Z -basis for the orders generated by the conjugates of algebraic - PDF document

Z -basis for the orders generated by the conjugates of algebraic integers St´ ephane R. LOUBOUTIN Aix Marseille Universit´ e, CNRS, Centrale Marseille, I2M, Marseille, FRANCE stephane.louboutin@univ-amu.fr June 26, 2018 Contents 1 Abstract 2 2 A Z -generating system of M α 5 3 A Z -basis of M α in the worst case 6 4 Remark on the Z -basis of M α 7 5 A Z -basis of M α in the cyclic cubic case 8 6 When is the order Z [ ε ] Galois-invariant? 10 7 On the orders Z [ α n ] and M α n , n � 1 13 7.1 The case that α is an algebraic unit . . . . . . . . . . . . . . . 13 7.2 The non-normal cubic case . . . . . . . . . . . . . . . . . . . . 14 7.3 The cyclic cubic case . . . . . . . . . . . . . . . . . . . . . . . 15 1

2 Z -basis for the orders generated by the conjugates of algebraic integers 1 Abstract Let ( α j � α i ) 2 2 Z \ { 0 } Y D α := 1  i<j  n be the discriminant of the minimal polynomial Π α ( X ) = X n � a n � 1 X n � 1 + · · · + ( � 1) n a 0 2 Z [ X ] of an algebraic integer α of degree n , where α 1 , · · · , α n are the n distinct complex roots of Π α ( X ). We consider M α = Z [ α 1 , · · · , α n ] , and order of L α = Q ( α 1 , · · · , α n ) be the normal closure of Q ( α ). It is a free Z -module of rank r = ( L α : Q ) � ( Q ( α ) : Q ) = n. If M is an order of a number field, let D M 2 Z denote its discriminant. Notice that D Z [ α ] = D α . The goal is to determine a Z -basis and the discrimi- nant D M α of M α and to give various applications of these determinations.

3 Z -basis for the orders generated by the conjugates of algebraic integers Let us explain how one usually constructs parametrized families of number fields of known discriminants and reg- ulators. One usually starts from explicit parametrized families of monic polynomials with integral coe ffi cients and constant coe ffi cient equal to ± 1, so that their com- plex roots are algebraic units. Let us for example consider the simplest cubic fields Q ( α ), where Π α ( X ) = X 3 � aX 2 +( a � 3) X +1, a � 2. Since D α = ( a 2 � 3 a + 9) 2 is a square, Q ( α ) / Q is a Galois cyclic cubic extension and since Π α ( X ) = ( X � α ) ( X � ( � α 2 + ( a � 1) α + 2))( X � ( α 2 � a α + a � 2)), the order Z [ α ] is Galois invariant. Moreover, the three conjugates α , α 0 and α 00 of α are algebraic units. Since Q ( α ) / Q is of prime degree, any 2 of these conjugates are multiplicatively independent in the group of units Z [ α ] ⇥ of the order Z [ α ]. In fact, for the simplest cubic fields we have Z [ α ] ⇥ = h� 1 , α , α 0 i . Hence, in the cases that Z [ α ] is equal to the ring of algebraic integers Z K of the number field K , we end up with cyclic cubic fields of known discriminants and regulators. Now, since D α = ( Z K : Z [ α ]) 2 d K and d K > 1, it follows that if a 2 � 3 a +9 = p is prime then Z K = Z [ α ]. Since the class number of K divides the class number h + p of the real cyclotomic field Q ( ζ p ) + , we easily end up with with examples of prime numbers p > 3 for which h + p > 1 (see [CW]). However, still assuming that α is an algebraic unit such that Q ( α ) / Q Galois cyclic of prime degree p � 3,

4 Z -basis for the orders generated by the conjugates of algebraic integers whereas the order Z [ α ] is not generally Galois invari- ant, the order M α is always Galois invariant. Hence it would be much more satisfactory to have families of parametrized polynomials for which D M α would be known and for which any p � 1 of the p conjugates of α would form a system of fundamental units of the order M α . In this respect we proved: Let ε , ε 0 Theorem 1 (See [LL14, Theorem 1.2]). and ε 00 be the three real roots of any one of the follow- ing parametrized families of Q -irreducible cubic poly- nomials of discriminants a square: Φ n ( X ) = X 3 � n ( n 2 + n + 3)( n 2 + 2) X 2 � ( n 3 + 2 n 2 + 3 n + 3) X � 1 n 2 Z , Ξ n ( X ) = X 3 � ( n 3 � 2 n 2 + 3 n � 3) X 2 � n 2 X � 1 1 , 2 6 = n 2 Z , Ψ n ( X ) = X 3 +( n 8 +2 n 6 � 3 n 5 +3 n 4 � 4 n 3 +5 n 2 � 3 n +3) X 2 � ( n 3 � 2) n 2 X � 1 n 2 Z . Then, { 1 , ε , ε 2 ε 0 } is a Z -basis of the totally real cubic order Z [ ε , ε 0 ] and { ε , ε 0 } is a system of fundamental units of this cubic order Z [ ε , ε 0 ] . Indeed, in these three cases 3 ac � b 2 divides D α and 3 b � a 2 . Hence the results on the Z -basis follow from Theorem 7.

5 Z -basis for the orders generated by the conjugates of algebraic integers 2 A Z -generating system of M α Proposition 2 (See [Lou16b]). The set Ω α := { α e 1 1 · · · e e n n ; 0  e k  n � k } is a Z -generating system (with n ! elements) of M α . Proof. Let us prove Proposition 2 for n = 3. We must show that Ω α = { 1 , α , α 2 , α 0 , αα 0 , α 2 α 0 } is a Z -generating system of the order M α = Z [ α , α 0 , α 00 ], where α , α 0 and α 00 are the complex roots of the minimal polynomial Π α ( X ) = X 3 � aX 2 + bX � c 2 Z [ X ]. Since a = α + α 0 + α 00 2 Z , we have M α = Z [ α , α 0 ] . Now, α 0 being a root of Π α ( X ) X � α = X 2 � ( a � α ) X + ( α 2 � a α + b ) 2 Z [ α ][ X ] , it is integral over Z [ α ] and we have M α = Z [ α , α 0 ] = Z [ α ][ α 0 ] = Z [ α ] + α 0 Z [ α ] . Since Z [ α ] = Z + Z α + Z α 2 , the desired result follows. •

6 Z -basis for the orders generated by the conjugates of algebraic integers 3 A Z -basis of M α in the worst case Theorem 3 (See [Lou16b]). Assume that Gal( L α / Q ) is isomorphic to the symmetric group S n . Then M α is a free Z -module of rank n ! , Ω α := { α e 1 1 · · · α e n n ; 0  e k  n � k } is a Z -basis of M α and the discriminant of M α is D M α = D n ! / 2 . α Now, of particular interest is the case where Q ( α ) is a normal number field. In that case M α is a Galois in- variant order of Q ( α ) and r = n . The matrix M α of the coordinates of the n ! elements of Ω α in the canonical Q -basis B α = { 1 , α , · · · , α n � 1 } of Q ( α ) is in M n,n ! ( Q ). Consequently, it is rather easy to develop and algorithm for constructing a Z -basis of M α and to compute D M α . However, from a theorytical point of view, in the Galois case, we present the only cases where a Z -basis of M α has been obtained: the quadratic and cubic cases.

7 Z -basis for the orders generated by the conjugates of algebraic integers 4 Remark on the Z -basis of M α Lemma 4 Let { ω 1 , · · · , ω r } be a Z -basis of a free Z - module M of rank r � 1 . There exists a Z -basis of M containing a given ω = a 1 ω 1 + · · · + a r ω r 2 M if and only if gcd( a 1 , · · · , a r ) = 1 . Consequently, if 1 2 M and M \ Q = Z , e.g. if M is an order of a number field of degree r , then there exists a Z -basis of M of the form { 1 , ω 2 , · · · , ω r } . Proof. Clearly, the condition is necessary. Conversely, assume that gcd( a 1 , · · · , a r ) = 1. Let u 1 , · · · , u r 2 Z be such that a 1 u 1 + · · · + a r u r = 1 (B´ ezout). Define a Z -linear map φ : M � ! Z by x = x 1 ω 1 + · · · + x r ω r 2 M 7! φ ( x ) = u 1 x 1 + · · · + u r x r 2 Z . Since x = φ ( x ) ω + ( x � φ ( x ) ω ) for x 2 M , we have M = Z ω � ker φ . Hence, there exist ω 0 2 , · · · , ω 0 r 2 M such that { ω , ω 0 2 , · · · , ω 0 r } is a Z -basis of M . • Proposition 5 There exists a Z -basis of M α of the form { 1 , α , ω 3 , · · · , ω r } .

8 Z -basis for the orders generated by the conjugates of algebraic integers 5 A Z -basis of M α in the cyclic cubic case Corollary 6 Let α ba a cubic algebraic integer. As- sume that Q ( α ) / Q is Galois, i.e. assume that D α = D 2 is a square. Then D M α = ( D/ ( M α : Z [ α ]) 2 and the index ( M α : Z [ α ]) is equal to the least d � 1 such that d M α ✓ Z [ α ] , i.e. is equal to the least common multiple of the denominators of the entries of the matrix M α of the coordinates in the Q -basis B α = { 1 , α , α 2 } of any Z -generating system of M α . We may assume that Ω α = { 1 , α , α 2 , α 0 , αα 0 , α 2 α 0 } with α 0 such that its coordinates in the canonical Q -basis B α = { 1 , α , α 2 } of Q ( α ) are given by the fourth column of the following matrix M α (see [Lou12, Proposition 10]): (2 a 2 � 6 b ) c 0 a 2 b +3 ac � 4 b 2 + Da 1 ( ab � 9 c � D ) c 1 0 0 2 D 2 D 2 D � 2 a 3 +7 ab � 9 c � D � a 2 b +3 ac +2 b 2 + aD 2 a 2 c � ab 2 +3 bc + bD 0 1 0 B C 2 D 2 D 2 D @ A 2 a 2 � 6 b � 6 ac +2 b 2 ab � 9 c � D 0 0 1 2 D 2 D 2 D Letting n i,j denote the numerator of the coe ffi cient ( i, j ) of M α , we have d := ( M α : Z [ α ]) = 2 D/ gcd( n i,j ) and ◆ 2 ✓ 1 D M α = ( D/d ) 2 = 2 gcd ( n 3 ,j ) , 4  j  6 D α , ( a 2 � 3 b ) 2 , ( b 2 � 3 ac ) 2 � � D M α = gcd .