Algebraic Construction of Quasi-split Algebraic Tori
aa r X i v : . [ m a t h . AG ] J a n Algebraic Construction of Quasi-split Algebraic Tori
Armin Jamshidpey , Nicole Lemire , and ´Eric Schost David Cheriton School of Computer Science, University of Waterloo Department of Mathematics, University of Western OntarioJanuary 30, 2018
Abstract
The main purpose of this work is to give a constructive proof for a particular caseof the no-name lemma. Let G be a finite group, K be a field, L be a permutation G -lattice and K [ L ] be the group algebra of L over K . The no-name lemma asserts that theinvariant field of the quotient field of K [ L ], K ( L ) G is a purely transcendental extensionof K G . In other words, there exist y , . . . , y n which are algebraically independent over K G such that K ( L ) G ∼ = K G ( y , . . . , y n ). We define elements { y , . . . , y n } ⊂ K [ L ] G with the desired properties, in the case when G is the Galois group of a finite extensionGal( K/F ), and L is a sign permutation G -lattice. An algebraic F -torus T is an algebraic group defined over a field F which splits over analgebraic closure ¯ F of F , that is, which is isomorphic to a torus (a finite product of copiesof the multiplicative group G m ) over ¯ F . In general, ¯ F is not the smallest field over which T splits: it is known that an algebraic F -torus T splits over a finite Galois extension of F .There is a unique minimal such extension, say K ; if G = Gal( K/F ), then G is called thesplitting group of T . For more details, see [16, p. 27].For a finite group G , a G -lattice L is a free Z -module A of finite rank, together with agroup homomorphism G −→ Aut( A ) (the group of automorphisms of A ). Given a modulebasis of A , any group homomorphism G −→ GL( n, Z ), with n = rank( A ), gives such anaction. If K is a field, the group algebra K [ L ] of L over K is isomorphic to the K -algebra ofLaurent polynomials K [ x ± , . . . , x ± n ], for some indeterminates x , . . . , x n . If K is equippedwith an action of G (that is, a G -field), we can extend the action of G on lattice L to anaction on K [ L ]; the ring K [ L ] G of multiplicative invariants consists of those elements in K [ L ] invariant under the action of G . The fraction field K ( L ) of K [ L ] is isomorphic to K ( x , . . . , x n ), and the subfield of invariants under the action of G is written K ( L ) G .1t is known that there is a duality between the category of algebraic tori with splittinggroup G and G -lattices. For a given algebraic torus T with splitting group G , its charactermodule Hom( T, G m ) is a G -lattice. Conversely, if L is a G -lattice, with G = Gal( K/F )for some finite Galois extension
K/F , then T = Spec( K [ L ] G ) is an algebraic F -torus withsplitting group G , coordinate ring K [ L ] G and function field K ( L ) G .A G -lattice L is called permutation (resp. sign permutation ) if it has a Z -basis which ispermuted (resp. up to sign changes) by G . In particular, an algebraic torus whose correspond-ing G -lattice is a permutation lattice is called a quasi-split torus . Quasi-split algebraic toriare characterized as being representable as a direct product of groups of the form R E/F ( G m ),where R E/F is the Weil restriction with respect to a finite separable field extension
E/F .Note that R E/F ( G m ) is a F -group scheme with F -points E × and character lattice Z [ G/H ],where
K/F is a Galois closure of
E/F , G = Gal( K/F ) and H is the subgroup of G whichfixes the subfield E . The Weil restriction here is not necessarily with respect to a finiteGalois extension K/F as this would only produce character lattices which are direct sumsof the group ring.A specific case of the no-name lemma asserts that if L is a permutation G -lattice and K is a G -field, then K ( L ) G is rational over K G [11, Chapter 9.4]; in particular, with G = Gal( K/F ), K ( L ) G is F -rational. The term ”no-name lemma” was first used by Dol-gachev in [3], expressing the fact that many researchers discovered the result independently.It is actually more general than the stated version here; see [3, p. 6], [1, Section 3.2], [10,Proposition 1.3], [4, Remark 2.4] and [5, Proposition 1.1]. In this paper, we give a con-structive proof of the particular case described above (and of a slight generalization thereof,using signed permutation matrices), by exhibiting a basis for such a field of invariants. Inconcrete terms, we start from a subgroup of GL( n, Z ) and describe the field of functions ofan associated torus. Definition 1.
Let G be a finite subgroup of GL( n, Z ) . The G -lattice L G corresponding to G is the rank n lattice Z n = { [ a , . . . , a n ] T : a i ∈ Z } on which G acts naturally by left-multiplication. Note that Z n has a standard basis { e i : i = 1 , . . . , n } , where e i is the columnvector [ δ i,j , i = 1 , . . . , n ] T . Suppose further that we are given an isomorphism ι : G → Gal(
K/F ), for some finiteGalois extension
K/F (in what follows, we simply say that
K/F has Galois group G ). Then,through the identification K [ x ± , . . . , x ± n ] ≃ K [ L G ], K [ x ± , . . . , x ± n ] is equipped with the G -action defined as follows: • for g in G and α in K , we write g ( α ) = ( ι ( g ))( α ) (so G acts as the Galois group on K ); • for g in G and j = 1 , . . . , n , g ( x i ) = Q nj =1 x g j,i j , where g j,i is the ( j, i )th -entry of g (sothat we also have g ( e i ) = P nj =1 g j,i e j ).If we let T G be the algebraic torus corresponding to L G , then T G is an algebraic F -toruswhich splits over K , with character lattice L G and function field K ( x , . . . , x n ) G .2onjugate subgroups of GL( n, Z ) correspond to isomorphic lattices, and isomorphic al-gebraic tori; in particular, for G a finite subgroup of GL( n, Z ), L G is a (signed) permutationlattice if and only if G is conjugate to a group of (signed) permutation matrices. Computa-tionally, we do not have an efficient algorithm at hand to decide whether a given lattice is(signed) permutation. Hence, in our main results, we will assume that G is a subgroup ofthe group S n of permutation matrices of size n , or more generally of the group B n of signedpermutation matrices of size n . In such a case, for i in { , . . . , n } and g in G , g ( e i ) = ± e j for some index j in { , . . . , n } (all signs being +1 if G is a subgroup of S n ), and the actionof g ∈ G on x i is given by g ( x i ) = ( x j if g ( e i ) = e j x − j if g ( e i ) = − e j . For such groups G , the F -rationality of the torus T G means that for K ( x , . . . , x n ),endowed with the G -action we just described, there exist algebraically independent y , . . . , y n in K ( x , . . . , x n ) such that K ( x , . . . , x n ) G = F ( y , . . . , y n ). However, the proofs of the no-name lemma we are aware of are nonconstructive. The goal of this paper is to exhibit suchan algebraically independent set { y i : i = 1 , . . . , n } ; we state two such results. Note that thefirst result is a special case of the general version of the no-name lemma. In particular, thiscan be applied to get an explicit transcendence basis of the function field of a quasi-splitalgebraic torus T G , provided G is given as a group of permutation matrices.In both our theorems, we rely on the notion of a normal element of a finite Galoisextension K/F with Galois group G ; we recall that α ∈ K is normal if α and all its Galoisconjugates form an F -basis of K . Any finite Galois extension admits a normal element [9,Theorem 6.13.1]; there exist algorithms to construct one, in characteristic zero [6] or inpositive characteristic [15, 12]. Theorem 2.
Let G be a subgroup of S n , let K/F be a finite Galois extension with Galoisgroup G , and let α ∈ K be a normal element for K/F . Then K ( x , . . . , x n ) G = F ( y , . . . , y n ) ,with y i = X g ∈ G g ( αx i ) , i = 1 , . . . , n. Our second statement is similar, but deals with the more general case of signed permu-tations (if the group G below happens to be a subgroup of S n , the construction is not thesame as in the previous theorem). Theorem 3.
Let G be a subgroup of B n , let K/F be a finite Galois extension with Galoisgroup G , and let α ∈ K be a normal element for K/F . Then K ( x , . . . , x n ) G = F ( y , . . . , y n ) , with y i = X g ∈ G g (cid:18) α x i (cid:19) , i = 1 , . . . , n. There are many algorithms for finding the invariant rings for polynomial invariants [14, 2].For multiplicative invariants, the algorithmic landscape is not developed to the same extent.3n [13] the author introduced an algorithm to compute a generating set for the ring ofmultiplicative invariants, when the acting group is a subgroup of a reflection group. A moregeneral algorithm for computing the ring of multiplicative invariants is given by Kemperin [8]. It is worth mentioning that although both above algorithms may be applied to ourproblem, they will not necessarily produce a transcendence basis of the invariant field.
Acknowledgements.
A large part of this work was done while the first author was a PhDstudent at the University of Western Ontario. The second and third authors are supportedby an NSERC Discovery Grant. We thank Gregor Kemper and Lex Renner for their feedbackon an earlier version of this work.
In what follows, we use the following notation: we still write S n , resp. B n , for the groupsof permutation, resp. signed permutation matrices of size n . Note that D n , the group ofdiagonal matrices in GL n ( Z ) (diagonal matrices with entries in {± } ) is a normal subgroupof B n such that B n = D n ⋊ S n is a semidirect product. We let S n be the symmetric groupof size n . Since S n is naturally isomorphic to S n , we see that there is a natural grouphomomorphism ρ : B n → S n obtained by mapping a signed permutation matrix to thepermutation ρ g such that ρ g ( i ) = j , where j is the index of the unique non-zero entry in the i th column of g . Note that the kernel of the group homomorphism is D n . Hence, in termsof the action defined in the previous section, for g in B n and for all i, j in { , . . . , n } , wehave g ( x i ) = x ± ρ g ( i ) . For i, j as above, we also denote by G ± i,j the set of all g in G such that g ( x i ) = x ± j , that is, such that ρ g ( i ) = j .We start with a lemma that generalizes known facts about Moore matrices over finitefields (see Example 5 below). Let G be a subgroup of B n and let K/F be a finite Galoisextension with Galois group isomorphic to G , as in the previous section. Through thisisomorphism, G acts on (column) vectors entrywise: for g ∈ G and a column vector in K m ,written as C = [ µ µ · · · µ m ] T , we define g ( C ) = [ g ( µ ) g ( µ ) · · · g ( µ m )] T .Let then M be in M n,n ( K ). For such a matrix, and for i = 1 , . . . , n , its i th column iswritten M i = [ µ ,i µ ,i · · · µ m,j ] T . We say that G permutes the columns of M up to sign iffor g in G and i in { , . . . , n } , we have g ( M i ) = ± M ρ g ( i ) . Lemma 4.
Let
K/F be a finite Galois extension with Galois group G . Let M be in M n,n ( K ) and assume that G permutes the columns of M up to sign. Assume also that the entries ofthe first column of M are F -linearly independent. Then M is invertible.Proof. Assume by contradiction that there is a non-zero vector in the left nullspace of M ;take x ∈ K n to be a vector with the minimum number of non-zero entries among the non-zero left nullspace elements (that is, such that x T M = 0). Let k ∈ { , . . . , n } be such that x k = 0 and let y = x k x ∈ K n , so that y k = 1, and y is still in the left nullspace of M .For i in { , . . . , n } , we have the equality y T M i = 0, where M i is the i th column of M .For g in G , we deduce g ( y T M i ) = g ( y ) T g ( M i ) = ± g ( y ) T M ρ g ( i ) = 0. Since this is true for4ll i , we obtain that g ( y ) is in the left nullspace of M as well. This further implies that y ′ = g ( y ) − y is in the left nullspace of M . However, since y k = 1, g ( y k ) = 1, so that y ′ k = 0.By construction of x , this implies that y ′ = 0, so that g ( y ) = y .Since this is true for all g , we deduce that y is in F n . Then, the relation y T M = 0implies that y = 0, a contradiction. Remark.
The hypotheses in the above lemma imply that the induced unsigned action of thegroup G on the columns of M is a transitive action. Let H = Stab G ( M ) . Then h ∈ H fixes M and so M ∈ ( K H ) n . But then the n entries of M are contained in K H and arealso F -linearly independent. The size of an F -linearly independent set in K H can be at most [ K H : F ] . Then n ≤ [ K H : F ] = [ K H : K G ] = [ G : H ] ≤ n implies that [ G : H ] = n . Butthen the orbit of M under the action of G has size [ G : H ] = [ G : Stab G ( M )] = n . Example 5.
Let F = F q , for some prime power q , and let K = F q n . The Galois group of K/F is cyclic of size n , generated by the Frobenius map x x q . Let then ( α , . . . , α n ) bein K , and consider the Moore matrix M = [ m i,j ] ≤ i,j ≤ n , with m i,j = α q j − i . The Frobeniusmap permutes the columns of M , and we recover the fact that if ( α , . . . , α n ) are F -linearlyindependent, M is invertible [7, Corollary 1.3.4]. We can now prove our first result.
Proof of Theorem 2.
We first prove the result under the additional assumption that G actstransitively on L G . The elements ( y , . . . , y n ), with y i = P g ∈ G g ( αx i ) as defined in thetheorem, are invariant under the action of G . We will show below that K ( x , . . . , x n ) = K ( y , . . . , y n ); this will prove that K ( x , . . . , x n ) G = F ( y , . . . , y n ), since K ( x , . . . , x n ) = K ( y , . . . , y n ) implies that K ( y , . . . , y n ) G = K G ( y , . . . , y n ) = F ( y , . . . , y n ).For i, j in { , . . . , n } , let G i,j = { g ∈ G : g ( x i ) = x j } , so that we can rewrite y i as y i = n X j =1 X g ∈ G i,j g ( α ) x j , i = 1 , . . . , n. Note that G i,i = Stab G ( x i ) and so is a subgroup of G . Since the action of G on L G istransitive, G i,j is non-empty for every 1 ≤ i, j ≤ n . Take such indices i, j , and fix some g i,j in G i,j . If g ∈ G i,j , then g − i,j g ( x i ) = x i shows that g is in g i,j G i,i . Since we also have g i,j G i,i ⊆ G i,j , we see that G i,j = g i,j G i,i .We now show that the matrix M with i th row the coordinate vector of y i with respectto the K -basis { x , . . . x n } is invertible. The matrix M has entries m i,j = P g ∈ G i,j g ( α ), i, j = 1 , . . . , n . We will apply Lemma 4 to show that M is invertible, which is sufficient toprove the theorem.We check the hypothesis of the lemma. First, let ρ : G → S n , ρ ( g ) = ρ g be the grouphomomorphism that corresponds to the action of G on ( x , . . . , x n ), so that ρ g ( i ) = j if andonly if g ( x i ) = x j for all 1 ≤ i, j ≤ n . We will show that the columns of M are permutedby the action of G . Let thus h be in G . Note that for g in G i,j , hg is in G i,ρ h ( j ) . Then since5 i,j = g i,j G i,i is a left coset of G i,i where g i,j is an arbitrary element of G i,j , we see that hG i,j = hg i,j G i,i = G i,ρ h ( j ) since hg ij ∈ G i,ρ h ( j ) . We then get h ( m i,j ) = X g ∈ G i,j hg ( α ) = X σ ∈ G i,ρh ( j ) σ ( α ) = m i,ρ h ( j ) . This shows that h ( M j ) = M ρ h ( j ) for all j = 1 , . . . , n , so that G permutes the columns of M .Finally, the first column M has entries P g ∈ G i, g ( α ), i = 1 , . . . , n . Since α is a normalelement of the Galois extension K/F with Galois group G , the set { g ( α ) : g ∈ G } is F -linearly independent. Since G = ⊔ ni =1 G i, is a disjoint union, and all G i, are non-empty, theset X g ∈ G i, g ( α ) , i = 1 , . . . , n is F -linearly independent as well. So Lemma 4 applies, and we conclude that M is invertible,as claimed.We can now give the proof of our claim in the general case. Let e , . . . , e n be thestandard basis of L G , and let { e j k | k = 1 , . . . , r } and correspondingly { x j k | k = 1 , . . . , r } bea complete set of G -orbit representatives among the basis vectors, and the indeterminates x , . . . , x n respectively. Then L k = ⊕ e i ∈ G e jk Z e i is a transitive permutation G -lattice for each k = 1 , . . . , r , and K ( L k ) = K ( x i | x i ∈ Gx j k ).The lattice L G = ⊕ rk =1 L k is a direct sum of transitive permutation G -lattices, so that K ( x , . . . , x n ) is the compositum of the fields K ( L k ), k = 1 , . . . , r . Thus, using the resultestablished in the transitive case, we obtain K ( x , . . . , x n ) G = F ( y i | x i ∈ Gx j k , k = 1 , . . . , r ),where for all k and for x i ∈ Gx j k , we have y i = P g ∈ G g ( αx i ). Example 6.
Let K be the splitting field of x − over F = Q . Then Gal( K/ Q ) ∼ = D , K = Q ( θ, i ) , with θ = √ , and { , θ, θ , θ , i, iθ, iθ , iθ } , is a Q -basis for K . Let n = 4 , let G ≤ GL(4 , Z ) be generated by r = and s = . and let ( x , . . . , x ) be new indeterminates, on which G acts as in Definition 1; this actionis transitive. One can verify that G is isomorphic to Gal( K/ Q ) ; through this isomorphism,the action of r and s on the generators of K is given by r ( i ) = i r ( θ ) = iθs ( i ) = − i s ( θ ) = θ. Now, define α = 1 + θ + θ + θ + i + iθ + iθ + iθ = ( X i =0 θ i )(1 + i );6 his is a normal element in K/ Q . Note that G , = Stab G ( x ) = h s i . Since r i ( x ) = x i +1 , i = 0 , , , , we see that G i,i = Stab G ( x i ) = r i Stab G ( x ) r − i and so G , = G , = h s i and G , = G , = h r s i . Also note that r j − i ∈ G i,j , where we may consider all exponents of r modulo 4. This shows that the elements ( y , . . . , y ) of Theorem 2, expressed on the basis ( x , . . . , x ) , are given by the coordinate matrix M = (1 + s )( α ) r (1 + s )( α ) r (1 + s )( α ) r (1 + s )( α ) r (1 + r s )( α ) (1 + r s )( α ) r (1 + r s )( α ) r (1 + r s )( α ) r (1 + s )( α ) r (1 + s )( α ) (1 + s )( α ) r (1 + s )( α ) r (1 + r s )( α ) r (1 + r s )( α ) r (1 + r s )( α ) (1 + r s )( α ) . Now since M j = r j − M for all j = 1 , , , , it is clear that the action of r permutes thecolumns. In particular, rM j = M j +1 (modulo 4). One can check that sM = M and since | Stab G ( M ) | = 8 / , this shows that Stab G ( M ) = h s i . Then sr j − M = r − j sM showsthat sM j = M − j , j = 1 , . . . , so s also permutes the columns of M . Remark.
With the assumptions of the previous theorem, we can actually compute the coor-dinate ring of the torus; we obtain K [ L G ] G ∼ = K [ x ± , x ± , . . . , x ± n ] G = F [ y , . . . , y n ] x ··· x n , for y , . . . , y n as in the theorem. Indeed, we have K [ x ± , x ± , . . . , x ± n ] = K [ x , . . . , x n ] x ··· x n . We are interested in K [ L G ] G , that is, (cid:0) K [ x ± , x ± , . . . , x ± n ] (cid:1) G = ( K [ x , . . . , x n ] x ··· x n ) G . Theproof of Theorem 2 shows that K [ x , . . . , x n ] = K [ y , . . . , y n ] . On the other hand since G permutes the x i ’s, x · · · x n is invariant under the action of G , and we can conclude ( K [ x , . . . , x n ] x ··· x n ) G = ( K [ y , . . . , y n ] x ··· x n ) G = K G [ y , . . . , y n ] x ··· x n = F [ y , . . . , y n ] x ··· x n . One could further rewrite x · · · x n as a degree n homogeneous polynomial in y , . . . , y n (butthe expression obtained this way is not particularly handy). We conclude with the proof of our second main result. The proof follows that of The-orem 2, the only difference being in the description of the coordinate matrix M . As inTheorem 2, we first prove the result under the extra assumption that G acts transitively upto sign on L G . Proof of Theorem 3.
Assume first that the action of G is transitive (up to sign). For i in { , . . . , n } , define z i = (1 + x i ) − . Then, for g ∈ G , g ( z i ) = ( z j if g ( x i ) = x j − z j if g ( x i ) = x − j , and K ( x , . . . , x n ) = K ( z , . . . , z n ) . The elements y i can be rewritten as y i = P g ∈ G g ( αz i ), for i in { , . . . , n } ; as before, in order to prove that K ( z , . . . , z n ) G = F ( y , . . . , y n ), it is enough7o prove that K ( y , . . . , y n ) = K ( z , . . . , z n ). This will be done by writing (1 , y , . . . , y n ) as K -linear combinations of (1 , z , . . . , z n ), and proving that the coordinate matrix is invertible.For i, j in { , . . . , n } , let G ± i,j be defined as in the preamble of this section, that is G ± i,j = { g ∈ G : g ( z i ) = z j or g ( z i ) = 1 − z j } . By the transitivity assumption, G ± i,j is non-empty forevery 1 ≤ i, j ≤ n . Let us further define G + i,j = { g ∈ G : g ( z i ) = z j } and G − i,j = { g ∈ G : g ( z i ) = 1 − z j } , so that G i,j = G + i,j ⊔ G − i,j .Note that G + i,i = Stab G ( z i ). Both G ± i,i and G + i,i are subgroups of G . One can show thatthe left cosets of G + i,i in G are the non-empty sets in the collection { G + i,j : 1 ≤ j ≤ n } ∪ { G − i,j : 1 ≤ j ≤ n } By the transitivity assumption, the sets G ± i,j = ∅ so one can guarantee that either G + i,j = ∅ or G − i,j = ∅ (although both are possible). If G + i,j = ∅ , and g + ij is an arbitrary element of G + i,j ,then G + i,j = g + ij G + i,j and if G − i,j = ∅ and g − ij is an arbitrary element of G − i,j , then G − i,j = g − ij G + i,j .Let M ∗ be the coordinate matrix of (1 , y , . . . , y n ) with respect to the K -basis (1 , z , . . . , z n );we have to show that det( M ∗ ) = 0. By definition, for i in { , . . . , n } , we have y i = X g ∈ G g ( αz i ) = n X j =1 (cid:16) X g ∈ G + i,j g ( α ) z j + X g ∈ G − i,j g ( α )(1 − z j ) (cid:17) = n X j =1 X g ∈ G − i,j g ( α ) + n X j =1 (cid:16) X g ∈ G + i,j g ( α ) − X g ∈ G − i,j g ( α ) (cid:17) z j . For i, j ∈ { , . . . , n } , define m i,j = P g ∈ G + i,j g ( α ) − P g ∈ G − i,j g ( α ) and c i = P nj =1 P g ∈ G − i,j g ( α ).The matrix M ∗ is then M ∗ = · · · c m , · · · m ,n ... ... ... c n m n, · · · m n,n . Let us write M = m , · · · m ,n ... ... m n, · · · m n,n . Since det( M ∗ ) = det( M ), it is enough to show that the determinant of M is non-zero; thiswill be done using Lemma 4. We now check that the hypotheses of the lemma are satisfied.As before, let ρ : G → S n , ρ ( g ) = ρ g be the group homomorphism that corresponds tothe action of G on { z , . . . , z n } , so that ρ g ( i ) = j if and only if g is in G ± i,j . We will showthat the columns M , . . . , M n of M are permuted up to sign by the action of G .Let h be in G and i, j be in { , . . . , n } . We can then write h ( m i,j ) = h (cid:16) X g ∈ G + i,j g ( α ) − X g ∈ G − i,j g ( α ) (cid:17) = X g ∈ G + i,j hg ( α ) − X g ∈ G − i,j hg ( α ) .
8s in the proof of Theorem 2, we have hG ± i,j = G ± i,ρ h ( j ) , but more precisely, we can write (cid:26) G + i,ρ h ( j ) = hG + i,j G − i,ρ h ( j ) = hG − i,j if h ∈ G + j,ρ h ( j ) and (cid:26) G + i,ρ h ( j ) = hG − i,j G − i,ρ h ( j ) = hG + i,j if h ∈ G − j,ρ h ( j ) . (1)In the first case, we deduce m i,ρ h ( j ) = X g ∈ G + i,ρh ( j ) g ( α ) − X g ∈ G − i,ρh ( j ) g ( α ) = X g ∈ G + i,j hg ( α ) − X g ∈ G − i,j hg ( α ) = h ( m i,j );in the second case, we get m iρ h ( j ) = X g ∈ G + i,ρh ( j ) g ( α ) − X g ∈ G − i,ρh ( j ) g ( α ) = X g ∈ G − i,j hg ( α ) − X g ∈ G + i,j hg ( α ) = − h ( m i,j ) . In other words, h ( M j ) = ± M ρ h ( j ) , so G permutes the columns of M up to sign. Secondly,the first column M has entries X g ∈ G + i, g ( α ) − X g ∈ G − i, g ( α ) , i = 1 , . . . , n. Since α is a normal element of the Galois extension K/F with Galois group G , and since G = ⊔ ni =1 G ± i,j = ⊔ ni =1 ( G + i, ⊔ G − i, ) is a disjoint union, with all G ± i,j non-empty (by the transitivityof the action), this set is F -linearly independent.So Lemma 4 applies, and we conclude that K ( y , . . . , y n ) = K ( z , . . . , z n ); this impliesthat K ( x , . . . , x n ) G = F ( y , . . . , y n ). This finishes the proof in the transitive case; the proofin the general case follows as in Theorem 2. Example 7.
Let K = Q ( ρ ) , where ρ is a primitive -th root of unity, so that K/ Q is Galois,with Gal( K/ Q ) ∼ = C . Take n = 3 , assume G ≤ GL(3 , Z ) is generated by σ = − − and let ( x , x , x ) be indeterminates over K , on which G acts as in Definition 1; this actionis not transitive. One can also verify that G is isomorphic to Gal( K/ Q ) ; σ ( ρ ) = ρ . Thisimplies that σ k ( ρ ) = ρ k , k = 0 , , , . In particular, σ ( ρ ) = ρ = ρ .We choose ρ as our normal element of the extension K/ Q . (A primitive p th root of unityis a normal element for the extension over Q that it generates.)In this example, note that G +1 , = G +2 , = { id } , G − , = G − , { σ } , G +1 , = G − , = { σ } ,and G − , = G +2 , = { σ } . Also, G +3 , = h σ i and G − , = σG +3 , = h σ, σ } . ith ( z , z , z ) and ( y , y , y ) defined as before, the matrix M ∗ giving the coordinates of (1 , y , y , y ) on the basis (1 , z , z , z ) is M ∗ = ρ + ρ ρ − ρ ρ − ρ ρ + ρ ρ − ρ ρ − ρ ρ + ρ ρ − ρ − ρ + ρ . Remark that due to the non-transitivity of the action of G , the bottom-right × submatrixof M ∗ , while invertible, does not satisfy the assumptions of Lemma 4 (this matrix is blockdiagonal, with blocks corresponding to K ( z , z ) and K ( z ) , for which the lemma applies). References [1] Jean-Louis Colliot-Th´el`ene and Jean-Jacques Sansuc. The rationality problem for fieldsof invariants under linear algebraic groups (with special regards to the Brauer group).In
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