Journal of Algebra 368 (2012) Contents lists available at SciVerse ScienceDirect. Journal of Algebra.

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1 Journal of Algebra ) Contents lists available at SciVerse ScienceDirect Journal of Algebra wwwelseviercom/locate/jalgebra The rationality problem for finite subgroups of GL 4 Q) Ming-chang Kang a,,1, Jian Zhou b,1, a Department of Mathematics and Taida Institute of Mathematical Sciences, National Taiwan University, Taipei, Taiwan b School of Mathematical Sciences, Peking University, Beijing, China article info abstract Article history: Received 6 May 010 Availableonline10July01 Communicated by Efim Zelmanov MSC: primary 13A50, 14E08, 14M0, 1F1 Keywords: Rationality problem Rationality Retract rationality Conic bundles Let G be a finite subgroup of GL 4 Q) The group G induces an action on Qx 1, x, x 3, x 4 ), the rational function field of four variables over Q Theorem The fixed subfield Qx 1, x, x 3, x 4 ) G := { f Qx 1, x, x 3, x 4 ): σ f = f for any σ G} is rational ie purely transcendental) over Q, except for two groups which are images of faithful representations of C 8 and C 3 C 8 into GL 4 Q) both fixed fields for these two exceptional cases are not rational over Q) There are precisely 7 such groups in GL 4 Q) up to conjugation; the answers to the rationality problem for most of them were proved by Kitayama and Yamasaki 009) [KY] except for four cases We solve these four cases left unsettled by Kitayama and Yamasaki; thus the whole problem is solved completely 01 Elsevier Inc All rights reserved 1 Introduction Let G be a finite subgroup of GL n Q) and Qx 1, x,,x n ) be the rational function field of n variables over Q The group G induces an action on Qx 1,,x n ) by Q-automorphisms defined as follows: For any σ = a ij ) 1 i, j n GL n Q), for any 1 j n, defineσ x j = 1 i n a ijx i In this article we will like to know whether the fixed subfield Qx 1,,x n ) G := { f Qx 1,,x n ): σ f = f for all σ G} is rational ie purely transcendental) over Q For any λ GL n Q), any finite subgroup G of GL n Q), if Qx 1,,x n ) G = Q f 1,, f N ), then Qx 1,,x n ) λ G λ = Qλ f 1 ),,λf N )) Thus the rationality of Qx 1,,x n ) G depends only on the conjugacy class of G within GL n Q) It is easy to see that Qx, x ) G is rational over Q for any finite * Corresponding author addresses: kang@mathntuedutw M-c Kang), zhjn@mathpkueducn J Zhou) 1 Both authors were partially supported by National Center for Theoretic Sciences Taipei Office) The work of this paper was finished when the second-named author visited National Taiwan University under the support by National Center for Theoretic Sciences Taipei Office) /$ see front matter 01 Elsevier Inc All rights reserved

2 54 M-c Kang, J Zhou / Journal of Algebra ) subgroup G of GL Q) By a result of Oura and Rikuna [OR], Qx 1, x, x 3 ) G is rational over Q for any finite subgroup G of GL 3 Q) also see Theorem A in Appendix A of this paper) The goal of this paper is to study the rationality of Qx 1, x, x 3, x 4 ) G where G is a finite subgroup of GL 4 Q) There are precisely 7 finite subgroups up to conjugation contained in GL 4 Q) A complete list of these subgroups can be found in the book of Brown, Bülow, Neubüser, Wondratschek and Zassenhaus [BBNWZ, pp 80 60] There these 7 groups are classified into 33 crystal systems Each crystal system contains one or more Q-classes; each Q-class is a conjugacy class of some finite subgroup of GL 4 Q) Since every finite subgroup of GL n Q) canberealizedasafinitesubgroupofgl n Z), each Q-class in [BBNWZ] contains one or more Z-classes; these Z-classes are not conjugate within GL n Z), but they are conjugate within GL n Q) and represent this Q-class A set of generators of each Z-class is exhibited in [BBNWZ] The notation 4, 6, 1) used in [KY] means the first Q-class in the 6-th crystal system of GL 4 Q) see [BBNWZ, p 3]) Note that these Z-classes can be found also in the data base of GAP at the command GeneratorsOfGroupMatGroupZClass4,33,3,1)) for the first Z-class in the Q-class 4, 33, 3) Wewillusethesamenotationasin[KY] We recall two known results of this question Theorem 11 See Kitayama [Ki]) For n = 4 or 5,ifHisafinite-group of Qx 1,,x n ),thenqx 1,,x n ) H is rational over Q if and only if H is not isomorphic to C 8, the cyclic group of order 8 Theorem 1 See Kitayama and Yamasaki [KY]) Let G be a finite subgroup of GL 4 Q) If G doesn t belong to the 6 conjugacy classes 4, 6, 1), 4, 33, ), 4, 33, 3), 4, 33, 6), 4, 33, 7), 4, 33, 11), then Qx 1, x, x 3, x 4 ) G is rational over Q If G is conjugate to 4, 6, 1) or 4, 33, ), thenqx 1, x, x 3, x 4 ) G is not rational over Q The main result of this paper is to solve the four case 4, 33, 3), 4, 33, 6), 4, 33, 7), 4, 33, 11) left unsettled in Theorem 1 Since the sets of generators of these four groups in[ky, pp ] are defined over Z[1/] see Section 3), we may consider similar rationality problems for a field k with char k Here is our result Theorem 13 1) Let k be a field with char k and G be a finite group belong to the conjugacy class 4, 33, 3) or 4, 33, 6), whichisdefinedon[ky, p 377] Thenbothkx 1, x, x 3, x 4 ) G and kx 1 /x 4, x /x 4, x 3 /x 4 ) G are rational over k ) Let k be a field with char k, 3 and G be a finite group belong to the conjugacy class 4, 33, 7) or 4, 33, 11) which is defined on [KY, p 378] Thenbothkx 1, x, x 3, x 4 ) G and kx 1 /x 4, x /x 4, x 3 /x 4 ) G are rational over k Combining Theorem 1 and Theorem 13, we obtain the following result Theorem 14 Let G be a finite subgroup of GL 4 Q) ThenQx 1, x, x 3, x 4 ) G is rational over Q if and only if G is not conjugate to the 4, 6, 1) or 4, 33, ) Note that the groups 4, 6, 1) and 4, 33, ) are images of faithful representations of C 8 and C 3 C 8 into GL 4 Q) respectively The fixed fields for these two groups are not rational over Q by Voskresenskii, Lenstra and Saltman by [Sa, Theorem 51, Theorem 511, Theorem 31], since Theorem 51 of [Sa] is valid for F V ) G where G GLV ) is any faithful representation of G The main idea of the proof of Theorem 13 is to enlarge the field k to K by adding, or 3tok so that the given representations of these groups will become more simple Then consider the fixed subfield of the projective part, ie K y 1 /y 4, y /y 4, y 3 /y 4 ) G, with the aid of Theorem 1 insectionsincewehaveenlargedk to K, we should descend the ground field from K to k, which leads to a field of the form of two successive conic bundles The method to show the rationality of the resulting field is elementary, but technical Moreover, in the proof of Theorem 13, lots of computations are necessary Some of them can be carried out by hands as most proof of the traditional mathematical theorems But we use the computer algebra package in some other computations for

3 M-c Kang, J Zhou / Journal of Algebra ) the sake of saving energy and keeping free from possible manual mistakes We emphasize that our use of the computer in this paper is limited only to the routine symbolic computation; no extra codes of data bases, eg GAP, are required We will note that the rationality of kx 1, x, x 3, x 4 ) G depends on the group G as an abstract group and also depends on the faithful representation of G into GL 4 Q) For example, both the groups 4, 3, 11) and 4, 33, 6) are isomorphic to GL F 3 ) as abstract groups see [KY, p 377]) The fixed field kx 1, x, x 3, x 4 ) G is rational for the group 4, 3, 11) by [Pl,Ri] Let kx 1, x, x 3, x 4 ) be the field with the action of G associated to 4, 33, 6) By applying Theorem in Section to kx 1, x, x 3, x 4 )x 1, x, x 3, x 4 ) and kx 1, x, x 3, x 4 )x 1, x, x 3, x 4 ), we find easily that kx 1, x, x 3, x 4 )G t 1, t, t 3, t 4 ) is rational over k where gt i ) = t i for any g G, any1 i 4) Thus kx 1, x, x 3, x 4 )G is stably rational over k But it is not obvious at all whether kx 1, x, x 3, x 4 )G is rational over k The same situation holds for the groups 4, 3, 5) and 4, 33, 3) We will organize this paper as follows We recall some preliminaries in Section, which will be used in the proof of Theorem 13 In Section 3, we give the sets of generators of the four groups 4, 33, 3), 4, 33, 6), 4, 33, 7) and 4, 33, 11) The rationality for the groups 4, 33, 3) and 4, 33, 6) will be given in Section 4 and Section 5 respectively The rationality of the groups 4, 33, 7) and 4, 33, 11) is given in Section 6 Thus the proof of Theorem 13 is finished Since the rationality of 4, 31, i) where 3 i 7in[KY, p 368, lines 6 8 from the bottom] was referred to an unpublished preprint of Yamasaki [Ya1], we include a proof of these cases in Appendix A for the convenience of the reader For the same reason we include a proof of Oura Rikuna s Theorem, ie the rationality for finite subgroups in GL 3 Q) in Appendix A, because Oura and Rikuna s paper [OR] hasn t been published when this paper was written although their preprint appeared already in 003) Since we don t have the preprints [OR] and [Ya1], we are not sure whether the proof in Theorem A1 and Theorem A are the same as theirs Notation and terminology The cyclic group of order n will be denoted by C n If G is a finite subgroup of GL n Z), foranyfieldk, we will say that G acts on the rational function field kx 1,,x n ) by monomial k-automorphisms, if for any σ = a ij ) 1 i, j n G GL n Z), wedefineσ x j = b j σ ) 1 i n xa ij i Preliminaries for some b j σ ) k\{0} We recall several results which will be used in tackling the rationality problem Theorem 1 See Ahmad, Hajja and Kang [AHK, Theorem 31]) Let L be any field, Lx) the rational function field of one variable over L and G a finite group acting on Lx) Suppose that, for any σ G, σ L) Land σ x) = a σ x + b σ where a σ, b σ Landa σ 0 ThenLx) G = L G f ) for some polynomial f L[x] Infact, if m = min{deg gx) 1: gx) L[x] G }, any polynomial f L[x] G with deg f = m satisfies the property Lx) G = L G f ) Theorem See Hajja and Kang [HK, Theorem 1]) Let G be a finite group acting on Lx 1,,x n ),the rational function field of n variables over a field L Suppose that i) for any σ G, σ L) L; ii) the restriction of the action of G to L is faithful; iii) for any σ G, σ x 1 ) x 1 σ x ) = Aσ ) x + Bσ ) σ x n ) x n where Aσ ) GL n L) and Bσ ) is an n 1 matrix over L

4 56 M-c Kang, J Zhou / Journal of Algebra ) Then there exist elements z 1,,z n Lx 1,,x n ) which are algebraically independent over L, and Lx 1,,x n ) = Lz 1,,z n ) so that σ z i ) = z i for any σ G, any 1 i n Theorem 3 See Kang [Ka, Theorem 4]) Let k be any field, σ be a k-automorphism of the rational function field kx, y) defined by σ x) = a/x, σ y) = b/y wherea k\{0}, b= c[x + a/x)]+dsuchthatc, d kand at least one of c and d is non-zero Then kx, y) σ = ku, v) where u and v are defined as u = x a/x) xy ab/xy), v = y b/y) xy ab/xy) Theorem 4 See Yamasaki [Ya]) Let k be a field with char k, a k\{0}, σ be a k-automorphism of the rational function field kx, y) defined by σ x) = a/x, σ y) = a/y Then kx, y) σ = ku, v) where u = x y)/a xy), v= x + y)/a + xy) Theorem 5 See Masuda [Ma, Theorem 3]; [HoK, Theorem ]) Let k be any field, σ be a k-automorphism of the rational function field kx, y, z) defined by σ : x y z x Then kx, y, z) σ = ks 1, u, v) = ks 3, u, v) where s 1,s,s 3 are the elementary symmetric functions of degree one, two, three in x, y, z, and u and v are defined as u = v = x y + y z + z x 3xyz x + y + z xy yz zx, xy + yz + zx 3xyz x + y + z xy yz zx Remark The formula of u and v was essentially due to Masuda [Ma, Theorem 3] with a misprint in the original expression The error was corrected by Rikuna [Ri] For the details, see the paragraph before Theorem of [HoK1] Theorem 6 See Hajja [Ha]) Let k be any field, and G be a finite group acting on the rational function field kx, y) by monomial k-automorphisms Then kx, y) G is rational over k 3 Sets of generators of the four groups In this section we recall the sets of generators of the groups 4, 33, 3), 4, 33, 6), 4, 33, 7) and 4, 33, 11) given in [KY, pp ] Note that various matrices in GL 4 Q) are defined on [KY, p 36] and groups of crystal systems 33 are defined on [KY, p 369] In particular, the group Q in the notation of [KY] is isomorphic to the quaternion group of order 8, and Q is generated by i and ij We write explicitly these matrices in GL 4 Q) as follows: 0 1 i = 0 0 1, ij = α = , α 0 = , 1,

5 M-c Kang, J Zhou / Journal of Algebra ) α 1 = , k 1α 0 = We will define a matrix T GL 4 Q) by 0 1 T = We define λ 1, λ, σ, τ, λ 3, λ 4 which are conjugates of i, ij, α, α 0, α 1, k 1 α 0 by the conjugation associated to T Explicitly, we get 0 0 λ 1 = T i T = , λ = T ij T = 0 10, σ = T α T = 1 1 1, τ = T α 0 T =, λ 3 = T α 1 T = 1 1 1, λ 4 = T k 1 α 0 T = Definition 31 The group 4, 33, 3) is generated by λ 1, λ, σ ; the group 4, 33, 6) is generated by λ 1, λ, σ, τ ; the group 4, 33, 7) is generated by λ 1, λ, σ, λ 3 ; the group 4, 33, 11) is generated by λ 1, λ, σ, λ 3, λ 4 Note that, as abstract groups, the group 4, 33, 3) is isomorphic to the group SL F 3 ), the group 4, 33, 6) is isomorphic to the group GL F 3 ), the group 4, 33, 7) is a central extension of the group SL F 3 ), while the group 4, 33, 11) is of order The rationality of the group 4, 33, 3) Throughout this section k is any field with char k, and G is the group 4, 33, 3) ThusG = λ 1,λ, σ by Definition 31 G acts on kx 1, x, x 3, x 4 ) by k-automorphisms defined as λ 1 : x 1 x, x x 1, x 3 x 4, x 4 x 3, λ : x 1 x 3 x 1, x x 4 x, σ : x 1 x 1 + x + x 3 + x 4 )/, x x 1 x + x 3 x 4 )/, x 3 x 1 x x 3 + x 4 )/, x 4 x 1 + x x 3 x 4 )/ 41) Step 1 Let π = Galk )/k) Note that, if k, thenπ ={1}; if / k, thenπ = ρ with ρ ) =

6 58 M-c Kang, J Zhou / Journal of Algebra ) We can extend the actions of G and π to k )x 1, x, x 3, x 4 ) by requiring that G acts trivially on k ) and π acts trivially on x 1, x, x 3, x 4 It follows that kx 1, x, x 3, x 4 ) G = {k )x 1, x, x 3, x 4 ) π } G = k )x 1, x, x 3, x 4 ) G,π Step Define y 1, y, y 3, y 4 by y 1 = x 1 x, y = x 3 x 4, y 3 = x 3 + x 4, y 4 = x 1 x 4) In other words, we define y 1, y, y 3, y 4 by y 1 y y 3 y = x 1 x x 3 x 4 It follows that k )x 1, x, x 3, x 4 ) = k )y 1, y, y 3, y 4 ) and the actions of G and π if / k) are given by λ 1 : y 1 y 1, y y, y 3 y 3, y 4 y 4, λ : y 1 y y 1, y 3 y 4 y 3, σ : y 1 )y 1 + y )/, y 1 )y 1 y )/, y 3 )y 3 + y 4 )/, y 4 1 )y 3 y 4 )/, ρ :, y 1 y 4, y y 3, y 4 y 1 43) Step 3 By the same arguments as in Step 1, it is easy to see that kx 1 /x 4, x /x 4, x 3 /x 4 ) G = k )y 1 /y, y 3 /y 4, y 1 /y 3 ) G,π Define z 1 = y 1 /y, z = y 3 /y 4, z 3 = y 1 /y 3 By Theorem 1, we find that k )y 1, y, y 3, y 4 ) G,π = k )z 1, z, z 3 )y 4 ) G,π = k )z 1, z, z 3 ) G,π z 0 ) where z 0 is fixed by the actions of G and π It remains to show that k )z 1, z, z 3 ) G,π is rational over k Step 4 Define u 1 = z 1 /z, u = z 1 z, u 3 = z 3 Then k )z 1, z, z 3 ) λ 1 = k )u 1, u, u 3 ) Moreover, λ acts on u 1, u, u 3 by λ : u 1 1/u 1, u 1/u, u 3 u 3 /u 1 Define v 3 = u /u 1 )) Then k )u 1, u, u 3 ) = k )u 1, u, v 3 ) By Theorem 4, we find that k )u 1, u, v 3 ) λ = k )v 1, v, v 3 ) where v 1 = u 1 u )/1 u 1 u ), v = u 1 + u )/1 + u 1 u ) In summary, we get k )z 1, z, z 3 ) λ 1,λ = k )v 1, v, v 3 ) Step 5 Note that the actions of σ and ρ if / k) are given by σ : v 1 v /v 1, v 1/v 1, v 3 v 3 v 1 + v )/ v v ) ), ρ :, v 1 1/v 1, v 1/v, v 3 + v 1 )1 + v )/ v 3 v 1 + v ) )

7 M-c Kang, J Zhou / Journal of Algebra ) Define X 3 = v v 1 + v )/1 + v 1 )1 + v )) Thenk )v 1, v, v 3 ) = k )v 1, v, X 3 ) and σ X 3 ) = X 3 Thus k )v 1, v, v 3 ) σ = k )v 1, v ) σ X 3 ) = k )X 1, X, X 3 ) by Theorem 5 regarding v 1, v /v 1,1/v as x, y, z in Theorem 5 and defining X 1 and X as u and v there) where X 1 and X are defined by X 1 = v 3 1 v3 + v3 1 + v3 ) 3v 1 v / v 4 1 v + v4 + v 1 v 1 v3 v 1v v3 1 ) v, X = v 1 v 4 + ) v 1v + v 4 1 v 3v 1 v / v 4 1 v + v4 + v 1 v 1 v3 v 1v v3 1 ) v In conclusion, k )z 1, z, z 3 ) G = k )X 1, X, X 3 ) Step 6 If k, we find that π ={1} Hencek )z 1, z, z 3 ) G = kz 1, z, z 3 ) G = kx 1, X, X 3 ) is rational over k From now on, we assume that / k and π = ρ We will show that k )X 1, X, X 3 ) ρ is rational over k We use the computer to perform the action of ρ on X 1, X, X 3 Weget ρ : X 1 X / X 1 X 1 X + X ), X X 1 / X 1 X 1 X + X ), X3 A/X 3 where A = g 1 g g 3 and g 1 = 1 + X 1 ) X 1 + X 1 ) + X, g = 1 + X ) X X ) + X 1, g 3 = 1 + X 1 + X + X X 3 + X 1 X 3X1 X X 1 X + ) + X X 4 Note that ρg 1 ) = g /X 1 X 1 X + X ) Define Y 1 = X 1 /X, Y = X 1, Y 3 = X 3 /g 1 Thenk )X 1, X, X 3 ) = k )Y 1, Y, Y 3 ) and ρ : Y 1 1/Y 1, Y Y 1 / Y 1 Y1 + Y 1)), Y3 B/Y 3 where B = Y 1 Y 1 Y 1 + Y 1 )/[Y Y 4 + Y 1Y Y 1 + Y ) + Y 1Y Y 3 1 Y 3 + Y 1 Y ) + 3Y 1 Y 4 + Y 3 1 Y 3 Y 1 Y ) + Y 4 1 Y 4 ] Note that k )Y 1, Y ) ρ is the function field of a conic bundle over k )Y 1 ) ρ which is the function field of P 1 k ), and k )Y 1, Y, Y 3 ) ρ is the function field of another conic bundle over some k-surface whose function field is k )Y 1, Y ) ρ Step 7 It is not difficult to verify that k )Y 1, Y, Y 3 ) ρ = ku 0, U 1, U, U 3, U 4 ) where U i s are defined by U 0 = 1 Y 1 )/1 + Y 1 ), U 1 = Y + Y 1 / Y 1 Y1 + Y 1)), U = [ Y Y 1 / Y 1 Y1 + Y 1))], U3 = Y 3 + B/Y 3, U 4 = Y 3 B/Y 3 ) with the relations U 1 + U = U 0 ) ) / 1 3U 0 and U 3 + U 4 = 4B 44) We will simplify the two relations in formula 44)

8 60 M-c Kang, J Zhou / Journal of Algebra ) For example, in the first relation of formula 44), multiply both sides by 1 + U 0 ) Use the identity a + b )c + d ) = ac bd) + ad + bc) to simplify the left-hand side of the resulting relation In other words, define V 1 = U 0 U 1 + U ) 1 3U 0 ) / + U 0), V = U 0 U U 1 ) ) 1 3U 0 / + U 0) We find the ku 0, U 1, U ) = ku 0, V 1, V ) and the relation becomes V 1 + V = 1 3U 0 The above relation can be written as [ V 1 /1 + V ) ] [ 1 V )/1 + V ) ] = 3U 0 /1 + V ) 45) Define W 1 = V 1 /1+ V ), W = U 0 /1+ V )Therelation45) guarantees that 1 V )/1+ V ) kw 1, W )ThusV kw 1, W ) It follows that ku 0, V 1, V ) = kw 1, W ) Now we rewrite the second relation U 3 + U 4 = 4B of formula 44) intermsofw 1, W Weget U 3 + U 4 = [ 4W 1 + W 1 + 3W 1) ] / [ W 1 W W 1 + 3W ) ] 46) Use the similar trick as above to simplify the relation 46) In short, define W 3 = [ U 3 W 1 +3W ) ][ + U 4 W 1 W 1 W ) W 1 +3W ] [ / 4W 1 + W 1 +3W ) ], W 4 = [ U 4 W 1 +3W ) ][ U 3 W 1 W 1 W ) W 1 +3W ] [ / 4W 1 + W 1 +3W ) ] We get kw 1, W, U 3, U 4 ) = kw 1, W, W 3, W 4 ) and the relation 46) becomes W 3 + W 4 W 1 + W + W 1 + 3W ) = 0 47) Define w 1 = W 1 /W 1 + 3W ), w = W /W 1 + 3W ), w 3 = W 3 /W 1 + 3W ), w 4 = W 4 / W 1 + 3W ) We find that kw 1, W, W 3, W 4 ) = kw 1, w, w 3, w 4 ) and the relation 47) becomes w 3 + w 4 w 1 + w + 1 = 0 The above relation can be written as w 3 + w = w 1 w )w 1 + w ) Thus w 1 + w kw 1 w, w 3, w 4 ) Hence w 1, w kw 1 w, w 3, w 4 ) It follows that kw 1, w, w 3, w 4 ) = kw 1 w, w 3, w 4 ) is rational over k

9 M-c Kang, J Zhou / Journal of Algebra ) The rationality of the group 4, 33, 6) Throughout this section k is any field with char k, and G is the group 4, 33, 6) ThusG = λ 1,λ, σ, τ by Definition 31 The actions of λ 1, λ, σ on kx 1, x, x 3, x 4 ) is the same as those given in formula 41) We record the action of τ as follows τ : x 1 x 3, x x, x 3 x 1, x 4 x 4 The method to prove that kx 1, x, x 3, x 4 ) G is k-rational is very similar to the method used in Section 4 In many situations, even the formulae of changing variables are identically the same Step 1 Let π = Galk, )/k) Note that π is isomorphic to {1}, C or C C In case [k, ) : k]=4, define ρ 1, ρ and ρ 3 by ρ 1 ) =, ρ 1 ) =, ρ ) =, ρ ) =, ρ 3 ) =, ρ 3 ) = We get π = ρ 1, ρ and ρ 3 = ρ 1 ρ in this situation In general, π ={1}, ρ 1, ρ, ρ 3 or ρ 1, ρ In the sequel, whenever we describe the action of ρ 1, we mean that / k and ρ 1 π ; similarly for ρ and ρ 3 As in Step 1 of Section 4, we may extend the actions of G and π to k, )x 1, x, x 3, x 4 ) The formula 4) shouldbemodifiedwedefiney 1, y, y 3, y 4 by y y = y x y x 0 1 x x 4 By the same arguments as in Step 1 of Section 4, we can show that kx 1, x, x 3, x 4 ) G = k, )y 1, y, y 3, y 4 ) G,π and kx 1 /x 4, x /x 4, x 3 /x 4 ) G = k, )y 1 /y, y 3 /y 4, y 1 /y 3 ) G,π Moreover, the actions of λ 1, λ, σ on y 1, y, y 3, y 4 are the same as those given in formula 43) The action of ρ 1 on y 1, y, y 3, y 4 is the same as the action of ρ on y 1, y, y 3, y 4 giveninformula 43); but remember that ρ 1 ) =, ρ 1 ) = We record the actions of τ and ρ as follows: τ : y 1 y 1 + y )/, y y 1 + y )/, y 3 y 3 + y 4 )/, y 4 y 3 y 4 )/, ρ : y 1 )y 3, y )y 4, y )y 1, y )y It remains to show that k, )y 1 /y, y 3 /y 4, y 1 /y 3 ) G,π is rational over k Step The substitution formulae for z 1, z, z 3, u 1, u, u 3, v 1, v, v 3, X 1, X, X 3 are completely the same as in Steps 3 5 of Section 4 Thus we get k, )y 1 /y, y 3 /y 4, y 1 /y 3 ) λ 1,λ,σ = k, )X 1, X, X 3 ) and k, )y 1, y, y 3, y 4 ) λ 1,λ,σ = k, )X 1, X, X 3, z 0 ) where z 0 is fixed by all elements of G and π Define Y 1 = X 1 /X, Y = X 1, Y 3 = X 1 X 3 /g 1 remember A = g 1 g g 3 and A is defined in Step 6 of Section 4) Note that k, )X 1, X, X 3 ) = k, )Y 1, Y, Y 3 ) and τ acts on X i, Y j as follows:

10 6 M-c Kang, J Zhou / Journal of Algebra ) τ : X 1 X 1 / X 1 X 1 X + X ), X X / X 1 X 1 X + X ), X3 X 3, Y 1 Y 1, Y Y 1 / Y 1 Y1 + Y 1)), Y3 Y 3 It is easy to verify that k, )Y 1, Y, Y 3 ) τ = k, )Z 1, Z, Z 3 ) where Z 1 = Y 1, Z = Y +[Y 1 /Y 1 Y 1 + Y 1 ))], Z 3 = Y 3 {Y [Y 1 /Y 1 Y 1 + Y 1 ))]} We conclude that k, )Y 1, Y, Y 3 ) τ is rational over k, ) If k, ) = k, ieπ ={1}, thenkx 1, x, x 3, x 4 ) G is k-rational In the remaining part of this section we will consider the situations when π = ρ 1, ρ, ρ 1, ρ or ρ 3 Step 3 We consider the case π = ρ 1, ρ first Using the computer for symbolic computation, we find that ρ 1 Z i ) = ρ Z i ), ρ 3 Z i ) = Z i for 1 i 3 and ρ 1 : Z 1 1/Z 1, Z Z /Z 1, Z 3 C/Z 3 where C = Z 1 4Z 1 + Z Z 1 Z + Z 1 Z )/[1 Z 1 + Z 1 ) Z 1 + Z 1 Z + Z + 4Z 3 1 Z 1 Z Z Z 1 Z + Z 4 1 Z Z 3 1 Z + Z 4 1 Z )] It follows that The action of ρ 1 is given by k, )Z 1, Z, Z 3 ) ρ 1,ρ = { k, )Z 1, Z, Z 3 ) ρ 1ρ } ρ 1 = k )Z 1, Z, Z 3 ) ρ 1 ρ 1 :, Z 1 1/Z 1, Z Z /Z 1, Z 3 C/Z 3 Step 4 It is not difficult to verify that k )Z 1, Z, Z 3 ) ρ 1 = ku 1, U, U 3, U 4 ) where U 1, U, U 3, U 4 are defined by U 1 = Z 1 + 1/Z1 ) ), U = Z 1 1/Z1 ) ), U 3 = Z 3 + C/Z 3 ), U 4 = Z 3 C/Z 3 ) ) with the relation U 3 + U 4 = U 1 3U ) U 1 + ) U / [ U 1 ) 3U 16U U U 4U 1U 1U 1 U + 9U 4 )] 51) We will simplify the relation 51) Use the same technique as Step 7 in Section 4 Define V 1 = 8U 1 / U 1 3U ), V = 4U / U 1 3U ), V 3 = αu 3, V 4 = αu 4 where α = 16U U U 4U 1U 1U 1 U + 9U 4 )/[4U 1 + U )U 1 3U )] It follows that ku 1, U ) = kv 1, V ), ku 1, U, U 3, U 4 ) = kv 1, V, V 3, V 4 ) and the relation 51) becomes V 3 + V 4 = 1 V 1 + 6V ) 1 + V 1 + 4V ) 5)

11 M-c Kang, J Zhou / Journal of Algebra ) Define w 1 = 1/1 + V 1 ), w = V /1 + V 1 ), w 3 = V 3 /1 + V 1 ), w 4 = V 4 /1 + V 1 ) Then kv 1, V, V 3, V 4 ) = kw 1, w, w 3, w 4 ) and the relation 5) becomes w 3 + w 4 = w 1 + 4w ) w w ) Thus we get a relation w 3 / w 1 + 4w )) + w4 / w 1 + 4w )) = w w ) / w1 + 4w ) 53) It follows that w w )/w 1 + 4w ) kw, w 3 /w 1 + 4w ), w 4/w 1 + 4w )) Hencew 1 belongs to this field We find that kw 1, w, w 3, w 4 ) = kw, w 3 /w 1 + 4w ), w 4/w 1 + 4w )) is rational over k Step 5 Consider the case π = ρ 3 ie k Hencek, )Z 1, Z, Z 3 ) ρ3 = kz 1, Z, Z 3 ) is k-rational Done Consider the case π = ρ 1,ie k Thenk, )Z 1, Z, Z 3 ) π = k )Z 1, Z, Z 3 ) ρ1 The action of ρ 1 is the same as that in the last line of Step 3 The proof of rationality of the present situation is completely the same as in Step 4 Done The case π = ρ,ie k, can be discussed in a similar way note that the actions of ρ 1 and ρ are the same on Z 1, Z, Z 3 ) The details of the proof is omitted 6 The rationality of the groups 4, 33, 7) and 4, 33, 11) In this section we assume that k is any field with char k, 3 Let G be the group 4, 33, 7) or 4, 33, 11) We will show that kx 1, x, x 3, x 4 ) G is rational over k By Definition 31, ifg is the group 4, 33, 7), theng = λ 1,λ, σ,λ 3 ;ifg is the group 4, 33, 11), then G = λ 1,λ, σ,λ 3,λ 4 As in Section 5, the proof in this section is very similar to that in Section 4 Step 1 Let π = Galk, 3 )/k) If[k, 3 ) : k] =4, define ρ 1, ρ and ρ 3 by ρ 1 ) =, ρ 1 3 ) = 3, ρ ) =, ρ 3 ) = 3, ρ 3 ) =, ρ 3 3 ) = 3 Then π = ρ 1, ρ In general, π ={1}, ρ 1, ρ, ρ 3 or ρ 1, ρ Throughout this section, whenever we describe the action of ρ 1, we mean that / k and ρ 1 π ; similarly for ρ and ρ 3 Asbefore,wemayextendtheactionsofG and π to k, 3 )x 1, x, x 3, x 4 ) The actions of λ 1, λ, σ on x 1, x, x 3, x 4 are the same as given in formula 41) The actions of λ 3 and λ 4 are given by λ 3 : x 1 x 1 x x 3 + x 4 )/, x x 1 x + x 3 + x 4 )/, x 3 x 1 x x 3 x 4 )/, x 4 x 1 x + x 3 x 4 )/, λ 4 : x 1 x 1, x x 4, x 3 x 3 Define y 1, y, y 3, y 4 by y 1 y y 3 y 4 = ) ) 0 + )1+ 3 ) 0 + )1+ 3 ) + 3 ) )

12 64 M-c Kang, J Zhou / Journal of Algebra ) x 1 x x x 4 By the same arguments as in Step 1 of Section 4, we can show that kx 1, x, x 3, x 4 ) G = k, 3 )y 1, y, y 3, y 4 ) G,π and kx 1 /x 4, x /x 4, x 3 /x 4 ) G = k, 3 )y 1 /y, y 3 /y 4, y 1 /y 3 ) G,π The actions of λ 1, λ, σ on y 1, y, y 3, y 4 are the same as those given in formula 43) We describe the actions of λ 3, λ 4, ρ 1, ρ and ρ 3 : ) λ 3 : y 1 ζ y 1, y ζ y, y 3 ζ y 3, y 4 ζ y 4 where ζ = + 3 )/, λ 4 : y 1 1 )y 3 + y 4 )/, y 1 )y 3 y 4 )/, y )y 1 + y )/, y )y 1 y )/, ρ 1 : y )y 4 /, y + 3 )y 3 /, y )y /, y )y 1 /, ρ : y )y 3 /, y + 3 )y 4 /, y )y 1 /, y )y /, ρ 3 : y ) + 3 )y /, y 1 + )1 3 )y 1 /, y 3 1 )1 3 )y 4 /, y 4 + )1 3 )y 3 / It remains to show that k, 3 )y 1 /y, y 3 /y 4, y 1 /y 3 ) G,π is rational over k Step The substitution formulae for z 1, z, z 3, u 1, u, u 3, v 1, v, v 3, X 1, X, X 3 are completely the same as in Step 3 Step 5 of Section 4 Thus we get k, 3 )y 1 /y, y 3 /y 4, y 1 /y 3 ) λ 1,λ,σ = k, 3 )X 1, X, X 3 ) and k, 3 )y 1, y, y 3, y 4 ) λ 1,λ,σ = k, 3 )X 1, X, X 3, z 0 ) where z 0 is fixed by all elements of G and π The action of λ 3 is given by λ 3 : X 1 X 1, X X, X 3 ζ X 3 Hence k, 3 )X 1, X, X 3 ) λ 3 = k, 3 )X 1, X, X 3 3 ) We will discuss the rationality problem for the group 4, 33, 7) inthenexttwostepsthediscussion for the group 4, 33, 11) will be postponed till Step 5 Step 3 Consider the group G = λ 1,λ, σ,λ 3 in this step We have shown that k, 3 )z 1, z, z 3 ) λ 1,λ,σ,λ 3 = k, 3 )X 1, X, X 3 3 )Ifπ ={1}, then kx 1, x, x 3, x 4 ) G is k-rational It remains to consider cases when π = ρ 1, ρ, ρ 1, ρ, ρ 3 We consider the case π = ρ 1, ρ first The action of ρ 1 on X 1, X, X 3 and on X 3 3 also) is given by ρ 1 : X 1 X / X 1 X 1 X + X ), X X 1 / X 1 X 1 X + X ), X3 A/X 3 61) where A = g 1 g g 3 and g 1, g, g 3 are the same polynomials defined in Step 6 of Section 4 Moreover, ρ X 1 ) = ρ 1 X 1 ), ρ X ) = ρ 1 X ), ρ X 3 ) = ρ 1 X 3 ) Define Y 1 = X 1 /X, Y = X 1, Y 3 = X 3 3 /g3 1

13 M-c Kang, J Zhou / Journal of Algebra ) The action of ρ 3 is given as ρ 3 : X 1 X 1, X X, X 3 X 3, Y 1 Y 1, Y Y, Y 3 Y 3 We find that k, 3 )Y 1, Y, Y 3 ) ρ 3 = k 3 )Y 1, Y, Y 3 + Y 3 ) 6) For simplicity, call Y 0 = Y 3 + Y 3 The action of ρ 1 is given by ρ 1 : 3 3, Y 1 1/Y 1, Y Y 1 / Y 1 Y1 + Y 1)), Y0 B 3 /Y 0 63) where B is defined in Step 6 of Section 4 The remaining proof is similar to Step 7 of Section 4, but the ground field is k 3 ) in the present situation It can be shown that k 3 )Y 1, Y, Y 0 ) ρ 1 = ku 0, U 1, U, U 3, U 4 ) where U 0 = 31 Y 1 )/1 + Y 1 ), U 1 = Y + Y 1 / Y 1 Y1 + Y 1)), U = 3 [ Y Y 1 / Y 1 Y1 + Y 1))], U3 = Y 0 + B 3 /Y 0, U 4 = Y 0 B 3 /Y 0 ), with two relations 3U 1 + U = U 0 ) ) / 1 U 0, 3U 3 + U 4 = 4B3 64) We will simplify the relations in 64) We use the identity 3a + b )3c + d ) = 3ad + bc) + bd 3ac) this time Define V 1 = U 0 U 1 + U ) 1 U 0 The first relation becomes ) / 6 + U 0), V = U 0 U 3U 1 ) ) 1 U 0 / 6 + U 0) 3V 1 + V = 1 U 0 65) Hence we get 3 [ V 1 /1 + V ) ] + [ U0 /1 + V ) ] = 1 V )/1 + V ) It is not difficult to show that ku 0, U 1, U ) = kw 1, W ) where W 1 = V 1 /1 + V ), W = U 0 /1 + V ),becauseofformula65) Now we will simplify the second relation 3U 3 + U 4 = 4B3 in 61) Express B 3 in terms of W 1, W This relation becomes 3U 3 + U 4 = 3/D3 66) where D = [9W 1 W 33W 1 + W ) ]/[33W W ) )3W W ) )] Regarding 3W W ) )3W W ) ) = 3a + b )3c + d ) and using the identity 3a + b )3c + d ) = 3ad + bc) + bd 3ac), we may change the variables by defining W 3 = D [ W 1 U 4 + 3W 1 + W 1) U 3 ], W 4 = D [ 6W 1 U 3 3W 1 + W 1) U 4 ]

14 66 M-c Kang, J Zhou / Journal of Algebra ) We get kw 1, W, U 3, U 4 ) = kw 1, W, W 3, W 4 ) with a relation 3W 3 + W 4 = [ 9W 1 W 3 3W 1 + W ) ] Define w i = W i /3W 1 + W ) for 1 i 4 We get kw 1, W, W 3, W 4 ) = kw 1, w, w 3, w 4 ) with arelation 3w 3 + w 4 9w 1 w ) + 6 = 0 67) Hence kw 1, w, w 3, w 4 ) = k3w 1 w, w 3, w 4 ) is rational over k, becauseofformula67) Done Step 4 Suppose π = ρ 1, ρ or ρ 3 for the group 4, 33, 7) If π = ρ 3,ie 3 k, the rationality was already proved in formula 6) ofthepreviousstep If π = ρ 1,ie 3 k, wegetk, 3 )Y1, Y, Y 3 ) π = k 3 )Y 1, Y, Y 0 ) ρ1 The proof is thesameasinstep3 The case π = ρ,ie k, is similar and the proof is omitted Step 5 Now we consider the group G = λ 1,λ, σ,λ 3,λ 4, ie the group 4, 33, 11) InStep,we have shown that k, 3 )z 1, z, z 3 ) λ 1,λ,σ,λ 3 = k, 3 )X 1, X, X 3 3 ) We will show that k, 3 )X 1, X, X 3 3 ) λ4,π is rational over k We modify the definition of Y 1, Y, Y 3 in Step 4 Now define Y 1 = X 1, Y = X, Y 3 = 1 + )X 3 3 / 1 + X 1 X + X 1 X 1 X + X ) 3 The actions of λ 4, ρ 1, ρ, ρ 3 on k, 3 )Y 1, Y, Y 3 ) are given by λ 4 : Y 1 Y, Y 3 8/ Y 3 D 3), ρ 1 : Y 1 Y / Y 1 Y 1Y + Y ), Y Y 1 / Y 1 Y 1Y + Y ), Y 3 8 Y 1 Y 1Y + Y ) 3/ Y3 D 3), ρ : Y 1 Y / Y 1 Y 1Y + Y ), Y Y 1 / Y 1 Y 1Y + Y ), Y 3 8 Y 1 Y 1Y + Y ) 3/ Y3 D 3), ρ 3 : Y 1 Y 1, Y Y, Y 3 Y 3, 68) where D = 1 + Y 1 + Y + Y 1 Y + Y Y 3 + Y 4 1 Y 3 1 Y + 3Y 1 Y Y 1Y 3 + Y 4 Define [ U 0 = Y 1 /Y, U 1 = Y / Y 1 Y 1Y + Y))], U = 3Y 1 [ 1 1/ Y 1 Y 1Y + Y ))], U3 = Y 3 [ 1 + Y 1 Y 1Y + Y ) 3 ] It is easy to verify that k, 3 )Y 1, Y, Y 3 ) = k, 3 )U 0, U 1, U, U 3 ) with a relation Define 3U 1 + U = 1U 0 / 1 U 0 + U 0) 69) V 0 = U 0 1, V 1 = U 1 1 U0 + U 0) /U0, V = U 1 U0 + U 0) /3U0 )

15 M-c Kang, J Zhou / Journal of Algebra ) Then k, 3 )U 0, U 1, U ) = k, 3 )V 0, V 1, V ) and the relation 69) becomes V 0 V 1 3V + 3 = 0 Since V 0 V 1 k, 3 )V 0 + V 1, V ) by the above relation, we find that k, 3 )V 0, V 1, V ) = k, 3 )V 0 + V 1, V ) Now define w 1 = V 0 + V 1, w = V, w 3 = U 3 Then k, 3 )Y 1, Y, Y 3 ) = k, 3 )U 0, U 1, U, U 3 ) = k, 3 )w 1, w, w 3 ) and the actions of λ 4, ρ 1, ρ, ρ 3 on w 1, w, w 3 are given by λ 4 : w 1 f 1 f f 3, w 4w 1 w f 3, w 3 h 1 h h 3 / w 3 w 3 1 4) h3, ρ 1 : w 1 f 1 f f 3, w 4w 1 w f 3, w 3 h 1 h h 3 / w 3 w 3 1 4) h3, ρ : w 1 f 1 f f 3, w 4w 1 w f 3, w 3 h 1 h h 3 / w 3 w 3 1 4) h3, ρ 3 : w 1 w 1, w w, w 3 w 3, 610) where f 1 = 3 + w 1 3w, f = 3 + w 1 + 3w, f 3 = 3 + w 1 + w 1 + 3w, h 1 = 3 + w 1 3w, h = 3 + w 1 6w 1 w 3w, h 3 = 3 + w 1 + 6w 1 w 3w, h 4 = 3 + 7w 1 + 5w 1 + w3 1 3w 9w 1 w Step 6 We claim that k, 3 )w 1, w, w 3 ) λ 4,π = kw 1, w, w 3 ) λ 4 no matter what π is For example, suppose π = ρ 1 By formula 610), the actions of λ 4 and ρ 1 on w 1, w, w 3 are the same Thus k, 3 )w 1, w, w 3 ) λ 4,ρ 1 ={k, 3 )w 1, w, w 3 ) λ 4ρ 1 } λ 4 = kw 1, w, w 3 ) λ 4 Similarly the other cases can be verified easily Step 7 We will prove that kw 1, w, w 3 ) λ 4 is k-rational Note that λ 4 kw 1, w )) = kw 1, w ) We will find an element t kw 1, w ) such that λ 4 t) = t But we simplify w 1, w, w 3 first Define W 1 = w 1 + 1, W = w, W 3 = h 1 h h 3 /w 3 w 1 h 4) Thenkw 1, w, w 3 ) = kw 1, W, W 3 ) and λ 4 acts on W 1, W, W 3 by λ 4 : W 1 4W 1 + W 1 6W ) / 4 + W 1 + 3W ), W 4W 1 W 4W )/ 4 + W 1 + ) 3W, W 3 4 W 1 + W 3 1 9W 1W + ) [ 6W / W W 1 + )] 3W We will find an element t kw 1, w ) = kw 1, W ) such that λ 4 t) = t We use a similar trick in [HoK1, Section ]: Examinetheeffectsofλ 4 on w 1, w, f 1, f, f 3 The action λ 4 acts on these five polynomials by monomial automorphisms but these five polynomial are not algebraically independent) It is not difficult to find a monomial fixed by λ 4 ;forexample,λ 4 f 1 / f ) = f 1 / f Define t 1 = f 1 / f = + W 1 3W )/ + W 1 + 3W )

16 68 M-c Kang, J Zhou / Journal of Algebra ) Then kw 1, W ) = kt 1, W 1 ) and λ 4 t 1 ) = t 1, λ 4 W 1 ) =[+t 1 t 1 )+ W 11+4t 1 +t 1 )]/[ 4t 1 t 1 ) + W 11 + t 1 + t 1 )] Note that λ 4W 1 ) is of the form a 1 + a W 1 )/a 3 + a 4 W 1 ) where a 1, a, a 3, a 4 kt 1 ) Thus we can bring it into the form a 0 /W for some W and some a 0 kt 1 ) Define E = t 1 /1 + t 1 + t 1 ), t = W 1 3E 1, t 3 = W 3 We find that kw 1, W, W 3 ) = kt 1, t, t 3 ) and λ 4 : t 1 t 1, t a/t, t 3 [ c t + a/t ) ) + d ] /t 3 where a = 9E1 + E), c =E, d = 41 9E 18E ) Thus we apply Theorem 3 to conclude that kt 1, t, t 3 ) λ 4 is rational over k Appendix A The following theorem is proved in [Ya1, Section ]; we provide a different proof here Bytheway,wenoteamisprintin[KY, p 378] There it was claimed that the rationality for the group 4, 33, 14) was treated already in [KY], while the proof for the group 4, 33, 15) would be postponed to [Ya1] This is not correct The correct version is that the group 4, 33, 15) was solved in [KY], and the group 4, 33, 14) was left to [Ya1] Theorem A1 LetGbeoneofthefivegroups4, 31, 3), 4, 31, 4), 4, 31, 5), 4, 31, 6), 4, 31, 7) in GL 4 Q) Then Qx 1, x, x 3, x 4 ) G is rational over Q Proof By [KY, 378], these groups are isomorphic to A 5, S 5, A 5 C or S 5 C as abstract groups where A 5 and S 5 are the alternating and symmetric group of degree five Since S 5 has two inequivalent irreducible representations of dimension four arising from the standard representation and its tensor product with the unique non-trivial linear character [FH, pp 7 9], it is easy to find these five groups as follows Define σ, τ 1, τ, τ 3,λ GL 4 Q) by σ =, τ = 10 10, τ = τ 3 = , λ= 1 0 0, Note that the matrix σ corresponds to the 5-cycle 1,, 3, 4, 5) S 5,thematrixτ 1 corresponds to the transposition 1, ) S 5,thematrixτ 3 corresponds to the 3-cycle 1,, 3) S 5 By comparing the character tables of these groups, we find that the group 4, 31, 3) is conjugate to the group σ, τ 3 which is the restriction of the standard representation to A 5 ), the group 4, 31, 4) conjugate to the group σ, τ 1 which is the standard representation of S 5 ), the group 4, 31, 5) conjugate to the group σ, τ, the group 4, 31, 6) conjugate to the group σ, τ 3,λ, and the group 4, 31, 7) conjugate to the group σ, τ 1,λ Suppose G is any one of the above five groups acting on Qx 1, x, x 3, x 4 ) By Theorem 1, we find that Qx 1, x, x 3, x 4 ) G = Qx 1 /x 4, x /x 4, x 3 /x 4 ) G x 0 ) where x 0 is fixed by all elements of G Note that Qx 1 /x 4, x /x 4, x 3 /x 4 ) is the function field of PV 0 ) where V 0 is the standard representation of S 5 see [HK, p 519]) By [HK, Lemma 1, Lemma 5], bothqx 1 /x 4, x /x 4, x 3 /x 4 ) S 5 and Qx 1 /x 4, x /x 4, x 3 /x 4 ) A 5 are Q-rational

17 M-c Kang, J Zhou / Journal of Algebra ) The following theorem was due to Oura and Rikuna [OR] It seems that [OR] has not been published in some journal Thus we include its proof here Theorem A Let G be a finite subgroup of GL 3 Q)ThenQx 1, x, x 3 ) G is rational over Q Proof Step 1 We look into the book [BBNWZ] There are 3 finite subgroups in GL 3 Q) up to conjugation Among them, there are subgroups in total, which are reducible ie suppose G acts on 1 i 3 Q x i; then without loss of generality we may assume that σ x 1, σ x Q x 1 Q x and σ x 3 Q x 3 for any σ G) The remaining 10 subgroups are M-groups in the sense that, if G acts on 1 i 3 Q x i,thenσ x i Q x σ i) for all σ G, for1 i 3see[CR, p 6] for details) The following 10 groups are M-groups: 3, 5, i), 3, 7, i) where 1 i 5 The remaining groups are reducible groups Be aware that there are more M-groups other than the 10 groups listed above But these extra groups are reducible groups also Step Suppose G is a reducible group We may assume that, for all σ G, σ x 1, σ x Q x 1 Q x, σ x 3 Q x 3 ThusQx 1, x, x 3 ) G = Qx 1, x ) G x 0 ) by Theorem 1 NowQx 1, x ) G = Qx 1 /x, x ) G = Qx 1 /x ) G y 0 ) by Theorem 1 again Since Qx 1 /x ) G is Q-rational by Lüroth s Theorem, we find that Qx 1, x, x 3 ) G is Q-rational Step 3 Suppose that G is an M-group acting on Qx 1, x, x 3 ) Then Qx 1, x, x 3 ) G = Qx 1 /x 3, x /x 3, x 3 ) G = Qx 1 /x 3, x /x 3 ) G x 0 ) by Theorem 1 The action of G on Qx 1 /x 3, x /x 3 ) are monomial actions in x 1 /x 3, x /x 3 see the last paragraph of Section 1 for the definition of monomial automorphisms) By Theorem 6, Qx 1 /x 3, x /x 3 ) G is rational over Q References [AHK] H Ahmad, M Hajja, M Kang, Rationality of some projective linear actions, J Algebra 8 000) [BBNWZ] H Brown, R Büllow, J Neubüser, H Wondratschek, H Zassenhaus, Crystallographic Groups of Four-Dimensional Spaces, John Wiley, New York, 1978 [CR] CW Curtis, I Reiner, Methods of Representation Theory, vol 1, John Wiley, New York, 1981 [FH] W Fulton, J Harris, Representation Theory: A First Course, Grad Texts in Math, vol 19, Springer-Verlag, Berlin, 1991 [Ha] M Hajja, Rationality of finite groups of monomial automorphisms of K x, y), J Algebra ) [HK] M Hajja, M Kang, Some actions of symmetric groups, J Algebra ) [HoK1] A Hoshi, M Kang, A rationality problem of some Cremona transformation, Proc Japan Academy Ser A ) [HoK] A Hoshi, M Kang, Twisted symmetric group actions, Pacific J Math, in press [Ka] M Kang, Rationality problem of GL 4 group actions, Adv Math ) [Ki] H Kitayama, Noether s problem for four- and five-dimensional linear actions, J Algebra ) [KY] H Kitayama, A Yamasaki, The rationality problem for four-dimensional linear actions, J Math Kyoto Univ ) [Ma] K Masuda, On a problem of Chevalley, Nagoya Math J ) [OR] M Oura, Y Rikuna, On three-dimensional linear Noether s problem, preprint, 003 [Pl] B Plans, Noether s problem for GL, 3), Manuscripta Math ) [Ri] Y Rikuna, The existence of generic polynomial for SL, 3) over Q, preprint [Sa] DJ Saltman, Generic Galois extensions and problems in field theory, Adv Math ) [Ya1] A Yamasaki, Some cases of four dimensional linear Noether s problem, J Math Soc Japan 6 010) [Ya] A Yamasaki, Negative solutions to three-dimensional monomial Noether problem, arxiv: v1