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On the Variety of Automorphisms of the Affine Plane
195, 604]623 Ž1997. JA977054
JOURNAL OF ALGEBRA ARTICLE NO.
On the Variety of Automorphisms of the Affine Plane Jean-Philippe Furter* U.M.P.A., E.N.S.L., 46 Allee ´ d’Italie, 69 007, Lyon, France Communicated by Peter Littelmann Received October 15, 1996
The main subject of our study is GA 2, n , the variety of automophisms of the affine plane of degree bounded by a positive integer n. After detailing some definitions and notations in Section 1, we give in Section 2 an algorithm to decide whether an endomorphism of the affine plane over an integral domain is a tame automorphism. Then, by applying this algorithm to the Nagata automorphism, we recover easily its known results. In Section 6, we compute the number of irreducible components of GA 2, n when n F 9 and we show that GA 2, n is reducible when n G 4. Our proofs are based on a precise decomposition theorem for automorphisms given in Section 3 and a characterization of length one automorphisms given in Section 5. Finally, in Section 7, we give some details on the case n s 4. Q 1997 Academic Press
1. DEFINITIONS AND NOTATIONS Let k be an integral domain. k w X, Y x is the polynomial algebra in the indeterminates X, Y Žendowed with the degree function and the valuation function at the origin, which are denoted by deg and val, and A2k s Spec k w X, Y x is the affine plane over k. A k-endomorphism f of A2k will be identified with its sequence f s Ž f 1 , f 2 . of coordinate functions f i g k w X, Y x Ž i s 1, 2.. If g is a k-endomorphism of A2k , we agree that fg will denote the k-endomorphism of A2k obtained by composition of f and g. We define: the degree of f by deg f s max deg f 1 , deg f 2 4 ; the bidegree of f by bidegŽ f . s Ždeg f 1 , deg f 2 .; the total degree of f by tdegŽ f . s deg f 1 q deg f 2 ; the Jacobian of f by JacŽ f . s Ž f 1 r X .Ž f 2 r Y . y Ž f 1 r Y .Ž f 2r X .. * E-mail address: [email protected]. 604 0021-8693r97 $25.00 Copyright Q 1997 by Academic Press All rights of reproduction in any form reserved.
AUTOMORPHISMS OF THE AFFINE PLANE
605
We introduce the following groups: GA 2 Ž k ., the group of all k-automorphisms of A2k ; Af 2 Ž k ., the subgroup of GA 2 Ž k . of affine automorphisms, i.e., automorphisms of the shape Ž aX q bY q c, dX q eY q f . where a, b, c, d, e, f are elements of k such that ae y bd is a unit of k; GL 2 Ž k ., the subgroup of Af 2 Ž k . of linear automorphisms; BA 2 Ž k ., the subgroup of GA 2 Ž k . of triangular or ‘‘de Jonquieres’’ ` automorphisms, i.e., automorphisms of the shape Ž aX q P Ž Y ., bY q c . where a, b are units of k, c is an element of k, and P Ž Y . is an element of k w Y x; TA 2 Ž k ., the subgroup of GA 2 Ž k . of tame automorphisms, i.e., the subgroup of GA 2 Ž k . generated by Af 2 Ž k . and BA 2 Ž k ..
2. AN ALGORITHM TO DECIDE WHETHER AN ENDOMORPHISM IS A TAME AUTOMORPHISM The following proposition is the key of the announced algorithm: if f g TA 2 Ž k . satisfies tdeg f G 3, it ensures the existence of a g GA 2 Ž k . of a very special shape such that tdegŽ ay1 f . - tdeg f. PROPOSITION 1. Let f s Ž f 1 , f 2 . g TA 2 Ž k . with bidegŽ f . s Ž d1 , d 2 .. Denote by g i the homogeneous component of degree d i of f i for i s 1, 2. Then d1 < d 2 or d 2 < d1. Moreo¨ er: Ži. If d1 - d 2 , then there exists l g k such that g 2 s l g 1d 2 r d1 ; Žii. If d 2 - d1 , then there exists l g k such that g 1 s l g 2d1 r d 2 ; Žiii. If d1 s d 2 ) 1, then there exists Ž a , h. g Af 2 Ž k . = TA 2 Ž k . such that f s a h and d1 s deg h1 ) deg h 2 where h s Ž h1 , h 2 .. Remarks. Ž1. Let us assume that degŽ f . ) 1. We set a s Ž X, Y q l X d 2 r d1 . in case Ži., a s Ž X q lY d1 r d 2 , Y . in case Žii., and if we are in case Žiii., we keep a as explained there. Then, it is clear that we have tdegŽ ay1 f . - tdeg f. Ž2. If d1 s d 2 , in general, there does not exist l g k such that g 2 s l g 1 or g 1 s l g 2 , as shown by the following example where k s Cw Z x: f s Ž Ž 1 y Z . X q ZY q Ž 1 y Z . Y 2 , yZX q Ž 1 q Z . Y y ZY 2 . .
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JEAN-PHILIPPE FURTER
It can be checked that f s ag where
½
a s Ž Ž 1 y Z . X q ZY , yZX q Ž 1 q Z . Y . g Af 2 Ž k . g g Ž X q Y 2 , Y . g BA 2 Ž k .
so that f g TA 2 Ž k .. However, bideg f s Ž2, 2., the homogeneous component of degree 2 of f is ŽŽ1 y Z .Y 2 , yZY 2 . and there exists no l g Cw Z x such that Ž1 y Z .Y 2 s lŽyZY 2 . or yZY 2 s lŽ1 y Z .Y 2 . Ž3. Nevertheless, still in the case d1 s d 2 , if k s C and deg f ) 1, the equality Jac f g C* implies that JacŽ g 1 , g 2 . s 0, which is well known to imply that g 1 and g 2 are proportional polynomials. The following group-theoretical lemma will be used in the proof of Proposition 1. LEMMA 1. If a group G is generated by two subgroups H and K, then each element g of G can be written g s h1 k 1 h 2 k 2 ??? h l k l h lq1 , where l is a non-negati¨ e integer and the h i Ž resp. k i . are elements of H Ž resp. K . such that h i f K for 2 F i F l Ž resp. k l f H for 1 F i F l .. Proof. If g is an element of G, we have g s h1 k 1 h 2 k 2 ??? h l k l h lq1 , where l is a non-negative integer and the h i Žresp. k i . are elements of H Žresp. K .. If the condition Ž h i f K for 2 F i F l and k i f H for 1 F i F l . is not satisfied, then we can obtain an expression for g of the same shape but where l is replaced by l y 1. Indeed, if, for example, h i 0 g K Ž2 F i 0 F l ., then we have g s h1 k 1 h 2 k 2 ??? h i 0y1 Ž k i 0y1 h i 0 k i 0 . h i 0q1 ??? h l k l h lq1 , where k i 0y1 h i 0 k i 0 g K. After a finite number of such reductions, we will necessarily obtain the desired expression for g. Proof of Proposition 1. By Lemma 1 applied with G s TA 2 Ž k ., H s Af 2 Ž k ., K s BA 2 Ž k ., and g s f, we obtain the existence of a non-negative integer l and of Ž a , b . g ŽAf 2 Ž k .. 2 , Ž a i .1 F i F ly1 g ŽAf 2 Ž k . _ BA 2 Ž k .. ly1, Žg i .1F i F l g ŽBA 2 Ž k . _ Af 2 Ž k .. l such that f s ag 1 a 1 g 2 ??? a ly2 g ly1 a ly1 g l b . We show by a decreasing induction on i, beginning with i s l and finishing with i s 1 that l
bideg Ž g i a i ??? g ly1 a ly1 g l b . s
ž
l
Ł deg g j , Ł jsi
jsiq1
/
deg g j .
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AUTOMORPHISMS OF THE AFFINE PLANE
The conclusion is clear for i s l. Let us suppose it is true for a given i with 2 F i F l. Let us write
a iy1 s Ž a iy1 X q biy1Y q c iy1 , d iy1 X q e iy1Y q f iy1 . . The relations a iy1 f BA 2 Ž k . and g iy1 f Af 2 Ž k . imply that d iy1 / 0 and degŽg iy1 . G 2. We now obtain bidegŽ a iy1 g i a i ??? g ly1 a ly1 g l b . s Ž pi , Ł ljs i deg g j . with pi F Ł ljs i deg g j and bidegŽ g iy 1 a iy 1 ??? g ly1 a py1 g l b . s ŽŁ ljsiy1 deg g j , Ł ljsi deg g j .. It is now proven that l
bideg Ž g 1 a 1 ??? g ly1 a ly1 g l b . s
ž
l
Ł deg g j , Ł deg g j js1
js2
/
.
Let us write a s Ž aX q bY q c, dX q eY q f .. To conclude, we just have to check that we are in case Ži. or Žii. or Žiii. according to whether a s 0 or d s 0 or ad / 0. We deduce at once from this last proposition an algorithm to decide if a k-endomorphism of A2k belongs to TA 2 Ž k .: ALGORITHM. Ž1. Enter a k-endomorphism f of A2k . Ž2. Let Ž d1 , d 2 . s bidegŽ f 1 , f 2 .. Ž3. If d1 s d 2 s 1, then goto 8. Ž4. If d1 / d 2 , then goto 6. Ž5. If there exists a g Af 2 Ž k . such that tdegŽ a f . - tdegŽ f ., then replace f by a f and goto 2, else STOP: f f TA 2 Ž k .. Ž6. If d 2 - d1 , then replace f s Ž f 1 , f 2 . by Ž f 2 , f 1 .. Ž7. If Ž d1 < d 2 and there exists l g k such that g 2 s l g 1d 2 r d1 ., then replace f by Ž X, Y y l X d 2 r d1 . f and goto 2, else STOP: f f TA 2 Ž k .. Ž8. If JacŽ f . is a unit of k, then STOP: f g TA 2 Ž k ., else STOP: f f TA 2 Ž k .. Application. Let us consider the Nagata automorphism Žcf. wNax. 2
f s X y 2Y Ž XZ q Y 2 . y Z Ž XZ q Y 2 . , Y q Z Ž XZ q Y 2 .
ž
/
g GA 2 Ž k . for k s Cw Z x and k s Cw Z, Zy1 x. By expanding f , we get
f s Ž X y 2 ZXY y Z 3 X 2 y 2Y 3 y 2 Z 2 XY 2 y ZY 4 , Y q Z 2 X q ZY 2 .
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JEAN-PHILIPPE FURTER
so that the homogeneous components of highest degree of f are yZY 4 and ZY 2 . Ž1. If k s Cw Z x, Ž ZY 2 . 2 does not divide yZY 4 and so f f TA 2 ŽCw Z x.. Ž2. If k s Cw Z, Zy1 x, the equation yZY 4 s yZy1 Ž ZY 2 . 2 allows us to obtain
f s Ž X y Zy1 Y 2 , Y . Ž X q Zy1 Y 2 , Y q Z 2 X q ZY 2 . and we easily see that
Ž X q Zy1 Y 2 , Y q Z 2 X q ZY 2 . s Ž X , Y q Z 2 X . Ž X q Zy1 Y 2 , Y . whence f s Ž X y Zy1 Y 2 , Y .Ž X, Y q Z 2 X .Ž X q Zy1 Y 2 , Y . and f g TA 2 ŽCw Z, Zy1 x.. 3. A PRECISE DECOMPOSITION THEOREM FOR AUTOMORPHISMS OF THE AFFINE PLANE From now on, we set k s C and we note GA 2 instead of GA 2 ŽC., Af 2 instead of Af 2 ŽC., etc. Let us recall that by a famous result of H.W.E. Jung and W. van der Kulk wvdKx, we have GA 2 s TA 2 . We give here a precise decomposition theorem for elements of GA 2 Žsee Theorem 1.. This result allows us Žsee Section 6, Proposition 10. to compute the dimension of the algebraic variety GA 2, n . Before stating the theorem, we need the following definitions: T2 is the subgroup of GA 2 of automorphisms of the shape Ž X q P Ž Y ., Y . where P Ž Y . is an element of Cw Y x satisfying valŽ P . G 1; UA 2 is the subgroup of T2 of elements of the shape Ž X q P Ž Y ., Y . where P Ž Y . is any element of Cw Y x satisfying valŽ P . G 2; for all Ž a, b . g C 2 , we set t Ž a, b. s Ž X q a, Y q b . g GA 2 and for all c g C, we set sc s Ž Y q cX, X . g GA 2 . We note s s s 0 s Ž Y, X .. THEOREM 1 ŽPrecise Decomposition Theorem.. _ Af 2 , then: Ži.
Let f s Ž f 1 , f 2 . g GA 2
if degŽ f 1 . ) degŽ f 2 ., '! Ž a, b . g C 2 , '! l g N, '! Ž g i . 1FiFl g Ž T2 _ GL 2 . = Ž UA 2 _ GL 2 . , '! b g GL 2
ly1
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AUTOMORPHISMS OF THE AFFINE PLANE
such that f s t Ž a, b. g 1 sg 2 s ??? sg ly1 sg l b ; Žii. if degŽ f 1 . F degŽ f 2 ., '! Ž a, b, c . g C 3 , '! l g N, '! Ž g i . 1FiFl g Ž T2 _ GL 2 .
ly1
= Ž UA 2 _ GL 2 . , '! b g GL 2 such that f s t Ž a, b. sc g 1 sg 2 s ??? sg ly1 sg l b . To prove this theorem, we will use the following three decomposition lemmas: LEMMA 2. Let f s Ž f 1 , f 2 . g GA 2 satisfy degŽ f 1 . ) degŽ f 2 . ) 1. Then '! Žg , p . g ŽT2 _ GL 2 . = GA 2 such that f s g p and p s Ž p1 , p 2 . satisfies degŽ p1 . - degŽ p 2 . s degŽ f 2 .. Proof. Existence. We begin by proving that for each automorphism f s Ž f 1 , f 2 . g GA 2 satisfying degŽ f 1 . G degŽ f 2 . ) 1 there exists Žg , p . g T2 = GA 2 such that f s g p and p s Ž p1 , p 2 . satisfies degŽ p1 . - degŽ f 1 .. For 1 F i F 2, let d i s deg f i and let g i denote the homogeneous component of degree d i of f i . From Proposition 1 and Remark 3 following it, we deduce that d 2 < d1 and that there exists l g C such that g 1 s l g 2d1 r d 2 . We can now just write f s Ž X q lY d1 r d 2 , Y . p where p g GA 2 and it is easy to check that p s Ž p1 , p 2 . satisfies p 2 s f 2 and degŽ p1 . - degŽ f 1 .. From this fact, we deduce by an immediate induction on deg f 1 that for each automorphism f s Ž f 1 , f 2 . g GA 2 satisfying degŽ f 1 . G degŽ f 2 . ) 1, there exists Žg , p . g T2 = GA 2 such that f s g p and p s Ž p1 , p 2 . satisfies degŽ p1 . - degŽ p 2 . s deg f 2 . This terminates the proof of the existence. Unicity. Let Žg , p . and Ž d , q . be as in Lemma 2 such that g p s d q. We can write g s Ž X q RŽ Y ., Y ., d s Ž X q S Ž Y ., Y . where RŽ Y . and SŽ Y . are elements of Cw Y x such that valŽ R . G 1 and valŽ S . G 1. Moreover, p s Ž p1 , p 2 . and q s Ž q1 , q2 . satisfy deg p1 - deg p 2 and deg q1 - deg q2 . We have Ž p1 q RŽ p 2 ., p 2 . s Ž q1 q SŽ q2 ., q2 . from which we deduce firstly that p 2 s q2 and secondly that p1 y q1 s S Ž p 2 . y RŽ p 2 .. In the last equality, the left hand term is of degree strictly less than deg p 2 which implies that S y R s 0 and finally g s d and p s q. LEMMA 3. Let f s Ž f 1 , f 2 . g GA 2 satisfy f Ž0, 0. s Ž0, 0. and degŽ f 1 . ) degŽ f 2 . s 1. Then '! Žg , b . g ŽUA 2 _ GL 2 . = GL 2 such that f s gb . Proof. Existence. Let f s Ž f 1 , f 2 . g GA 2 satisfy f Ž0, 0. s Ž0, 0. and degŽ f 1 . ) degŽ f 2 . s 1. We define b as the linear part of f. Then, if b s Ž b 1 , b 2 ., we have Cw b 1 , b 2 x s Cw X, Y x and so f can be written as Ž b 1 q g 1Ž b 1 , b 2 ., b 2 . where g 1 g Cw X, Y x and valŽ g 1 . G 2. Let us set g s Ž X q g 1 , Y .. On one hand, we have f s gb and on the other hand, g being an automorphism, we have necessarily g 1 g Cw Y x and finally g g UA 2 _ GL 2 .
610
JEAN-PHILIPPE FURTER
Unicity. If f s gb where Žg , b . g ŽUA 2 _ GL 2 . = GL 2 , then b is necessarily equal to the linear part of f and this gives us the uniqueness. LEMMA 4. Let f s Ž f 1 , f 2 . g GA 2 satisfy f Ž0, 0. s Ž0, 0. and degŽ f 1 . ) degŽ f 2 .. Then '! l g N*, '! Ž g i . 1FiFl g Ž T2 _ GL 2 .
ly1
= Ž UA 2 _ GL 2 . , '! b g GL 2
such that f s g 1 sg 2 s ??? sg ly1 sg l b . Proof. Existence. By induction on degŽ f 2 ., if degŽ f 2 . s 1, we are done by Lemma 3. If degŽ f 2 . ) 1, by Lemma 2, f can be written f s gs f˜ where Žg , f˜. g ŽT2 _ GL 2 . = GA 2 and f˜s Ž f˜1 , f˜2 . with degŽ f˜2 . - degŽ f˜1 . s degŽ f 2 .. The equality f Ž0, 0. s Ž0, 0. implies the equality f˜Ž0, 0. s Ž0, 0.. We have the ˜ result by applying the induction hypothesis to f. Uniqueness. This follows from the uniqueness in Lemmas 2 and 3 by noting that if l G 2, Žg i .1 F i F l g ŽT2 _ GL 2 . ly1 = ŽUA 2 _ GL 2 . and b g GL 2 , then the automorphism p s Ž p1 , p 2 . s sg 2 s ??? sg l b satisfies deg p1 - deg p 2 . Proof of Theorem 1. Let f g GA 2 _ Af 2 . If deg f 1 ) deg f 2 , the following applies. Existence. If we set Ž a, b . s f Ž0, 0., then the automorphism f˜s Ž f˜1 , f˜2 . ˜Ž . Ž . Ž ˜. Ž ˜. s ty1 Ž a, b. f satisfies f 0, 0 s 0, 0 and deg f 1 ) deg f 2 . We obtain the ˜ existence by Lemma 4 applied to f. Unicity. If f is written as in Ži. of Theorem 1, it is clear that we must have Ž a, b . s f Ž0, 0. and the unicity now follows from the unicity of Lemma 4 applied to f˜s ty1 Ž a, b. f. If deg f 1 F deg f 2 , we can see that there exist unique complex numbers ˜ ˆ ˆ a, b, and c such that fˆs Ž fˆ1 , fˆ2 . s scy1ty1 Ž a, b. f satisfies deg f 1 ) deg f 2 and fˆŽ0, 0. s Ž0, 0.. The result is now a consequence of case Ži. of the ˆ theorem applied to f.
4. ON THE MULTIDEGREE OF AN AUTOMORPHISM OF THE AFFINE PLANE We introduce here a notion of multidegree of automorphisms of the affine plane and obtain some basic properties. The multidegree of an automorphism is an element of the following set D: DEFINITIONS. D denotes the set of all finite sequences of integers greater than or equal to 2 and D* denotes the set D with the vacuum sequence omitted.
AUTOMORPHISMS OF THE AFFINE PLANE
611
We define the following partial order F on D: if d s Ž d1 , . . . , d l . g D and if e s Ž e1 , . . . , e m . g D, then we say that d F e if l F m and if there exist 1 F i1 - i 2 - ??? - i l F m such that ; j g 1, . . . , l 4 , d j F e i j . Also, we will denote the concatenation of d and e by de s Ž d1 , . . . , d l , e1 , . . . , e m .. We set: l Ž d . s l; < d < s d1 q d 2 q ??? qd l ; degŽ d . s d1 = d 2 = ??? = d l ; dy1 s Ž d l , . . . , d1 .. Let us set B 2 s Af 2 l BA 2 . W. van der Kulk proved that GA 2 is the amalgamated product of Af 2 and BA 2 over B 2 Žsee wvdKx.. The following theorem expresses this result: THEOREM ŽvdK.. Let f g GA 2 . Then there exist l g N, Ž a i .1 F i F lq1 g ŽAf 2 . lq1, and Žg i .1 F i F l g ŽBA 2 . l such that
¡f s a g a g ??? a g a ¢; i g 2, . . . , l 4 , a f BA .
~; i g 1, . . . , l 4 , g f Af 1
1
2
l
2
l
i
lq1 2
i
2
Moreo¨ er, if m g N, Ž a iX .1 F i F mq1 g ŽAf 2 . mq 1 , and Žg iX .1F i F m g ŽBA 2 . m are such that
¡f s a g a g ??? a g a ~; i g 1, . . . , m4 , g f Af ¢; i g 2, . . . , m4 , a f BA X 1
X 1
X 2
X 2
X X m m X i X i
X mq1 2
2
then l s m and there exist Ž bi .1 F i F l g ŽB 2 . l and Ž d i .1 F i F l g ŽB 2 . l such that
¡a s a b ~; i g 2, . . . , l 4 , a s d ¢a s d a X 1
X lq1
y1 1
1
X i
l
iy1 a i
by1 i
lq1
and ; i g 1, . . . , l 4 , g iX s bi g i dy1 i . Remarks. Ž1. If Ž a iX .1 F i F lq1 g ŽAf 2 . lq1 and Žg iX .1 F i F l g ŽBA 2 . l satisfy the latter relations, we easily check that X y1 y1 y1 a 1X g 1X a 2X g 2X ??? a lXg lX a lq1 s a 1 by1 1 b1g 1 d 1 d 1 a 2 b 2 b 2 g 2 d 2 d 2 y1 ??? d ly1 a l by1 l b l g l d l d l a lq1
s a 1 g 1 a 2 g 2 ??? a l g l a lq1
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JEAN-PHILIPPE FURTER
Ž2. If g g BA 2 , then g g B 2 if and only if deg g s 1. Furthermore, for all b , b 9 in Af 2 , we have degŽ bgb 9. s deg g . Thanks to the Theorem of W. van der Kulk and to the point Ž2. of the Remarks, we can define the multidegree dŽ f . of f and the length l Ž f . of f as follows: DEFINITION.
dŽ f . s Ždeg g 1 , . . . , deg g l . g D; l Ž f . s l.
Remark. The definition of the length l Ž f . of f that we use in this paper is different from the usual one which is equal to the length of f as an element of the amalgamated product Af 2 ) B 2 BA 2 . PROPOSITION 2. We ha¨ e deg f s degŽ dŽ f ... Proof. Let us write f s a 1 g 1 a 2 g 2 ??? a l g l a lq1 as in the last theorem. By the argument of the proof of Proposition 1, we obtain l
bideg Ž g 1 a 2 g 2 a 3 ??? g ly1 a l g l a lq1 . s
ž
Ł deg g j , js1
l
Ł deg g j js2
/
,
whence the result. The next result is obvious from the definitions. PROPOSITION 3.
dŽ fy1 . s dŽ f .y1 .
Remark. From the last two propositions, we get degŽ fy1 . s degŽ f ., which is nothing else than the n s 2 case of the following formula of Gabber: any automorphism f of AnC satisfies deg fy1 F Ždeg f . ny 1. PROPOSITION 4. If f and g are elements of GA 2 , then dŽ fg . F dŽ f . dŽ g . and we ha¨ e equality if and only if degŽ fg . s degŽ f .degŽ g .. Proof. Let us prove dŽ fg . F dŽ f . dŽ g .. If f g Af 2 , we have dŽ fg . s dŽ g . and dŽ f . s B so that dŽ fg . s dŽ f . dŽ g .. X X If f g BA 2 , let us write g s a 1X g 1X a 2X g 2X ??? a m gmX a mq1 as in the last X theorem. If a 1 f B 2 , it is clear that dŽ fg . s dŽ f . dŽ g .. So, let us suppose a 1X g B 2 , whence f a 1X g 1X g BA 2 . We now claim that: either f a 1X g 1X g B 2 , in which case dŽ fg . s Ždeg g 2X , . . . , deg gmX . so that dŽ fg . F dŽ g . F dŽ f . dŽ g ., either f a 1X g 1X f B 2 , in which case deg f a 1X g 1X F max deg f, deg g 1X 4 so that dŽ fg . s Ždeg f a 1X g 1X , deg g 2X , . . . , deg gmX . F dŽ f . dŽ g ..
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AUTOMORPHISMS OF THE AFFINE PLANE
Now that we have proven dŽ fg . F dŽ f . dŽ g . when f is either in Af 2 or in BA 2 , we can deduce the same inequality for any f by writing f s a 1 g 1 a 2 g 2 ??? a l g l a lq1 as in the last theorem. Indeed d Ž fg . s d Ž a 1 g 1 a 2 g 2 ??? a l g l a lq1 g . F d Ž a 1 . d Ž g 1 . d Ž a 2 . d Ž g 2 . ??? d Ž a l . d Ž g l . d Ž a lq1 . d Ž g . F d Ž g 1 . d Ž g 2 . ??? d Ž g l . d Ž g . F dŽ f . dŽ g . . If dŽ fg . s dŽ f . dŽ g ., it is clear that degŽ fg . s degŽ f .degŽ g . by Proposition 2. Conversely, if dŽ fg . / dŽ f . dŽ g ., then dŽ fg . - dŽ f . dŽ g . so that deg dŽ fg . - degŽ dŽ f . dŽ g .. and we obtain deg Ž fg . s deg Ž d Ž fg . . - deg Ž d Ž f . d Ž g . . s deg f deg g . For any d g D, we define sets Ud and Vd which will play a key role in Section 6. DEFINITION.
We set l
fd :
Af 2 =
Ł T2, d = GL 2
is1
i
Ž a , Ž g1 , . . . , gl . , b .
ª
GA 2, degŽ d.
¬
ag 1 sg 2 ??? sg l b
;
1 Ud s ImŽ f d .; cd s f d < Af 2=Ł ly ; and finally is 1 ŽT 2, d i _T 2, d iy 1 .=ŽUA 2, d l _UA 2, d ly 1 .=GL 2 Vd s ImŽ cd ..
In the present section, we will show that Ud and Vd have a very simple characterization in terms of the multidegree. PROPOSITION 5.
For any d g D, Vd s f g GA 2 such that dŽ f . s d4
Proof. For any d g D, let us set VdX s f g GA 2 such that dŽ f . s d4 . We have clearly Vd ; VdX . By Theorem 1, we have GA 2 s @ d g D Vd . The equality GA 2 s @ d g D VdX now shows that Vd s VdX . We will need the following lemma. LEMMA 5. Let f s Ž f 1 , f 2 . g GA 2 satisfy degŽ f 1 . ) degŽ f 2 . ) 1 and let a g Af 2 _ B 2 . Then there exists Žg , p . g ŽT2 _ GL 2 . = GA 2 such that f s ga p with deg p1 ) deg p 2 and deg p1 s deg f 2 .
614
JEAN-PHILIPPE FURTER
Remark. When a s s , Lemma 5 is the existence result of Lemma 2. Proof. Because a f B 2 , we have ay1 f B 2 so that ay1 s Ž aX q bY q c, dX q eY q f . where a, b, c, d, e, f are complex numbers with d / 0. On the one hand, we have
ay1 s aX q b y
ž
ž
ae d
/
Y q c, dX q f
/ž
Xq
e d
Y, Y
/
which is equivalent to
ž
Xy
e d
Y , Y s a aX q b y
/ ž
ž
ae d
/
Y q c, dX q f
/
and on the other hand, by Lemma 2, we know the existence of Žg 9, g . g ŽT2 _ GL 2 . = GA 2 such that g 9 f s g with deg g 1 - deg g 2 s deg f 2 . So, we have
ž
Xy
e d
Y , Y g 9 f s a ag 1 q b y
ž
/
ž
ae d
/
g 2 q c, dg 1 q f .
/
By setting
¡g s g 9
~
y1
¢p s ž ag
1
ž
Xq
e d
q by
ž
Y, Y ae d
/
/ g 2 q c, dg 1 q f
/
we have the desired result Žwe use the fact that b y aerd / 0, which is true because Ž aX q bY q c, dX q eY q f . is an automorphism.. PROPOSITION 6. Let f g GA 2 . Set dŽ f . s d and l Ž d . s l. Then, for any Ž a i .1 F i F ly1 g ŽGL 2 _ B 2 . ly1, there exist Ž a , b . g Af 2 = GL 2 and Žg i .1F i F l g ŽT2 . l such that f s ag 1 a 1 ??? g ly1 a ly1 g l b . Proof. If l s 0, there is nothing to do. So, let us suppose l G 1, so that deg f G 2. We know the existence of a g Af 2 such that g s ay1 f satisfies g Ž0, 0. s Ž0, 0. and degŽ g 1 . ) degŽ g 2 .. Now, the result comes from the fact that if Ž a i . i g N* g ŽGL 2 _ B 2 .N* and if g g GA 2 satisfies g Ž0, 0. s Ž0, 0. and degŽ g 1 . ) degŽ g 2 ., then there exist m g N*, Žg i .1 F i F m g ŽT2 _ GL 2 . m and b g GL 2 such that g s g 1 a 1 ??? gmy1 a my1 gm b . This result is proved in the same way as Lemma 4 Žof course using Lemma 5 instead of Lemma 2..
AUTOMORPHISMS OF THE AFFINE PLANE
PROPOSITION 7.
615
For any e in D, we ha¨ e Ue s @ d F e Vd .
Proof. By Proposition 4, it is clear that Ue ; @ d F e Vd . To prove the reverse inclusion, it is sufficient to prove that if e s Ž e1 , . . . , e m ., k g 1, . . . , m4 , and d s Ž e1 , . . . , ˆ e k , . . . , e m . then Vd ; Ue . Let f g Vd . By definition, there exist Ž a , b . g Af 2 = GL 2 and Žg i .1, . . . , ˆk, . . . , m g ŽT2 . my 1 such that f s ag 1 sg 2 s ??? sg ky1 sg kq1 s ??? sgm b , i.e., f s f d Ž a , Žg 1 , . . . , g ˆk , . . . , gm ., b .. If k s 1, we have f s f e Ž as , Ž Id, g 2 , . . . , gm . , b . and if k s m, we have f s f e Ž a , Ž g 1 , . . . , gmy1 , Id . , sb . so that in these two cases we have f g Ue . Therefore, we can suppose that k g 2, . . . , m y 14 . We set u s Ž X q Y, Y . g T2 , so that sus s Ž X, X q Y . g GL 2 _ B 2 . Define Ž a i .1 F i F my1 g ŽGL 2 _ B 2 . my 1 by a i s s if i / k and a k s sus . By Proposition 6, we deduce the existence of Ž a 9, b 9. g Af 2 = GL 2 and Žg iX .1, . . . , ˆk, . . . , m g ŽT2 . my 1 such that X X f s a 9g 1X sg 2X s ??? sg ky1 susg kq1 s ??? sgmX b 9,
i.e., X X f s f e Ž a 9, Ž g 1X , . . . , g ky1 , u , g kq1 , . . . , gmX . , b 9 . .
So f g Ue . COROLLARY 1. Ud s f g GA 2 , such that dŽ f . F d4 . COROLLARY 2.
d F e is equi¨ alent to Ud ; Ue .
5. AUTOMORPHISMS OF LENGTH AT MOST ONE The next proposition gives us some simple characterizations for an automorphism to be of length at most one: PROPOSITION 8. Let f s Ž f 1 , f 2 . g GA 2 . Write f s Ý`is0 Ž Pi , Q i ., where the Ž Pi , Q i . are homogeneous C-endomorphisms of A2C of degree i and set Ž P G 2 , Q G 2 . s ŽÝ`is2 Pi Ý`is2 Q i .. Then, the six following assertions are equi¨ -
616
JEAN-PHILIPPE FURTER
alent: Ži. P G 2 and Q G 2 are linearly dependent; Žii. l Ž f . F 1; Žiii. ; i, j G 2, JacŽ Pi , Q j . s 0; Živ. JacŽ P G 2 , Q G 2 . s 0; Žv. P G 2 and Q G 2 are algebraically dependent o¨ er C; Žvi. ' Ž u, ¨ . g Cw t x, ' R g Cw X, Y x such that Ž P G 2 , Q G 2 . s Ž uŽ R ., ¨ Ž R ... We use the following lemma: LEMMA 6. Let Ž u, ¨ . g Cw t x 2 _ Ž0, 0.4 and R g Cw X, Y x such that uŽ R .Ž Rr X . q ¨ Ž R .Ž Rr Y . s 0. Then there exist Ž a, b . g C 2 and w g Cw t x such that R s w Ž aX q bY .. Proof. We can obviously suppose that R is not a constant. Let us write
½
u s u 0 q u1 t q ??? qu m t m ¨ s ¨ 0 q ¨ 1 t q ??? q¨ m t m ,
where m is a positive integer and the u i , ¨ i are complex numbers. We can suppose that u and ¨ are relatively prime, so that u 0 / 0 or ¨ 0 / 0. The relation uŽ R .Ž Rr X . q ¨ Ž R .Ž Rr Y . s 0 can be written
u0
R X
q ¨0
R Y
s y u1 R q ??? u m R m
R X
y ¨ 1 R q ??? ¨ m R m
R Y
.
So we see that R divides u 0 Ž Rr X . q ¨ 0 Ž Rr Y .. Furthermore deg u 0 Ž Rr X . q ¨ 0 Ž Rr Y . - deg R so that u 0 Ž Rr X . q ¨ 0 Ž Rr Y . s 0. By making a linear change of coordinates in this last equality, we obtain the existence of w g Cw t x such that R s w Ž ¨ 0 X y u 0 Y .. Proof of Proposition 8. Ži. « Žii.. If Ži. is true, there exists a in Af 2 such that bidegŽ a f . s Ždeg f, 1. and Lemma 3 shows that l Ž a f . F 1, whence Žii.. Žii. « Žiii.. If Žii. is true, there exists a linear form l in Cw X, Y x such that for all i G 2, Pi and Q i are proportional to l i, whence Žiii.. Žiii. « Živ.. This is clear. The equivalence of Živ., Žv., and Žvi. comes from the equivalence of the following three assertions Žsee wNox., where k is any field of characteristic
AUTOMORPHISMS OF THE AFFINE PLANE
617
zero and g 1 , g 2 are any elements of k w X, Y x: Ž1. JacŽ g 1 , g 2 . s 0. Ž2. g 1 and g 2 are algebraically dependent over k. Ž3. ' Ž u, ¨ . g k w t x, ' R g k w X, Y x such that Ž g 1 , g 2 . s Ž uŽ R ., ¨ Ž R ... Let us now prove that Žvi. « Ži.. We can suppose that f s Ž X q uŽ R ., Y q ¨ Ž R .. where uŽ0. s ¨ Ž0. s RŽ0, 0. s 0, valŽ uŽ R .. G 2, and valŽ ¨ Ž R .. G 2. Since f is an automorphism, we necessarily have JacŽ f . s 1, which implies that u9Ž R .Ž Rr X . q ¨ 9Ž R .Ž Rr Y . s 0. By Lemma 6, if Ž u9, ¨ 9. / Ž0, 0., there exist Ž a, b . g C 2 and w g Cw t x such that R s w Ž aX q bY .. So, we can suppose that R s aX q bY. The condition u9Ž R .Ž Rr X . q ¨ 9Ž R .Ž Rr Y . s 0 implies that u9Ž R . and ¨ 9Ž R . are linearly dependent. It follows easily that uŽ R . and ¨ Ž R . are also linearly dependent.
6. ON THE IRREDUCIBLE COMPONENTS OF GA 2, n If n is a positive integer, let E 2, n be the vector space of C-endomorphisms f of A2C satisfying degŽ f . F n. Let us set GA 2, n s GA 2 l E 2, n , T2, n s T2 l E 2, n and UA 2, n s UA 2 l E 2, n . We set J2, n s f g E 2, n , such that Jac f s 14 and G2, n s GA 2, n l J2, n . E 2, n is an affine space and J2, n is a ŽZariski. closed subset of E 2, n . H. Bass, E. H. Connell, and D. Wright show in the paper wB-C-Wx that G2, n is also a closed subset of E 2, n . By slightly modifying their proof, we obtain that GA 2, n is a locally closed subset of E 2, n . In particular, GA 2, n is an algebraic variety. We have GA 2, n , C* = G2, n , via f s Ž f 1 , f 2 . ¬ ŽJacŽ f ., Ž f 1rJacŽ f ., f 2 .., so that the irreducible components of GA 2, n are in one to one correspondence with the irreducible components of G2, n . Our interest in the irreducible components of G2, n is motivated by the Jacobian Conjecture in dimension 2 and degree n asserting that an element of E 2, n is invertible if and only if its Jacobian is a nonzero constant. This conjecture is equivalent to the equality J2, n s G2, n . In Remark Ž1.7. of the last quoted paper, the authors note that to show this equality, it would suffice to show, if possible, that dim J2, n s dim G2, n and that the algebraic variety J2, n is irreducible Žbecause we have of course J2, n ; G2, n .. We answer here negatively to this hope. Indeed, on one hand J2, n s G2, n when n F 100, by T. T. Moh wMox and on the other hand we show that G2, n is reducible when n G 4 ŽProposition 11 below.. We see thus that the variety J2, n is reducible when 4 F n F 100.
618
JEAN-PHILIPPE FURTER
The results of Section 3 lead us to believe that we can get some insight into the decomposition in irreducible components of GA 2, n via purely combinatorial methods Žsee the Conjecture formulated in this section.. Let us recall that for any d g D, we introduced in Section 4 the subsets Ud and Vd of GA 2 which turned out to be
½
Ud s f g GA 2 such that d Ž f . F d 4 Vd s f g GA 2 such that d Ž f . s d 4
.
DEFINITION. We note that by the definition of Ud and Vd Žsee Section 4., their Zariski closures in GA 2, degŽ d. are equal. We set Wd s Uds Vd ŽZariski closure in GA 2, degŽ d. .. When l Ž d . s 1, the next proposition shows that Wd s Ud : PROPOSITION 9. If n is an integer greater than or equal to two, then UŽ n. is a closed sub¨ ariety of GA 2, n . Proof. UŽ n. s f g GA 2, n such that l Ž f . F 14 and condition Žiii. Žor Živ.. of Proposition 8 shows us that this condition is closed. LEMMA 7. < d < q 6.
If d g D, then Wd is an irreducible ¨ ariety of dimension
Proof. If d s B, the lemma is true. Let us now suppose that d g D*. We have ly1
Wd s cd Af 2 =
ž
Ł Ž T2, d _ T2, d y1 . = Ž UA 2, d _ UA 2, d y1 . = GL 2
is1
i
i
l
l
/
,
Ž . Ž . where Af 2 = Ł ly1 is1 T2, d i _ T2, d iy1 = UA 2, d p _ UA 2, d ly1 = GL 2 is an irreducible variety Žas a product of irreducible varieties. so that Wd is an irreducible variety. Let us set: 1 c 1, d s cd < A 1=Ł ly is 1 ŽT 2, d i _T 2, d iy 1 .=ŽUA 2, d l _UA 2, d ly 1 .=GL 2 group of translations of A2C ;
where A 1 is the
1 c 2, d s cd < A 2=Ł ly where A 2 is the set is 1 ŽT 2,d i _T 2,d iy 1 .=ŽUA 2, d l _UA 2,d ly 1 .=GL 2 of automorphisms of A2C of the shape Ž Y q a, X q cY q b . where a, b, c are elements of C;
V1, d s ImŽ c 1, d .; V2, d s ImŽ c 2, d ..
AUTOMORPHISMS OF THE AFFINE PLANE
619
We have Vd s V1, d " V2, d Žby Theorem 1.. c 1, d and c 2, d being injective morphisms Žagain by Theorem 1., we obtain dimŽV1, d. s < d < q 5 and dimŽV2, d. s < d < q 6. Finally, dim Wd s dim Vd s dim Ž V1, d j V2 , d . s max dimV1, d , dimV2, d 4 s < d < q 6. DEFINITION. For n g N*, we set Dn s d g D, degŽ d . F n4 and Cn denotes the set of maximal elements of Dn . For example, C1 s B 4 ; C2 s Ž 2 . 4 ; C3 s Ž 3 . 4 ; C4 s Ž 4 . , Ž 2, 2 . 4 ; C5 s Ž 5 . , Ž 2, 2 . 4 ; C6 s Ž 6 . , Ž 2, 3 . , Ž 3, 2 . 4 ; C 7 s Ž 7 . , Ž 2, 3 . , Ž 3, 2 . 4 ; C8 s Ž 8 . , Ž 2, 4 . , Ž 4, 2 . , Ž 2, 2, 2 . 4 ; C9 s Ž 9 . , Ž 2, 4 . , Ž 3, 3 . , Ž 4, 2 . , Ž 2, 2, 2 . 4 . The next lemma follows immediately from Corollary 2 to Proposition 7: LEMMA 8.
If n is a positi¨ e integer, then we ha¨ e GA 2, n s
D
dgC n
Ud s
@ Vd s D
dgDn
Wd .
dgC n
PROPOSITION 10. If n is a positi¨ e integer, then GA 2, n is an algebraic ¨ ariety of dimension 6 if n s 1 and of dimension n q 6 otherwise. Proof. By Lemmas 7 and 8, we have dim GA 2, n s max d g D n < d < q 6. If n s 1, D 1 s B4 and the proposition is clear.
620
JEAN-PHILIPPE FURTER
If n ) 1, ; d s Ž d1 , . . . , d l . g Dn , we have < d< s
l
l
Ý di F Ł di F n is1
is1
hence, max d g D n < d < is obtained for d s Ž n. where it is equal to n. Being now familiarized with the definitions, we give the announced conjecture: Conjecture. If n is a positive integer, then GA 2, n s D d g C n Wd is the decomposition in irreducible components of GA 2, n . This conjecture is equivalently formulated as 2
; Ž d, e . g Ž Cn . , d / e « Wd o We . We will prove this conjecture for n F 9 by using the next lemma: LEMMA 9. If n is an integer greater than or equal to two and if d g Cn with d / Ž n., then we ha¨ e Wd o WŽ n.
and
WŽ n. o Wd .
Proof. Ud contains an automorphism f such that l Ž f . G 2 and so Ud o UŽ n. . The equality UŽ n. s WŽ n. shows that Wd o WŽ n. . We deduce at once from this that WŽ n. o Wd . Otherwise the inequality dim Wd F dim WŽ n. together with the irreducibility of Wd , would imply WŽ n. s Wd , which is not the case. THEOREM 2.
If 1 F n F 9, then the Conjecture is true.
Proof. If n s 1, then GA 2, n s WŽB. . For 2 F n F 3, Cn s Ž n.4 and hence GA 2, n s WŽ n. is irreducible. Let us now suppose that 4 F n F 9. By Lemma 9, it is sufficient to prove that if d and e are two elements of Cn different from Ž n., then Wd o We . But the condition n F 9 implies the equality dim Wd s dim We Žby the list following the definition of Cn .. It is then sufficient to show that Wd / We . Now, Vd is a dense constructible subset of Wd and Ve is a dense constructible subset of We . Furthermore, Vd l Ve s B. This proves that we cannot have Wd s We . Even if we cannot yet compute the number of irreducible components of GA 2, n , we can however decide whether this variety is irreducible: PROPOSITION 11. If n is a positi¨ e integer, then the ¨ ariety GA 2, n is irreducible if and only if n F 3.
AUTOMORPHISMS OF THE AFFINE PLANE
621
Proof. It is sufficient to prove that GA 2, n is reducible for n G 4. Now, WŽ n. is an irreducible component of GA 2, n Žbecause it is an irreducible variety of the same dimension as GA 2, n .. The element Ž X q Y 2 , Y . s Ž X q Y 2 , Y . is in WŽ2, 2. ; GA 2, n but not in WŽ n. . COROLLARY OF PROPOSITION 11. reducible.
If 4 F n F 100, then the ¨ ariety J 2, n is
Remark. It is not true that the mapping d: GA 2 ª D f ¬ dŽ f . is lower-continuous, i.e., if d g D, it is not necessarily true that Ud is a closed subset of GA 2, degŽ d. . To check this, we consider the family of GA 2 given by the Nagata automorphism,
x : C ª GA 2 , 4 Z ¬ x Ž Z. , where x Ž Z . s Ž X y 2Y Ž XZ q Y 2 . y ZŽ XZ q Y 2 . 2 , Y q ZŽ XZ q Y 2 ... Then, if Z g C*, we have x Ž Z . s g 1 a 1 g 2 where
¡g
~a
¢g
1
s Ž X y Zy1 Y 2 , Y . g BA 2 _ B 2
1
s Ž X , Y q Z 2 X . g Af 2 _ B 2
2
s Ž X q Zy1 Y 2 , Y . g BA 2 _ B 2
so that dŽ x Ž Z .. s Ž2, 2.. However, dŽ x Ž0.. s Ž3. so that x Ž0. g UŽ2, 2. _ UŽ2, 2. . 7. MORE INSIGHT INTO THE IRREDUCIBLE COMPONENTS OF GA 2, 4 It results from Theorem 2 that WŽ4. o WŽ2, 2. . The next proposition gives us an explicit equation satisfied by all elements of WŽ2, 2. but not necessarily by elements of WŽ4. . PROPOSITION 12. Let f s Ž f 1 , f 2 . g WŽ2, 2. . Write f s Ý4is0 Ž Pi , Q i ., where Ž Pi , Q i . are C-endomorphisms of A2C which are homogeneous of degree i. Then, P2 P3 y 2 P1 P4 and Q2 Q3 y 2 Q1 Q4 are proportional polynomials.
622
JEAN-PHILIPPE FURTER
This is equi¨ alent to the algebraic condition Jac Ž P2 P3 y 2 P1 P4 , Q2 Q3 y 2 Q1 Q4 . s 0. Proof. Let f s Ž f 1 , f 2 . g VŽ2, 2. . Let us note that deg f s 4 by Proposition 5 so that deg f 1 s 4 or deg f 2 s 4. By Proposition 1, we have deg f 1 s 1, 2, or 4. Furthermore, it is impossible to have deg f 1 s 1 because otherwise we would have l Ž f . F 1 by Proposition 8. In the same way, we obtain deg f 2 s 2 or 4. Let us write f s Ý4is0 Ž Pi , Q i . where Ž Pi , Q i . are C-endomorphisms of 2 A C which are homogeneous of degree i. We begin by proving that there exists a complex number l such that 2 2 Ž Q2 Q3 y 2 Q1 Q4 . s lŽ P1 Q2 y P2 Q1 . Q4 .
If Q4 s 0, we have Q3 s 0 and it is sufficient to take l s 0. Otherwise, we have deg f 2 s 4 and we claim that there exists a complex number a such that degŽ f 1 y a f 2 . s 2. Indeed, if deg f 1 s 2, take a s 0. If deg f 1 s 4, by Remark 3 following Proposition 1, we know the existence of a such that degŽ f 1 y a f 2 . - 4. However, Ž f 1 y a f 2 , f 2 . g VŽ2, 2. so that degŽ f 1 y a f 2 . s 2 and the claim is proven. We set g s Ž g 1 , g 2 . s Ž f 1 y a f 2 , f 2 . s ŽÝ2is0 Ž Pi y a Q i ., Ý4is0 Q i . g GA 2 . We have bideg g s Ž2, 4.. Still by Proposition 1, we deduce the existence of a complex number b such that degŽ g 2 y b g 12 . - 4. Because Ž g 1 , g 2 y b g 12 . is an automorphism, we have in fact degŽ g 2 y b g 12 . F 2 whence
½
Q3 s 2 b Ž P1 y a Q1 . Ž P2 y a Q2 . Q4 s b Ž P2 y a Q2 .
2
which give us successively Q2 Q3 y 2 Q1 Q4 s 2 b Ž P2 y a Q2 . Ž Ž P1 y a Q1 . Q2 y Ž P2 y a Q2 . Q1 . Q2 Q3 y 2 Q1 Q4 s 2 b Ž P2 y a Q2 . Ž P1 Q2 y P2 Q1 . 2 2 2 Ž Q2 Q3 y 2 Q1 Q4 . s 4b 2 Ž P2 y a Q2 . Ž P1 Q2 y P2 Q1 . 2 2 Ž Q2 Q3 y 2 Q1 Q4 . s 4b Ž P1 Q2 y P2 Q1 . Q4 ,
whence the existence of l. By the same argument, there exists m such that 2 2 Ž P2 P3 y 2 P1 P4 . s m Ž P1 Q2 y P2 Q1 . P4 .
AUTOMORPHISMS OF THE AFFINE PLANE
623
Since P4 and Q4 are proportional polynomials, the same holds for Ž P2 P3 y 2 P1 P4 . 2 and Ž Q2 Q3 y 2 Q1 Q4 . 2 and hence for P2 P3 y 2 P1 P4 and Q2 Q3 y 2 Q1 Q 4 . We have thus proved the relation Jac Ž P2 P3 y 2 P1 P4 , Q2 Q3 y 2 Q1 Q4 . s 0 and this relation remains true if we only suppose that f g WŽ2, 2. s VŽ2 , 2.. Proposition 12 is enough to prove that g s Ž X q Y 4 , X q Y q Y 4 . g WŽ4. _ WŽ2 , 2. . Indeed, the homogeneous components Ž Pi , Q i . of degree i of g are Ž P1 , Q1 . s Ž X, X q Y ., Ž P2 , Q2 . s Ž0, 0., Ž P3 , Q3 . s Ž0, 0., and Ž P4 , Q4 . s Ž Y 4 , Y 4 . so that P2 P3 y 2 P1 P4 and Q2 Q3 y 2 Q1 Q4 are equal to y2 XY 4 and y2Ž X q Y .Y 4 which are not proportional polynomials.
ACKNOWLEDGMENTS I would like to thank Jose ´ Bertin for his advice and Michel Brion for his constant help. Note added in proof. The author afterward noticed that the analogous notion of ‘‘multidegree’’ has already been introduced in wF-Mx, where the authors also compute the dimension of the manifold of automorphisms with the prescribed multidegree. I am indebted to E. Ghys for pointing out this reference to me.
REFERENCES wB-C-Wx H. Bass, E. H. Connell, and D. Wright, The Jacobian conjecture: Reduction of degree and formal expansion of the inverse, Bull. Amer. Math. Soc. Ž1982., 287]330. wF-Mx S. Friedland and J. Milnor, Dynamical properties of plane polynomial automorphisms, Ergodic Th. & Dynam. Systems 9 Ž1989., 67]99. wvdKx W. van der Kulk, On polynomial rings in two variables, Nieuw. Arch. Wisk. (3) 1 Ž1953., 33]41. wMox T. T. Moh, On the global Jacobian conjecture and the configuration of roots, J. Reine Angew. Math. 340 Ž1983., 140]212. wNax M. Nagata, On automorphisms group of k w x, y x, in ‘‘Lecture Notes in Math.,’’ Vol. 5, Kyoto University, 1972. wNox A. Nowicki, On the Jacobian equation J Ž f, g . s 0 for polynomials in k w x, y x, Nagoya Math. J. 109 Ž1988., 151]157.
JOURNAL OF ALGEBRA ARTICLE NO.
On the Variety of Automorphisms of the Affine Plane Jean-Philippe Furter* U.M.P.A., E.N.S.L., 46 Allee ´ d’Italie, 69 007, Lyon, France Communicated by Peter Littelmann Received October 15, 1996
The main subject of our study is GA 2, n , the variety of automophisms of the affine plane of degree bounded by a positive integer n. After detailing some definitions and notations in Section 1, we give in Section 2 an algorithm to decide whether an endomorphism of the affine plane over an integral domain is a tame automorphism. Then, by applying this algorithm to the Nagata automorphism, we recover easily its known results. In Section 6, we compute the number of irreducible components of GA 2, n when n F 9 and we show that GA 2, n is reducible when n G 4. Our proofs are based on a precise decomposition theorem for automorphisms given in Section 3 and a characterization of length one automorphisms given in Section 5. Finally, in Section 7, we give some details on the case n s 4. Q 1997 Academic Press
1. DEFINITIONS AND NOTATIONS Let k be an integral domain. k w X, Y x is the polynomial algebra in the indeterminates X, Y Žendowed with the degree function and the valuation function at the origin, which are denoted by deg and val, and A2k s Spec k w X, Y x is the affine plane over k. A k-endomorphism f of A2k will be identified with its sequence f s Ž f 1 , f 2 . of coordinate functions f i g k w X, Y x Ž i s 1, 2.. If g is a k-endomorphism of A2k , we agree that fg will denote the k-endomorphism of A2k obtained by composition of f and g. We define: the degree of f by deg f s max deg f 1 , deg f 2 4 ; the bidegree of f by bidegŽ f . s Ždeg f 1 , deg f 2 .; the total degree of f by tdegŽ f . s deg f 1 q deg f 2 ; the Jacobian of f by JacŽ f . s Ž f 1 r X .Ž f 2 r Y . y Ž f 1 r Y .Ž f 2r X .. * E-mail address: [email protected]. 604 0021-8693r97 $25.00 Copyright Q 1997 by Academic Press All rights of reproduction in any form reserved.
AUTOMORPHISMS OF THE AFFINE PLANE
605
We introduce the following groups: GA 2 Ž k ., the group of all k-automorphisms of A2k ; Af 2 Ž k ., the subgroup of GA 2 Ž k . of affine automorphisms, i.e., automorphisms of the shape Ž aX q bY q c, dX q eY q f . where a, b, c, d, e, f are elements of k such that ae y bd is a unit of k; GL 2 Ž k ., the subgroup of Af 2 Ž k . of linear automorphisms; BA 2 Ž k ., the subgroup of GA 2 Ž k . of triangular or ‘‘de Jonquieres’’ ` automorphisms, i.e., automorphisms of the shape Ž aX q P Ž Y ., bY q c . where a, b are units of k, c is an element of k, and P Ž Y . is an element of k w Y x; TA 2 Ž k ., the subgroup of GA 2 Ž k . of tame automorphisms, i.e., the subgroup of GA 2 Ž k . generated by Af 2 Ž k . and BA 2 Ž k ..
2. AN ALGORITHM TO DECIDE WHETHER AN ENDOMORPHISM IS A TAME AUTOMORPHISM The following proposition is the key of the announced algorithm: if f g TA 2 Ž k . satisfies tdeg f G 3, it ensures the existence of a g GA 2 Ž k . of a very special shape such that tdegŽ ay1 f . - tdeg f. PROPOSITION 1. Let f s Ž f 1 , f 2 . g TA 2 Ž k . with bidegŽ f . s Ž d1 , d 2 .. Denote by g i the homogeneous component of degree d i of f i for i s 1, 2. Then d1 < d 2 or d 2 < d1. Moreo¨ er: Ži. If d1 - d 2 , then there exists l g k such that g 2 s l g 1d 2 r d1 ; Žii. If d 2 - d1 , then there exists l g k such that g 1 s l g 2d1 r d 2 ; Žiii. If d1 s d 2 ) 1, then there exists Ž a , h. g Af 2 Ž k . = TA 2 Ž k . such that f s a h and d1 s deg h1 ) deg h 2 where h s Ž h1 , h 2 .. Remarks. Ž1. Let us assume that degŽ f . ) 1. We set a s Ž X, Y q l X d 2 r d1 . in case Ži., a s Ž X q lY d1 r d 2 , Y . in case Žii., and if we are in case Žiii., we keep a as explained there. Then, it is clear that we have tdegŽ ay1 f . - tdeg f. Ž2. If d1 s d 2 , in general, there does not exist l g k such that g 2 s l g 1 or g 1 s l g 2 , as shown by the following example where k s Cw Z x: f s Ž Ž 1 y Z . X q ZY q Ž 1 y Z . Y 2 , yZX q Ž 1 q Z . Y y ZY 2 . .
606
JEAN-PHILIPPE FURTER
It can be checked that f s ag where
½
a s Ž Ž 1 y Z . X q ZY , yZX q Ž 1 q Z . Y . g Af 2 Ž k . g g Ž X q Y 2 , Y . g BA 2 Ž k .
so that f g TA 2 Ž k .. However, bideg f s Ž2, 2., the homogeneous component of degree 2 of f is ŽŽ1 y Z .Y 2 , yZY 2 . and there exists no l g Cw Z x such that Ž1 y Z .Y 2 s lŽyZY 2 . or yZY 2 s lŽ1 y Z .Y 2 . Ž3. Nevertheless, still in the case d1 s d 2 , if k s C and deg f ) 1, the equality Jac f g C* implies that JacŽ g 1 , g 2 . s 0, which is well known to imply that g 1 and g 2 are proportional polynomials. The following group-theoretical lemma will be used in the proof of Proposition 1. LEMMA 1. If a group G is generated by two subgroups H and K, then each element g of G can be written g s h1 k 1 h 2 k 2 ??? h l k l h lq1 , where l is a non-negati¨ e integer and the h i Ž resp. k i . are elements of H Ž resp. K . such that h i f K for 2 F i F l Ž resp. k l f H for 1 F i F l .. Proof. If g is an element of G, we have g s h1 k 1 h 2 k 2 ??? h l k l h lq1 , where l is a non-negative integer and the h i Žresp. k i . are elements of H Žresp. K .. If the condition Ž h i f K for 2 F i F l and k i f H for 1 F i F l . is not satisfied, then we can obtain an expression for g of the same shape but where l is replaced by l y 1. Indeed, if, for example, h i 0 g K Ž2 F i 0 F l ., then we have g s h1 k 1 h 2 k 2 ??? h i 0y1 Ž k i 0y1 h i 0 k i 0 . h i 0q1 ??? h l k l h lq1 , where k i 0y1 h i 0 k i 0 g K. After a finite number of such reductions, we will necessarily obtain the desired expression for g. Proof of Proposition 1. By Lemma 1 applied with G s TA 2 Ž k ., H s Af 2 Ž k ., K s BA 2 Ž k ., and g s f, we obtain the existence of a non-negative integer l and of Ž a , b . g ŽAf 2 Ž k .. 2 , Ž a i .1 F i F ly1 g ŽAf 2 Ž k . _ BA 2 Ž k .. ly1, Žg i .1F i F l g ŽBA 2 Ž k . _ Af 2 Ž k .. l such that f s ag 1 a 1 g 2 ??? a ly2 g ly1 a ly1 g l b . We show by a decreasing induction on i, beginning with i s l and finishing with i s 1 that l
bideg Ž g i a i ??? g ly1 a ly1 g l b . s
ž
l
Ł deg g j , Ł jsi
jsiq1
/
deg g j .
607
AUTOMORPHISMS OF THE AFFINE PLANE
The conclusion is clear for i s l. Let us suppose it is true for a given i with 2 F i F l. Let us write
a iy1 s Ž a iy1 X q biy1Y q c iy1 , d iy1 X q e iy1Y q f iy1 . . The relations a iy1 f BA 2 Ž k . and g iy1 f Af 2 Ž k . imply that d iy1 / 0 and degŽg iy1 . G 2. We now obtain bidegŽ a iy1 g i a i ??? g ly1 a ly1 g l b . s Ž pi , Ł ljs i deg g j . with pi F Ł ljs i deg g j and bidegŽ g iy 1 a iy 1 ??? g ly1 a py1 g l b . s ŽŁ ljsiy1 deg g j , Ł ljsi deg g j .. It is now proven that l
bideg Ž g 1 a 1 ??? g ly1 a ly1 g l b . s
ž
l
Ł deg g j , Ł deg g j js1
js2
/
.
Let us write a s Ž aX q bY q c, dX q eY q f .. To conclude, we just have to check that we are in case Ži. or Žii. or Žiii. according to whether a s 0 or d s 0 or ad / 0. We deduce at once from this last proposition an algorithm to decide if a k-endomorphism of A2k belongs to TA 2 Ž k .: ALGORITHM. Ž1. Enter a k-endomorphism f of A2k . Ž2. Let Ž d1 , d 2 . s bidegŽ f 1 , f 2 .. Ž3. If d1 s d 2 s 1, then goto 8. Ž4. If d1 / d 2 , then goto 6. Ž5. If there exists a g Af 2 Ž k . such that tdegŽ a f . - tdegŽ f ., then replace f by a f and goto 2, else STOP: f f TA 2 Ž k .. Ž6. If d 2 - d1 , then replace f s Ž f 1 , f 2 . by Ž f 2 , f 1 .. Ž7. If Ž d1 < d 2 and there exists l g k such that g 2 s l g 1d 2 r d1 ., then replace f by Ž X, Y y l X d 2 r d1 . f and goto 2, else STOP: f f TA 2 Ž k .. Ž8. If JacŽ f . is a unit of k, then STOP: f g TA 2 Ž k ., else STOP: f f TA 2 Ž k .. Application. Let us consider the Nagata automorphism Žcf. wNax. 2
f s X y 2Y Ž XZ q Y 2 . y Z Ž XZ q Y 2 . , Y q Z Ž XZ q Y 2 .
ž
/
g GA 2 Ž k . for k s Cw Z x and k s Cw Z, Zy1 x. By expanding f , we get
f s Ž X y 2 ZXY y Z 3 X 2 y 2Y 3 y 2 Z 2 XY 2 y ZY 4 , Y q Z 2 X q ZY 2 .
608
JEAN-PHILIPPE FURTER
so that the homogeneous components of highest degree of f are yZY 4 and ZY 2 . Ž1. If k s Cw Z x, Ž ZY 2 . 2 does not divide yZY 4 and so f f TA 2 ŽCw Z x.. Ž2. If k s Cw Z, Zy1 x, the equation yZY 4 s yZy1 Ž ZY 2 . 2 allows us to obtain
f s Ž X y Zy1 Y 2 , Y . Ž X q Zy1 Y 2 , Y q Z 2 X q ZY 2 . and we easily see that
Ž X q Zy1 Y 2 , Y q Z 2 X q ZY 2 . s Ž X , Y q Z 2 X . Ž X q Zy1 Y 2 , Y . whence f s Ž X y Zy1 Y 2 , Y .Ž X, Y q Z 2 X .Ž X q Zy1 Y 2 , Y . and f g TA 2 ŽCw Z, Zy1 x.. 3. A PRECISE DECOMPOSITION THEOREM FOR AUTOMORPHISMS OF THE AFFINE PLANE From now on, we set k s C and we note GA 2 instead of GA 2 ŽC., Af 2 instead of Af 2 ŽC., etc. Let us recall that by a famous result of H.W.E. Jung and W. van der Kulk wvdKx, we have GA 2 s TA 2 . We give here a precise decomposition theorem for elements of GA 2 Žsee Theorem 1.. This result allows us Žsee Section 6, Proposition 10. to compute the dimension of the algebraic variety GA 2, n . Before stating the theorem, we need the following definitions: T2 is the subgroup of GA 2 of automorphisms of the shape Ž X q P Ž Y ., Y . where P Ž Y . is an element of Cw Y x satisfying valŽ P . G 1; UA 2 is the subgroup of T2 of elements of the shape Ž X q P Ž Y ., Y . where P Ž Y . is any element of Cw Y x satisfying valŽ P . G 2; for all Ž a, b . g C 2 , we set t Ž a, b. s Ž X q a, Y q b . g GA 2 and for all c g C, we set sc s Ž Y q cX, X . g GA 2 . We note s s s 0 s Ž Y, X .. THEOREM 1 ŽPrecise Decomposition Theorem.. _ Af 2 , then: Ži.
Let f s Ž f 1 , f 2 . g GA 2
if degŽ f 1 . ) degŽ f 2 ., '! Ž a, b . g C 2 , '! l g N, '! Ž g i . 1FiFl g Ž T2 _ GL 2 . = Ž UA 2 _ GL 2 . , '! b g GL 2
ly1
609
AUTOMORPHISMS OF THE AFFINE PLANE
such that f s t Ž a, b. g 1 sg 2 s ??? sg ly1 sg l b ; Žii. if degŽ f 1 . F degŽ f 2 ., '! Ž a, b, c . g C 3 , '! l g N, '! Ž g i . 1FiFl g Ž T2 _ GL 2 .
ly1
= Ž UA 2 _ GL 2 . , '! b g GL 2 such that f s t Ž a, b. sc g 1 sg 2 s ??? sg ly1 sg l b . To prove this theorem, we will use the following three decomposition lemmas: LEMMA 2. Let f s Ž f 1 , f 2 . g GA 2 satisfy degŽ f 1 . ) degŽ f 2 . ) 1. Then '! Žg , p . g ŽT2 _ GL 2 . = GA 2 such that f s g p and p s Ž p1 , p 2 . satisfies degŽ p1 . - degŽ p 2 . s degŽ f 2 .. Proof. Existence. We begin by proving that for each automorphism f s Ž f 1 , f 2 . g GA 2 satisfying degŽ f 1 . G degŽ f 2 . ) 1 there exists Žg , p . g T2 = GA 2 such that f s g p and p s Ž p1 , p 2 . satisfies degŽ p1 . - degŽ f 1 .. For 1 F i F 2, let d i s deg f i and let g i denote the homogeneous component of degree d i of f i . From Proposition 1 and Remark 3 following it, we deduce that d 2 < d1 and that there exists l g C such that g 1 s l g 2d1 r d 2 . We can now just write f s Ž X q lY d1 r d 2 , Y . p where p g GA 2 and it is easy to check that p s Ž p1 , p 2 . satisfies p 2 s f 2 and degŽ p1 . - degŽ f 1 .. From this fact, we deduce by an immediate induction on deg f 1 that for each automorphism f s Ž f 1 , f 2 . g GA 2 satisfying degŽ f 1 . G degŽ f 2 . ) 1, there exists Žg , p . g T2 = GA 2 such that f s g p and p s Ž p1 , p 2 . satisfies degŽ p1 . - degŽ p 2 . s deg f 2 . This terminates the proof of the existence. Unicity. Let Žg , p . and Ž d , q . be as in Lemma 2 such that g p s d q. We can write g s Ž X q RŽ Y ., Y ., d s Ž X q S Ž Y ., Y . where RŽ Y . and SŽ Y . are elements of Cw Y x such that valŽ R . G 1 and valŽ S . G 1. Moreover, p s Ž p1 , p 2 . and q s Ž q1 , q2 . satisfy deg p1 - deg p 2 and deg q1 - deg q2 . We have Ž p1 q RŽ p 2 ., p 2 . s Ž q1 q SŽ q2 ., q2 . from which we deduce firstly that p 2 s q2 and secondly that p1 y q1 s S Ž p 2 . y RŽ p 2 .. In the last equality, the left hand term is of degree strictly less than deg p 2 which implies that S y R s 0 and finally g s d and p s q. LEMMA 3. Let f s Ž f 1 , f 2 . g GA 2 satisfy f Ž0, 0. s Ž0, 0. and degŽ f 1 . ) degŽ f 2 . s 1. Then '! Žg , b . g ŽUA 2 _ GL 2 . = GL 2 such that f s gb . Proof. Existence. Let f s Ž f 1 , f 2 . g GA 2 satisfy f Ž0, 0. s Ž0, 0. and degŽ f 1 . ) degŽ f 2 . s 1. We define b as the linear part of f. Then, if b s Ž b 1 , b 2 ., we have Cw b 1 , b 2 x s Cw X, Y x and so f can be written as Ž b 1 q g 1Ž b 1 , b 2 ., b 2 . where g 1 g Cw X, Y x and valŽ g 1 . G 2. Let us set g s Ž X q g 1 , Y .. On one hand, we have f s gb and on the other hand, g being an automorphism, we have necessarily g 1 g Cw Y x and finally g g UA 2 _ GL 2 .
610
JEAN-PHILIPPE FURTER
Unicity. If f s gb where Žg , b . g ŽUA 2 _ GL 2 . = GL 2 , then b is necessarily equal to the linear part of f and this gives us the uniqueness. LEMMA 4. Let f s Ž f 1 , f 2 . g GA 2 satisfy f Ž0, 0. s Ž0, 0. and degŽ f 1 . ) degŽ f 2 .. Then '! l g N*, '! Ž g i . 1FiFl g Ž T2 _ GL 2 .
ly1
= Ž UA 2 _ GL 2 . , '! b g GL 2
such that f s g 1 sg 2 s ??? sg ly1 sg l b . Proof. Existence. By induction on degŽ f 2 ., if degŽ f 2 . s 1, we are done by Lemma 3. If degŽ f 2 . ) 1, by Lemma 2, f can be written f s gs f˜ where Žg , f˜. g ŽT2 _ GL 2 . = GA 2 and f˜s Ž f˜1 , f˜2 . with degŽ f˜2 . - degŽ f˜1 . s degŽ f 2 .. The equality f Ž0, 0. s Ž0, 0. implies the equality f˜Ž0, 0. s Ž0, 0.. We have the ˜ result by applying the induction hypothesis to f. Uniqueness. This follows from the uniqueness in Lemmas 2 and 3 by noting that if l G 2, Žg i .1 F i F l g ŽT2 _ GL 2 . ly1 = ŽUA 2 _ GL 2 . and b g GL 2 , then the automorphism p s Ž p1 , p 2 . s sg 2 s ??? sg l b satisfies deg p1 - deg p 2 . Proof of Theorem 1. Let f g GA 2 _ Af 2 . If deg f 1 ) deg f 2 , the following applies. Existence. If we set Ž a, b . s f Ž0, 0., then the automorphism f˜s Ž f˜1 , f˜2 . ˜Ž . Ž . Ž ˜. Ž ˜. s ty1 Ž a, b. f satisfies f 0, 0 s 0, 0 and deg f 1 ) deg f 2 . We obtain the ˜ existence by Lemma 4 applied to f. Unicity. If f is written as in Ži. of Theorem 1, it is clear that we must have Ž a, b . s f Ž0, 0. and the unicity now follows from the unicity of Lemma 4 applied to f˜s ty1 Ž a, b. f. If deg f 1 F deg f 2 , we can see that there exist unique complex numbers ˜ ˆ ˆ a, b, and c such that fˆs Ž fˆ1 , fˆ2 . s scy1ty1 Ž a, b. f satisfies deg f 1 ) deg f 2 and fˆŽ0, 0. s Ž0, 0.. The result is now a consequence of case Ži. of the ˆ theorem applied to f.
4. ON THE MULTIDEGREE OF AN AUTOMORPHISM OF THE AFFINE PLANE We introduce here a notion of multidegree of automorphisms of the affine plane and obtain some basic properties. The multidegree of an automorphism is an element of the following set D: DEFINITIONS. D denotes the set of all finite sequences of integers greater than or equal to 2 and D* denotes the set D with the vacuum sequence omitted.
AUTOMORPHISMS OF THE AFFINE PLANE
611
We define the following partial order F on D: if d s Ž d1 , . . . , d l . g D and if e s Ž e1 , . . . , e m . g D, then we say that d F e if l F m and if there exist 1 F i1 - i 2 - ??? - i l F m such that ; j g 1, . . . , l 4 , d j F e i j . Also, we will denote the concatenation of d and e by de s Ž d1 , . . . , d l , e1 , . . . , e m .. We set: l Ž d . s l; < d < s d1 q d 2 q ??? qd l ; degŽ d . s d1 = d 2 = ??? = d l ; dy1 s Ž d l , . . . , d1 .. Let us set B 2 s Af 2 l BA 2 . W. van der Kulk proved that GA 2 is the amalgamated product of Af 2 and BA 2 over B 2 Žsee wvdKx.. The following theorem expresses this result: THEOREM ŽvdK.. Let f g GA 2 . Then there exist l g N, Ž a i .1 F i F lq1 g ŽAf 2 . lq1, and Žg i .1 F i F l g ŽBA 2 . l such that
¡f s a g a g ??? a g a ¢; i g 2, . . . , l 4 , a f BA .
~; i g 1, . . . , l 4 , g f Af 1
1
2
l
2
l
i
lq1 2
i
2
Moreo¨ er, if m g N, Ž a iX .1 F i F mq1 g ŽAf 2 . mq 1 , and Žg iX .1F i F m g ŽBA 2 . m are such that
¡f s a g a g ??? a g a ~; i g 1, . . . , m4 , g f Af ¢; i g 2, . . . , m4 , a f BA X 1
X 1
X 2
X 2
X X m m X i X i
X mq1 2
2
then l s m and there exist Ž bi .1 F i F l g ŽB 2 . l and Ž d i .1 F i F l g ŽB 2 . l such that
¡a s a b ~; i g 2, . . . , l 4 , a s d ¢a s d a X 1
X lq1
y1 1
1
X i
l
iy1 a i
by1 i
lq1
and ; i g 1, . . . , l 4 , g iX s bi g i dy1 i . Remarks. Ž1. If Ž a iX .1 F i F lq1 g ŽAf 2 . lq1 and Žg iX .1 F i F l g ŽBA 2 . l satisfy the latter relations, we easily check that X y1 y1 y1 a 1X g 1X a 2X g 2X ??? a lXg lX a lq1 s a 1 by1 1 b1g 1 d 1 d 1 a 2 b 2 b 2 g 2 d 2 d 2 y1 ??? d ly1 a l by1 l b l g l d l d l a lq1
s a 1 g 1 a 2 g 2 ??? a l g l a lq1
612
JEAN-PHILIPPE FURTER
Ž2. If g g BA 2 , then g g B 2 if and only if deg g s 1. Furthermore, for all b , b 9 in Af 2 , we have degŽ bgb 9. s deg g . Thanks to the Theorem of W. van der Kulk and to the point Ž2. of the Remarks, we can define the multidegree dŽ f . of f and the length l Ž f . of f as follows: DEFINITION.
dŽ f . s Ždeg g 1 , . . . , deg g l . g D; l Ž f . s l.
Remark. The definition of the length l Ž f . of f that we use in this paper is different from the usual one which is equal to the length of f as an element of the amalgamated product Af 2 ) B 2 BA 2 . PROPOSITION 2. We ha¨ e deg f s degŽ dŽ f ... Proof. Let us write f s a 1 g 1 a 2 g 2 ??? a l g l a lq1 as in the last theorem. By the argument of the proof of Proposition 1, we obtain l
bideg Ž g 1 a 2 g 2 a 3 ??? g ly1 a l g l a lq1 . s
ž
Ł deg g j , js1
l
Ł deg g j js2
/
,
whence the result. The next result is obvious from the definitions. PROPOSITION 3.
dŽ fy1 . s dŽ f .y1 .
Remark. From the last two propositions, we get degŽ fy1 . s degŽ f ., which is nothing else than the n s 2 case of the following formula of Gabber: any automorphism f of AnC satisfies deg fy1 F Ždeg f . ny 1. PROPOSITION 4. If f and g are elements of GA 2 , then dŽ fg . F dŽ f . dŽ g . and we ha¨ e equality if and only if degŽ fg . s degŽ f .degŽ g .. Proof. Let us prove dŽ fg . F dŽ f . dŽ g .. If f g Af 2 , we have dŽ fg . s dŽ g . and dŽ f . s B so that dŽ fg . s dŽ f . dŽ g .. X X If f g BA 2 , let us write g s a 1X g 1X a 2X g 2X ??? a m gmX a mq1 as in the last X theorem. If a 1 f B 2 , it is clear that dŽ fg . s dŽ f . dŽ g .. So, let us suppose a 1X g B 2 , whence f a 1X g 1X g BA 2 . We now claim that: either f a 1X g 1X g B 2 , in which case dŽ fg . s Ždeg g 2X , . . . , deg gmX . so that dŽ fg . F dŽ g . F dŽ f . dŽ g ., either f a 1X g 1X f B 2 , in which case deg f a 1X g 1X F max deg f, deg g 1X 4 so that dŽ fg . s Ždeg f a 1X g 1X , deg g 2X , . . . , deg gmX . F dŽ f . dŽ g ..
613
AUTOMORPHISMS OF THE AFFINE PLANE
Now that we have proven dŽ fg . F dŽ f . dŽ g . when f is either in Af 2 or in BA 2 , we can deduce the same inequality for any f by writing f s a 1 g 1 a 2 g 2 ??? a l g l a lq1 as in the last theorem. Indeed d Ž fg . s d Ž a 1 g 1 a 2 g 2 ??? a l g l a lq1 g . F d Ž a 1 . d Ž g 1 . d Ž a 2 . d Ž g 2 . ??? d Ž a l . d Ž g l . d Ž a lq1 . d Ž g . F d Ž g 1 . d Ž g 2 . ??? d Ž g l . d Ž g . F dŽ f . dŽ g . . If dŽ fg . s dŽ f . dŽ g ., it is clear that degŽ fg . s degŽ f .degŽ g . by Proposition 2. Conversely, if dŽ fg . / dŽ f . dŽ g ., then dŽ fg . - dŽ f . dŽ g . so that deg dŽ fg . - degŽ dŽ f . dŽ g .. and we obtain deg Ž fg . s deg Ž d Ž fg . . - deg Ž d Ž f . d Ž g . . s deg f deg g . For any d g D, we define sets Ud and Vd which will play a key role in Section 6. DEFINITION.
We set l
fd :
Af 2 =
Ł T2, d = GL 2
is1
i
Ž a , Ž g1 , . . . , gl . , b .
ª
GA 2, degŽ d.
¬
ag 1 sg 2 ??? sg l b
;
1 Ud s ImŽ f d .; cd s f d < Af 2=Ł ly ; and finally is 1 ŽT 2, d i _T 2, d iy 1 .=ŽUA 2, d l _UA 2, d ly 1 .=GL 2 Vd s ImŽ cd ..
In the present section, we will show that Ud and Vd have a very simple characterization in terms of the multidegree. PROPOSITION 5.
For any d g D, Vd s f g GA 2 such that dŽ f . s d4
Proof. For any d g D, let us set VdX s f g GA 2 such that dŽ f . s d4 . We have clearly Vd ; VdX . By Theorem 1, we have GA 2 s @ d g D Vd . The equality GA 2 s @ d g D VdX now shows that Vd s VdX . We will need the following lemma. LEMMA 5. Let f s Ž f 1 , f 2 . g GA 2 satisfy degŽ f 1 . ) degŽ f 2 . ) 1 and let a g Af 2 _ B 2 . Then there exists Žg , p . g ŽT2 _ GL 2 . = GA 2 such that f s ga p with deg p1 ) deg p 2 and deg p1 s deg f 2 .
614
JEAN-PHILIPPE FURTER
Remark. When a s s , Lemma 5 is the existence result of Lemma 2. Proof. Because a f B 2 , we have ay1 f B 2 so that ay1 s Ž aX q bY q c, dX q eY q f . where a, b, c, d, e, f are complex numbers with d / 0. On the one hand, we have
ay1 s aX q b y
ž
ž
ae d
/
Y q c, dX q f
/ž
Xq
e d
Y, Y
/
which is equivalent to
ž
Xy
e d
Y , Y s a aX q b y
/ ž
ž
ae d
/
Y q c, dX q f
/
and on the other hand, by Lemma 2, we know the existence of Žg 9, g . g ŽT2 _ GL 2 . = GA 2 such that g 9 f s g with deg g 1 - deg g 2 s deg f 2 . So, we have
ž
Xy
e d
Y , Y g 9 f s a ag 1 q b y
ž
/
ž
ae d
/
g 2 q c, dg 1 q f .
/
By setting
¡g s g 9
~
y1
¢p s ž ag
1
ž
Xq
e d
q by
ž
Y, Y ae d
/
/ g 2 q c, dg 1 q f
/
we have the desired result Žwe use the fact that b y aerd / 0, which is true because Ž aX q bY q c, dX q eY q f . is an automorphism.. PROPOSITION 6. Let f g GA 2 . Set dŽ f . s d and l Ž d . s l. Then, for any Ž a i .1 F i F ly1 g ŽGL 2 _ B 2 . ly1, there exist Ž a , b . g Af 2 = GL 2 and Žg i .1F i F l g ŽT2 . l such that f s ag 1 a 1 ??? g ly1 a ly1 g l b . Proof. If l s 0, there is nothing to do. So, let us suppose l G 1, so that deg f G 2. We know the existence of a g Af 2 such that g s ay1 f satisfies g Ž0, 0. s Ž0, 0. and degŽ g 1 . ) degŽ g 2 .. Now, the result comes from the fact that if Ž a i . i g N* g ŽGL 2 _ B 2 .N* and if g g GA 2 satisfies g Ž0, 0. s Ž0, 0. and degŽ g 1 . ) degŽ g 2 ., then there exist m g N*, Žg i .1 F i F m g ŽT2 _ GL 2 . m and b g GL 2 such that g s g 1 a 1 ??? gmy1 a my1 gm b . This result is proved in the same way as Lemma 4 Žof course using Lemma 5 instead of Lemma 2..
AUTOMORPHISMS OF THE AFFINE PLANE
PROPOSITION 7.
615
For any e in D, we ha¨ e Ue s @ d F e Vd .
Proof. By Proposition 4, it is clear that Ue ; @ d F e Vd . To prove the reverse inclusion, it is sufficient to prove that if e s Ž e1 , . . . , e m ., k g 1, . . . , m4 , and d s Ž e1 , . . . , ˆ e k , . . . , e m . then Vd ; Ue . Let f g Vd . By definition, there exist Ž a , b . g Af 2 = GL 2 and Žg i .1, . . . , ˆk, . . . , m g ŽT2 . my 1 such that f s ag 1 sg 2 s ??? sg ky1 sg kq1 s ??? sgm b , i.e., f s f d Ž a , Žg 1 , . . . , g ˆk , . . . , gm ., b .. If k s 1, we have f s f e Ž as , Ž Id, g 2 , . . . , gm . , b . and if k s m, we have f s f e Ž a , Ž g 1 , . . . , gmy1 , Id . , sb . so that in these two cases we have f g Ue . Therefore, we can suppose that k g 2, . . . , m y 14 . We set u s Ž X q Y, Y . g T2 , so that sus s Ž X, X q Y . g GL 2 _ B 2 . Define Ž a i .1 F i F my1 g ŽGL 2 _ B 2 . my 1 by a i s s if i / k and a k s sus . By Proposition 6, we deduce the existence of Ž a 9, b 9. g Af 2 = GL 2 and Žg iX .1, . . . , ˆk, . . . , m g ŽT2 . my 1 such that X X f s a 9g 1X sg 2X s ??? sg ky1 susg kq1 s ??? sgmX b 9,
i.e., X X f s f e Ž a 9, Ž g 1X , . . . , g ky1 , u , g kq1 , . . . , gmX . , b 9 . .
So f g Ue . COROLLARY 1. Ud s f g GA 2 , such that dŽ f . F d4 . COROLLARY 2.
d F e is equi¨ alent to Ud ; Ue .
5. AUTOMORPHISMS OF LENGTH AT MOST ONE The next proposition gives us some simple characterizations for an automorphism to be of length at most one: PROPOSITION 8. Let f s Ž f 1 , f 2 . g GA 2 . Write f s Ý`is0 Ž Pi , Q i ., where the Ž Pi , Q i . are homogeneous C-endomorphisms of A2C of degree i and set Ž P G 2 , Q G 2 . s ŽÝ`is2 Pi Ý`is2 Q i .. Then, the six following assertions are equi¨ -
616
JEAN-PHILIPPE FURTER
alent: Ži. P G 2 and Q G 2 are linearly dependent; Žii. l Ž f . F 1; Žiii. ; i, j G 2, JacŽ Pi , Q j . s 0; Živ. JacŽ P G 2 , Q G 2 . s 0; Žv. P G 2 and Q G 2 are algebraically dependent o¨ er C; Žvi. ' Ž u, ¨ . g Cw t x, ' R g Cw X, Y x such that Ž P G 2 , Q G 2 . s Ž uŽ R ., ¨ Ž R ... We use the following lemma: LEMMA 6. Let Ž u, ¨ . g Cw t x 2 _ Ž0, 0.4 and R g Cw X, Y x such that uŽ R .Ž Rr X . q ¨ Ž R .Ž Rr Y . s 0. Then there exist Ž a, b . g C 2 and w g Cw t x such that R s w Ž aX q bY .. Proof. We can obviously suppose that R is not a constant. Let us write
½
u s u 0 q u1 t q ??? qu m t m ¨ s ¨ 0 q ¨ 1 t q ??? q¨ m t m ,
where m is a positive integer and the u i , ¨ i are complex numbers. We can suppose that u and ¨ are relatively prime, so that u 0 / 0 or ¨ 0 / 0. The relation uŽ R .Ž Rr X . q ¨ Ž R .Ž Rr Y . s 0 can be written
u0
R X
q ¨0
R Y
s y u1 R q ??? u m R m
R X
y ¨ 1 R q ??? ¨ m R m
R Y
.
So we see that R divides u 0 Ž Rr X . q ¨ 0 Ž Rr Y .. Furthermore deg u 0 Ž Rr X . q ¨ 0 Ž Rr Y . - deg R so that u 0 Ž Rr X . q ¨ 0 Ž Rr Y . s 0. By making a linear change of coordinates in this last equality, we obtain the existence of w g Cw t x such that R s w Ž ¨ 0 X y u 0 Y .. Proof of Proposition 8. Ži. « Žii.. If Ži. is true, there exists a in Af 2 such that bidegŽ a f . s Ždeg f, 1. and Lemma 3 shows that l Ž a f . F 1, whence Žii.. Žii. « Žiii.. If Žii. is true, there exists a linear form l in Cw X, Y x such that for all i G 2, Pi and Q i are proportional to l i, whence Žiii.. Žiii. « Živ.. This is clear. The equivalence of Živ., Žv., and Žvi. comes from the equivalence of the following three assertions Žsee wNox., where k is any field of characteristic
AUTOMORPHISMS OF THE AFFINE PLANE
617
zero and g 1 , g 2 are any elements of k w X, Y x: Ž1. JacŽ g 1 , g 2 . s 0. Ž2. g 1 and g 2 are algebraically dependent over k. Ž3. ' Ž u, ¨ . g k w t x, ' R g k w X, Y x such that Ž g 1 , g 2 . s Ž uŽ R ., ¨ Ž R ... Let us now prove that Žvi. « Ži.. We can suppose that f s Ž X q uŽ R ., Y q ¨ Ž R .. where uŽ0. s ¨ Ž0. s RŽ0, 0. s 0, valŽ uŽ R .. G 2, and valŽ ¨ Ž R .. G 2. Since f is an automorphism, we necessarily have JacŽ f . s 1, which implies that u9Ž R .Ž Rr X . q ¨ 9Ž R .Ž Rr Y . s 0. By Lemma 6, if Ž u9, ¨ 9. / Ž0, 0., there exist Ž a, b . g C 2 and w g Cw t x such that R s w Ž aX q bY .. So, we can suppose that R s aX q bY. The condition u9Ž R .Ž Rr X . q ¨ 9Ž R .Ž Rr Y . s 0 implies that u9Ž R . and ¨ 9Ž R . are linearly dependent. It follows easily that uŽ R . and ¨ Ž R . are also linearly dependent.
6. ON THE IRREDUCIBLE COMPONENTS OF GA 2, n If n is a positive integer, let E 2, n be the vector space of C-endomorphisms f of A2C satisfying degŽ f . F n. Let us set GA 2, n s GA 2 l E 2, n , T2, n s T2 l E 2, n and UA 2, n s UA 2 l E 2, n . We set J2, n s f g E 2, n , such that Jac f s 14 and G2, n s GA 2, n l J2, n . E 2, n is an affine space and J2, n is a ŽZariski. closed subset of E 2, n . H. Bass, E. H. Connell, and D. Wright show in the paper wB-C-Wx that G2, n is also a closed subset of E 2, n . By slightly modifying their proof, we obtain that GA 2, n is a locally closed subset of E 2, n . In particular, GA 2, n is an algebraic variety. We have GA 2, n , C* = G2, n , via f s Ž f 1 , f 2 . ¬ ŽJacŽ f ., Ž f 1rJacŽ f ., f 2 .., so that the irreducible components of GA 2, n are in one to one correspondence with the irreducible components of G2, n . Our interest in the irreducible components of G2, n is motivated by the Jacobian Conjecture in dimension 2 and degree n asserting that an element of E 2, n is invertible if and only if its Jacobian is a nonzero constant. This conjecture is equivalent to the equality J2, n s G2, n . In Remark Ž1.7. of the last quoted paper, the authors note that to show this equality, it would suffice to show, if possible, that dim J2, n s dim G2, n and that the algebraic variety J2, n is irreducible Žbecause we have of course J2, n ; G2, n .. We answer here negatively to this hope. Indeed, on one hand J2, n s G2, n when n F 100, by T. T. Moh wMox and on the other hand we show that G2, n is reducible when n G 4 ŽProposition 11 below.. We see thus that the variety J2, n is reducible when 4 F n F 100.
618
JEAN-PHILIPPE FURTER
The results of Section 3 lead us to believe that we can get some insight into the decomposition in irreducible components of GA 2, n via purely combinatorial methods Žsee the Conjecture formulated in this section.. Let us recall that for any d g D, we introduced in Section 4 the subsets Ud and Vd of GA 2 which turned out to be
½
Ud s f g GA 2 such that d Ž f . F d 4 Vd s f g GA 2 such that d Ž f . s d 4
.
DEFINITION. We note that by the definition of Ud and Vd Žsee Section 4., their Zariski closures in GA 2, degŽ d. are equal. We set Wd s Uds Vd ŽZariski closure in GA 2, degŽ d. .. When l Ž d . s 1, the next proposition shows that Wd s Ud : PROPOSITION 9. If n is an integer greater than or equal to two, then UŽ n. is a closed sub¨ ariety of GA 2, n . Proof. UŽ n. s f g GA 2, n such that l Ž f . F 14 and condition Žiii. Žor Živ.. of Proposition 8 shows us that this condition is closed. LEMMA 7. < d < q 6.
If d g D, then Wd is an irreducible ¨ ariety of dimension
Proof. If d s B, the lemma is true. Let us now suppose that d g D*. We have ly1
Wd s cd Af 2 =
ž
Ł Ž T2, d _ T2, d y1 . = Ž UA 2, d _ UA 2, d y1 . = GL 2
is1
i
i
l
l
/
,
Ž . Ž . where Af 2 = Ł ly1 is1 T2, d i _ T2, d iy1 = UA 2, d p _ UA 2, d ly1 = GL 2 is an irreducible variety Žas a product of irreducible varieties. so that Wd is an irreducible variety. Let us set: 1 c 1, d s cd < A 1=Ł ly is 1 ŽT 2, d i _T 2, d iy 1 .=ŽUA 2, d l _UA 2, d ly 1 .=GL 2 group of translations of A2C ;
where A 1 is the
1 c 2, d s cd < A 2=Ł ly where A 2 is the set is 1 ŽT 2,d i _T 2,d iy 1 .=ŽUA 2, d l _UA 2,d ly 1 .=GL 2 of automorphisms of A2C of the shape Ž Y q a, X q cY q b . where a, b, c are elements of C;
V1, d s ImŽ c 1, d .; V2, d s ImŽ c 2, d ..
AUTOMORPHISMS OF THE AFFINE PLANE
619
We have Vd s V1, d " V2, d Žby Theorem 1.. c 1, d and c 2, d being injective morphisms Žagain by Theorem 1., we obtain dimŽV1, d. s < d < q 5 and dimŽV2, d. s < d < q 6. Finally, dim Wd s dim Vd s dim Ž V1, d j V2 , d . s max dimV1, d , dimV2, d 4 s < d < q 6. DEFINITION. For n g N*, we set Dn s d g D, degŽ d . F n4 and Cn denotes the set of maximal elements of Dn . For example, C1 s B 4 ; C2 s Ž 2 . 4 ; C3 s Ž 3 . 4 ; C4 s Ž 4 . , Ž 2, 2 . 4 ; C5 s Ž 5 . , Ž 2, 2 . 4 ; C6 s Ž 6 . , Ž 2, 3 . , Ž 3, 2 . 4 ; C 7 s Ž 7 . , Ž 2, 3 . , Ž 3, 2 . 4 ; C8 s Ž 8 . , Ž 2, 4 . , Ž 4, 2 . , Ž 2, 2, 2 . 4 ; C9 s Ž 9 . , Ž 2, 4 . , Ž 3, 3 . , Ž 4, 2 . , Ž 2, 2, 2 . 4 . The next lemma follows immediately from Corollary 2 to Proposition 7: LEMMA 8.
If n is a positi¨ e integer, then we ha¨ e GA 2, n s
D
dgC n
Ud s
@ Vd s D
dgDn
Wd .
dgC n
PROPOSITION 10. If n is a positi¨ e integer, then GA 2, n is an algebraic ¨ ariety of dimension 6 if n s 1 and of dimension n q 6 otherwise. Proof. By Lemmas 7 and 8, we have dim GA 2, n s max d g D n < d < q 6. If n s 1, D 1 s B4 and the proposition is clear.
620
JEAN-PHILIPPE FURTER
If n ) 1, ; d s Ž d1 , . . . , d l . g Dn , we have < d< s
l
l
Ý di F Ł di F n is1
is1
hence, max d g D n < d < is obtained for d s Ž n. where it is equal to n. Being now familiarized with the definitions, we give the announced conjecture: Conjecture. If n is a positive integer, then GA 2, n s D d g C n Wd is the decomposition in irreducible components of GA 2, n . This conjecture is equivalently formulated as 2
; Ž d, e . g Ž Cn . , d / e « Wd o We . We will prove this conjecture for n F 9 by using the next lemma: LEMMA 9. If n is an integer greater than or equal to two and if d g Cn with d / Ž n., then we ha¨ e Wd o WŽ n.
and
WŽ n. o Wd .
Proof. Ud contains an automorphism f such that l Ž f . G 2 and so Ud o UŽ n. . The equality UŽ n. s WŽ n. shows that Wd o WŽ n. . We deduce at once from this that WŽ n. o Wd . Otherwise the inequality dim Wd F dim WŽ n. together with the irreducibility of Wd , would imply WŽ n. s Wd , which is not the case. THEOREM 2.
If 1 F n F 9, then the Conjecture is true.
Proof. If n s 1, then GA 2, n s WŽB. . For 2 F n F 3, Cn s Ž n.4 and hence GA 2, n s WŽ n. is irreducible. Let us now suppose that 4 F n F 9. By Lemma 9, it is sufficient to prove that if d and e are two elements of Cn different from Ž n., then Wd o We . But the condition n F 9 implies the equality dim Wd s dim We Žby the list following the definition of Cn .. It is then sufficient to show that Wd / We . Now, Vd is a dense constructible subset of Wd and Ve is a dense constructible subset of We . Furthermore, Vd l Ve s B. This proves that we cannot have Wd s We . Even if we cannot yet compute the number of irreducible components of GA 2, n , we can however decide whether this variety is irreducible: PROPOSITION 11. If n is a positi¨ e integer, then the ¨ ariety GA 2, n is irreducible if and only if n F 3.
AUTOMORPHISMS OF THE AFFINE PLANE
621
Proof. It is sufficient to prove that GA 2, n is reducible for n G 4. Now, WŽ n. is an irreducible component of GA 2, n Žbecause it is an irreducible variety of the same dimension as GA 2, n .. The element Ž X q Y 2 , Y . s Ž X q Y 2 , Y . is in WŽ2, 2. ; GA 2, n but not in WŽ n. . COROLLARY OF PROPOSITION 11. reducible.
If 4 F n F 100, then the ¨ ariety J 2, n is
Remark. It is not true that the mapping d: GA 2 ª D f ¬ dŽ f . is lower-continuous, i.e., if d g D, it is not necessarily true that Ud is a closed subset of GA 2, degŽ d. . To check this, we consider the family of GA 2 given by the Nagata automorphism,
x : C ª GA 2 , 4 Z ¬ x Ž Z. , where x Ž Z . s Ž X y 2Y Ž XZ q Y 2 . y ZŽ XZ q Y 2 . 2 , Y q ZŽ XZ q Y 2 ... Then, if Z g C*, we have x Ž Z . s g 1 a 1 g 2 where
¡g
~a
¢g
1
s Ž X y Zy1 Y 2 , Y . g BA 2 _ B 2
1
s Ž X , Y q Z 2 X . g Af 2 _ B 2
2
s Ž X q Zy1 Y 2 , Y . g BA 2 _ B 2
so that dŽ x Ž Z .. s Ž2, 2.. However, dŽ x Ž0.. s Ž3. so that x Ž0. g UŽ2, 2. _ UŽ2, 2. . 7. MORE INSIGHT INTO THE IRREDUCIBLE COMPONENTS OF GA 2, 4 It results from Theorem 2 that WŽ4. o WŽ2, 2. . The next proposition gives us an explicit equation satisfied by all elements of WŽ2, 2. but not necessarily by elements of WŽ4. . PROPOSITION 12. Let f s Ž f 1 , f 2 . g WŽ2, 2. . Write f s Ý4is0 Ž Pi , Q i ., where Ž Pi , Q i . are C-endomorphisms of A2C which are homogeneous of degree i. Then, P2 P3 y 2 P1 P4 and Q2 Q3 y 2 Q1 Q4 are proportional polynomials.
622
JEAN-PHILIPPE FURTER
This is equi¨ alent to the algebraic condition Jac Ž P2 P3 y 2 P1 P4 , Q2 Q3 y 2 Q1 Q4 . s 0. Proof. Let f s Ž f 1 , f 2 . g VŽ2, 2. . Let us note that deg f s 4 by Proposition 5 so that deg f 1 s 4 or deg f 2 s 4. By Proposition 1, we have deg f 1 s 1, 2, or 4. Furthermore, it is impossible to have deg f 1 s 1 because otherwise we would have l Ž f . F 1 by Proposition 8. In the same way, we obtain deg f 2 s 2 or 4. Let us write f s Ý4is0 Ž Pi , Q i . where Ž Pi , Q i . are C-endomorphisms of 2 A C which are homogeneous of degree i. We begin by proving that there exists a complex number l such that 2 2 Ž Q2 Q3 y 2 Q1 Q4 . s lŽ P1 Q2 y P2 Q1 . Q4 .
If Q4 s 0, we have Q3 s 0 and it is sufficient to take l s 0. Otherwise, we have deg f 2 s 4 and we claim that there exists a complex number a such that degŽ f 1 y a f 2 . s 2. Indeed, if deg f 1 s 2, take a s 0. If deg f 1 s 4, by Remark 3 following Proposition 1, we know the existence of a such that degŽ f 1 y a f 2 . - 4. However, Ž f 1 y a f 2 , f 2 . g VŽ2, 2. so that degŽ f 1 y a f 2 . s 2 and the claim is proven. We set g s Ž g 1 , g 2 . s Ž f 1 y a f 2 , f 2 . s ŽÝ2is0 Ž Pi y a Q i ., Ý4is0 Q i . g GA 2 . We have bideg g s Ž2, 4.. Still by Proposition 1, we deduce the existence of a complex number b such that degŽ g 2 y b g 12 . - 4. Because Ž g 1 , g 2 y b g 12 . is an automorphism, we have in fact degŽ g 2 y b g 12 . F 2 whence
½
Q3 s 2 b Ž P1 y a Q1 . Ž P2 y a Q2 . Q4 s b Ž P2 y a Q2 .
2
which give us successively Q2 Q3 y 2 Q1 Q4 s 2 b Ž P2 y a Q2 . Ž Ž P1 y a Q1 . Q2 y Ž P2 y a Q2 . Q1 . Q2 Q3 y 2 Q1 Q4 s 2 b Ž P2 y a Q2 . Ž P1 Q2 y P2 Q1 . 2 2 2 Ž Q2 Q3 y 2 Q1 Q4 . s 4b 2 Ž P2 y a Q2 . Ž P1 Q2 y P2 Q1 . 2 2 Ž Q2 Q3 y 2 Q1 Q4 . s 4b Ž P1 Q2 y P2 Q1 . Q4 ,
whence the existence of l. By the same argument, there exists m such that 2 2 Ž P2 P3 y 2 P1 P4 . s m Ž P1 Q2 y P2 Q1 . P4 .
AUTOMORPHISMS OF THE AFFINE PLANE
623
Since P4 and Q4 are proportional polynomials, the same holds for Ž P2 P3 y 2 P1 P4 . 2 and Ž Q2 Q3 y 2 Q1 Q4 . 2 and hence for P2 P3 y 2 P1 P4 and Q2 Q3 y 2 Q1 Q 4 . We have thus proved the relation Jac Ž P2 P3 y 2 P1 P4 , Q2 Q3 y 2 Q1 Q4 . s 0 and this relation remains true if we only suppose that f g WŽ2, 2. s VŽ2 , 2.. Proposition 12 is enough to prove that g s Ž X q Y 4 , X q Y q Y 4 . g WŽ4. _ WŽ2 , 2. . Indeed, the homogeneous components Ž Pi , Q i . of degree i of g are Ž P1 , Q1 . s Ž X, X q Y ., Ž P2 , Q2 . s Ž0, 0., Ž P3 , Q3 . s Ž0, 0., and Ž P4 , Q4 . s Ž Y 4 , Y 4 . so that P2 P3 y 2 P1 P4 and Q2 Q3 y 2 Q1 Q4 are equal to y2 XY 4 and y2Ž X q Y .Y 4 which are not proportional polynomials.
ACKNOWLEDGMENTS I would like to thank Jose ´ Bertin for his advice and Michel Brion for his constant help. Note added in proof. The author afterward noticed that the analogous notion of ‘‘multidegree’’ has already been introduced in wF-Mx, where the authors also compute the dimension of the manifold of automorphisms with the prescribed multidegree. I am indebted to E. Ghys for pointing out this reference to me.
REFERENCES wB-C-Wx H. Bass, E. H. Connell, and D. Wright, The Jacobian conjecture: Reduction of degree and formal expansion of the inverse, Bull. Amer. Math. Soc. Ž1982., 287]330. wF-Mx S. Friedland and J. Milnor, Dynamical properties of plane polynomial automorphisms, Ergodic Th. & Dynam. Systems 9 Ž1989., 67]99. wvdKx W. van der Kulk, On polynomial rings in two variables, Nieuw. Arch. Wisk. (3) 1 Ž1953., 33]41. wMox T. T. Moh, On the global Jacobian conjecture and the configuration of roots, J. Reine Angew. Math. 340 Ž1983., 140]212. wNax M. Nagata, On automorphisms group of k w x, y x, in ‘‘Lecture Notes in Math.,’’ Vol. 5, Kyoto University, 1972. wNox A. Nowicki, On the Jacobian equation J Ž f, g . s 0 for polynomials in k w x, y x, Nagoya Math. J. 109 Ž1988., 151]157.