Quadratic homogeneous Keller maps of rank two

Quadratic homogeneous Keller maps of rank two

Linear Algebra and its Applications 476 (2015) 16–27 Contents lists available at ScienceDirect Linear Algebra and its Applications www.elsevier.com/...

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Linear Algebra and its Applications 476 (2015) 16–27

Contents lists available at ScienceDirect

Linear Algebra and its Applications www.elsevier.com/locate/laa

Quadratic homogeneous Keller maps of rank two Kevin Pate, Charles Ching-An Cheng ∗ Department of Mathematics and Statistics, Oakland University, Rochester, MI, United States

a r t i c l e

i n f o

Article history: Received 13 June 2014 Accepted 6 February 2015 Available online 10 March 2015 Submitted by J.M. Landsberg MSC: 14R10 14R20

a b s t r a c t Let H be a quadratic homogeneous polynomial map of dimension n over an infinite field in which 2 is invertible such that its Jacobian JH is nilpotent. Meisters and Olech have shown that JH is strongly nilpotent if n ≤ 4. They also proved that it is not true when n = 5. We show that if rank JH ≤ 2 and n arbitrary, then JH is strongly nilpotent. We also give examples to show that this is no longer true for any rank and dimension as long as the rank is greater than 2. © 2015 Elsevier Inc. All rights reserved.

Keywords: Quadratic Homogeneous Polynomial map Keller map Nilpotent Strongly nilpotent Linearly triangularizable

1. Introduction Let k be a field and H = (H1 , . . . , Hn ) : kn → kn a polynomial map, i.e. H(x) = (H1 (x), . . . , Hn (x)) where Hi ∈ k[X] = k[X1 , . . . , Xn ]. The degree of H is defined by deg H = max{deg Hi |i = 1, 2, . . . , n}. H is homogeneous of degree d if each Hi is either homogeneous of degree d or equal to zero. A matrix M ∈ Mn (k[X1 , . . . , Xs ]), is strongly * Corresponding author. E-mail address: [email protected] (C.C.-A. Cheng). http://dx.doi.org/10.1016/j.laa.2015.02.014 0024-3795/© 2015 Elsevier Inc. All rights reserved.

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nilpotent if M (v1 )M (v2 ) · · · M (vn ) = 0 for any v1 , v2 , . . . , vn ∈ ks . Meisters and Olech [8] proved for any quadratic homogeneous map H : Rn → Rn that if JH is nilpotent then JH is strongly nilpotent if n ≤ 4 and the statement is false if n ≥ 5. In this paper we prove the following. Theorem 1. Suppose k is an infinite field with 2 invertible and H : kn → kn is a quadratic homogeneous polynomial map such that JH is nilpotent. If JH ∈ Mn (k(X)) has rank at most 2 then JH is strongly nilpotent. This extends the result of Meisters and Olech [8] when n = 3. When the field k has characteristic 0, an anonymous referee has pointed out that this result can be derived from [3] and a proof is given in Appendix A. A polynomial map F is invertible if there exists a polynomial map G such that F G = GF = (X1 , . . . , Xn ). The map F is Keller if the determinant of its Jacobian, JF = (∂Fi /∂Xj ), is a nonzero element of k. By the chain rule for Jacobians, invertible polynomial maps are Keller maps. The famous Jacobian Conjecture states that if k has characteristic 0 then any Keller map is invertible. Wang [11] proved it if deg F ≤ 2. Bass, Connel and Wright [1] reduced the conjecture to the case where F = X + H with H homogeneous of degree 3. (Druzkowski [5] has further reduced it to where H is cubic linear.) When H is homogeneous of degree ≥ 2, it can be shown that F = X + H is Keller if and only if JH is nilpotent (see [1] or [12]). A polynomial map of the form (X1 + p1 , . . . , Xn + pn ) is lower triangular if each pi is a polynomial in k[X1 , . . . , Xi−1 ]. It is upper triangular if each pi is a polynomial in k[Xi+1 , Xi+2 , . . . , Xn ]. An upper triangular map can be turned into a lower triangular map, and vice versa, by conjugating it with the invertible linear map (Xn , . . . , X1 ). A polynomial map F is triangular if it is either upper or lower triangular. It is linearly triangularizable (LT) if it is linearly conjugate to a triangular map, i.e., there exists an invertible linear map T such that T −1 F T is triangular. When k has characteristic 0, van den Essen and Hubbers [7] proved that JH is strongly nilpotent if and only if the polynomial map X + H is LT. Thus we may deduce from Theorem 1 the following. Corollary 2. Suppose k has characteristic 0 and F = X + H is a quadratic homogeneous Keller polynomial map with rank JH ≤ 2. Then F is LT. This extends a result of Cheng [4] that all quadratic linear maps of rank 2 are LT. In Section 2 we prove Theorem 1. In Section 3 we prove a technical lemma which is needed for the proof of Theorem 1. In Section 4 we exhibit examples to show that the conclusion of Theorem 1 is no longer true if the rank of JH is greater than 2. In Appendix A we provide another proof of Theorem 1 when the field k is algebraically closed with characteristic 0.

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2. Proof of Theorem 1 Proposition 3. ([8]) Suppose k is an arbitrary field and H : kn → kn a quadratic homogeneous polynomial map. Then, for any x, y ∈ kn , (i) JH(αx + βy) = αJH(x) + βJH(y) for α, β ∈ k (linearity), (ii) JH(x)y = JH(y)x (symmetry). Proof. (i) follows from the hypothesis that H is quadratic. Since, for each s,  ∂Hs = aij Xi , ∂Xj i=1 n

[JH(x)y]s = JHs (x)y =

n  ∂Hs j=1

=

n 

n 

(

i=1 j=1

∂Xj

aij yj )xi =

(x)yj =

n  i=1

n  n  ( aij xi )yj j=1 i=1

∂Hs (y)xi = JHk (y)x = [JH(y)x]s . ∂Xi

This proves (ii). 2 Lemma 4. (Meisters, Olech [8]) Suppose k is a field in which 2 is invertible. If H : kn → kn is a quadratic homogeneous map such that JH 2 = 0, then JH is strongly nilpotent. Proof. By Proposition 3, we have 0 = JH(x +y)2 = [JH(x) +JH(y)]2 = JH(x)JH(y) + JH(y)JH(x). So 0 = JH(x)JH(y)y + JH(y)JH(x)y = JH(x)JH(y)y + JH(y)2 x = JH(x)JH(y)y = JH(x)(2H(y)) by Euler’s rule. Taking Jacobians with respect to y gives JH(x)JH(y) = 0. Thus JH is strongly nilpotent. 2 The proof of Theorem 1 is by induction and relies on the following result. Lemma 5. (Sun [10]) Suppose k is an arbitrary field and H : kn → k n is a quadratic polynomial map such that JH is nilpotent. If the columns of JH are dependent over k then there exists an invertible linear map T such that JH  has its last column zero where ˜ is strongly nilpotent where H ˜ = (H  , H  , · · · , H  ), H  = T −1 HT . Furthermore, if J H 1 2 n−1 then JH is also strongly nilpotent. Proof. Since the columns of JH are dependent over k, there exists v ∈ kn such that JH · v = 0. Let T ∈ GLn (k) with its last column v and let H  = T −1 HT . Then JH  · en = T −1 JH(T X)T · en = T −1 JH(T X) · v = 0. Hence the last column of JH  is zero. When 2 is invertible in k and H is quadratic, ˜ is strongly nilpotent, then so is JH  and thus JH. 2 H  does not involve Xn . If J H

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˜ above is a Remark 1. If 2 is invertible in k and H is quadratic homogeneous then H n n polynomial map k → k . Proof of Theorem 1. If rank JH < 2 or if n < 3 then JH 2 = 0 and so, by Lemma 4, JH is strongly nilpotent. Hence Theorem 1 is true for n < 3 or rank JH < 2. Henceforth we will assume rank JH = 2 and n ≥ 3 and will prove by induction on n. Since JH is nilpotent, JH = S −1 N S for some S ∈ GLn (k(X)) where ⎛

J2 N =⎝ 0 0

0 J2 0

⎞ 0 0 ⎠

(1)

0n−4

or  N=

J3 0



0

(2)

0n−3

and where  J2 =

0 0 1 0





0 ⎝ and J3 = 1 0

0 0 1

⎞ 0 0⎠. 0

In case (1), N 2 = 0 and so JH 2 = 0. By Lemma 4, JH is strongly nilpotent. Therefore we may assume (2), so JH 3 = 0 and JH 2 = 0. Lemma 6 below shows that, since n ≥ 3, the third column of JH can be made zero by linear conjugation, i.e. there ¯ has its third column zero. So, by ¯ = T −1 HT1 then J H exists T1 ∈ GLn (k) such that if H 1 ¯ then JH  has Lemma 5, there is an invertible linear map T such that if H  = T −1 HT ˜ is strongly nilpotent where H ˜ = (H  , H  , · · · , H  ) then so last column zero and if J H 1 2 n−1  ¯ ˜ is J H and thus JH. Since JH is nilpotent of rank 2, J H is nilpotent of rank at most 2. ˜ is strongly nilpotent. Therefore JH is also strongly nilpotent. Hence, by induction, J H This completes the proof. 2 Lemma 6. Suppose 2 is invertible in k and H is a quadratic homogeneous polynomial map of dimension n ≥ 3 such that JH 3 = 0 and JH 2 = 0. If rank JH = 2 then there ¯ = T −1 HT1 then J H ¯ has its third column zero. exists T1 ∈ GLn (k) such that if H 1 3. Proof of Lemma 6 Lemma 7. Suppose A = (aij ) ∈ Matn (k) is nilpotent where k is a field. If there exists i such that every entry of the ith row (resp. column) is zero except possibly the ith, then this entry is also zero. Proof. The result follows from the fact that the ith row (resp. column) of As is zero except possibly the ith entry (aii )s . 2

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Proof of Lemma 6. Since JH 2 = 0, there exists p such that JH(p)2 = 0. Thus rank JH(p)2 ≥ 1. The nilpotence of JH(p) implies that rank JH(p) > rank JH(p)2 . But 2 = rank JH ≥ rank JH(p). So rank JH(p) = 2 and rank JH(p)2 = 1. Hence dim ker JH(p) = n − 2 and dim ker JH(p)2 = n − 1. Moreover, ker JH(p) ⊂ ker JH(p)2 . There are two cases to consider: either JH 2 X = 0 or JH 2 X = 0. Case 1: JH 2 X = 0. Since k is infinite, the intersection of {p | H(p) = 0} and {p | JH(p)2 = 0} is nonempty. Hence we may choose p such that JH(p)p = 2H(p) = 0 and JH(p)2 = 0. So there exists u such that JH(p)2 u = 0. Let p1 = u and p2 = p. Then p1 ∈ / ker JH(p)2 , and 2 2 2 p2 ∈ ker JH(p) because JH X = 0 implies JH(p) p = 0. Next define p3 = JH(p)p ∈ ker JH(p). Note that p3 = JH(p)p2 . Since p1 , p2 , p3 are linearly independent, p4 , . . . , pn can be chosen in ker JH(p) so that {p1 , . . . , pn } is a basis of kn . Let P be the invertible matrix [p1 , . . . , pn ], let P¯ = p1 X1 + · · · + pn Xn be the corre¯ = P¯ −1 H P¯ . For 1 ≤ i ≤ n, let Ai = J H(e ¯ i ). sponding invertible linear map and let H Thus Ai = P −1 JH(P ei )P = P −1 JH(pi )P. ¯ = A1 X1 + · · · + An Xn , so J H ¯ has its third column zero if By the linearity of JH, J H and only if all the Ai have theirs zero also. By symmetry, Ai ej = P −1 JH(P ei )P ej = P −1 JH(P ej )P ei = Aj ei for all i, j. Denoting the (i, j)th entry of Ak by akij , we have akij = eti Ak ej = eti Aj ek = ajik

(3)

for all indices i, j and k. Since JH(p)p1 ∈ ker JH(p)2 = p2 , . . . , pn , we have JH(p)p1 = c2 p2 + · · · + cn pn for some ci ∈ k. Also c2 = 0, otherwise JH(p)p1 ∈ p3 , . . . , pn = ker JH(p). Replacing p1 by (1/c2 )p1 , we may assume c2 = 1. Since JH(p)p2 = p3 , and p3 , . . . , pn ∈ ker JH(p), we have A2 = P −1 JH(p2 )P = P −1 [JH(p)p1 , . . . , JH(p)pn ] = P −1 [p2 + c3 p3 + · · · + cn pn , p3 , 0, . . . , 0] = [e2 + c3 e3 + · · · + cn en , e3 , 0, . . . , 0]. Hence, by (3), ak12 = a21k = 0 for all k.

(4)

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3 ¯ ¯ i ) ± J H(e ¯ j ))3 = (J H(e ¯ i ± ej ))3 = 0. Hence Since J H(x) = 0, (Ai ± Aj )3 = (J H(e 0 = (Ai + Aj )3 − (Ai − Aj )3 = 2(A2i Aj + Aj A2i + Ai Aj Ai ). Taking i = 2 and j = k, we have

A22 Ak + Ak A22 + A2 Ak A2 = 0.

(5)

It is not difficult to check that ⎛

0 ⎜ 0 ⎜ ⎜ ak ⎜ 11 2 A2 Ak = ⎜ ⎜ 0 ⎜ .. ⎝ . 0

0 0 ak12 0 .. .

0 0 ak13 0 .. .

0

0



⎞ ··· 0 ··· 0 ⎟ ⎟ · · · ak1n ⎟ ⎟ , ··· 0 ⎟ ⎟ .. ⎟ ··· . ⎠ ···

0

ak13 ⎜ ak23 ⎜ ⎜ k Ak A22 = ⎜ a33 ⎜ . ⎝ ..

0 0 0 .. .

⎞ ··· 0 ··· 0⎟ ⎟ ··· 0⎟ ⎟, .. ⎟ ··· . ⎠

akn3

0

··· 0

and ⎛

0 ⎜ (ak12 + c3 ak13 + · · · + cn ak1n ) ⎜ ⎜ ∗ ⎜ A2 Ak A2 = ⎜ c (ak + c ak + · · · + c ak ) 3 13 n 1n ⎜ 4 12 ⎜ .. ⎝ .

0 ak13 (c3 ak13 + ak23 ) c4 ak13 .. .

cn (ak12 + c3 ak13 + · · · + cn ak1n )

cn ak13

0⎞ 0⎟ ⎟ 0⎟ ⎟ . 0⎟ ⎟ .. ⎟ ··· . ⎠ 0 ··· 0 0 0 0 0 .. .

··· ··· ··· ···

Thus Eq. (5) becomes ⎛

ak13 ⎜ ak23 + (ak12 + c3 ak13 + · · · + cn ak1n ) ⎜ ⎜ ∗ ⎜ ⎜ ak + c (ak + c ak + · · · + c ak ) 4 12 3 13 n 1n ⎜ 43 ⎜ .. ⎝ . akn3 + cn (ak12 + c3 ak13 + · · · + cn ak1n )

0 ak13 k a12 + (c3 ak13 + ak23 ) c4 ak13 .. .

0 0 ak13 0 .. .

cn ak13

0

This implies ak13 = ak14 = · · · = ak1n = 0 for all k,

··· 0 ⎞ ··· 0 ⎟ ⎟ · · · ak1n ⎟ ⎟ = 0. ··· 0 ⎟ ⎟ ⎟ .. .. ⎠ . . ··· 0

K. Pate, C.C.-A. Cheng / Linear Algebra and its Applications 476 (2015) 16–27

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reducing it to ⎛

0 ⎜ ak23 + ak12 ⎜ ⎜ ∗ ⎜ ⎜ ak + c (ak ) 4 12 ⎜ 43 ⎜ .. ⎝ . akn3 + cn (ak12 )

0 0 ak12 + ak23 0 .. . 0

0⎞ 0⎟ ⎟ 0⎟ ⎟ = 0. 0⎟ ⎟ .. ⎟ .⎠ 0 ··· 0

0 0 0 0 .. .

··· ··· ··· ··· .. .

(6)

But ak12 = 0 by (4), so (6) gives ak23 = ak43 = · · · = akn3 = 0 for all k. Thus ak13 = ak23 = ak43 = · · · = akn3 = 0 for all k. So the only possible nonzero entry of the third column of Ak is the third, so by Lemma 7, ak33 = 0 for all k. Case 2: JH 2 X = 0. This is similar to Case 1. First choose a basis for kn . Since JH 2 X = 0, p can be chosen so that JH(p)2 p = 0. Let p1 = p. Then JH(p)2 p1 = 0. Let p2 = JH(p)p ∈ ker JH(p)2 and p3 = JH(p)2 p ∈ ker JH(p). Note that p2 = JH(p)p1 and p3 = JH(p)p2 . Finally, p4 , . . . , pn can be chosen in ker JH(p) so that {p1 , . . . , pn } is a basis of kn . Take P to be ¯ and the A s as in Case 1. the invertible matrix [p1 , . . . , pn ], and define H i Since JH(p)p1 = p2 , JH(p)p2 = p3 , and p3 , . . . , pn ∈ ker JH(p), we have A1 = P −1 [JH(p1 )p1 , . . . , JH(p1 )pn ] = P −1 [p2 , p3 , 0, . . . , 0] = [e2 , e3 , 0, . . . , 0]. But this is precisely the matrix A2 from Case 1 if we take c3 = · · · = cn = 0, so the computations for Eqs. (5) and (6) are still valid, only replacing A2 from Case 1 by A1 . Eq. (4) no longer holds, instead we have ak11 = 0 for all k. As before, we have ak13 = ak14 = · · · = ak1n = 0 for all k. Symmetry gives a31k = a41k = · · · = an1k = 0 for all k. Therefore the first rows of A3 , A4 , . . . , An are zero.

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Since ci = 0, 3 ≤ i ≤ n, we also obtain from Eq. (6) ak43 = · · · = akn3 = 0 for all k. This together with ak13 = 0 implies that in the third column of each Ak , all entries except possibly the second and third, ak23 and ak33 , are zero. By (3) ⎛

0 ⎜ a123 ⎜ ⎜ a1 ⎜ 33 A3 = ⎜ ⎜ 0 ⎜ .. ⎝ .

⎞ ··· 0 · · · an23 ⎟ ⎟ · · · an33 ⎟ ⎟ . ··· 0 ⎟ ⎟ .. ⎟ ··· . ⎠

0 a223 a233 0 .. .

0

···

0

0

By (6), ak23 = −ak12 for each k. Thus ⎛

0 ⎜ −a112 ⎜ ⎜ a1 ⎜ 33 A3 = ⎜ ⎜ 0 ⎜ .. ⎝ .

⎞ ··· 0 · · · −an12 ⎟ ⎟ · · · an33 ⎟ ⎟ . ··· 0 ⎟ ⎟ .. ⎟ ··· . ⎠

0 −a212 a233 0 .. .

0

···

0

0

Since the first rows of A1 , A3 , A4 , . . . , An are zero, −ak12 = 0 for all k = 2. Therefore ⎛

0 ⎜ 0 ⎜ ⎜ a1 ⎜ 33 A3 = ⎜ ⎜ 0 ⎜ .. ⎝ . 0

0 −a212 a233 0 .. .

0 0 a333 0 .. .

0

0

⎞ ··· 0 ··· 0 ⎟ ⎟ · · · an33 ⎟ ⎟ . ··· 0 ⎟ ⎟ .. ⎟ ··· . ⎠ ···

0

Applying Lemma 7 to the second row, −a212 = 0. Thus ak23 = 0 for all k; hence every entry of the third column of Ak is zero except possibly the third entry. By Lemma 7, ak33 = 0 for all k, and the third column of every Ak is zero. This completes the proof of Lemma 6. 2 4. Examples Meisters and Olech [8] show that in dimension 5, Vasyunin has produced a quadratic homogeneous polynomial map F4 with Jacobian matrix of rank 4 which is nilpotent but not strongly nilpotent where

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1 1 F4 = (0, X1 X3 , X1 X4 + X22 , X1 X5 − X2 X3 , X32 ). 2 2 Using the following we can conclude that when rank JH is 4, JH being nilpotent does not always imply it being strongly nilpotent regardless of what dimension H has. Proposition 8. Suppose H is a polynomial map over k of dimension n and suppose JH ¯ = (H1 , . . . , Hn , 0, . . . , 0) be of dimension is nilpotent but not strongly nilpotent. Let H ¯ is nilpotent and not strongly nilpotent. s > n. Then J H Proof. It is easy to check that  ¯ = JH

JH 0

0 0

 .

Then  ¯ (1) ) · · · J H(v ¯ (k) ) = J H(v

JH(v (1) ) · · · JH(v (k) ) 0 0 0

 .

2

What if rank JH ≥ 5? The following proposition combined with Proposition 8 concludes that nilpotence does not imply strong nilpotence when the rank is at least 5 in any dimension. Proposition 9. Suppose H = (0, X1 X3 , X1 X4 + 12 X22 , X1 X5 − X2 X3 , 12 X32 , 12 X52 , 12 X62 , . . . , 1 2 2 Xn−1 ) is a polynomial map over k of dimension n ≥ 5. Then JH is nilpotent but not strongly nilpotent and rank JH = n − 1. Proof. Note that the first five components of H are the components of the example F4 of Vasyunin. In this case JH = A + B where ⎛

0 0 0 ⎜X 0 X 3 1 ⎜ ⎜ X 0 X ⎜ 4 2 ⎜ ⎜ X5 −X3 −X2 JF4 0 ⎜ =⎜ 0 A= 0 X3 ⎜ 0 0 ⎜ ⎜ 0 0 0 ⎜ .. .. ⎜ .. ⎝ . . . 0 0 0 and

0 0 X1 0 0

0 0 0 X1 0

0 0 0 0 0

··· ··· ··· ··· ···

0 .. . 0

0 .. . 0

0 .. . 0

··· .. . ···

⎞ 0 0⎟ ⎟ ⎟ 0⎟ ⎟ 0⎟ ⎟ 0⎟ ⎟ ⎟ 0⎟ ⎟ .. ⎟ .⎠ 0

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0 ⎜. ⎜ .. ⎜ ⎜ ⎜0 ⎜ ⎜ B = ⎜0 ⎜ ⎜ ⎜0 ⎜ ⎜ ⎝0 0

··· .. . ···

0 .. . 0

0 .. . 0

0 ··· .. .. . . 0 ···

0 .. . 0

· · · 0 X5 0 · · · 0 .. . · · · 0 0 X6 . . . .. . 0 ··· 0 0 0 · · · 0 0 0 0 Xn−1

25

⎞ 0 .. ⎟ .⎟ ⎟ ⎟ 0⎟ ⎟ 0⎟ ⎟. .. ⎟ ⎟ .⎟ ⎟ ⎟ 0⎠ 0

It is not hard to check that rank JH = n − 1. Since A5 = 0, B n−4 = 0 and AB = 0, JH n = (A + B)n = B n−3 A3 + B n−2 A2 + B n−1 A + B n = 0, i.e. JH is nilpotent. It is easy to see that JH(e1 ) = [0, 0, e2 , e3 , e4 , 0, . . . , 0] and JH(e2 ) = [0, e3 , −e4 , 0, . . . , 0]. Hence JH(e1 )JH(e2 ) = [0, e2 , −e3 , 0, . . . , 0] and so JH is not strongly nilpotent. 2 It remains to investigate the case where rank JH = 3. For dimension 4, Meisters and Olech [8] have proved that nilpotence does imply strong nilpotence. But for higher dimensions, the answer is negative as the following example together with Proposition 8 concludes. This example is a modification of that of Vasyunin, H = (0, X1 X3 , X1 X4 + X2 X5 , X1 X5 − X3 X5 , 0). The details are left for the reader. It is worth noting that de Bondt [2] has computed all homogeneous Keller maps of dimensions 4 and 5. Appendix A As pointed out earlier, when the field k has characteristic 0, Theorem 1 can be deduced from the following result of de Bondt and van den Essen [3, Theorem 2.1]. Their proof uses a theorem of Bertini [9, Theorem 18]. Theorem 10 (de Bondt, van den Essen). Suppose k is algebraically closed with characteristic 0 and H = (H1 , . . . , Hn ) : kn → k n is a homogeneous polynomial map of degree at least two. If rank JH ≤ 2 then Hi = ghi (p, q) for all i where g = gcd(Hi ), hi ∈ k[t1 , t2 ] are homogeneous of the same degree or 0 and p, q ∈ k[X1 , . . . , Xn ] are homogeneous of the same degree. Lemma 11. Suppose a2 + bc = 0 where a, b, c ∈ k[X1 , . . . , Xn ] and k is a field. If a = a1 X1 + · · · + an Xn , b = b1 X1 + · · · + an Xn and c = c1 X1 + · · · + cn Xn with ai , bi , ci ∈ k then the following holds for all i, j. (i) bi cj = bj ci ,

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(ii) ai aj + bi cj = 0, (iii) ai bj = aj bi , (iv) ai cj = aj ci . Proof. The hypothesis is equivalent to 2ai aj + bi cj + bj ci = 0.

(∗)

In particular, when i = j, a2i + bi ci = 0. √ Therefore, in the algebraic closure k of k, ai = ± −bi ci for all i. Hence we obtain from (∗)

bi cj + bj ci = −2(± −bi ci )(± −bj cj ). Squaring both sides we obtain (i). Using (i) and (∗), (ii) follows. By (i) and (ii) ai aj = −bi bj = −bj bi = aj ai establishing (iii). (iv) is symmetrical to (iii). 2 Proposition 12. Suppose H = (H1 , H2 ) : k n → k2 where k is a field and H1 , H2 ∈ k[X1 , . . . , Xn ] are quadratic homogeneous with n ≥ 2. If J = JX1 ,X2 (H1 , H2 ) is nilpotent then it is strongly nilpotent.   Proof. Since J = ac db is nilpotent, d = −a. Let a = a1 X1 + · · · + an Xn , b = b1 X1 + · · · + an Xn and c = c1 X1 + · · · + cn Xn with ai , bi , ci ∈ k. Then 

  a(X) b(X) a(Y ) b(Y ) J(X)J(Y ) = c(X) −a(X) c(Y ) −a(Y )   a(X)a(Y ) + b(X)c(Y ) a(X)b(Y ) − b(X)a(Y ) . = c(X)a(Y ) − a(X)c(Y ) c(X)b(Y ) + a(X)a(Y ) The reader may check that the coefficients of Xi Yj in the above matrix entries are all zero by Lemma 11 showing that J is strongly nilpotent. 2 Corollary 13. Suppose H = (H1 , H2 , 0, · · · , 0) : kn → kn is a nilpotent quadratic homogeneous polynomial map with k a field. Then JH is strongly nilpotent.  k k−1    k N N M N = J (H , H ). Since JH = where Proof. Note that JH = N0 M X ,X 1 2 1 2 0 0 0 and JH is nilpotent, N is also nilpotent. By Proposition 12, N is also strongly nilpotent. Hence JH is strongly nilpotent. 2 Proof of Theorem 1 when k is algebraically closed with characteristic 0. By Theorem 10, Hi = ghi (p, q). If deg g = 2 then hi ∈ k. If deg g = 1 then Hi = gli where li is a linear

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form. If deg g = 0 and deg hi = 1 then Hi = ci p + di q with ci , di ∈ k. In all these cases, H is conjugate to a quadratic homogeneous polynomial map (F1 , F2 , 0, · · · , 0) and hence JH is strongly nilpotent by Corollary 13. The only remaining case is when deg g = 0 and deg hi = 2. In this case, Hi = hi (p, q) where p, q are linear forms. If p, q are linearly dependent, then rank H ≤ 1 and H is conjugate to a quadratic homogeneous polynomial map (F1 , 0, . . . , 0) so, as before, JH is strongly nilpotent. If p and q are linearly independent then, without loss of generality, we may assume that p = X1 , q = X2 . Then Hi ∈ k[X1 , X2 ]. Since JH is nilpotent, J = JX1 ,X2 (H1 H2 ) is nilpotent. By Proposition 12, J is strongly nilpotent. It follows that JH is also strongly nilpotent. This proves Theorem 1 when k is algebraically closed with characteristic 0. Using the well-known technique of Lefschetz principle (see, e.g. [6]), one can remove the condition that k is algebraically closed. 2 References [1] H. Bass, E. Connell, D. Wright, The Jacobian conjecture: reduction of degree and formal expansion of the inverse, Bull. Amer. Math. Soc. 7 (1982) 287–330. [2] M. de Bondt, Homogeneous Keller Maps, Ph.D. thesis, Radboud University, Nijmegen, Netherlands, July 2009. [3] M. de Bondt, A. van den Essen, The Jacobian conjecture: linear triangularization for homogeneous polynomial maps in dimension three, J. Algebra 294 (2005) 294–306. [4] C.C. Cheng, Quadratic linear Keller maps, Linear Algebra Appl. 348 (2002) 203–207. [5] L.M. Drużkowski, The Jacobian conjecture in case of rank or corank less than three, J. Pure Appl. Algebra 85 (1993) 233–244. [6] A. van den Essen, Polynomial Automorphisms and the Jacobian Conjecture, Progr. Math., vol. 190, Birkhäuser, 2000. [7] A. van den Essen, E. Hubbers, Polynomial maps with strongly nilpotent Jacobian matrix and the Jacobian conjecture, Linear Algebra Appl. 247 (1996) 121–132. [8] G.H. Meisters, C. Olech, Strong nilpotence holds in dimension up to five only, Linear Multilinear Algebra 30 (1991) 231–255. [9] A. Schinzel, Selected Topics on Polynomials, The University of Michigan Press, Ann Arbor, 1982. [10] X. Sun, On quadratic homogeneous quasi-translations, J. Pure Appl. Algebra 214 (2010) 1962–1972. [11] S.S.S. Wang, A Jacobian criterion for separability, J. Algebra 65 (1980) 453–494. [12] D. Wright, Formal inverse expansion and the Jacobian conjecture, J. Pure Appl. Algebra 48 (1987) 199–219.