JOURNAL OF ALGEBRA 179, 838᎐851Ž. 1996 ARTICLE NO. 0039
On Invariants of F4 and the Center of the Albert Algebra
A. V. Iltyakov and I. P. Shestakov
Institute of Mathematics, No¨osibirsk, 630090, Russia* Communicated by Georgia Benkart
Received October 26, 1994
1. INTRODUCTION
It is known that some simple algebraic groups can be represented as groups of automorphisms of simple algebras. For example, the group PSLk is isomorphic to AutŽ.Mkk, where M is the algebra of k = k-matrices; the exceptional groups G24and F can be represented as AutŽ.O and Aut Ž.A , respectively, where O is a split Cayley algebra and A is a split Albert algebra, i.e., the algebra of Hermitian 3 = 3-matrices over O, with the symmetric multiplication a( b s 1r2Ž.ab q ba . Let G s AutŽ.A for some algebra A over a field F. Consider the diagonal action of G on the direct sum nA of n copies of A, which can be nm considered as an affine space F , m s dimŽ.A . It induces naturally an action of AutŽ.A on the field FnA Ž .of rational functions on nA, i.e., the field of rational functions on A in n vector variables. One of the main problems concerning the structure of the invariant field FnAŽ.Gis whether this field is purely transcendental over F wx1.If F is an algebraically closed field of characteristic zero and A is a central simple algebra, then the field G FnAŽ.coincides with the quotient field KAnŽ.of the centroid of the algebra FnŽ.A generated over F by n generic elements of A wx4Ž a ‘‘free affine algebra of type A,’’ in notations ofwx 8. , so, the problem of rational- G ity of FnAŽ.reduces to that of KAnŽ.. Ž. Ž. It is known that the field KAnkis rational over F if A s M s MFk, kF42,3,orwxAsO wx8 ; so the fields of rational invariants of corre- sponding representations of PSLk , k F 4, and G2 are rational. The main result of this paper is that K nŽ.A is also rational over F in the case of the Albert algebra A for any field F of characteristic / 2, 3.
* During 1993r94 Becarios de FICYT at the University of Oviedo, Asturias, Spain.
838
0021-8693r96 $12.00 Copyright ᮊ 1996 by Academic Press, Inc. All rights of reproduction in any form reserved. INVARIANTS OF F4 839
In particular, if F is algebraically closed field of characteristic zero, the Ž.G Ž. field of invariants FnA is rational over F for G s Aut A of type F4. Moreover, the subspace A 0 of zero trace elements from A is known to be an irreducible G-module of minimal dimensionwx 6, p. 382 , and we obtain as a corollary of the above result that the field of rational invariants of a minimal representation of a simple algebraic group of type F4 in several variables is rational over F. Ž. Ž Ž .. The field K113A , KMF is generated by traces of a generic ele- ment and, therefore, it is rational; the rationality of K2Ž.A is proved inwx 9 . Moreover, there it is also proved that K nŽ.A is purely transcendental over K33Ž.A, so it suffices to prove the rationality of K Ž.A . We do it by giving an explicit list of free generators for this field.
2. PRELIMINARY RESULTS
Let us recall some properties of Cayley᎐Dickson and Albert algebras Žseewx 6, 11. . The elements of a split Cayley᎐Dickson algebra O can be written in the form
␣ ¨ , ž/¨*
3 where ␣,  g F and ¨, ¨* g F s F = F = F. The multiplication is defined by the rule
␣11¨ ␣ 22¨ UUи ž/ž/¨11 ¨ 22
U U U ␣␣12qŽ.¨ 1,¨ 2 ␣12¨ q 21¨ y¨ 1=¨ 2 s UU U , ž/␣21¨q 12¨q¨ 1=¨¨ 2Ž. 1,¨ 2q 1 2
3 where Ž.¨, w , ¨ = w are the scalar and vector product in F , respectively. The algebra O is equipped with the involution
␣ ¨  y¨ s . ž/ž/¨* y¨*␣
Observe that the trace tŽ.x of a matrix x g O equals x q x g 1 и F , F. Let O O F be the algebra of 3 3-matrices over O.Itis 3s mF 3 = equipped with the involution a m b ª a m bii, where b ª b is the matrix transposition. The subspace HŽ.O3 of symmetric elements with the 840 ILTYAKOV AND SHESTAKOV multiplication a( b s 1r2Ž.ab q ba is a central simple Jordan algebra A of dimension 27, which is called aŽ. split Albert algebra. Ž Ä4.Ž Let eiiiijs e , ei,jg1, 2, 3 , i / j be matrix units of O3we identify . 1mF33with F . Then an arbitrary element of A can be written in the form
3 csÝÝeii␣q wxc ijk , is1 Ž.ijk Ž. where ␣ig F, c i g O, wxc jk means c и ejkq c и e k j, and ijk runs over the subgroup A3 of even permutations. Denote by trŽ.c the trace of c, that is ␣123q ␣ q ␣ . We frequently use the following multiplication rules Ž.i, j, k are distinct :
11 wxaij( wxb jk s22 wab xik; wxa ij( wxb ij s tŽ.Žab eijq e ..
It is easy to check that the trace trŽ.Rc of the linear transformation Ž. Rc :xªx(cof the vector space A equals to 9 и tr c for any c g A. A permutation of the idempotents e123, e , e induces an automorphism of A. The set of such automorphisms is a subgroup of G s AutŽ.A , which Ž. is isomorphic to S3. Further, an automorphism g Aut O defines an Ž. automorphism of O3 by a и eij ª a и eij. Its restriction on A is an automorphism of A. So, we can assume that AutŽ.O is embedded in AutŽ.A under this map. Let : Ž.Ž␣ ¨  y¨ * .. It is easy to see that AutŽ.O . We set ¨ *  ª y¨ ␣ g Ä4 Hsgr S33, . Since s for each g S , this group is isomorphic to S32= Z . In particular, < 27 X a x A FYŽ.Ž n. ,1Ž. jiis Ým jFg m is1 Žn.Ä4 where a127,...,a is a basis of A and Y s xij N i s 1, . . . , 27; j s 1,...,n is a set of variables. The algebra A is generated by three elementswx 11, p. 58, Ex. 12 . Hence, by Proposition 2.1 inwx 8Ž see also Lemma 2 inwx 4. the center Cn of the Žn. algebra Bnnnlies in FYŽ .. Besides, if K is the quotient field of C , then BKnnis a central simple algebra over K nof dimension 27. Note that the algebra BKnnhas unit and, therefore, K ncoincides with the quotient field of the centroid of Bn. Let b B . Then, since it is an element of the Albert algebra A g n m F Ž Žn..ŽŽn..Ž.ŽŽn..Ž. FY over FY , the trace tr b g FY . It turns out that tr b g K n INVARIANTS OF F4 841 wx9, Section 1.2 . Moreover, in the same way one can check that for any central simple algebra B A FYŽ Žn..Žover a subfield K FYŽn..of : m F : dimension 27 the trace trŽ.b of an element b g B lies in K. We will show that X12, X can be transformed to a special form. Let us take a set of variables U X s Ä4xiiiijijikik, y , z ; y , z ; ¨ , ¨ N i, j, k s 1,2,3; i / j Ž.UŽUUU. and put ¨iis ¨ 1, ¨i2, ¨i3, ¨iis ¨ 1, ¨i2, ¨i3. Consider the elements from A A FXŽ., smF 3 xsÝexii, is1 3 ysÝÝeyiiq wxy ijk , is1 Ž.ijk 3 zsÝÝeziiq wxzijk , is1 Ž.ijk yjk 0 zjk¨ i where y Ž., z Ž.U are from O FXŽ.. These elements iis 0 ykj s ¨ ikjz m F turn to be free generators of B3 Žcf.wx 2. . LEMMA 1. The algebra alg F Ä4x, y, z is isomorphic to B3. Proof. Let K FYŽ Ž2.., consider the subalgebra B и K A A s 2 : KFs m 3Ž.2 Ž. Ž. Ž. K. Since X111121131q tXXqtXXqtXs0, where tXi1gKand tX31Ž./0wx 6, p. 232 , the unit of A lies in this subalgebra. Hence, by w 6, p. 363, Ex. 2xwBK2 is a central simple algebra of dimension 9. From 8, Lemma 1.3x it follows that BK22is a central simple Jordan algebra over K22of dimension 9 and K : K. ŽŽ3.. Let P be the algebraic closure of the field FY . Then B3 : A P s A P and K P. The algebra BK Pis a central simple one mmF22: 2K2 Ž.Žq. over P wx11, p. 137 ; therefore, it is isomorphic to BP23,MP s MFŽ.Žq. P A 6, p. 204 . 3 m FP: wx Ž. Ž . Ž.Žq. Bywx 6, p. 389, Ex.3 there is g Aut A P such that BP23sMP and Ž.X1 has a diagonal form. Hence, there is an epimorphism from Ä4Ž. Ž.Ž.Ž. alg F x, y, z to B3, which sends x ª X 123, y ª X , z ª X . Ž. Besides, since B33, B is a free algebra of rank 3 in its variety, this map is invertible. This just proves the lemma. Ž. Let us define an action of H on FX: for any ␦ g S3 we set ␦ и ti s t␦ Ži., Ä4 ŽÄ4. where t is a letter from x, y, z , ␦ и tij s tt␦Ži.␦Žj. gy,z, and Ž. U Ž.U Ž Ä ␦и¨ij ssgn ␦ ¨␦Ži.jij, ␦ и ¨ s sgn ␦ ¨␦Ži.jii. Further, и t s ttgx,y, 842 ILTYAKOV AND SHESTAKOV 4.ŽÄ4. UU z,иtijstt ji gy,zand, finally, и ¨ijsy¨ ij, и¨ ijsy¨ ij.We identify FXŽ.with 1 m FX Ž., where 1 is the unit of A. Ž. Ž.H LEMMA 2. The field K33s K A lies in the fixed field F X . Proof. An element g g G defines an automorphism of A over FXŽ., Ž.g gid:a␣Ž.aиg␣. 1 s mF mª m Ž. On the other hand, every g g Aut F FX defines an automorphism of A over F, Ž.g id g : a ␣ a Ž.g и ␣ . 2 s mF m ª m Ä4 Ž. Žy1. Note that for any t g x, y, z and g g H we have t и 12g s t и g . Therefore, it holds for any t g B3. Ž. Ž. It is known that K33is generated by the set tr B wx9 . Let T s 1g , where g g G. For any b g B3 we have 11y1 trŽ.b и T s 99tr Ž.RbиTbs trŽ.TRTstrŽ.b . Ž. Ž. Hence, if f g K31, then f и g s f. Therefore, f s f и 2g for any H ggH, and this means f g FXŽ.. Ž. We set ␣ s yyy12 23 31 and K˜3s Kx 3 1,x 2,x 3,␣. LEMMA 3. The field K˜3 contains the following set M: Ž. Ž . 1 xiii,y,zis1, 2, 3 ; Ž.2 yy12 21,yy 23 32,yy 31 13; Ž.3 ␣; Ž. Ž . 4 zyij ji i,js1, 2, 3, i / j ; Ž. Ž U.Ž . 5 ¨ii,¨ is1, 2, 3 ; Ž. Ž U.Ž . 6 ¨ij,¨yi ij,js1, 2, 3, i / j ; Ž. Ž . 7 ⌬sdet ¨ 123, ¨ , ¨ . Proof. Let B˜˜˜BK K BиK A. It is a central simple 333s m K3333, : Ž. algebra over K˜˜33of dimension 27. Hence, for any b g B we get tr b g K˜3. Let us check that M : K˜3. At the same time we prove the following assertion. LEMMA 4. The algebra B˜3 contains the elements: Ž.a e123,e,e; Ž. yij 0 Ä4 b,wx00ij where i, j g 1, 2, 3 , i / j; 0 00 Ž.c,,¨i U whereŽ. ijk A ; wxw00 jk¨ i 0 x jk g 3 00 Ž. 0¨jjiy U Ž. d,,wxwx00 jk ¨ jijy 0 jk ijk g S3. INVARIANTS OF F4 843 First, since x g B˜31and the roots x , x2, x3of its characteristic polyno- mial lie in K˜˜31, we have e , e2, e33g B . Applying Peirce decomposition, we Ž. get yii, z g K˜˜3for any i and wxyijk, wxz ijkg B33for any ijk g A . There- Ž2. fore, yyjk k jstr wxy i jk g K˜3. Let ␣ s ␣ и s yyy13 32 21syy 12 21иyy 23 32иyy 31 13r␣gK˜3. Hence, y 0 1 ij ␣y yy y y s ž/wxkiij y2Ž.jjiijŽ. wxjk( wxki 00ij ␣y␣ ˜˜yji 0 yi j 0 belongs to B3. Also, wx00jis wxwxy k i jyg00 i j B3. Further, zyij ji s Žz yji 0 .Ž.Ž.˜˜ tr wxkij( w00 x ji g K3; therefore, zijry ijs zy ij jiryy ij ji gK3 and 0, ¨i zzjk yjk 000k j Uszi jk yy 0 wx 0 ykj ¨i jk yyjk00jk k j jk U 1 0 ¨ i 2 lies in B˜˜. Hence, Ž., tr ŽU .K . Next, 3 ¨ii¨ sy 2 wx¨ i 0 jkg 3 0 00 ¨ j yij 0 U ˜ y 0 s2 U ( gK3, y¨jij ¨j 0 00 jkki i j Ž. Ž U. and in a similar way one can get all elements d . Now, ¨ jkjiik, ¨ yy s 0y 00 trŽ¨jji U.Ž.K˜˜and since yy y ␣ yy K,we ywxwx00 jk ( ¨kiky 0 jk g 3 ji ikr jk s r jk k j g 3 Ž U. get ¨ jkjk, ¨ y to belong to this field. Taking an appropriate pair of elements fromŽ. d we prove all elementsŽ. 6 to lie in K˜3. Finally, we have yy 0 ¨iikyy ki00y ik ki ¨ i ( s , 00ij 00ki 2 00jk 00 So, 0¨i B˜. Moreover, 2 0¨jk0¨ is an element of wx00 jk g 3 w¨ jk= ¨ 0 xjksy wxwx00 ki( 00 i j B˜3, and 00 0¨i ⌬sŽ.¨jki=¨,¨sytr g K˜3. ¨jk=¨0( ž/jk 00jk Lemma 3 and Lemma 4 are proved. LEMMA 5. The field K˜3 is generated o¨er F by M. Proof. Let V be the subspace over FMŽ.spanned by the elements Ž.Ž.a᎐d of Lemma 4. We prove that it is an algebra over FM Ž .. It suffices to show a( b g V for any a, b fromŽ.Ž. a ᎐ d. 844 ILTYAKOV AND SHESTAKOV If a s ei, this is evident. Define a representation : H ª Aut F A by Ž. Ž. Ž. Ž. Ž. putting g s 12g и g for g g H. Since g NsFŽX.2g, the field FMŽ.is invariant under this action of H. Actually, K3 is fixed under Hby Lemma 2, and vect F Ä4␣, ␣, x123, x , x is invariant under H. Moreover, H transposes the elementsŽ.Ž. a ᎐ d up to sign. Hence, V is invariant under H and it suffices to show that a( b g V only for pairs a, b from different H-orbits. Note that the first and the second elements inŽ. c andŽ. d are conjugate under , so we need only consider the cases where a is one of the following elements: 0 y 00y Ž.1,2,3.¨ 12Ž. 3Ž. ¨221 0023 0 023 0 023 In the first case we have a и Ž.23 sya. Hence, we need to take b from differentŽ. 23 -orbits: 0 ¨ 120 ¨ 00 00 bg,,,,UU ½¨120¨0 0023 0031 23 31 y 0 y 0 y 0 23 ,,12 21 5 0023 0012 0021 or 0 ¨ 221y 0 ¨ 223y 0 ¨ 113y bg½,,, 0023 0012 0012 00 00 00 UUU,, .2Ž. ¨212y 0 ¨232y 0 ¨131y 0 5 23 12 12 In caseŽ. 2 we have a и Ž.23 s a; so, similarly, y23 0 y32 0 y12 0 y21 0 b g ½5,,, 0023 0032 0012 0021 or b is as inŽ. 2 . Finally, in caseŽ. 3 b runs over the elements Ž. d . For all these a, b one can show a( b g V in a straightforward way. We consider only several typical cases. For example, ␣ y23 0 y12 0 y13 0 ( sgV. 0023 0012 2yy13 31 0013 INVARIANTS OF F4 845 Next, Ž., U y и yy 0¨13100 ¨ 1¨ 2 12 23 32 y0 (Uy0sgV. 0023 ¨232 12 ␣ 0 031 Finally, consider 1 0 ¨ 120 ¨ 00 (sy = 0 . 0023 0031 2¨12¨ 12 UU U Ž. Let ¨ 12= ¨ s ␣ 1311313232¨ q ␣ ¨ y q ␣ ¨ y , where ␣ig FX. Then the Ž.Ž.Ž. conditions ¨ 121= ¨ , ¨ s ¨ 122= ¨ , ¨ s 0, ¨ 123= ¨ , ¨ s ⌬ yield eas- Ž. Ž. Ž. ily that ␣i g FM. Notice that B3 : V. Moreover, tr V : FM. Hence, trŽ.B31is contained in this subfield. Since x , x2, x3and ␣ also lie there, Ž. we just get K˜3s FM. Ž.H LEMMA 6. K3s FM . Ž.H Ž. Ž.H Proof. First, K3 : FX lFM sFM . Like in Lemma 3 we Ž. obtain ␣␣ g Kx31,x 2,x 3and ␣q␣strŽ.Ž.wxy3112( wxy 23( wxy 2 31 g Kx 3123Ž.,x,x; Ž.Ž. hence, < Kx123,x,x,␣:Kx 123,x,x << H On the other hand, by Artin’s theorem HH H <<<<<<< 3. LIST OF GENERATORS Now we are going to give an explicit list of algebraically independent Ž.H generators for the field K3 s FM . First, let us find generators of Ž.Z2 Ä4 FM , where Z2 s 1, . Let MЈ be a set obtained from M by Ž.UUŽU replacing the elements yy12 21 and ¨ 1, ¨ 1 by ␣ and ⌬ sydet ¨ 1, ¨ 2 , U .Ž.Ž. ¨3,respectively.Weclaim FM sFMЈ.Infact, yy12 21 s Ž.Ž . ŽŽU . .Ž ␣␣ r yyyy23 32 31 13 . Besides, ⌬⌬ sydet ¨ij, ¨ N i, j s 1, 2, 3 s ␣11¨ , 846 ILTYAKOV AND SHESTAKOV U .ŽÄŽU.4. ¨12q␣, where ␣ 12, ␣ g FM_ ¨ 11,¨ and ␣ 1/ 0. It proves the claimed statement. We break down MЈ into two subsets: U U U M0s Ä4xiii, y , z ; yy23 32,yy 31 13;Ž.Ž.¨ 2,¨ 2,¨ 3,¨ 3 ; X U M1s ½␣, ␣, ijkkjiis zy,sи;␥ isŽ.¨ jkjki,¨ y,␥ s␥iи;⌬,⌬NŽ.ijk g A35. Ž. LEMMA 7. Let K˜s Kt11,t,...,tmm, t be a purely transcendental exten- sion of a field K of degree 2m and the group Z2 acts on K˜ o¨er K by Z 2 tllи s t . Then the fixed field K˜ is generated o¨er K by the elements 2 h1,...,hm, s121, s rs ,...,smrs1, where hlllllls t q t , s s t y t . Proof. Let KЈ be the subfield generated by these elements. It is evident ZZ22 that KЈ : K˜˜. Now let frg g K , where f, g g R s Ktwx11,t,...,tmm, t . Z Then frg s 1r2 и Ž.Ž.fg q fg r gg . Hence, it suffices to show that R 2: KЈ. Let f be a monomial in the free generators. By induction on degŽ.f we prove f q f g KЈ. If degŽ.f s 1, this is trivial. Otherwise, f s ab¨, where Ä4 Ž. a,bgtll,tNls1,...,m and ¨ is a monomial of degree deg f y 2 Ž. maybe, it is the unit . Also, notice that ssijgKЈ for any i, j. Hence, modulo KЈ we have 11 fqfs22Ž.Ž.aqabŽ.¨qb¨q Ž.ayab Ž.¨yb¨ ' Ž.ayab Ž.¨yb¨ 1 s4Ž.Ž.Ž.Ž.Ž.Ž.Ž.ayabyb¨q¨qbqbaya¨y¨ '0. The lemma is proved. X X Ž .Ž.Ž.Z2 Now let us put K s FM01. Then Lemma 7 holds for KM ,soFM X is generated over F by a set N which consists of M0 and the elements ␣ q ␣ , iiiiq  , ␥ q ␥ , ⌬ q ⌬, 2 Ž.␣y␣,Ž.iiyr Ž.␣y␣,Ž.␥iiy␥r Ž.␣y␣, Ž.Ž.⌬y⌬r␣y␣. INVARIANTS OF F4 847 Ž.H Ž.S3 Construct now a transcendence base of K3 s FM sFN . Denote Ž U .Ž U.Ž U. d11221233231132syyqyyqyy and d s ¨ 11, ¨ q ¨ 22, ¨ q ¨ 33, ¨ , Ž. then d12, d g FN. We need to change the set of generators N a bit. Let 2 ␣q␣␣Ž.Ž.y␣ ⌬q⌬⌬y⌬ N0s ,,,2 ; ½5dd122d1Ž.␣y␣d Ž. moreover, let N1 consist of the elements i s 1, 2, 3 iiy  t1iisx,t2iisy,t3iisz,t4iiisq,t5is , ␣y␣ ␥iiq␥␥iiy␥yyjkkj t6is ,t7is ,t8isŽ.Ž.ijk g A3, d22Ž.␣y␣dd1 U Ž.¨ii,¨ t9is . d2 Ž. It is evident, that NЈ s N01j N lies in FN. LEMMA 8. The set NЈ generates FŽ. N . Ž. Ž. Proof. We need only show N : FNЈ. It holds, if d12, d g FNЈ. 3 Ž. First, ␣␣rd1818283s ttt gFNЈand 2 2 ␣␣ 1 ␣ q ␣␣Ž.y␣ 22sygFNŽ.Ј. dd114ž/ž/d1 Ž. Ž .2 Ž. Therefore, d1g FNЈ. It also implies ␣ y ␣ g FNЈ. 2221ŽŽ..Ž. 2 Ž Next, ␥␥iird26s4t iyt7i␣y␣ gFNЈ. In a similar way see the proof of Lemma 7. one may check that ␥␥␥1231 ␥␥␥ 1231 q ddd222␣ ddd 222␣ Ž. 3 ŽŽ belongs to FNЈ too. Now direct calculations shows ⌬⌬rd2 sydet ¨i, U . .32 Ž. Ž. ¨j Ni,js1, 2, 3 rd22g FNЈ. Besides, ⌬⌬rd also lies in FNЈ. Thus, Ž. d2 gFNЈand the lemma is proved. 848 ILTYAKOV AND SHESTAKOV Ž.S3 Note, that the set N03: FN sK. Consider now the elements 3 pjjsÝ ti,js1,...,7; is1 3 q1111212131311stt qttqtt, qjsÝ tt1iji, js2,...,9; is1 3 2 r1111213sttt, rjsÝ tt1iji, js2,...,9. is1 Ä4 THEOREM 1. The set T s N01j p ,..., p 71919; q ,...,q ; r ,...,r forms a rationality base for K3 o¨er F. Ž. Ž.S3 Ž 4S3 Let F10s FN , then K 3s FNЈ sFt1ij Njs1,...,9; is1, 2, 3 . Ž.Ž.S3 We have F21s Fpjjj,q,r:FN . Besides, by the Vandermonde Ž.Ž.Ž determinant argument we get Fx21,x 2,x 3sFNЈ note that pjmakes . sense fo rj s 8, 9 and equals 1 . Further, xi are the roots of the polyno- 32 Ž. mial t y pt111qqtyr. Hence, THEOREM 2. Let F be a field of characteristic / 2, 3; let Bn be the algebra generated o¨er F by n generic elements X1,..., Xn of a split Albert Ž. algebra A s H O3 and let Knn be the quotient field of the center of B . Then Kn is a rational function field o¨er F. For n ) 3 a rationality base for K n over K3 may be obtained as follows Ž. wx8 . Let u127,...,u be a basis for B333и K over K . Then the 27 n y 3 polynomials trŽ.Xuij,1FjF27; 4 F i F n, form a rationality base for K n over K3. In particular, tr и deg FnK s 27n y52Ž this equality is the main result ofwx 9. . Rationality of K2 over F is proved inwx 9 ; in this case tr и deg F K2 s 10 and a rationality base is given by j 2 2 2 2 ÄtrŽ.Xi; i s 1,2; j s 1,2,3; trŽ.XX12,trŽX 12X .Ž,tr XX 12 .Ž,tr X 12X .4. 2 3 Finally, K11is generated by trŽ.ŽX ,tr X1.Ž,tr X1.and is rational over F, toowx 6 . INVARIANTS OF F4 849 Apply now these results to the structure of the field of invariants of AutŽ.A . THEOREM 3. Let A be a split Albert algebra o¨er an algebraically closed G field F of characteristic 0 and G s AutŽ.A . Then for any n the field F Ž nA . of rational in¨ariants of G on n ¨ariables is a rational function field o¨er F. Ž.G Proof. If n G 3, then FnA sKn bywx 4 . For n s 1 the same state- ment follows fromwx 6, p. 389 . Finally, for n s 2 it suffices to prove the following resultŽ seewx 10, Section 2.1. . LEMMA 9. There is a finite subset S of the subalgebra of Fwx2A , generated by traces of elements of B2 , which separates orbits of general position in 2A, i.e., there is an open set U of the affine space 2A such that for any two points X,XЈgU with different G-orbits there exists h g S such that hŽ. X / hX ŽЈ .. Ž. Ä Proof. For any X s X12, X g 2A let AXFbe the subalgebra alg X1, 4 X2 :A. We shall prove that for a point X of general position this Ž.Žq. Ž2. subalgebra is isomorphic to MF3 . Actually, let R s FYw xbe the algebra of polynomials in Y Ž2. and let P be the algebraic closure of the Ž Ž2..Ž.Žq. field K s FY . Then, like in Lemma 1 we have BP23,MP :AP AP. smF Choose a basis b192,...,b gB of the vector space BP2over P. Then Ž. the matrix units eij, i, j s 1, 2, 3, are polynomials ij b19,...,b over P. Let ⌳ be a finite subset of P which includes the coefficients of all ij and the structure constants of the algebra BP2 with respect to the basis Ž. b19,...,b . Then K ⌳ is a finite extension Kwx␣ of K, where ␣ g P. Ž. Every g ⌳ equals h ␣ , where h is a polynomial over K. Denote by g the common denominator of the coefficients of h, g ⌳, and the Ä r Ä44 minimal polynomial for ␣. Put R1 s frg N f g R; r g N j 0 . Then ␣ is integral over R11and the R wx␣ -module generated by b19,...,b is an R1wx␣-algebra, which contains the matrix units eij. Hence, this algebra Ž.Žq. equals B221s BRwx␣ and includes an F-subalgebra C , MF3 . Ž. Ž. Suppose, gX/0. Then the homomorphism : B211ª A, X s X , Ž. X22sXcan be extended to a homomorphism : B2ª A. It is clear that the images of and coincide. Recall that the dimension of Ž.B2 Ž.Žq. Ž. is not bigger than 9wx 6, p. 382 ; besides, it contains MF3 sC. Ž. Ž.Žq. Hence, AX s B23, MF as claimed. Remark. As the referee noted, using results ofwx 7 one can find a basis ࠻࠻ of BP21over P in an explicit formŽ 1, X , X1, X2, X2, X1= X2, ࠻࠻࠻࠻ X121212=X,X=X,X=X., which provides another proof of this statement. 850 ILTYAKOV AND SHESTAKOV Žq. Restriction of the Killing form trŽ.xy on MF3 Ž. is nondegenerate. Let Ž.Žq. us take two points XЈ, XЉ g 2A such that AX Ј , AX Љ , MF3 and Ž. Ž . Ž. Ž . fXЈsfXЉ for any fXsfX12,X gtrwx 2A . We will show that X YX Y X1122иgsX,XиgsXfor some g g G. Ž.Žq. First, we may assume that AX Ј s AX Љ s MF3 wx6, p. 389, Ex.3 . Ž. Ž Ž.. Later on, for any f g B2 if fXЈs0, then 0 s tr AfXXЈ Јs ŽŽ.. Ž. tr AfXX Љ Љ and, therefore, fXЉs0. Hence, we can define a map Ž.Žq. Ž.Žq. ŽŽ .. Ž . g:MF33ªMF by gfXЈ sfXЉ for every f g B2.Itis Žq. XY evidently an automorphism of MF3Ž. that sends Xiito X , and it can be lifted up to an automorphism of A wx6, p. 370, Theorem 3 . So, the lemma Ž.G is proved and bywx 10, Section 2.1 we get K2 s F 2A . COROLLARY. Let the ground field be algebraically closed of characteristic 0. Then the field of rational in¨ariants of a minimal irreducible representation of a simple algebraic group of type F4 in se¨eral ¨ariables is a rational function field. Proof. It is known that the subspace A 0 of elements with zero trace in Ais an irreducible G-module of minimal dimensionwx 6, p. 382 . We have Ž.G Ž. Ž. FnA001:FnA . Let X ,..., Xnnbe free generators of B 1 , where Ž.Ž. a122s1 and a ,...,a 7is a basis of A 0cf.wx 8 . Observe that tr yjs 3 x1j. Let us set 27 trŽ.Xi ZiisXy1и sÝaiijmx. 3 is2 We will obtain the set of algebraically independent generators of FnŽ.AG X Ä Ž. 4 X in the form Tnns T j tr X iN i s 1,...,n, where elements from Tn X depend only on Z1,...,Znn. Then T is just a transcendence base of G FnŽ.A0 . Ž.G Ž i.Ž . X The generators of F A have the form tr Xi11s1, 2, 3 . Hence, T Ži.Ž . consists of tr Zi1 s2, 3 . In case n s 2 the fixed field is generated by Ž j.Ž .Ž.Ž 2 .Ž 2.Ž22. tr Xii s1, 2, j s 1, 2, 3 and tr XX12,tr XX 12,tr XX 12,tr XX 12 X wx9 . Again, replacing Xiiby Z we get T n. Let us consider the case n s 3. By Lemma 1 we can assume that Ž. XX X123sx,Xsy,Xsz. Note that T includes tXkksp. Denote by q jj, r Ž. Ž. js1, . . . , 9 the elements qjj, r , where ti ik,ks1, 2, 3 is replaced by 1 Ä4X tiky3p k. It is easy to check that the set T31233s p , p , p j T , where XÄ4ÄXX 4 T3 spjjjNjs4,...,7 j q, r Njs1,...,9 jN03, generates K . Be- Ž.XX Ž.G Ž.G sides, pjjjG4,q,rjbelong to FnA000, therefore, N : FnA . So, it completes the proof for n s 3. G G Finally, for n ) 3 the rationality base for FnŽ.A00over FŽ.3A may be obtained just as in the case of K n; one should note only that A can be generated by three elements with zero tracewx 11, p. 58, Ex.12 . INVARIANTS OF F4 851 ACKNOWLEDGMENTS The authors thank the referee for his useful remarks. REFERENCES 1. I. V. Dolgachev, Rationality of fields of invariants, ‘‘Proc. Symp. Pure Math.,’’ Vol. 46, Part 2, Amer. Math. Soc., Providence, Rhode Island, 1987. 2. E. Formanek, The center of the ring of 3 = 3 generic matrices, Linear and Multilinear Algebra 7 Ž.1979 , 203᎐212. 3. E. Formanek, The center of the ring of 4 = 4 generic matrices, J. Algebra 62 Ž.1980 , 304᎐319. 4. A. V. Iltyakov, Trace polynomials and invariant theory, Geom. Dedicata, to appear. 5. N. Jacobson, Some groups of transformation defined by Jordan algebras. II. Groups of type F4 , J. Reine Angew. Math. 204 Ž.1960 , 74᎐98. 6. N. Jacobson, ‘‘Structure and Representations of Jordan Algebras,’’ Amer. Math. Soc. Colloq. Publ., Vol. 39, Amer. Math. Soc., Providence, Rhode Island, 1968. 7. K. McCrimmon, The Freudenthal᎐Springer᎐Tits constructions revisited, Trans. Amer. Math. Soc. 148 Ž.1970 , 293᎐314. 8. S. V. Polikarpov and I. P. Shestakov, Nonassociative affine algebras, Algebra i Logika 29 Ž.1990 , 709᎐723. 9. S. V. Polikarpov, Free affine Albert algebras, Sibirsk. Mat. Zh. 32, No. 6Ž. 1991 , 131᎐141. 10. V. L. Popov and E. B. Vinberg, Invariant theory, in ‘‘Encyclopaedia of Mathematical Sciences. Algebraic Geometry IV, Vol. 55, pp. 123᎐284, Springer-Verlag, New YorkrBerlin, 1994. 11. K. A. Zhevlakov, A. M. Slinko, I. P. Shestakov, and A. I. Shirshov, ‘‘Rings That Are Nearly Associative,’’ Academic Press, New York, 1982.