ) r K we define A ◦ g = ag + bcg + d , which lies in K ( X ) r K . As in the case of C r K we have the relation A ◦ A ◦ . . . ◦ A ◦ g = A n ◦ g, A occurs n times on the left hand side. Our candidate for f is a numerator polynomial f of the rational function F = X + A ◦ X + A ◦ X + . . . + A n − ◦ X + C = f f , (4)where f , f ∈ K [ X ] are co-prime polynomials. The constant C ∈ K allows a variation of F .Such a variation can be necessary for making f irreducible. The polynomial f is uniquelydetermined only up to a factor in K r { } . Example.
Let A be as in the above example. We obtain, with C = 1, F = X + X + 1 − X + 2 + 3 − X + 3 + − X + 6 − X + 3 − − X + 99 X + − X + 9 − X − , whose numerator polynomial f = 2 X + 2 X − X + 40 X + 5 X − X + 2is irreducible over Q . Indeed, it is a polynomial with Galois group h ( x x x x x x ) i , where x is an arbitrary zero of f and x j = A j − ◦ x for j = 2 , . . . ,
6. This follows from thefollowing more general result.
Theorem 1.
Let A ∈ GL (2 , K ) have the above properties, in particular, A satisfies (3) .Let n ≥ , F be defined as in (4) and f a numerator polynomial of F . Suppose that f is irreducible and has degree n . For an arbitrary zero x of f in C , put x j = A j − ◦ x , j = 2 , . . . , n . Then the numbers x , . . . , x n are the complex zeros of f . Further, K [ x ] is aGalois extension of K with Galois group h ( x x . . . x n ) i .Proof. The numbers x , . . . , x n are zeros of f . Indeed, for x = A ◦ x we have F ( x ) = A ◦ x + A ◦ x + . . . + A n ◦ x + C. However, since A n ◦ x = x , this is the same as F ( x ), which is zero (observe that f ( x ) = 0for the denominator polynomial f of F since f and f are co-prime). This argument alsoworks for x , . . . , x n .Suppose that the numbers x , . . . , x n are not pairwise different, so x j = x k for 1 ≤ j Let n = N = 5 and ζ = ζ . We put r = 1. Our matrix A = (cid:18) a bc d (cid:19) hasto satisfy a + d = ζ + ζ − = 2 cos(2 π/ 5) = ( − √ / ad − bc = 1. Hence we put K = Q [ √ 5] and A = (cid:18) − √ / 21 ( − √ / (cid:19) . If we choose C = 0, a numerator polynomial of F is f = (3 − √ X + ( − √ X + ( − 80 + 32 √ X +(300 − √ X + ( − 395 + 175 √ X + 178 − √ . (5)MAPLE is able to factorize f over the field K . It says that f is irreducible (the proof ofTheorem 2 yields a different method to show this). Starting with the zero x ≈ − . f , we obtain x = A ◦ x ≈ . x = A ◦ x ≈ − . x = A ◦ x ≈ . x = A ◦ x ≈ . f .In Section 4 we will see that f gives rise to an irreducible polynomial in Q [ X ] of degree5 whose splitting field has a Galois group isomorphic to the dihedral group D of order 10. Remark. In the setting of Theorem 1 it is easy to find a polynomial P ∈ K [ X ] suchthat x = P ( x ). We may assume that c = 0, since otherwise F (and f ) is a polynomialof degree ≤ 1. As f and cX + d are co-prime, the extended Euclidean algorithm yieldspolynomials g and h in K [ X ] such that f g + ( cX + d ) h = 1 . Because f ( x ) = 0, we have ( cx + d ) h ( x ) = 1 and A ◦ x = ( ax + b ) h ( x ). The remainderof ( aX + b ) h modulo f is a polynomial P of degree ≤ n − P = − X + ( − √ X / − √ X +( − 61 + 15 √ X/ − √ . Next we show how Theorem 1 works for n = 8, 10 and 12 provided that K is a suitablequadratic extension of Q . In all of these cases N = 2 n . Here ζ + ζ − does not lie in thequadratic extension K of Q , but ( ζ + ζ − ) does. Accordingly, we put r = sζ + ζ − for a number s ∈ K . Then condition (3) reads a + d = s, ad − bc = s ( ζ + ζ − ) . (6)For n = 8 we have, with ζ = ζ , ( ζ + ζ − ) = ζ + 2 + ζ − = 2 + 2 cos( π/ 4) = 2 + √ 2. Then(6) says a + d = s, ad − bc = s √ s (2 − √ . Hence we put K = Q [ √ s = 2, say, but s = 2 + √ s / (2 + √ 2) = 2 + √ 2. A matrix A with this property is A = (cid:18) − √ √ (cid:19) . 5f we choose C = 1, we obtain a numerator polynomial f ∈ K [ X ] with the properties ofTheorem 1.In the case n = 10 we choose ζ = ζ , which gives a nicer result than ζ itself. We have( ζ + ζ − ) = 2 + 2 cos(3 π/ 5) = (5 − √ / 2. Then (6) reads a + d = s, ad − bc = s (5 + √ / . If we choose s = 10, a suitable matrix A is A = (cid:18) − − √ (cid:19) , and on putting C = 1 we obtain a polynomial f of the desired kind.In the case n = 12 we choose ζ = ζ and have ( ζ + ζ − ) = 2 + 2 cos( π/ 6) = 2 + √ a + d = s, ad − bc = s (2 − √ . If we choose s = 1, a suitable matrix is A = (cid:18) − √ − (cid:19) With C = 1 this gives a polynomial f of the desired kind. Remark. In the setting of (3) suppose that K is a real field. Hence the determinant r of A is positive. This, however, means that the M¨obius transform z A ◦ z maps the upperhalf-plane { u + iv : u, v ∈ R , v > } into itself; and the same holds for the lower half-plane.Therefore, all complex zeros of f must be real. Indeed, suppose that x R . Since thecomplex-conjugate x of x is a zero of f , we have x = x k for some k . But x k = A k − ◦ x lies in the same half-plane as x , which is impossible for x . In particular, we have only realzeros in the above cases n = 8 , , f in the above way. The parameter r has only the purpose ofdiminishing the degree of K over Q . If we multiply this parameter by a non-vanishingelement of K , we obtain the same set of polynomials F . Hence r cannot be considered asa free parameter. By (3), however, two of the four parameters a, b, c, d are free, and theconstant C is also free. Therefore, we can vary three quantities in K in order to obtain anirreducible polynomial f of degree n . In the above cases n = 5 , , , 12 we do not find polynomials f with rational coefficientsbut with coefficients in a quadratic extension K of Q . How can we obtain a polynomialin Q [ X ] from such an f ? Here the multiplication with the conjugate polynomial f ′ is anobvious way.In order to make this idea precise, we suppose that K equals Q [ √ D ] for D ∈ { , , } .Let f ∈ K [ X ] be given and( ) ′ : K → K : u + v √ D u − v √ D be the non-trivial field automorphism of K . The map ( ) ′ has a natural extension to thepolynomial ring K [ X ]. It maps each coefficient α of a polynomial to the respective coefficient α ′ = ( ) ′ ( α ). We denote this extension also by ( ) ′ . For example, the polynomial f of (5)is mapped to the conjugate polynomial f ′ = (3 + √ X + ( − − √ X + ( − − √ X +(300 + 128 √ X + ( − − √ X + 178 + 80 √ . f = f · f ′ . Then this polynomial lies in Q [ X ]. In our case, we obtain f = 4( X − X − X + 170 X + 35 X − X +3753 X − X + 1825 X − X − . The polynomial f is irreducible. What is the Galois group of the splitting field of f over Q ?It turns out that it is isomorphic to the wreath product C ≀ C . Roughly speaking, this meansthe following: Let C denote a cyclic group of order 5. Then C ≀ C contains the cartesianproduct C × C as a normal subgroup of index 2. In particular, | C ≀ C | = 5 · · C ≀ C contains a cyclic group C of order 2 that acts on this normal subgroup in a certainway. We will explain this action in greater detail in Theorem 3.The proof that the Galois group has this structure is fairly easy: One factorizes f moduloa number of primes until the factorization type (5 , , , , , 1) occurs, i. e., there is an irre-ducible factor of degree 5 and five factors of degree 1, each of which occurs with multiplicity1. In our case 29 is such a prime.But all these things will be explained in the next two theorems. In these theorems weneed not assume that f or f ′ is irreducible in K [ X ]. It suffices that f f ′ is irreducible in Q [ X ]. Theorem 2. Let n ∈ { , , , } and K the corresponding quadratic extension of Q .Suppose that the polynomial f ∈ K [ X ] of Theorem 1 has degree n . Let f ′ ∈ K [ X ] be theconjugate polynomial of f and f = f f ′ be irreducible in Q [ X ] . Let m be a natural numbersuch that mf lies in Z [ X ] . Suppose that, for some prime number p , mf has a factorizationof type ( n, , . . . , modulo p . Then L = K [ x , y ] is the splitting field of f over Q , where f ( x ) = 0 = f ′ ( y ) . The degree of L over Q is [ L : Q ] = 2 n . Proof. Since f is irreducible over Q , f must be irreducible in K [ X ]. Otherwise, we have anon-trivial factorization f = g · h in K [ X ], which produces the corresponding factorization f ′ = g ′ h ′ of the conjugate polynomial. But then gg ′ and hh ′ lie in Q [ X ] and f has a non-trivial factorization over Q , which we have excluded. In the same way f ′ is irreducible in K [ X ].As we assume, x ∈ C is a zero of f , so x = A ◦ x , . . . , x n = A n − ◦ x are the remainingzeros of f , by Theorem 1. By assumption, y ∈ C is a zero of f ′ . Now f ′ is a numeratorpolynomial of F ′ = X + A ′ ◦ X + A ′ ◦ X + . . . + A ′ n − ◦ X + C ′ , where A ′ arises from A if we apply the map ( ) ′ to the entries of A . As in the case of f and x , we see that y = A ′ ◦ y , . . . , y n = A ′ n − ◦ y are the zeros of f ′ . Hence L = K [ x , y ] isthe splitting field of f over K .We show that L equals Q [ x , . . . , x n , y , . . . , y n ], the splitting field of f over Q . For thispurpose we assume that f is monic (recall that it is unique only up to a factor in K r { } ).Then f does not lie in Q [ X ]. Indeed, if f were in Q [ X ], then f ′ would be equal to f , and f = f would be reducible. So f must have a coefficient in K rQ . This coefficient generatesthe quadratic field K over Q , and it has the form P ( x , . . . , x n ) for a (symmetric) polynomial P in n variables with coefficients in Q . Accordingly, K [ x ] = Q [ x , . . . , x n ] and, in the sameway, K [ y ] = Q [ y , . . . , y n ]. Altogether, L = K [ x , y ] = Q [ x , . . . , x n , y , . . . , y n ].Next we consider the tower of fields Q ⊆ K ⊆ K [ x ] ⊆ K [ x , y ] = L and obtain the degrees [ K : Q ] = 2, [ K [ x ] : K ] = n and [ K [ x , y ] : K [ x ]] ≤ [ K [ y ] : K ] = n . Therefore, [ L : K ] ≤ · n · n = 2 n . 7ow mf ∈ Z [ X ] has a factorization of type ( n, , . . . , 1) mod p for some prime p . Thismeans that there exists an automorphism σ of order n in the Galois group of L over Q thatfixes n zeros of f (the existence of σ is a consequence of Tchebotarev’s density theorem, see[5] or [1, p. 402]). Let z be such a zero. Then σ also fixes Q [ z ], and [ L : Q [ z ]] must be ≥ n ,the order of σ . On the other hand, [ Q [ z ] : Q ] = 2 n , since f is irreducible. Altogether, weobtain [ L : Q ] ≥ n · n = 2 n . Both estimates together show [ L : Q ] = 2 n .The field L is the composite of the Galois extensions K = K [ x ] and K = K [ y ] of K ,whose Galois groups are h ( x . . . x n ) i and h ( y . . . y n ) i (see [3, Sect. 2, F11]. It is well-knownthat the map σ ( σ | K , σ | K )embeds the Galois group of L over K in the cartesian product h ( x . . . x n ) i × h ( y . . . y n ) i (see [3, Sect. 12, F2]). Since both groups have the same number of elements (which equals n = [ L : K ]), this embedding must be an isomorphism. Hence the Galois group of L over K contains an element σ such that σ | K = ( x . . . x n ), σ | K = id , and an element τ suchthat τ | K = id and τ | K = ( y . . . y n ).Let G be the Galois group of L over Q . For the sake of simplicity we denote σ ∈ G simply by ( x . . . x n ) and τ ∈ G by ( y . . . y n ). The group h σ, τ i of order n is just the Galoisgroup of L over K . Let ρ ∈ G be such that ρ | K = ( ) ′ . Such a ρ exists, since K is a Galoisextension of Q , whose Galois group arises from G by restriction: If we consider µ | K for all µ ∈ G , we must obtain all elements of the Galois group of K over Q (see [3, Sect. 8, F4]).Note that ρ ( x ) must be one of y , . . . , y n , since0 = f ( x ) = ρ ( f )( ρ ( x )) = f ′ ( ρ ( x )) . As above, ρ ( f ) means that ρ is applied to all coefficients of f . The same argument showsthat ρ ( y ) is one of x , . . . , x n . This implies that, for suitably chosen exponents j , k , theautomorphism σ j τ k ρ maps x to y and y to x . For reasons of simplicity we write ρ insteadof σ j τ k ρ . So ρ exchanges x and y . But then it also exchanges x j and y j for j = 2 , . . . , n ,since we have ρ ( x j ) = ρ ( A j − ◦ x ) = ρ ( A j − ) ◦ ρ ( x ) = A ′ j − ◦ y = y j . Altogether, we may write ρ = ( x y )( x y ) . . . ( x n y n ). Hence we have shown the followingtheorem. Theorem 3. Under the assumptions of Theorem , the Galois group G of L over K isgenerated by σ = ( x . . . x n ) , τ = ( y . . . y n ) , and ρ = ( x y ) . . . ( x n y n ) . It contains thenormal subgroup h σ, τ i of order n and index 2. Further G = { σ j τ k ρ l : j, k = 0 , . . . , n − , l =0 , } .Remark. In the context of the theorem, the groups h σ i and h τ i are cyclic subgroups of G , isomorphic to the abstract cyclic group C n of order n . Then h σ, τ i is a group isomorphicto the cartesian product C n × C n . Moreover, it is a normal subgroup of index 2 in G , and h ρ i is isomorphic to C . In addition, we have the composition rules ρσ = τ ρ, ρτ = σρ. (7)This is what the notation G ∼ = C n ≀ C (wreath product of C n by C ) says.The examples of the foregoing section for n = 5 , , , 12 all lead to polynomials over Q with Galois group C n ≀ C . For instance, if n = 10, we obtain f = X + 2 X − X − X + 28905 X + 26568 X − X − X + 5691930 X + 2011900 X − X − X + 28459650 X + 4137000 X − X − X + 3613125 X + 191250 X − X − X + 3125 , where we have divided our original f by − · .8 Dihedral groups Let C n = h σ i be a cyclic group of order n . The dihedral group D n of order 2 n can be definedas the group generated by σ and an element ρ of order 2 such that σρ = ρσ − . (8)Following [4], we construct a polynomial g ∈ Q [ X ] of degree n whose Galois group is D n for n = 5 , , 10 and 12.Let f = f f ′ be as in the foregoing section. In particular, x , . . . , x n are the zeros of f , y , . . . , y n the zeros of f ′ . Moreover, L = Q [ x , . . . , x n , y , . . . , y n ] is the splitting fieldof f over Q , whose Galois group G is generated by σ = ( x . . . x n ), τ = ( y . . . y n ) and ρ = ( x y ) . . . ( x n y n ). Let H = h στ i , a cyclic subgroup of G of order n . By (7), ρσρ = τ and ρτ ρ = σ , so we have ρστ ρ = στ. (9)Further, σ and τ commute with στ . These facts show that H is a normal subgroup of G ,the factor group G/H being of order 2 n .Observe that σ ( στ ) − = τ − , so we obtain σ = τ − for the corresponding residue classesin G/H . From (7) we see that σρ = ρτ , hence σ ρ = ρ σ − . In other words, σ and ρ satisfy the relation (8). Accordingly, G/H is (isomorphic to) thegroup D n . Since H is normal in G , the fixed field M of H is a Galois extension of Q withGalois group isomorphic to G/H . In particular, [ M : Q ] = 2 n .For our purpose, however, we need a subfield M ′ of M with [ M ′ : Q ] = n . Therefore,we look at the relation (9), which shows that the group H ′ = h H, ρ i = h στ, ρ i is abelian oforder 2 n . However, it is not a normal subgroup of G , since σρσ − = ρτ σ − = ρσ − τ H ′ .Hence the fixed field of H ′ is a subfield M ′ of M that is not Galois over Q . Since [ M : M ′ ] =[ H ′ : H ] = 2, the Galois closure of M ′ must be M .In order to generate the field M ′ , we look for an element z ∈ L that remains fixed under H ′ . Such an element is, for instance, z = x y + x y + . . . + x n y n . (10)But this element generates M ′ only if H ′ is the set of all elements of the Galois group thatfix z . In other words, for an element µ ∈ G r H we should have µ ( z ) = z . This assertioncan be phrased in a slightly different way: If R is a system of representatives of G/H ′ , thenumbers µ ( z ) should be distinct for µ ∈ R . In our case the group h σ i itself is a suitableset R . So if the numbers z , σ ( z ), . . . , σ n − ( z ) are pairwise different, we have M ′ = Q [ z ].Moreover, g = n − Y j =0 ( X − σ j ( z ))is the minimal polynomial of z over Q , a polynomial whose splitting field has the Galoisgroup D n .Possibly the simplest way to check whether z is a suitable choice consists in approximatingthe elements z, σ ( z ) , σ n − ( z ) numerically. Provided that the monic polynomial that belongsto f has integer coefficients, the numbers z , σ ( z ) , . . . , σ n − ( z ) are algebraic integers. Thisimplies that g is a polynomial with integer coefficients. Hence a sufficiently precise numericalapproximation of these coefficients exhibits g . Examples. Let n = 5 and f be as in the example preceding Theorem 2. Obviously, thecorresponding monic polynomial has integer coefficients. With z as in (10), we obtain9 = X − X − X − X + 24775 X − f . The numerical result for g is nearly as precise. To be sure, we determinethe Galois group of the splitting field of g with the help of MAPLE. It turns out that it is D .In the case n = 10 we look at the example at the end of the foregoing section. We find g = X − X − X + 6120 X + 166823200 X − X − X + 257616832000 X + 23507471680000 X +13131916800000 X − z . MAPLE does not supply the Galois group of polynomialsof a degree ≥ 10, but factorizing g modulo a number of primes produces only the types(10) , (5 , , (2 , , , , , (2 , , , , , 1) and (1 , . . . , D (recallthe proof of Theorem 2 and the references [5] and [1, p. 402]). Hence it is at least plausiblethat our computations are correct.What can we do if the number z of (10) fails to generate M ′ since the conjugates σ j ( z ), j = 0 , . . . , n − x y + . . . + x n y n . Remark. The method presented here is not restricted to the ground fields Q or Q [ √ D ]for D = 2 , , 5. In the case n = 7, for instance, a possible ground field K is Q [ v ] for v = ζ + ζ − = 2 cos(2 π/ X + X − X − 1. MAPLE allowscomputations in the cubic field K . For instance, if we take A = (cid:18) v − v − (cid:19) and C = 1, we arrive at an irreducible polynomial in f ∈ K [ X ] of degree 7. By Theorem 1,we obtain an extension K [ x ] of K with cyclic Galois group of order 7. Acknowledgment. The author wishes to thank Gerhard Kirchner for a helpful discussion. References [1] David A. Cox, Galois Theory , Wiley, Hoboken (NJ), 2004.[2] Ralf Dentzer, Polynomials with cyclic Galois groups, Comm. Alg. (1995), 1593–1603.[3] Falko Lorenz, Algebra, Volume I: Fields and Galois Theory , Springer, New York, 2006.[4] Dominique Martinais, Leila Schneps, Polynˆomes `a groupe de Galois di´edral,