aa r X i v : . [ m a t h . A C ] D ec A characterization of Keller maps
Piotr J ֒ edrzejewicz Faculty of Mathematics and Computer ScienceNicolaus Copernicus UniversityToru´n, Poland
October 30, 2018
Abstract
Let k be a field of characteristic zero. Let ϕ be a k -endomorphismof the polynomial algebra k [ x , . . . , x n ]. It is known that ϕ is an auto-morphism if and only if it maps irreducible polynomials to irreduciblepolynomials. In this paper we show that ϕ satisfies the jacobian con-dition if and only if it maps irreducible polynomials to square-freepolynomials. Therefore, the Jacobian Conjecture is equivalent to thefollowing statement: every k -endomorphism of k [ x , . . . , x n ], mappingirreducible polynomials to square-free polynomials, maps irreduciblepolynomials to irreducible polynomials. Throughout this article k is a field of characteristic zero. By k [ x , . . . , x n ]we denote the k -algebra of polynomials in n variables. If ϕ is a k -endo-morphism of k [ x , . . . , x n ], then by Jac ϕ we denote the jacobian determinantof the polynomials ϕ ( x ), . . . , ϕ ( x n ) with respect to the variables x , . . . , x n . If Jac ϕ ∈ k \ { } , then we say that ϕ satisfies the jacobian condition.In this case the respective polynomial map F : k n → k n , F ( x , . . . , x n ) =( ϕ ( x ) , . . . , ϕ ( x n )), is called a Keller map. The famous Jacobian Conjecture,stated by Keller in [11], asserts that every k -endomorphism of k [ x , . . . , x n ]satisfying the jacobian condition is an automorphism of k [ x , . . . , x n ]. Keywords: jacobian conjecture, Keller map, jacobian determinant.2010 Mathematics Subject Classification: Primary 13F20, Secondary 14R15, 13N15. C -endo-morphism of C [ x , . . . , x n ] mapping variables to variables is an automorphism(recall that by a variable we mean an element of any set of n generators).The affirmative answer was given by Jelonek in [7], Theorem 2. It was notedthat this fact holds for arbitrary algebraically closed field of characteristiczero ([3], a comment to Theorem 10.5.9 on p. 273). Another characterizationof polynomial automorphisms was obtained by Bakalarski in [1], Theorem 3.7(see also a remark at the end of [1]). He showed that a C -endomorphism of C [ x , . . . , x n ] mapping irreducible polynomials to irreducible polynomials isan automorphism. We present another proof of Bakalarski’s theorem for anarbitrary field k of characteristic zero (Theorem 5.2).The aim of this paper is to obtain a characterization of polynomial en-domorphisms satisfying the jacobian condition as those mapping irreduciblepolynomials to square-free polynomials (Theorem 5.1). Hence, using theresult of Bakalarski, we have the following equivalent formulation of theJacobian Conjecture: every k -endomorphism of k [ x , . . . , x n ] mapping ir-reducible polynomials to square-free polynomials maps irreducible polyno-mials to irreducible polynomials (Theorem 5.3). Our characterization of k -endomorphisms satisfying the jacobian condition is an immediate conse-quence of Theorem 4.1, where we prove that an irreducible polynomial g di-vides the jacobian of given polynomials f , . . . , f n if and only if there exists anirreducible polynomial w ∈ k [ x , . . . , x n ] such that g divides w ( f , . . . , f n ).Basic definitions and facts are presented in Section 2. In Section 3 weprove preparatory lemmas, which will be useful in the proof of Theorem 4.1,given in Section 4. In Section 5 we obtain a characterization of endomor-phisms satisfying the jacobian condition and the equivalent formulation ofthe Jacobian Conjecture. Some conclusions and comments are presented inSection 6. By a ring we mean a commutative ring with unity. Let A be a ring. Anadditive map d : A → A such that d ( ab ) = d ( a ) b + ad ( b ) for a, b ∈ A is calleda derivation of A . The set A d = { a ∈ A ; d ( a ) = 0 } is called the ring ofconstants of d . If K is a subring of A , then a derivation d of A is K -linearif and only if K ⊂ A d . In this case we call d a K -derivation. Hence, if A isa k -algebra, where k is a field, and a , . . . , a m ∈ A , then a map d : A → A isa k [ a , . . . , a m ]-derivation if and only if d is a k -derivation and d ( a i ) = 0 for i = 1 , . . . , m . 2f d is a k -derivation of a k -algebra A , where k is a field, then for anelement a ∈ A and a polynomial w ( x ) ∈ k [ x ] we have d ( w ( a )) = w ′ ( a ) d ( a ).More generally, for a , . . . , a m ∈ A and a polynomial w ( x , . . . , x m ) ∈ k [ x ,. . . , x m ] the following holds: d ( w ( a , . . . , a m )) = ∂w∂x ( a , . . . , a m ) d ( a ) + . . . + ∂w∂x m ( a , . . . , a m ) d ( a m ) . In particular, if d is a k -derivation of k [ x , . . . , x n ], then d ( f ) = ∂f∂x d ( x ) + . . . + ∂f∂x n d ( x n ) for every polynomial f ∈ k [ x , . . . , x n ].Let k be a field of characteristic zero, let A be a finitely generated k -domain (that is, a commutative k -algebra with unity, without zero divisors)and let R be a k -subalgebra of A . Denote by R the field of fractions of R .Nowicki ([13], Theorem 5.4; [12], Theorem 4.1.4) proved that the followingconditions are equivalent:(1) R = A d for some k -derivation d of A ,(2) R is integrally closed in A and R ∩ A = R .Daigle observed in ”Locally nilpotent derivations” (unpublished lecturenotes, available on his website) that the condition (2) means that R is alge-braically closed in A as a subring. The present author noted in [9] that thischaracterization holds also for K -derivations, where K is a subring of A . Inthis case we have the following corollary from [9], Theorem 3.1. Corollary 2.1.
Let A be a finitely generated K -domain of characteristiczero, where K is a subring of A . An element b ∈ A belongs to the ring ofconstants of every K -derivation of A if and only if b is algebraic over thefield K . Given n polynomials f , . . . , f n ∈ k [ x , . . . , x n ], by Jac( f , . . . , f n ) wedenote the jacobian determinant of f , . . . , f n with respect to x , . . . , x n .Note that the map d ( f ) = Jac( f , . . . , f n − , f ) for f ∈ k [ x , . . . , x n ] is a k -derivation of k [ x , . . . , x n ], such that d ( f i ) = 0 for i = 1 , . . . , n −
1. Moregenerally, given m polynomials f , . . . , f m ∈ k [ x , . . . , x n ], where 1 m n ,and given arbitrary j , . . . , j m ∈ { , . . . , n } , by Jac f ,...,f m j ,...,j m we denote thejacobian determinant of f , . . . , f m with respect to x j , . . . , x j m . The map d ( f ) = Jac f ,...,f i − ,f,f i +1 ,...,f m j ,...,j m for f ∈ k [ x , . . . , x n ] is also a k -derivation of k [ x , . . . , x n ], and we have d ( f j ) = 0 for j = i .Following [10], we introduce the notion of a differential gcd of polynomials:dgcd( f , . . . , f m ) = gcd (cid:16) Jac f ,...,f m j ,...,j m ; 1 j , . . . , j m n (cid:17) f , . . . , f m ∈ k [ x , . . . , x n ]. Of course, dgcd is defined with respect to ascalar multiple. For a single polynomial f ∈ k [ x , . . . , x n ] we have dgcd( f ) = c · gcd (cid:16) ∂f∂x , . . . , ∂f∂x n (cid:17) , where c ∈ k \ { } . For n polynomials f , . . . , f n ∈ k [ x , . . . , x n ] we have dgcd( f , . . . , f n ) = c · Jac( f , . . . , f n ), where c ∈ k \ { } . Recall that k is a field of characteristic zero. In Lemmas 3.1 – 3.3 belowwe consider: arbitrary polynomials f , . . . , f n ∈ k [ x , . . . , x n ], an irreduciblepolynomial g ∈ k [ x , . . . , x n ] and the factor algebra A = k [ x , . . . , x n ] / ( g ).By f we denote the respective class in A of a polynomial f ∈ k [ x , . . . , x n ],that is, f = f + ( g ). Lemma 3.1.
For a given i ∈ { , . . . , n } consider the following condition: ( ∗ ) there exist s , . . . , s n ∈ k [ x , . . . , x n ] , where g ∤ s i , such that g | s d ( f ) + . . . + s n d ( f n ) for every k -derivation d of k [ x , . . . , x n ] . a) The jacobian determinant
Jac( f , . . . , f n ) is divisible by g if and only ifthe condition ( ∗ ) holds for some i ∈ { , . . . , n } . b) If, for a given i ∈ { , . . . , n } , the condition ( ∗ ) holds, then f i is algebraicover the field k ( f , . . . , f i − , f i +1 , . . . , f n ) .Proof. a) The jacobian determinant Jac( f , . . . , f n ) is divisible by g if andonly if the determinant of the matrix ∂f ∂x ∂f ∂x · · · ∂f ∂x n ∂f ∂x ∂f ∂x · · · ∂f ∂x n ... ... ... ∂f n ∂x ∂f n ∂x · · · ∂f n ∂x n equals 0 in A . If we consider this matrix over the field A , the last conditionis equivalent to the linear dependence over A of the rows of this matrix.This condition can be written with coefficients in A : there exist polynomials s , . . . , s n ∈ k [ x , . . . , x n ], where s i = 0 for some i , such that s h ∂f ∂x , . . . , ∂f ∂x n i + . . . + s n h ∂f n ∂x , . . . , ∂f n ∂x n i = (cid:2) , . . . , (cid:3) . h = s ∂f ∂x + . . . + s n ∂f n ∂x , . . . , h n = s ∂f ∂x n + . . . + s n ∂f n ∂x n are divisible by g .Now, observe that, for an arbitrary k -derivation d of k [ x , . . . , x n ], wehave s d ( f ) + . . . + s n d ( f n )= s (cid:16) ∂f ∂x d ( x ) + . . . + ∂f ∂x n d ( x n ) (cid:17) + . . . + s n (cid:16) ∂f n ∂x d ( x ) + . . . + ∂f n ∂x n d ( x n ) (cid:17) = (cid:16) s ∂f ∂x + . . . + s n ∂f n ∂x (cid:17) d ( x ) + . . . + (cid:16) s n ∂f ∂x n + . . . + s n ∂f n ∂x n (cid:17) d ( x n )= h d ( x ) + . . . + h n d ( x n ) . Hence, if the polynomials h , . . . , h n are divisible by g , then g | s d ( f ) + . . . + s n d ( f n ). On the other hand, if the polynomial s d ( f ) + . . . + s n d ( f n )is divisible by g for every k -derivation d , then, in particular, for the partialderivatives d = ∂∂x j , j = 1 , . . . , n , we obtain that the polynomials h , . . . , h n are divisible by g . b) Assume that the condition ( ∗ ) holds for some i ∈ { , . . . , n } . Let δ be anarbitrary k [ f , . . . , f i − , f i +1 , . . . , f n ]-derivation of the factor algebra A , thatis, a k -derivation such that δ ( f j ) = 0 for each j ∈ { , . . . , i − , i + 1 , . . . , n } .Consider a k -derivation d of k [ x , . . . , x n ] such that δ ( f ) = d ( f ) for every f ∈ k [ x , . . . , x n ] ([8], Lemma 3.2). We have d ( f j ) = δ ( f j ) = 0, that is, g | d ( f j ), for each j = i . Hence, the condition ( ∗ ) yields that g | s i d ( f i ), so g | d ( f i ), because g ∤ s i . This means that δ ( f i ) = d ( f i ) = 0. By Corollary2.1, since δ ( f i ) = 0 for an arbitrary k [ f , . . . , f i − , f i +1 , . . . , f n ]-derivation δ of A , f i is algebraic over the field k ( f , . . . , f i − , f i +1 , . . . , f n ).Note the following easy observation. Lemma 3.2.
Let m ∈ { , . . . , n } . a) The elements f , . . . , f m ∈ A are algebraically dependent over k if andonly if g | w ( f , . . . , f m ) for some nonzero polynomial w ∈ k [ x , . . . , x m ] . b) Let i ∈ { , . . . , m } . The element f i ∈ A is algebraic over the field k ( f , . . . , f i − , f i +1 , . . . , f n ) if and only if g | w ( f , . . . , f m ) for some nonzeropolynomial w ∈ k [ x , . . . , x m ] of positive degree with respect to x i . In the case of n polynomials in n variables we have the following. Lemma 3.3.
There exists an irreducible polynomial w ∈ k [ x , . . . , x n ] suchthat g | w ( f , . . . , f n ) . roof. The ideal ( g ) has height 1, so the Krull dimension of A equals n − f , . . . , f n are algebraically dependent over k . Then,by Lemma 3.2.a, there exists a nonzero polynomial u ∈ k [ x , . . . , x n ] suchthat g | u ( f , . . . , f n ). The polynomial u is obviously non-constant. Then,for some irreducible factor w of u , the polynomial w ( f , . . . , f n ) is divisibleby g . Lemma 3.4.
Assume that n > and r n − . Consider polyno-mials u ∈ k [ x , . . . , x r , x r +1 ] \ k [ x , . . . , x r ] and u ∈ k [ x , . . . , x r , x r +2 ] \ k [ x , . . . , x r ] . If the degrees of u with respect to x r +1 and of u with re-spect to x r +2 are relatively prime, then there exist nonzero polynomials w ∈ k [ x , . . . , x r ] , w ∈ k [ x , . . . , x r , x r +1 , x r +2 ] \ k [ x , . . . , x r ] such that w is ir-reducible and u + u = w w .Proof. Consider a decomposition of u + u into irreducible factors in k [ x ,. . . , x r , x r +1 , x r +2 ]: u + u = v . . . v s v s +1 . . . v t , where v , . . . , v s ∈ k [ x , . . . ,x r ] and v s +1 , . . . , v t k [ x , . . . , x r ], 0 s t . Observe that s < t , because u + u k [ x , . . . , x r ].Now, consider the field L = k ( x , . . . , x r ). Since the degrees of the poly-nomials: u in L [ x r +1 ] and u in L [ x r +2 ] are positive and relatively prime,the polynomial u + u is irreducible in L [ x r +1 , x r +2 ], by Corollary 3 to Theo-rem 21 in [14], p. 94 (see also [2]). Hence t = s + 1. Finally, put w = v . . . v s ( w = 1 if s = 0) and w = v s +1 . Lemma 3.5.
Let w ∈ k [ x , . . . , x m ] be an irreducible polynomial such that ∂w∂x i = 0 for some i ∈ { , . . . , m } . Then there exist polynomials v , v ∈ k [ x , . . . , x m ] and v ∈ k [ x , . . . , x i − , x i +1 , . . . , x m ] \ { } such that v w + v ∂w∂x i = v. Proof.
Consider the field L = k ( x , . . . , x i − , x i +1 , . . . , x m ). The polynomial w is irreducible in L [ x i ], and the polynomial ∂w∂x i is nonzero, so they arerelatively prime in L [ x i ]. Hence there exist polynomials u , u ∈ L [ x i ] suchthat u w + u ∂w∂x i = 1 . Let v ( v ∈ k [ x , . . . , x i − , x i +1 , . . . , x m ]) be the least common denominator ofthe coefficients of polynomials u , u . Multiplying the above equality by v and denoting v = u v , v = u v we get the lemma.6 Irreducible factors of jacobians
Theorem 4.1.
Let k be a field of characteristic zero, let f , . . . , f n ∈ k [ x ,. . . , x n ] be arbitrary polynomials, and let g ∈ k [ x , . . . , x n ] be an irreduciblepolynomial. The following conditions are equivalent: ( i ) g divides Jac( f , . . . , f n ) , ( ii ) g divides w ( f , . . . , f n ) for some irreducible polynomial w ∈ k [ x , . . . , x n ] .Proof. ( i ) ⇒ ( ii ) Assume that g | Jac( f , . . . , f n ).By Lemma 3.3, g | w ( f , . . . , f n ) for some irreducible polynomial w ∈ k [ x , . . . , x n ], so(1) w ( f , . . . , f n ) = gh for some h ∈ k [ x , . . . , x n ]. Without loss of generality, we may assume that w is of positive degree with respect to x n , so f n is algebraic over the field k ( f , . . . , f n − ), by Lemma 3.2.b. Assume that g ∤ w ( f , . . . , f n ), that is, g ∤ h .Consider the k -derivation d n of k [ x , . . . , x n ] defined by d n ( f ) = Jac( f , . . . , f n − , f )for f ∈ k [ x , . . . , x n ]. Observe that d n ( f i ) = 0 for i = 1 , . . . , n − d n ( f n ) = Jac( f , . . . , f n ). Applying the derivation d n to both sides of (1) weobtain ∂w∂x n ( f , . . . , f n ) d n ( f n ) = d n ( g ) h + gd n ( h ) . Since g | d n ( f n ) and g ∤ h , we have g | d n ( g ), that is, g | Jac( f , . . . , f n − , g ).From Lemma 3.1 we obtain that there exist polynomials s , . . . , s n ∈ k [ x , . . . , x n ], where g ∤ s i for some i ∈ { , . . . , n } , such that g | s d ( f ) + . . . + s n − d ( f n − ) + s n d ( g )for every k -derivation d of k [ x , . . . , x n ]. Note that the polynomials s , . . . ,s n − can not all together be divisible by g . Indeed, in this case we wouldhave g ∤ s n and g | s n d ( g ), so g | d ( g ) for every k -derivation d , what is nottrue for d = ∂∂x j such that ∂g∂x j = 0.Thus g ∤ s i for some i ∈ { , . . . , n − } ; we may assume that g ∤ s n − . ByLemma 3.1, f n − is algebraic over the field k ( f , . . . , f n − , g ) = k ( f , . . . ,f n − ). Recall that f n is algebraic over k ( f , . . . , f n − ), so if we denote7 = tr deg k k ( f , . . . , f n ), we have r n −
2. Hence, we may assume that f , . . . , f r are algebraically independent over k .Let L = k ( f , . . . , f r ). Since f r +1 and f r +2 are algebraic over L , thereexist nonzero polynomials v ∈ k [ x , . . . , x r , x r +1 ], v ∈ k [ x , . . . , x r , x r +2 ]of positive degrees t , t with respect to x r +1 , x r +2 , respectively, such thatthe polynomials v ( f , . . . , f r , f r +1 ) and v ( f , . . . , f r , f r +2 ) are both divis-ible by g (Lemma 3.2.b). Put u = v x t +1 r +1 , u = v x t r +2 . Then thepolynomials u ( f , . . . , f r , f r +1 ) and u ( f , . . . , f r , f r +2 ) are both divisible by g and, by Lemma 3.4, there exist nonzero polynomials w ∈ k [ x , . . . , x r ], w ∈ k [ x , . . . , x r , x r +1 , x r +2 ] such that w is irreducible and u + u = w w .We obtain that g | w ( f , . . . , f r ) w ( f , . . . , f r , f r +1 , f r +2 ) , but g ∤ w ( f , . . . , f r ) by Lemma 3.2.a, because f , . . . , f r are algebraicallyindependent over k . Finally, g | w ( f , . . . , f r , f r +1 , f r +2 ).( ii ) ⇒ ( i ) We will show by induction on m ∈ { , . . . , n } that for m arbitrary polynomials f , . . . , f m ∈ k [ x , . . . , x n ] and an irreducible polyno-mial g ∈ k [ x , . . . , x n ], if g | w ( f , . . . , f m ) for some irreducible polynomial w ∈ k [ x , . . . , x m ], then g | dgcd( f , . . . , f m ). Recall that dgcd( f , . . . , f m ) =gcd (cid:16) Jac f ,...,f m j ,...,j m ; 1 j , . . . , j m n (cid:17) .Let m = 1. Assume that g | w ( f ), where w ∈ k [ x ] is an irreduciblepolynomial, so w ( f ) = g h for some h ∈ k [ x , . . . , x n ]. Applying the partialderivative with respect to x i for i ∈ { , . . . , n } we obtain w ′ ( f ) ∂f ∂x i = 2 g ∂g∂x i h + g ∂h∂x i , so g | w ′ ( f ) ∂f ∂x i . Since w is irreducible, w ′ is relatively prime to w , so uw + vw ′ = 1 for some polynomials u, v ∈ k [ x ]. This yields u ( f ) w ( f ) + v ( f ) w ′ ( f ) = 1, so g ∤ w ′ ( f ). Therefore g | ∂f ∂x i for each i , so g | dgcd( f ).Now, let m ∈ { , . . . , n } . Assume that the induction hypothesis holdsfor m −
1. Assume that g | w ( f , . . . , f m ) for some irreducible polynomial w ∈ k [ x , . . . , x m ]:(2) w ( f , . . . , f m ) = g h, where h ∈ k [ x , . . . , x n ].First, consider the case when g | u ( f , . . . , f i − , f i +1 , . . . , f m ) for some i ∈{ , . . . , m } and some irreducible polynomial u ∈ k [ x , . . . , x i − , x i +1 , . . . , x m ].In this case, by the induction hypothesis, g | dgcd( f , . . . , f i − , f i +1 , . . . , f m ),so every jacobian determinant of f , . . . , f i − , f i +1 , . . . , f m is divisible by g .Then, for arbitrary j , . . . , j m ∈ { , . . . , n } , from the Laplace expansion with8espect to i -th row, we see that the determinant Jac f ,...,f m j ,...,j m is divisible by g ,so g | dgcd( f , . . . , f m ).Now, assume that g ∤ u ( f , . . . , f i − , f i +1 , . . . , f m ) for each i ∈ { , . . . , m } and every irreducible polynomial u ∈ k [ x , . . . , x i − , x i +1 , . . . , x m ]. Hence, inparticular, ∂w∂x i = 0. Suppose that g ∤ dgcd( f , . . . , f m ), that is, g ∤ Jac f ,...,f m j ,...,j m for some j , . . . , j m ∈ { , . . . , n } . Denote by d i , for i = 1 , . . . , m , the k -derivation of k [ x , . . . , x n ] defined by d i ( f ) = Jac f ,...,f i − ,f,f i +1 ,...,f m j ,...,j m for f ∈ k [ x , . . . , x n ]. Observe that d i ( f j ) = 0 for j = i and d i ( f i ) =Jac f ,...,f m j ,...,j m . Applying the derivation d i to both sides of (2) we have ∂w∂x i ( f , . . . , f m ) Jac f ,...,f m j ,...,j m = 2 gd i ( g ) h + g d i ( h ) , so g | ∂w∂x i ( f , . . . , f m ).From Lemma 3.5 we obtain that g | v ( f , . . . , f i − , f i +1 , . . . , f m ) for somenonzero polynomial v ∈ k [ x , . . . , x i − , x i +1 , . . . , x m ]. The polynomial v isobviously non-constant. Then there exists an irreducible polynomial u i ∈ k [ x , . . . , x i − , x i +1 , . . . , x m ] such that the polynomial u i ( f , . . . , f i − , f i +1 , . . . ,f m ) is divisible by g , that is,(3) u i ( f , . . . , f i − , f i +1 , . . . , f m ) = gs i for some s i ∈ k [ x , . . . , x n ]. By the assumption, the left side of (3) is notdivisible by g , so g ∤ s i . Applying the derivation d i to both sides of (3) weobtain 0 = d i ( g ) s i + gd i ( s i ), so g | d i ( g ) (for arbitrary i ∈ { , . . . , m } ).Now, consider arbitrary j ∈ { , . . . , m } and apply the derivation d j toboth sides of (3) for i ∈ { , . . . , m } , i = j : ∂u i ∂x j ( f , . . . , f i − , f i +1 , . . . , f m ) Jac f ,...,f m j ,...,j m = d j ( g ) s i + gd j ( s i ) . Since g | d j ( g ) and g ∤ Jac f ,...,f m j ,...,j m , we have g | ∂u i ∂x j ( f , . . . , f i − , f i +1 , . . . , f m ).On the other hand, applying the derivation ∂∂x j to both sides of (3) we obtain ∂u i ∂x ( f , . . . , f i − , f i +1 , . . . , f m ) ∂f ∂x j + . . . + ∂u i ∂x m ( f , . . . , f i − , f i +1 , . . . , f m ) ∂f m ∂x j = ∂g∂x j s i + g ∂s i ∂x j . Recall that g ∤ s i , so g | ∂g∂x j , that is, ∂g∂x j = 0 for each j ∈ { , . . . , m } , acontradiction. 9ote the following immediate consequence of Theorem 4.1. Corollary 4.2.
For arbitrary polynomials f , . . . , f n ∈ k [ x , . . . , x n ] the fol-lowing conditions are equivalent: ( i ) Jac( f , . . . , f n ) ∈ k \ { } , ( ii ) for every irreducible polynomial w ∈ k [ x , . . . , x n ] the polynomial w ( f ,. . . , f n ) is square-free. If ϕ is a k -endomorphism of k [ x , . . . , x n ], then by Jac ϕ we denote thejacobian determinant of the polynomials ϕ ( x ), . . . , ϕ ( x n ) with respect to x , . . . , x n : Jac ϕ = Jac( ϕ ( x ) , . . . , ϕ ( x n )) . We obtain the following characterization of k -endomorphisms satisfying thejacobian condition. Theorem 5.1.
Let k be a field of characteristic zero. Let ϕ be a k -endo-morphism of the polynomial algebra k [ x , . . . , x n ] . The following conditionsare equivalent: ( i ) Jac ϕ ∈ k \ { } , ( ii ) for every irreducible polynomial w ∈ k [ x , . . . , x n ] the polynomial ϕ ( w ) is square-free.Proof. Put f = ϕ ( x ), . . . , f n = ϕ ( x n ). Since ϕ is a k -endomorphism, forevery polynomial w ∈ k [ x , . . . , x n ] we have ϕ ( w ( x , . . . , x n )) = w ( ϕ ( x ) , . . . , ϕ ( x n )) = w ( f , . . . , f n ) . The rest follows from Corollary 4.2.The following theorem was obtained by Bakalarski in [1] (Theorem 3.7)under an additional assumption, but it was noted in a remark added in theproof that this assumption is not necessary. Here we present another proofof Bakalarski’s theorem, based on our Lemma 3.3.
Theorem 5.2 (Bakalarski) . Let k be a field of characteristic zero and let ϕ be a k -endomorphism of the polynomial algebra k [ x , . . . , x n ] . The followingconditions are equivalent: i ) ϕ is a k -automorphism of k [ x , . . . , x n ] , ( ii ) for every irreducible polynomial w ∈ k [ x , . . . , x n ] the polynomial ϕ ( w ) is irreducible.Proof. Every automorphism of a ring maps irreducible elements into irre-ducible elements, so it is enough to prove the implication ( ii ) ⇒ ( i ). Assumethat ϕ ( w ) is an irreducible polynomial for every irreducible w ∈ k [ x , . . . , x n ].Observe that ϕ is a monomorphism: if ϕ ( f ) = 0 and f = g . . . g r is a de-composition into irreducible factors, then ϕ ( g i ) = 0 for some i , contrary tothe assumption.Now we will prove that ϕ is surjective. Put f i = ϕ ( x i ) for i = 1 , . . . , n .Suppose that there exists a polynomial g ∈ k [ x , . . . , x n ], such that g k [ f , . . . , f n ]. In this case at least one of irreducible factors of g does not be-long to k [ f , . . . , f n ], so we may assume that g is irreducible. Then, by Lemma3.3, g | w ( f , . . . , f n ) for some irreducible polynomial w ∈ k [ x , . . . , x n ], thatis, w ( f , . . . , f n ) = gh , where h ∈ k [ x , . . . , x n ]. However, h k , because g k [ f , . . . , f n ], so w ( f , . . . , f n ) is a reducible polynomial.By Theorems 5.1 and 5.2 we have. Theorem 5.3.
Let k be a field of characteristic zero, let n be a positiveinteger. The following conditions are equivalent: ( i ) every k -endomorphism ϕ of k [ x , . . . , x n ] such that Jac ϕ ∈ k \ { } is anautomorphism of k [ x , . . . , x n ] (the Jacobian Conjecture), ( ii ) every k -endomorphism of k [ x , . . . , x n ] mapping irreducible polynomialsto square-free polynomials maps irreducible polynomials to irreducible poly-nomials. Remark 1.
From Theorem 5.1 we know that a k -endomorphism ϕ of k [ x , . . . , x n ] satisfies the jacobian condition if and only if it maps irreduciblepolynomials to square-free polynomials. It is natural to ask if there existsa non-trivial example of such a k -endomorphism; non-trivial in the follow-ing sense: for some irreducible polynomial w ∈ k [ x , . . . , x n ] the polynomial ϕ ( w ) is reducible. From Theorem 5.3 we know that such an example wouldbe a counter-example to the Jacobian Conjecture, and if such an exampledoes not exist, the Jacobian Conjecture is true.11 emark 2. It may be interesting to consider the following property of agiven ring (a commutative ring with unity):( ∗ ) every endomorphism mapping irreducible elements to square-freeelements maps irreducible elements to irreducible elements. Question 6.1.
Let R be a unique factorization domain satisfying the con-dition ( ∗ ) . Does the ring R [ x ] of polynomials in one variable over R alsosatisfy the condition ( ∗ ) ? If the answer to this question is positive, then the Jacobian Conjecture istrue. Namely, in this case, by an obvious induction, every ring endomorphismof k [ x , . . . , x n ] mapping irreducible polynomials to square-free polynomialsmaps irreducible polynomials to irreducible polynomials. And then, in par-ticular, every k -endomorphism of k [ x , . . . , x n ] mapping irreducible polyno-mials to square-free polynomials maps irreducible polynomials to irreduciblepolynomials. Remark 3.
Theorem 4.1 is a multi-dimensional version of the followinglemma of Freudenburg ([6]): if an irreducible polynomial g ∈ C [ x, y ] dividesboth partial derivatives ∂f∂x , ∂f∂y of a given polynomial f ∈ C [ x, y ], then g divides f + c for some c ∈ C . Van den Essen, Nowicki and Tyc ([4]) generalizedthis lemma for n variables over an algebraically closed field of characteristic 0.In [8] the author obtained the following generalization for an arbitrary field k of characteristic 0 (not necessarily algebraically closed): an irreduciblepolynomial g ∈ k [ x , . . . , x n ] divides all partial derivatives ∂f∂x , . . . , ∂f∂x n of agiven polynomial f ∈ k [ x , . . . , x n ] if and only if g divides W ( f ) for someirreducible polynomial W ( T ) ∈ k [ T ].Let us take a closer look at a very specific analogy between the cases ofa single polynomial and of n polynomials. In fact, comparing the proofs, wemay argue that there is no real analogy here. The only crucial implicationfor a single polynomial f ∈ k [ x , . . . , x n ] is the following: if an irreduciblepolynomial g ∈ k [ x , . . . , x n ] divides ∂f∂x , . . . , ∂f∂x n , then g divides W ( f ) forsome irreducible polynomial W ( T ) ∈ k [ T ]. For n polynomials f , . . . , f n ∈ k [ x , . . . , x n ] and an irreducible polynomial g ∈ k [ x , . . . , x n ], without anyassumptions, there always exists an irreducible polynomial w ∈ k [ x , . . . , x n ]such that g divides w ( f , . . . , f n ). We have established this fact in Lemma 3.3.Now, for a single polynomial f ∈ k [ x , . . . , x n ] it is easy to show that ifan irreducible polynomial g ∈ k [ x , . . . , x n ] divides ∂f∂x , . . . , ∂f∂x n and W ( f ),where W ( T ) ∈ k [ T ] is an irreducible polynomial, then g divides W ( f ). Theanalog of this fact for n polynomials is, in general, not true, as the followingexample shows. 12 xample 6.2 (Gwo´zdziewicz, Jelonek) . Consider the following polynomialsin k [ x, y ] : f = x , f = xy , g = x and w = x . Then Jac( f , f ) = x and w ( f , f ) = x are divisible by g , but w ( f , f ) is not divisible by g . However, we still can prove that if g divides the jacobian of f , . . . , f n ,then there exists an irreducible polynomial w ∈ k [ x , . . . , x n ] such that g divides w ( f , . . . , f n ). Finally, the reverse implication is also not easy to beproved, in contrast to the case of a single polynomial. It is an easy exerciseto show for f, g ∈ k [ x , . . . , x n ], where g is irreducible, that if g divides W ( f ) for some irreducible polynomial W ( T ) ∈ k [ T ], then g divides ∂f∂x i for i = 1 , . . . , n . Acknowledgements.
The author would like to thank Dr. Janusz Gwo´z-dziewicz and Prof. Zbigniew Jelonek for Example 6.2. The author would alsolike to thank Prof. Ludwik M. Dru˙zkowski for the correction of Lemma 3.5from the first version of this paper.
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