arXiv:1711.00904v1 [] 2 Nov 2017

Quadratic homogeneous polynomial maps H and
Keller maps x + H with 3 ≤ rk J H ≤ 4
Michiel de Bondt
November 6, 2017
Abstract
We compute by hand all quadratic homogeneous polynomial maps H
and all Keller maps of the form x + H, for which rk J H = 3, over a field
of arbitrary characteristic.
Furthermore, we use computer support to compute Keller maps of the
form x + H with rk J H = 4, namely:
• all such maps in dimension 5 over fields with 21 ;
• all such maps in dimension 6 over fields without 12 .
We use these results to prove the following over fields of arbitrary characteristic: for Keller maps x + H for which rk J H ≤ 4, the rows of J H
are dependent over the base field.

1

Introduction

Let n be a positive integer and let x = (x1 , x2 , . . . , xn ) be an n-tuple of variables.
We write a|b=c for the result of substituting b by c in a.
Let K be any field. In the scope of this introduction, denote by L an
unspecified (but big enough) field, which contains K or even K(x).
For a polynomial or rational map H = (H1 , H2 , . . . , Hm ) ∈ Lm , write J H
or Jx H for the Jacobian matrix of H with respect to x. So




J H = Jx H = 


∂
∂x2 H1
∂
∂x2 H2

∂
∂x1 H1
∂
∂x1 H2

..
.
∂
∂x1 Hm

..
.
∂
∂x2 Hm

···
···
.. .. ..
...
···

∂
∂xn H1
∂
∂xn H2






..

.
∂
∂xn Hm

Denote by rk M the rank of a matrix M , whose entries are contained in L,
and write trdegK L for the transcendence degree of L over K. It is known that
rk J H ≤ trdegK K(H) for a rational map H of any degree, with equality if
K(H) ⊆ K(x) is separable, in particular if K has characteristic zero. This is
proved in [dB2, Th. 1.3], see also [PSS, Ths. 10, 13].

1

Let a Keller map be a polynomial map F ∈ K[x]n , for which det J F ∈
K \ {0}. If H ∈ K[x]n is homogeneous of degree at least 2, then x + H is a
Keller map, if and only if J H is nilpotent.
We say that a matrix M ∈ Matn (L) is similar over K to a matrix M̃ ∈
Matn (L), if there exists a T ∈ GLn (K) such that M̃ = T −1 M T . If J H
is similar over K to a triangular matrix, say that T −1 (J H)T is a triangular
matrix, then

J T −1 H(T x) = T −1 (J H)|x=T x T

is triangular as well.
Section 2 is about quadratic homogeneous maps H in general, with the focus
on compositions with invertible linear maps, and the invariant r = rk J H. In
section 3, a classification is given for the case r = 3.
In section 4, a classification is given for the case r = 3, combined with the
nilpotency of J H. Nilpotency is an invariant of conjugations with invertible
linear maps, but is not an invariant of compositions with invertible linear maps
in general.
In section 5, we compute all Keller maps x + H with H quadratic homogeneous, for which rk J H = 4 in dimension 5 over fields with 21 , and in dimension
6 over fields without 21 . We use these results to prove the following over fields of
arbitrary characteristic: for Keller maps x + H with H quadratic homogeneous,
for which rk J H ≤ 4, the rows of J H are dependent over the base field.

2

rank r

Theorem 2.1. Let H ∈ K[x]m be a quadratic homogeneous polynomial map,
and r := rk J H.
Then there are S ∈ GLm (K) and T ∈ GLn (K), such that for H̃ := SH(T x),
only the first 12 r2 + 21 r rows of J H̃ may be nonzero, and one of the following
statements holds:
(1) Only the first 21 r2 − 12 r + 1 rows of J H̃ may be nonzero;
(2) char K 6= 2 and only the first r columns of J H̃ are nonzero;
(3) char K = 2 and only the first r + 1 columns of J H̃ are nonzero.
Conversely, rk J H̃ ≤ r if either H̃ is as in (2) or (3), or 1 ≤ r ≤ 2 and H̃ is
as in (1).
Corollary 2.2. Let H ∈ K[x]m be a quadratic homogeneous polynomial map.
Suppose that r := rk J H ≤ 4.
Then there are S ∈ GLm (K) and T ∈ GLn (K), such that for H̃ := SH(T x),
only the first 21 r2 + 21 r of J H̃ may be nonzero, and one of the following statements holds:
(1) Only the first r + 1 rows of J H̃ may be nonzero;

2

(2) r = 4, H̃i ∈ K[x1 , x2 , x3 ] for all i ≥ 2, and char K 6= 2;
(3) r = 4, H̃i ∈ K[x1 , x2 , x3 , x4 , x25 , x26 , . . . , x2n ] for all i ≥ 2, and char K = 2;
(4) Only the first r + 1 columns of J H̃ may be nonzero;
(5) r = 4, only the leading principal minor matrix of size 6 of J H̃ is nonzero,
and char K = 2.
Lemma 2.3. If H̃ is as in (2) or (3) of theorem 2.1, then theorem 2.1 holds for
H.
Proof. The number of terms of degree d in x1 , x2 , . . . , xr is


r−1+d
d
and the number of square-free terms of degree d in x1 , x2 , . . . , xr , xr+1 is


r+1
d
which is both 12 r2 + 21 r if d = 2.
If H̃ is as in (2) of theorem 2.1, then char K 6= 2 and the rows of J H̃ are
dependent over K on the 12 r2 + 21 r rows of
J (x21 , x1 x2 , . . . , x2r )
If H̃ is as in (3) of theorem 2.1, then char K = 2 and the rows of J H̃ are
dependent over K on the 21 r2 + 21 r rows of
J (x1 x2 , x1 x3 , . . . , xr xr+1 )
So at most 12 r2 + 12 r rows of J H̃ as in theorem 2.1 need to be nonzero.
If H̃ is as in (2) of theorem 2.1, then rk J H̃ ≤ r is direct. If H̃ is as in
(3) of theorem 2.1, then rk J H̃ ≤ r follows as well, because J H̃ · x = 0 if
char K = 2.
Lemma 2.4. Let M be a nonzero matrix whose entries are linear forms in
K[x]. Suppose that r := rk M does not exceed the cardinality of K.
Then there are invertible matrices S and T over K, such that for M̃ :=
SM T ,
M̃ = M̃ (1) L1 + M̃ (2) L2 + · · · + M̃ (n) Ln
where M̃ (i) is a matrix with coefficients in K for each i, L1 , L2 , . . . , Ln are
independent linear forms, and


Ir 0
M̃ (1) =
0 0
3

Proof. Since rk M = r ≥ 1, M has a minor matrix of size r × r whose determinant is nonzero. Assume without loss of generality that the determinant of
the leading principal minor matrix of size r × r of M is nonzero. Then this
determinant f is a homogeneous polynomial of degree r. From [dB1, Lemma
5.1 (ii)], it follows that there exists a v ∈ K n such that f (v) 6= 0.
Take independent linear forms L1 , L2 , . . . , Ln such that Li (v) = 0 for all
i ≥ 2. Then L1 (v) 6= 0, and we may assume that L1 (v) = 1. We can write
M = M (1) L1 + M (2) L2 + · · · + M (n) Ln
where M (i) is a matrix with coefficients in K for each i. If we substitute x = v
on both sides, we see that rk M (1) ≤ r and that the leading principal minor
matrix of size r × r of M (1) has a nonzero determinant.
So rk M (1) = r, and we can choose invertible matrices S and T over K, such
that


Ir 0
(1)
SM T =
0 0
So we can take M̃ (i) = SM (i) T for each i.
Suppose that M̃ is as in lemma 2.4. Write


A B
M̃ =
C D

(2.1)

where A ∈ Matr (K). If we extend A with one row and one column of J H,
we get an element of Matr+1 K(x) from which the determinant is zero. If we
focus on the coefficients of Lr1 and Lr−1
of this determinant, we see that
1
D=0

and

C ·B =0

(2.2)

and

C ·B =0

(2.3)

respectively. In particular,
D=0

Lemma 2.5. Let H̃ ∈ K[x]m , such that J H̃ is as M̃ in lemma 2.4.
Suppose that either char K 6= 2 or the rows of B are dependent over K.
Then the following hold.
(i) The columns of C in (2.1) are dependent over K.
(ii) If C 6= 0, then there exists a v ∈ K n of which the first r coordinates are
not all zero, such that


Ir 0
·x
(J H̃) · v =
0 0
Proof. The claims are direct if C = 0, so assume that C 6= 0. Then there exists
an i > r, such that H̃i 6= 0.
4

(i) Take v as in (ii). From D = 0, we deduce that C · v ′ = (C|D) · v = 0 for
some nonzero v ′ ∈ K r , which yields (i).
(ii) Take v as in lemma 2.4, and write v = (v ′ , v ′′ ), such that v ′ ∈ K r and
v ′′ ∈ K n−r . Since H̃ is quadratic homogeneous, we have


Ir 0
(J H̃) · v = (J H̃)|x=v · x = M̃ (1) · x =
·x
0 0
So it remains to show that v ′ 6= 0. We distinguish two cases:
• the rows of B are dependent over K.
Take w ∈ K r nonzero, such that wt B = 0. Then
t

′

t

′

t

t

′′

w A v = w A v + w B v = w (A|B) v = w

t

x1
x2

..
.

xr

!

=

r
X

wi xi 6= 0

i=1

so v ′ 6= 0.
• char K 6= 2.
From CB = 0, we deduce that
CA v ′ = CA v ′ + CB v ′′ = C (A|B) v = C

x1
x2

..
.

xr

!

 H̃

r+1

H̃



.. 
= 2  r+2
.
H̃m

As H̃i 6= 0 for some i > r, the right-hand side is nonzero, so v ′ 6=
0.
Lemma 2.6. Let H̃ ∈ K[x]m , such that J H̃ is as M̃ in lemma 2.4.
Suppose that rk B + rk C = r and that the columns of C are dependent over
K. Then the column space of B contains a nonzero constant vector.
Proof. From rk C + rk B = r and CB = 0, we deduce that ker C is equal to the
column space of B. Hence any v ′ ∈ K r such that Cv ′ = 0 is contained in the
column space of B.
Proof of theorem 2.1. From lemma 2.7 below, it follows that we may assume
that K has at least r elements. Let M = J H and take S and T as in lemma
2.4. Then S(J H)T is as M̃ in lemma 2.4. Let H̃ := SH(T x). Then J H̃ =
S(J H)|x=T x T is as M̃ in lemma 2.4 as well, but for different linear forms Li .
Take M̃ = J H̃ and take A, B, C, D as in (2.1). We distinguish four cases:
• The column space of B contains a nonzero constant vector.
Then there exists an U ∈ GLm (K), such that the column space of U M̃
contains e1 , because D = 0. Consequently, the matrix which consists of
the last m − 1 rows of J (U H̃) = U M̃ has rank r − 1. By induction on r,
it follows that we can choose U such that only
1
2 (r

− 1)2 + 21 (r − 1) = ( 12 r2 − r + 12 ) + ( 21 r − 21 ) = 12 r2 − 12 r
5

rows of J (U H̃) are nonzero besides the first row of J (U H̃). So U H̃ is as
H̃ in (1) of theorem 2.1.
• The rows of B are dependent over K in pairs.
If B 6= 0, then the column space of B contains a nonzero constant vector,
and the case above applies because D = 0.
So assume that B = 0. Then only the first r columns of J H̃ may be
nonzero, because D = 0. Since rk J H̃ = r, the first r columns of J H̃
are indeed nonzero. Furthermore, it follows from J H̃ · x = 2H̃ that
char K 6= 2. So H̃ is as in (2) of theorem 2.1, and the result follows from
lemma 2.3.
• char K = 2 and rk B ≤ 1.
If the rows of B are dependent over K in pairs, then the second case above
applies, so assume that the rows of B are not dependent over K in pairs.
On account of [dB4, Theorem 2.1], the columns of B are dependent over
K in pairs. As D = 0, there exists an U ′′ ∈ GLn−r (K) such that only the
first column of
 
B
U ′′
D
may be nonzero. Hence there exists an U ∈ GLn (K) such that only the
first r + 1 columns of (J H̃) U may be nonzero. Consequently, H̃(U x) is
as H̃ in (3) of theorem 2.1, and the result follows from lemma 2.3.
• None of the above.
We first show that rk C ≤ r − 2. So assume that rk C ≥ r − 1. From
rk C + rk B ≤ r, it follows that rk B ≤ 1. As the last case above does
not apply, char K 6= 2. From (i) of lemma 2.5, it follows that the columns
of C are dependent over K. As the first case above does not apply, it
follows from lemma 2.6 that rk C + rk B < r. So rk B = 0 and the rows
of B are dependent over K in pairs, which is the second case above, and
a contradiction. So rk C ≤ r − 2 indeed.
By induction on r, it follows that C needs to have at most
1
2 (r

− 2)2 + 21 (r − 2) = ( 21 r2 − 2r + 2) + ( 12 r − 1) = 21 r2 − 32 r + 1

nonzero rows. As A has r rows, there exists an U ∈ GLm (K) such that
U H̃ is as H̃ in (1) of theorem 2.1.
The last claim of theorem 2.1 follows from lemma 2.3 and the fact that 21 r2 −
1
2 r + 1 = r if 1 ≤ r ≤ 2.
Lemma 2.7. Let L be an extension field of K. If theorem 2.1 holds for L
instead of K, then theorem 2.1 holds.

6

Proof. We only prove lemma 2.7 for the first claim of theorem 2.1, because the
second claim can be treated in a similar manner, and the last claim does not
depend on the actual base field.
Suppose H satisfies the first claim of theorem 2.1, but with L instead of K.
If m ≤ 21 r2 + 21 r, then only the first 12 r2 + 21 r rows of J H may be nonzero, and
H satisfies the first claim of theorem 2.1.
So assume that m > 21 r2 + 12 r. Then the rows of J H are dependent over
L. Since L is a vector space over K, the rows of J H are dependent over
K. So we may assume that the last row of J H is zero. By induction on m,
(H1 , H2 , . . . , Hm−1 ) satisfies the first claim for H in in theorem 2.1. As Hm = 0,
we conclude that H satisfies the first claim of theorem 2.1.
Proof of corollary 2.2. If (2) or (3) of theorem 2.1 applies, then (4) of corollary
2.2 follows. So assume that (1) of theorem 2.1 applies. Then only the first
1 2
1
2 r − 2 r + 1 rows of J H̃ may be nonzero.
Suppose first that r ≤ 3. Then
1 2
2r

− 21 r + 1 ≤ 32 r − 21 r + 1 = r + 1

and corollary 2.2 follows.
Suppose next that r = 4. Take M̃ = J H̃ and take A, B, C, D as in (2.1).
We distinguish three cases.
• The column space of B contains a nonzero constant vector.
Then there exists an U ∈ GLm (K), such that the column space of U M̃
contains e1 . So the matrix which consists of the last m − 1 rows of
J (U H̃) = U M̃ has rank r − 1.
Make Ũ from U by replacing its first row by the zero row. Then rk J (Ũ H̃) =
r − 1, and we can apply theorem 2.1 to Ũ H̃.
– If case (1) of theorem 2.1 applies for Ũ H̃, then case (1) of corollary
2.2 follows, because
1
2 (r

− 1)2 − 21 (r − 1) + 1 =

9
2

−

3
2

+1=4=r

– If case (2) of theorem 2.1 applies for Ũ H̃, then case (2) of corollary
2.2 follows.
– If case (3) of theorem 2.1 applies for Ũ H̃, then case (3) of corollary
2.2 follows.
• rk B ≤ 1.
If the columns of B are dependent over K in pairs, then (4) of corollary
2.2 is satisfied. So assume that the columns of B are not dependent over
K in pairs. Then B 6= 0, and from [dB4, Theorem 2.1], it follows that
the rows of B are dependent over K in pairs. Hence the first case above
applies.

7

• rk C ≤ 1.
Then it follows from theorem 2.1 that at most one row of C needs to be
nonzero. So (1) of corollary 2.2 is satisfied.
• None of the above.
We first show that rk C = rk B = 2 and that the columns of C are independent over K. Since rk C + rk B ≤ r = 4, we deduce from rk B > 1 and
rk C > 1 that rk C = rk B = 2. So rk C + rk B = 4 = r. As the first case
above does not apply, it follows from lemma 2.6 that the columns of C are
independent over K.
From (i) of lemma 2.5, we deduce that char K = 2 and that the rows of B
are independent over K. Since the rk C + 2 columns of C are independent
over K, it follows from theorem 2.1 that C needs to have at most
1
2

· 22 −

1
2

·2+1 =2−1+1 = 2

nonzero rows. Since the rk B + 2 rows of B are independent over K, it
follows from [dB4, Theorem 2.3] that B needs to have at most 2 nonzero
columns, because the first case above does not apply. So (5) of corollary
2.2 is satisfied.
The last case in corollary 2.2 is indeed necessary, e.g. J H̃ = H(x1 x2 x3 +
x4 x5 x6 ), or J H̃ = H(x1 x2 x3 + x1 x5 x6 + x4 x2 x6 + x4 x5 x3 ).

3

rank 3

Theorem 3.1. Let H ∈ K[x]m be a quadratic homogeneous polynomial map,
such that r := rk J H = 3. Then we can choose S ∈ GLm (K) and T ∈ GLn (K),
such that for H̃ := SH(T x), one of the following statements holds:
(1) Only the first 3 rows of J H̃ may be nonzero;
(2) Only the first 4 rows of J H̃ may be nonzero, and
(H̃1 , H̃2 , H̃3 , H̃4 ) = (H̃1 , 12 x21 , x1 x2 , 12 x22 )
(in particular, char K 6= 2);
(3) Only the first 4 rows of J H̃ may be nonzero,
J (H̃1 , H̃2 , H̃3 , H̃4 ) = J (H̃1 , x1 x2 , x1 x3 , x2 x3 )
and char K = 2;
(4) H̃ is as in (2) or (3) of theorem 2.1;
(5) Only the first 4 rows of J H̃ may be nonzero, and


H̃1 , H̃2 , H̃3 , H̃4 = x1 x3 + cx2 x4 , x2 x3 − x1 x4 , 12 x23 + 2c x24 , 21 x21 + 2c x22
for some nonzero c ∈ K (in particular, char K 6= 2).
8

Conversely, rk J H̃ ≤ 3 in each of the five statements above.
Corollary 3.2. Let H ∈ K[x]m be a quadratic homogeneous polynomial map,
such that rk J H ≤ 3. If char K 6= 2, then rk J H = trdegK K(H).
Proof. Since rk J H ≤ trdegK K(H), it suffices to show that trdegK K(H) ≤ 3
if char K 6= 2. In (5) of theorem 3.1, we have trdegK K(H) ≤ 3 because
H̃12 + cH̃22 − 4H̃3 H̃4 = 0
In the other cases of theorem 3.1 where char K 6= 2, trdegK K(H) ≤ 3 follows
directly.
Lemma 3.3. Let H̃ ∈ K[x]m , such that J H̃ is as M̃ in lemma 2.4.
If rk C = 1 in (2.1) and r is odd, then the columns of C in (2.1) are dependent over K.
Proof. The case where char K 6= 2 follows from (i) of lemma 2.5, so assume that
char K = 2. Since rk C = 1 = 21 · 12 + 21 · 1, we deduce from theorem 2.1 that the
rows of C are dependent over K in pairs. Say that the first row of C is nonzero.
As r is odd, it follows from proposition 3.4 below that rk HH̃r+1 < r. Hence
there exists a v ′ ∈ K r such that (HH̃r+1 ) v ′ = 0, and
(J H̃r+1 ) v ′ = xt (HH̃r+1 ) = 0
The row space of C is spanned by J H̃r+1 , so C v ′ = 0.
Proposition 3.4. Let M ∈ Matn,n (K) be either a symmetric matrix, or an
anti-symmetric matrix with zeroes on the diagonal.
Then there exists a lower triangular matrix T ∈ Matn,n (K) with ones on the
diagonal, such that T t M T is the product of a symmetric permutation matrix
and a diagonal matrix.
In particular, rk M is even if M is an anti-symmetric matrix with zeroes on
the diagonal.
Proof. We describe an algorithm to transform M to the product of a symmetric
permutation matrix and a diagonal matrix. We distinguish three cases.
• The last column of M is zero.
Advance with the principal minor of M that we get by removing row and
column n.
• The entry in the lower right corner of M is nonzero.
Use Mnn as a pivot to clean the rest of the last column and the last row
of M . Advance with the principal minor of M that we get by removing
row and column n.

9

• None of the above.
Let i be the index of the lowest nonzero entry in the last column of M .
Use Min and Mni as pivots to clean the rest of columns i and n of M and
rows i and n of M . Advance with the principal minor of M that we get
by removing rows and columns i and n.
If M is M is an anti-symmetric matrix with zeroes on the diagonal, then so
is T t M T . By combining this with that T t M T is the product of a symmetric
permutation matrix and a diagonal matrix, we infer the last claim.
Notice that over characteristic 2, any Hessian matrix of a polynomial is
(anti-)symmetric with zeroes on the diagonal, which has even rank by extension
of scalars. Furthermore, one can show that any nondegenerate quadratic form
in odd dimension r over a perfect field of characteristic 2 is equivalent to
x1 x2 + x3 x4 + · · · + xr−2 xr−1 + x2r
Corollary 3.5. Let char K 6= 2 and M ∈ Matn,n (K) be a symmetric matrix.
Then there exists a T ∈ GLn (K), such that T t M T is a diagonal matrix.
Proof. Notice that the permutation matrix P in proposition 3.4 only has cycles
of length 1 and 2, because just like M , the product of P and the diagonal
matrix D is symmetric. Furthermore, for every cycle of length 2, the entries
on the diagonal of D which correspond to the two coordinates of that cycle are
equal. Since



 

1 1
0 c
1 −1
2c
0
=
−1 1
c 0
1 1
0 −2c
we can get rid of the cycles of length 2 in P .
Lemma 3.6. Suppose that H ∈ K[x1 , x2 , x3 , x4 ]4 , such that
 x1 
J H4 = ( x1 cx2 0 0 )
and
J H · v = xx23
0

for some nonzero c ∈ K, and a v ∈ K 4 of which the first 3 coordinates are not
all zero.
Suppose in addition that det J H = 0 and that the last column of J H does
not generate a nonzero constant vector.
Then there are S, T ∈ GL4 (K), such that for H̃ := SH(T x), J H̃ is of the
form


A11 A12 x1 cx2
 A21 A22 x2 −x1 

J H̃ = 
 A31 A32 x3 c̃x4 
x1 cx2 0
0
where the coefficients of x1 in A11 , A21 , A21 are zero.

10

Proof. Since the last row of J H is ( x1 cx2 0 0 ) and the last coordinate of
J H̃ · v is zero, we deduce that v1 = v2 = 0. As the first 3 coordinates of v are
not all zero, we conclude that v3 6= 0.
Make S from v3 I4 by changing column 4 to e4 Make T from I4 by changing
column 3 to v3−1 v, i.e. replacing the last entry of column 3 by v3−1 v4 , and by
replacing the last column by an arbitrary scalar multiple of e4 . Then

(J H̃) · e3 = S (J H)|x=T x · T e3 = S (J H) · v3−1 v x=T x
 x1 
 x1 
x2
−1
= v3 S x3
= xx23
0

x=T x

0

and
H̃4 = H4 (T x) = H4
So the third column of J H̃ is as claimed. It follows that J H̃ is as M̃ in lemma
2.4, with L1 = x3 , L2 = x2 , L3 = x1 , and L4 = x4 . Define A, B, C and D as in
(2.1).
Just like the last column of J H, the last column of J H̃ does not generate
a nonzero constant vector. Consequently, B11 and B21 are not both zero. From
(2.2), it follows that C11 B11 + C12 B21 = 0. So we can choose the last column
of T , such that B11 = C12 = cx2 and B21 = −C11 = −x1 .
The coefficient of x3 in B31 is zero, and by changing the third row of I4 on
the left of the diagonal in a proper way, we can get an U ∈ GL4 such that
! 
 
B11
cx2 
B
B21
3
U
= c̃x4 = −x
c̃x4
D
0
D11
for some c̃ ∈ K. Since U −1 can be obtained by changing the third row of I4 on
the left of the diagonal in a proper way as well, we infer that

J U −1 H̃(U x) · e3 = U −1 (J H̃)|x=Ux U · e3 = U −1 (J H̃)|x=Ux · e3
 x1 
 x1 
x2
−1
= xx23
=U
x3
0

x=Ux

0

and
(U −1 )4 H̃(U x) = H̃4 (U x) = H̃4
So if we replace S and T by U −1 S and T U respectively, then H̃ will be replaced
by U −1 H̃(U x).
So we can get B31 of the form c̃x4 for some c̃ ∈ K. Finally, we can clean the
coefficients of x1 in A11 , A21 , A21 with row operations, using C11 as a pivot. So
we can get the coefficients of x1 in A11 , A21 , A21 equal to zero.
Lemma 3.7. Suppose that det J H̃ = 0

A11 A12
 A21 A22

 A31 A32
x1 cx2

and J H̃ is of the form

x1 cx2
x2 −x1 

x3 c̃x4 
0
0

11

where c, c̃ ∈ K, such that c 6= 0. If the coefficients of x1
zero, then



x1 x3 + cx2 x4
x3
 x2 x3 − x1 x4 
 −x4

H̃ = 
and
J H̃ = 
 1 x23 + c x24 
 0
2
2
1 2
c 2
x1
x
+
x
2 1
2 2

Proof. Let ai and bi be the coefficients of x1 and x2 in Ai2
i ≤ 3. Then


a1 x1 x2 + 12 b1 x22 + x1 x3 + cx2 x4
 a2 x1 x2 + 1 b2 x22 + x2 x3 − x1 x4 
2

H̃ = 
 a3 x1 x2 + 1 b3 x22 + 1 x23 + c̃ x24 
2
2
2
1 2
c 2
2 x1 + 2 x2

a1 x2 + x3 a1 x1 + b1 x2 + cx4 x1
 a2 x2 − x4 a2 x1 + b2 x2 + x3 x2
J H̃ = 

a3 x2
a3 x1 + b3 x2
x3
x1
cx2
0

in A11 , A21 , A21 are
cx4
x3
0
cx2

x1
x2
x3
0


cx2
−x1 

cx4 
0

respectively, for each

and

cx2
−x1 

c̃x4 
0

Consequently, it suffices to show that ai = bi = 0 for each i ≤ 3, and that c̃ = c.
Suppose that the coefficients of x41 in det J H̃ are zero. Then we see by
expansion along rows 3, 4, 1, in that order, that a3 x1 = 0. Hence the third row
of J H̃ reads
J H̃3 = ( 0 b3 x2 x3 c̃x4 )
Since the coefficients of x31 x2 and x31 x3 in det J H̃ are zero, we see by expansion
along rows 3, 4, 1, in that order, that b3 x2 = a1 x1 = 0. Hence the third row of
J H̃ reads
J H̃3 = ( 0 0 x3 c̃x4 )
Since the coefficient of x32 x3 in det J H̃ is zero,
3, 4, 2, in that order, that a2 x2 = 0. So

x3 b1 x2 + cx4
 −x4 b2 x2 + x3

J H̃ = 
0
0
x1
cx2

we see by expansion along rows
x1
x2
x3
0


cx2
−x1 

c̃x4 
0

Since the coefficient of x21 x3 x4 in det J H̃ is zero, we see by expansion along row
3, and columns 2 and 1, in that order, that c̃x4 = cx4 .
Since the coefficient of x1 x22 x3 in det J H̃ is zero, we see by expansion along
row 3, and columns 1, 4, in that order, that b2 x2 = 0. Using that and that the
coefficient of x21 x2 x3 in det J H̃ is zero, we see by expansion along row 3, and
columns 1, 2, in that order, that b1 x2 = 0. So H̃ is as claimed.
Proof of theorem 3.1. Take M̃ = J H̃ and take A, B, C, D as in (2.1). We
distinguish three cases:
12

• The column space of B contains a nonzero constant vector.
Take U and Ũ as in the corresponding case in the proof of corollary 2.2.
– If case (1) of theorem 2.1 applies for Ũ H̃, then case (1) of theorem
3.1 follows, because
1
2 (r

− 1)2 − 21 (r − 1) + 1 =

4
2

−

2
2

+1=2=r−1

– If case (2) of theorem 2.1 applies for Ũ H̃, then case (1) or case (2)
of theorem 3.1 follows.
– If case (3) of theorem 2.1 applies for Ũ H̃, then case (1) or case (3)
of theorem 3.1 follows.
• The columns of B are dependent over K in pairs.
If char K = 2, then H̃ is as in (3) of theorem 2.1, which is included in (4)
of theorem 3.1. So assume that char K 6= 2. If B = 0, then H̃ is as in (2)
of theorem 2.1, which is included in (4) of theorem 3.1 as well. So assume
that B 6= 0.
From (i) of lemma 2.5, it follows that the columns of C are dependent
over K. If rk C ≥ 2, then rk C = 2 because rk B + rk C ≤ r = 3 and
B 6= 0, and B contains a nonzero constant vector because of lemma 2.6.
If rk C = 0, then (1) of theorem 3.1 follows. So assume that rk C = 1.
Since rk C = 1 = 12 · 12 + 21 · 1, we deduce from theorem 2.1 that the rows
of C are dependent over K in pairs. So we can choose S such that only
the first row of C is nonzero.
From corollary 3.5, it follows that there exists an U ′ ∈ GLr (K), such that

Hx1 ,x2 ,...,xr H̃r+1 (U ′ x) = (U ′ )−1 (HH̃r+1 ) U ′

is a diagonal matrix. Furthermore, the first entry on the diagonal is
nonzero because HH̃r+1 6= 0, and the last entry on the diagonal is zero
because the columns of C are dependent over K. By adapting S, we can
obtain that the entries on the diagonal of Hx1 ,x2 ,...,xr (H̃r+1 (U ′ x) are 1, c
and 0, in that order.
By adapting S and T , we can replace H̃ by
  ′

0
U
(U ′ )−1
H̃
0
0
Im−3

0
In−3

 
x

Then the first row of C becomes ( x1 cx2 0 ). We distinguish two cases.
– c = 0.
Notice that
J (H̃|x1 =1 ) = (J H̃)|x1 =1 · (J (1, x2 , x3 , . . . , xm ))
so rk J (H̃|x1 =1 ) = r − 1 = 2, and we can apply [dB4, Theorem 2.3].
13

∗ In the case of [dB4, Theorem 2.3] (1), case (2) of theorem 2.1
follows, which yields (4) of theorem 3.1.
∗ In the case of [dB4, Theorem 2.3] (2), case (1) of theorem 3.1
follows, because any linear combination of the rows of J H̃ which
is dependent on e1 , is linearly dependent on C1 .
∗ In the case of [dB4, Theorem 2.3] (3), case (1) or case (2) of
theorem 3.1 follows.
– c 6= 0.
From (ii) of lemma 2.5 and lemma 3.6, it follows that we can choose
S and T , such that the leading principal minor matrix of size 4 of
J H̃ is of the form


A11 A12 x1 cx2
 A21 A22 x2 −x1 


 A31 A32 x3 c̃x4 
x1 cx2 0
0
where the coefficients of x1 in A11 , A21 , A21 are zero. From lemma
3.7, case (5) of theorem 3.1 follows.

• None of the above.
We first show that rk B ≥ 2. For that purpose, assume that rk B ≤ 1.
Since the columns of B are not dependent over K in pairs, we deduce
that B 6= 0, and it follows from [dB4, Theorem 2.1] that the rows of B
are dependent over K in pairs. This contradicts the fact that the column
space of B does not contain a nonzero constant vector. So rk B ≥ 2 indeed.
From rk B + rk C = r, we deduce that rk C ≤ 1. If C = 0, then (1) of
theorem 3.1 follows, so assume that C 6= 0. Then rk C = 1 and rk B =
r − 1 = 2. From lemmas 3.3 and 2.6, we deduce that the column space of
B contains a nonzero constant vector, which is a contradiction.
So it remains to prove the last claim. In case of (5) of theorem 3.1, the last claim
follows from the proof of corollary 3.2. Otherwise, the last claim of theorem 3.1
follows in a similar manner as the last claim of theorem 2.1.
Lemma 3.8. Let K be a field of characteristic 2 and L be an extension field of
K. If theorem 3.1 holds, but for L instead of K, then theorem 3.1 holds.
Proof. Take H as in theorem 3.1. Then H is as in (1), (3) or (4) of theorem 3.1,
but with L instead of K. We assume that H is as in (3) of theorem 3.1 with L
instead of K, because the other case follows in a similar manner as lemma 2.7.
Notice that

0 x3 x2 x1 y4 y5 · · · ym · J H̃ = 0
Since J H̃ = S −1 (J H)|T x T , it follows that


0 x3 x2 x1 y4 y5 · · · ym · S · J H = 0
14

as well.
Suppose first that m = 4. As rk J H = m − 1, there exists a nonzero
v ∈ K(x)m such that ker v1 v2 v3 v4 is equal to the column space
 of J H.
Since the column space of J H is contained in ker ( 0 x3 x2 x1 ) · S , it follows
that v1 v2 v3 v4 is dependent on ( 0 x3 x2 x1 ) · S. So
v1 (S −1 )11 + v2 (S −1 )21 + v3 (S −1 )31 + v4 (S −1 )41 = 0

and the components of v are dependent over L. Consequently, the components
of v are dependent over K. So ker v1 v2 v3 v4 ) contains a nonzero vector
over K, and so does the column space of J H. Now we can follow the same
argumentation as in the first case in the proof of theorem 3.1.
Suppose next that m > 4. Then the rows of J H are dependent over L.
Hence the rows of J H are dependent over K as well. So we may assume that
the last row of J H is zero. By induction on m, (H1 , H2 , . . . , Hm−1 ) is as H in
theorem 3.1. As Hm = 0, we conclude that H satisfies theorem 3.1.

4

rank 3 with nilpotency

Theorem 4.1. Let H ∈ K[x]n be quadratic homogeneous, such that J H is
nilpotent and rk J H ≤ 3. Then there exists a T ∈ GLn (K), such that for
H̃ := T −1 H(T x), one of the following statements holds:
(1) J H̃ is lower triangular with zeroes on the diagonal;
(2) n ≥ 5, rk J H = 3, and J H̃ is of the form

0 x5
0
∗
 x4 0 −x5
∗

 0 x4
0
∗


 0
0
0
0

 .
..
..
..
 ..
.
.
.
0
0
0
0

···
···
···
···
···


∗
∗ 

∗ 


0 

.. 
. 
0

(3) n ≥ 6, rk J H = 3, char K = 2 and J H̃ is of the form

0 x6
0
0 0 x2
0 ···
 x5 0 −x6
0
∗
∗
∗
···

 0 x5
0
0
x
0
0
···
2



0
0
0 x6 x5
0 ···
 0

0
0
0 0
0
0 ···
 0

0
0
0 0
0
0 ···
 0


 0
0
0
0 0
0
0 ···

 .
.
.
.
.
.
..
..
..
.. ..
..
 ..
.
0
0
0
0 0
0
0 ···
15

0
∗
0
0
0
0
0
..
.
0




















Furthermore, x + H is tame if char K 6= 2, and there exists a tame invertible
map x + H̄ such that H̄ is quadratic homogeneous and J H̄ = J H in general.
Conversely, J H̃ is nilpotent in each of the three statements above. Furthermore, rk J H̃ = 3 in (2), rk J H̃ = 4 in (3) if char K 6= 2, and rk J H̃ = 3 in (3)
if char K = 2.
Lemma 4.2. Let H ∈ K[x]3 be homogeneous of degree 2, such that Jx1 ,x2 ,x3 H
is nilpotent. If Jx1 ,x2 ,x3 H is not similar over K to a triangular matrix, then
Jx1 ,x2 ,x3 H is similar over K to a matrix of the form


0 f 0
 b 0 f 
0 −b 0

where f and b are independent linear forms in K[x4 , x5 , . . . , xn ].

Proof. Suppose that Jx1 ,x2 ,x3 H is not similar over K to a triangular matrix.
Take i such that the coefficient matrix of xi of Jx1 ,x2 ,x3 H is nonzero, and define
N := Jx1 ,x2 ,x3 (H|xi =xi +1 ) = (Jx1 ,x2 ,x3 H)|xi =xi +1
Then N is nilpotent, and N is not similar over K to a triangular matrix. N (0)
is similar over K to the matrix N (0) in either (iii) or (iv) of [dB4, Lemma 3.1],
so it follows from [dB4, Lemma 3.1] that N is similar over K to a matrix of the
form


0 f +1
0
 b
0
f +1 
0
−b
0

where b and f are linear forms. b and f are independent, because the coefficients
of xi in b and f are 0 and 1 respectively. So Jx1 ,x2 ,x3 H is similar over K to a
matrix of the form


0 f 0
 b 0 f 
0 −b 0

where b and f are independent linear forms.
Let H̄ be the quadratic part with respect to x1 , x2 , x3 of H. Then Jx1 ,x2 ,x3 H̄
is nilpotent. By showing that H̄ = 0, we prove that b and f are contained in
K[x4 , x5 , . . . , xn ].
So assume that H̄ 6= 0. From [dB4, Theorem 3.2], it follows that we may
assume that Jx1 ,x2 ,x3 H̄ is lower triangular. If the coefficient matrix of x2 of
Jx1 ,x2 ,x3 H̄ is nonzero, then it has rank 1 because only the last row is nonzero,
and so has the coefficient matrix of x2 of Jx1 ,x2 ,x3 H. Otherwise, the coefficient
matrix of x1 of Jx1 ,x2 ,x3 H̄ and Jx1 ,x2 ,x3 H has rank 1, because only the first
column is nonzero.
So we could have chosen i, such that the coefficient matrix of xi of Jx1 ,x2 ,x3 H
would have rank 1. From [dB4, Lemma 3.1], it follows that Jx1 ,x2 ,x3 H is similar
over K to a triangular matrix. Contradiction, so H̄ = 0 indeed.
16

Lemma 4.3. Suppose that H ∈ K[x]n , such that J H is nilpotent.
(i) Suppose that J H may only be nonzero in the first row and the first 2
columns. Then there exists a T ∈ GLn (K) such that for H̃ := T −1 H(T x),
the following holds.
(a) J H̃ may only be nonzero in the first row and the first 2 columns (just
like J H).
(b) The Hessian matrix of the leading part with respect to x2 , x3 , . . . , xn of
H̃1 is the product of a symmetric permutation matrix and a diagonal
matrix.
(c) Every principal minor determinant of the leading principal minor matrix of size 2 of J H̃ is zero.
(ii) Suppose that char K = 2 and that J H may only be nonzero in the first
row and the first 3 columns. Then there exists a T ∈ GLn (K) such that
for H̃ := T −1 H(T x), the following holds.
(a) J H̃ may only be nonzero in the first row and the first 3 columns (just
like J H).
(b) The Hessian matrix of the leading part with respect to x2 , x3 , . . . , xn of
H̃1 is the product of a symmetric permutation matrix and a diagonal
matrix.
(c) Every principal minor determinant of the leading principal minor matrix of size 3 of J H̃ is zero.
Proof. From proposition 3.4, it follows that there exists a T ∈ Matn,n (K) of the
form


1 0 0 ··· 0
 0 1 0 ··· 0 



.. 
..
 0 ∗ 1
. . 



 . . .
.. ... 0 
 .. ..
0 ∗ ···
∗ 1

such that the Hessian matrix of the leading part with respect to x2 , x3 , . . . , xn of
H̃1 = H1 (T x) is the product of a symmetric permutation matrix and a diagonal
matrix.
Furthermore, J H̃ may only be nonzero in the first row and the first 2 or 3
columns respectively (just like J H), because of the form of T .
(i) Let N be the principal minor matrix of size 2 of J H̃.
Suppose first that J H̃ may only be nonzero in the first 2 columns. From
[dB4, Lemma 1.2], it follows that N is nilpotent. On account of [dB4,
Theorem 3.2], N is similar over K to a triangular matrix.
Hence the rows of N are dependent over K. If the second row of N is zero,
then (i) follows. If the second row of N is not zero, then we may assume
that the first row of N is zero, and (i) follows as well.
17

Suppose next that J H̃ may only be nonzero in the first row and the first
2 columns, but not just the first 2 columns. We distinguish two cases.
•

∂
∂x2 H̃1

∈ K[x1 , x2 ].
∂
Since tr J H̃ = 0, we deduce that ∂x
H̃1 ∈ K[x1 , x2 ] as well. Let G :=
1
H̃(x1 , x2 , 0, 0, . . . , 0). Then J G = (J H̃)|x3 =x4 =···=xn =0 . Consequently, the nonzero part of J G is restricted to the first two columns.
So the leading principal minor matrix of size 2 of J G is nilpotent.
But this minor matrix is just N , and just as for H̃ before, we may
assume that only one row of N is nonzero. This gives (i).
∂
H̃1 ∈
/ K[x1 , x2 ].
• ∂x
2
Since H(H̃1 |x1 =0 ) is the product of a permutation matrix and a diag∂
onal matrix, it follows that ∂x
H̃1 is a linear combination of x1 and
2
xi , where i ≥ 3, such that xi does not occur in any other entry of
J H̃. Looking at the coefficient of x1i in the sum of the determinants
∂
H̃2 = 0.
of the principal minors of size 2, we infer that ∂x
1
t
So the second row of J H̃ is dependent on e2 . From a permuted
∂
version of [dB4, Lemma 1.2], we infer that ∂x
H̃2 is nilpotent. Hence
2
the second row of J H̃ is zero. Since tr J H̃ = 0, we infer (i).

(ii) Let N be the principal minor matrix of size 3 of J H̃.
Suppose first that J H̃ may only be nonzero in the first 3 columns. From
[dB4, Lemma 1.2], it follows that the leading principal minor of size 3 of
J H̃ is nilpotent. On account of [dB4, Theorem 3.2], N is similar over K
to a triangular matrix. But for a triangular nilpotent Jacobian matrix of
size 3 over characteristic 2, the rank cannot be 2. So rk N ≤ 1.
Hence the rows of N are dependent over K in pairs. If the second and the
third row of N are zero, then (ii) follows. If the second or the third row
of N is not zero, then we may assume that the first 2 rows of N are zero,
and (ii) follows as well.
Suppose next that J H̃ may only be nonzero in the first row and the first
3 columns, but not just the first 3 columns. We distinguish three cases.
•

∂
∂
∂x2 H̃1 , ∂x3 H̃1

∈ K[x1 , x2 , x3 ].
Using techniques of the proof of (i), we can reduce to the case where
J H̃ may only be nonzero in the first 3 columns.
∂
∂
H̃1 , ∂x
H̃1 ∈
/ K[x1 , x2 , x3 ].
• ∂x
2
3
∂
H̃2 =
Using techniques of the proof of (i), we can deduce that ∂x
1
∂
(
H̃
,
H̃
)
is
nilpotent.
H̃
=
0,
and
that
J
2
3
x2 ,x3
∂x1 3
On account of [dB4, Theorem 3.2], Jx2 ,x3 (H̃2 , H̃3 ) is similar over
K to a triangular matrix. But for a triangular nilpotent Jacobian
matrix of size 2 over characteristic 2, the rank cannot be 1. So
rk Jx2 ,x3 (H̃2 , H̃3 ) = 0. Consequently, the last two rows of N are
zero, and (ii) follows.

18

• None of the above.
∂
Assume without loss of generality that ∂x
H̃1 ∈ K[x1 , x2 , x3 ] and
2
∂
/ K[x1 , x2 , x3 ]. Since H(H̃1 |x1 =0 ) is the product of a permu∂x3 H̃1 ∈
∂
tation matrix and a diagonal matrix, it follows that ∂x
H̃1 is a linear
3
combination of x1 and xi , where i ≥ 4, such that xi does not occur
in any other entry of J H̃.
Looking at the coefficient of x1i in the sum of the determinants of the
∂
∂
principal minors of size 2, we infer that ∂x
H̃3 = 0. If ∂x
H̃2 = 0 as
1
1
∂
well, then we can advance as above, so assume that ∂x1 H̃2 6= 0.
Looking at the coefficient of x1i in the sum of the determinants of the
∂
principal minors of size 3, we infer that ∂x
H̃3 = 0. So the third row
2
t
of J H̃ is dependent on e3 . From a permuted version of [dB4, Lemma
∂
H̃3 is nilpotent. Hence the third row of J H̃ is
1.2], we infer that ∂x
3
zero.
∂
∂
From tr J H̃ = 0, we deduce that ∂x
H̃1 = − ∂x
H̃2 . We show that
1
2
∂
∂
H̃1 =
H̃2 = 0
∂x1
∂x2

(4.1)

∂
∂
For that purpose, suppose that ∂x
H̃1 6= 0. Since ∂x
x2 = 0, the
1
1 1
∂
H̃1 is zero. Similarly, the coefficient of x2 in
coefficient of x1 in ∂x
1
∂
H̃
is
zero.
As
H̃
∈
K[x1 , x2 , x3 ], we infer that
2
∂x2 2

∂
∂
H̃1 = −
H̃2 ∈ Kx3 \ {0}
∂x1
∂x2
Looking at the coefficient of x23 in the sum of the determinants of
the principal minors of size 2, we deduce that the coefficient of x23 in
∂
∂
( ∂x
H̃1 ) · ( ∂x
H̃i ) is nonzero.
i
1
∂
∂
∂
Consequently, the coefficient of x33 in ( ∂x
H̃1 ) · ( ∂x
H̃2 ) · ( ∂x
H̃i ) ∈
i
2
1
3
Kx3 \ {0} is nonzero. This contributes to the coefficient of x33 in the
sum of the determinants of the principal minors of size 3. Contradiction, because this contribution cannot be canceled.
So (4.1) is satisfied. We show that in addition,
∂
H̃1 = 0
∂x2

(4.2)

∂
The coefficient of x1 of ∂x
H̃1 is zero, because of (4.1). The coefficient
2
∂
∂
x2 = 0. The coefficient of x3 of
of x2 of ∂x2 H̃1 is zero, because ∂x
2 2
∂
∂
∂x2 H̃1 is zero, because the coefficient of x2 of ∂x3 H̃1 ∈ Kx1 + Kxi is
zero. So (4.2) is satisfied as well.
Recall that the third row of N is zero. From (4.1) and (4.2), it follows
that the diagonal and the second column of N are zero as well. Hence
every principal minor determinant of N is zero, which gives (ii).

19

Lemma 4.4. Let H̃ be as in lemma 4.3. Suppose that J H̃ has a principal
minor matrix M of which the determinant is nonzero. Then
(a) H̃ is as in (ii) of lemma 4.3;
(b) rows 2 and 3 of J H̃ are zero;
(c) M has size 2 and x2 x3 | det M ;
(d) Besides M , there exists exactly one principal minor matrix M ′ of size 2 of
J H, such that det M ′ = − det M .
Proof. Take N as in (i) or (ii) of lemma 4.3 respectively. Then M is not a
principal minor matrix of N . So if M does not contain the upper left corner of
J H̃, then the last column of M is zero. Hence M does contain the upper left
corner of J H̃.
If M has two columns outside the column range of N , then both columns
are dependent on e1 . So M has exactly one column outside the column range
of N , say column i.
(i) Suppose first that H̃ is as in (i) of lemma 4.3. Then either M has size 2
with row and column indices 1 and i, or M has size 3 with row and column
indices 1, 2 and i.
The coefficient of x1 in the upper right corner of M is zero, because N11 =
0. Hence the upper right corner of M is of the form cxj for some nonzero
c ∈ K and a j ≥ 2. If j ≥ 3, then det M is equal to the sum of the terms
which are divisible by xj in the sum of the determinants of the principal
minors of the same size as M , which is zero.
So j = 2. Now M is the only principal minor matrix of its size, of which
the determinant is nonzero, because if we would take another i, then j
changes as well. Contradiction, because the sum of the determinants of
the principal minors of the same size as M is zero.
(ii) Suppose next that H̃ is as in (ii) of lemma 4.3. If the second row of J H̃
is nonzero, then the coefficient of x1 x3 in H̃2 is nonzero, because N22 = 0.
If the third row of J H̃ is nonzero, then the coefficient of x1 x2 in H̃3 is
nonzero, because N33 = 0. Since every principal minor determinant of N
is zero, we infer that N23 N32 = 0, so either the second or the third row of
J H̃ is zero.
Assume without loss of generality that the second row of J H̃ is zero. Then
either M has size 2 with row and column indices 1 and i, or M has size
3 with row and column indices 1, 3 and i. The upper right corner of M
is of the form cxj for some nonzero c ∈ K, and with the techniques in (i)
above, we see that 2 ≤ j ≤ 3.
Furthermore, we infer with the techniques in (i) above that J H̃ has another principal minor matrix M ′ of the same size as M ′ , of which the

20

determinant is nonzero as well. The upper right corner of M ′ can only be
of the form c′ x5−j for some nonzero c′ ∈ K.
It follows that N12 6= 0 and N13 6= 0. Consequently, N21 = N31 = 0. This
is only possible if both the second and the third row of J H̃ are zero. So
M has size 2, and claims (c) and (d) follow.
Lemma 4.5. Let n = 4,

x1 x3 + cx2 x4
 x2 x3 − x1 x4
H̃ = 
 1 x23 + c x24
2
2
1 2
c 2
2 x1 + 2 x2








x3
 −x4
J H̃ = 
 0
x1

and

cx4
x3
0
cx2

x1
x2
x3
0


cx2
−x1 

cx4 
0

as in lemma 3.7 (with c 6= 0).

Let M ∈ Mat4,4 (K), such that deg det J H̃ + M ≤ 2. Then there exists a
translation G, such that


H̃ G(x) − H̃ + M x ∈ K 4


In particular, det J H̃ + M = det J H̃ + M x = 0.

Proof. Since the quartic part of det(J H̃+M ) is zero, we deduce that det(J H̃) =
0. By way of completing the squares, we can choose a translation G such that
the linear part of F := H̃(G−1 (x)) + M G−1 (x) is of the form


a1 x1 + b1 x2 + c1 x3 + d1 x4
 a2 x1 + b2 x2 + c2 x3 + d2 x4 


 a3 x1 + b3 x2

c4 x3 + d4 x4

Notice that deg det J F ≤ 2. Looking at the coefficients of x31 , x32 , x33 , and x34
of det J F , we see that b3 = a3 = d4 = c4 = 0. Looking at the coefficients of
x21 x3 , x1 x23 , x22 x4 , and x2 x24 of det J F , we see that b1 = d1 = a1 = c1 = 0.
Looking at the coefficients of x21 x4 , x1 x24 , x22 x3 , and x2 x23 of det J F , we see that
b2 = c2 = a2 = d2 = 0.
So F has trivial linear part, and H̃ − F ∈ K 4 . Hence H̃(G) − F (G) ∈ K 4 ,
as claimed. The last claim follows from det(J H̃) = 0.
Proof of theorem 4.1. From [dB4, Theorem 3.2], it follows that (1) is satisfied
if rk J H ≤ 2. So assume that rk J H = 3. Then we can follow the cases of
theorem 3.1.
• H is as in (1) of theorem 3.1.
Let H̃ = SH(S −1 x). Then only the first 3 rows of J H̃ may be nonzero.
If the leading principal minor matrix N of size 3 of J H̃ is similar over
K to a triangular matrix, then so is J H̃ itself. So assume that N is not

21

similar over K to a triangular matrix. From lemma 4.2, it follows that N
is similar over K to a matrix of the form


0 f 0
 b 0 f 
(4.3)
0 −b 0
where f and b are independent linear forms in K[x4 , x5 , . . . , xn ].

Consequently, we can choose S, such that for H̃ := SH(S −1 x), only the
first 3 rows of J H̃ are nonzero, and the leading principal minor matrix of
size 3 is as in (4.3). If we negate the third row and the third column of
(4.3), and replace b by x4 and f by x5 , then we get


0 x5
0
 x4 0 −x5 
0 x4
0
We can even choose S such that the leading principal minor matrix of size
3 of J H̃ is as above. So (2) of theorem 4.1 is satisfied for T = S −1 .

If we replace H̃2 by 0, then all principal minor determinants of J H̃ become
zero. On account of [Dru, Lemma 1.2], J H̃ becomes permutation similar
to a triangular matrix. From proposition 4.6 below, we infer that x + H
is tame if char K = 0.
• H is as in (2) of theorem 3.1 or as in (2) of theorem 2.1.
Let H̃ = T −1 H(T x). Then the rows of Jx3 ,x4 ,...,xn H̃ are dependent over
K in pairs. Suppose first that the first 2 rows of Jx3 ,x4 ,...,xn H̃ are zero.
Then we can choose T such that only the last row of Jx3 ,x4 ,...,xn H̃ may
be nonzero.
From [dB4, Lemma 1.2], it follows that leading principal minor matrix
N of size 2 of J H̃ is nilpotent as well. On account of [dB4, Theorem
3.2], N is similar over K to a triangular matrix. From [dB4, Corollary
1.4], we deduce that J H̃ is similar over K to a triangular matrix. So
we can choose T such that J H̃ is lower triangular, and (1) is satisfied.
Furthermore, x + H is tame if char K = 0.
Suppose next that the first 2 rows of Jx3 ,x4 ,...,xn H̃ are not both zero.
Then we can choose T such that only the first row of Jx3 ,x4 ,...,xn H̃ may
be nonzero. From lemma 4.3 (i) and lemma 4.4, we infer that we can
choose T such that every principal minor of J H̃ has determinant zero.
From [Dru, Lemma 1.2], it follows that J H̃ is permutation similar to
a triangular matrix. So (1) is satisfied. Furthermore, x + H is tame if
char K = 0.
• H is as in (3) of theorem 3.1 or as in (3) of theorem 2.1.
Let H̃ = T −1 H(T x). Then the rows of Jx4 ,x5 ,...,xn H̃ are dependent over
K in pairs. Suppose first that the first 3 rows of Jx4 ,x5 ,...,xn H̃ are zero.
22

Then we can choose T such that only the last row of Jx4 ,x5 ,...,xn H̃ may be
nonzero, and just as above, (1) is satisfied. Furthermore, x + H is tame if
char K = 0.
Suppose next that the first 3 rows of Jx4 ,x5 ,...,xn H̃ are not all zero. Then
we can choose T such that only the first row of Jx4 ,x5 ,...,xn H̃ may be
nonzero. If we can choose T such that every principal minor of J H̃ has
determinant zero, then (1) is satified, just as before. Furthermore, x + H
is tame if char K = 0.
So assume that we cannot choose T such that every principal minor of
J H̃ has determinant zero. From lemma 4.3 (ii) and lemma 4.4, we infer
that we can choose T such that H̃ is as in lemma 4.4. More precisely, we
can choose T such that J H̃ is of the form


0 x4 −x5 x2 −x3 ∗ ∗ · · ·
 0
0
0
0
0
0 0 ··· 


 0
0
0
0
0
0 0 ··· 


 x3 ∗
∗
0
0
0 0 ··· 


(4.4)
 x2 ∗
∗
0
0
0 0 ··· 


 ∗

∗
∗
0
0
0 0 ··· 

 ∗
∗
∗
0
0
0 0 ··· 


..
..
..
..
..
.. .. . .
.
.
.
.
.
.
. .

∂
x2 = 0, the coefficients of x1 of the starred entries in the first
Since ∂x
1 1
column of (4.4) are zero. Hence we can clean these starred entries by way
of row operations with rows 4 and 5.

We can also clean them by way of a linear conjugation, because if a starred
entry in the first column is nonzero, the transposed entry in the first row
is zero, so the corresponding column operations which are induced by the
linear conjugation will not have any effect.
If rows 1, 4, and 5 remain as the only nonzero rows, then H is as in
(1) of theorem 3.1, which is the first case. So assume that another row
remains nonzero. By way of additional row operations and associated
column operations, we can get J H̃ of the form


0 x4 −x5 x2 −x3 ∗ ∗ · · ·
 0
0
0
0
0
0 0 ··· 


 0
0
0
0
0
0 0 ··· 


 x3 0
x1
0
0
0 0 ··· 


 x2 x1
0
0
0
0 0 ··· 
(4.5)


 0 x3 x2
0
0
0 0 ··· 


 0
0
0
0
0
0 0 ··· 


 0
0
0
0
0
0 0 ··· 


.. .. . .
..
..
..
..
..
.
. .
.
.
.
.
.
23

where we do not maintain (b) of lemma 4.3 (ii) any more. Hence P −1 H̃(P x)
is as H̃ in (3) for a suitable permutation matrix P .
In a similar manner as in the case where H is as in (1) of theorem 3.1, we
infer that x + H is tame if char K = 0.
• H is as in (5) of theorem 3.1.
Then only the first 4 columns of H̃ = T −1 H(T x) may be nonzero. Hence
the leading principal minor matrix N of size 4 of J H̃ is nilpotent. We
distinguish two subcases.
– The rows of N are linearly independent over K.
Then there exists an U ∈ GL4 (K), such that U N is as J H̃ in lemma
4.5. Furthermore,
det(U N + U ) = det U det(N + I4 ) = det U ∈ K ∗
So det(U N + U ) 6= 0 and deg det(U N + U ) ≤ 2. This contradicts
lemma 4.5.
– The rows of N are linearly independent over K.
Then we can apply the first case above on the map (H̃1 , H̃2 , H̃3 , H̃4 ).
Since N has less than 5 columns, the case where N is not similar
over K to a triangular matrix cannot occur in the case above. So N
is similar over K to a triangular matrix, and so are J H̃ and J H.
Furthermore, x + H is tame if char K = 0.
So it remains to prove the last claim in the case char K 6= 0. This follows by
way of techniques in the proof of theorem 5.1 in the next section.
Proposition 4.6. Let R be an integral domain of characteristic zero, and let
F ∈ R[x]n be a polynomial map which is invertible over R. Let F̃ ∈ R[x]n ,
such that only the ith component of F̃ is different from that of F , and such that
det J F̃ = det J F .
Then F̃ (F −1 ) is an elementary invertible polynomial map over R. In particular, F̃ is invertible over R.
Proof. Assume without loss of generality that i = n. Let K be the fraction field
of R. Then rk J H = trdegK K(H), because K has characteristic zero. Since
rk J (F1 , F2 , . . . , Fn−1 , 0) = n − 1 = rk J (F1 , F2 , . . . , Fn−1 , F̃n − Fn )
it follows that F̃n − Fn is algebraically dependent over K on F1 , F2 , . . . , Fn−1 .
As F is invertible over R,
F̃n − Fn ∈ R[F1 , F2 , . . . , Fn−1 ]
So F̃n (F −1 ) − xn ∈ R[x1 , x2 , . . . , xn−1 ]. Furthermore, F̃i (F −1 ) = Fi (F −1 ) = xi
for all i < n by assumption, so F̃ (F −1 ) is elementary.

24

5

rank 4 with nilpotency

Let M ∈ Matn (K) be nilpotent and v ∈ K n be nonzero. Define the image
exponent of v with respect to M as
IE(M, v) = IEK (M, v) := max{i ∈ N | M i v 6= 0}
and the preimage exponent of v with respect to M as
PE(M, v) = PEK (M, v) := max{i ∈ N | M i w = v for some w ∈ K n }
Theorem 5.1. Let char K 6= 2 and n = 5. Suppose that H ∈ K[x]5 , such
that J H is nilpotent and rk J H ≥ 4. Then rk J H = 4, and there exists a
T ∈ GL5 (K), such that J T −1 H(T x) is either triangular with zeroes on the
diagonal, or of one of the following forms.

 

0
0
0
0
0
0
0
0
0
0
 ∗

0
0
0
0 
0
0
x4
0 

  ∗

 0


x4
0 x2
0   x2 x1 0 −x5 −x4 


 x3 −x5 x1 0 −x2   x3 0 x1
0
x5 
∗
∗
0 x1
0
∗
0
0
x1
0
Furthermore, x + H is tame.

Proof. Suppose first that K is infinite. From [dB3, Theorem 4.4], it follows that
PEK(x) (J H, x) = 0. As rk J H = n − 1, IEK(x) (J H, x) + PEK(x) (J H, x) =
n − 1, so IEK(x) (J H, x) = n − 1. From [dB3, Corollary 4.3], it follows that there
exists a T ∈ GL5 (K), such that for H̃ := T −1 H(T x), we have


0 0 0 0 0
 1 0 0 0 0 



(J H̃)|x=e1 = 
 0 1 0 0 0 
 0 0 1 0 0 
0 0 0 1 0

Using this, one can compute all solutions, and match them against the given
classification. We did this with Maple 8, see dim5qdr.pdf.
To prove that x + H is tame, we show that the maps of the classification are
tame. If J H is similar over K to a triangular matrix, then x + H is tame. So
assume that J H is not similar over K to a triangular matrix. If




0
0


 z1 x21 + 1 x24 
z1 x21
2




 or G =  x1 x2 − x4 x5 
x2 x4
G=






 x1 x3 − 1 x25 
x1 x3 − x2 x5
2
z2 x21 + x1 x4
z2 x21 + z3 x1 x2 + z4 x22 + x1 x4

then we can apply proposition 4.6 with R = Z[z1 , z2 , z3 , z4 ] and i = 4, to obtain
that x + G is the composition of an elementary map and a map x + G̃ for which
25

J G̃ is permutation similar to a triangular matrix with zeroes on the diagonal.
So x + G is tame over R. Hence x + G modulo I is tame over R/I for any ideal
I of R. Since x + H has this form up to conjugation with a linear map, we infer
that x + H is tame.
Suppose next that K is finite, and let L be an infinite extension field of K.
If J H is similar over L to a triangular matrix, then by [dB4, Proposition 1.3],
J H is similar over K to a triangular matrix as well. So assume that J H is
not similar over K to a triangular matrix. Then one can check for the solutions
over L that [dB3, Theorem 5.2] is satisfied, which we did. So [dB3, Corollary
4.3] holds over K as well, and so does this theorem.
Theorem 5.2. Let char K = 2 and n = 6. Suppose that H ∈ K[x]6 , such
that J H is nilpotent and rk J H ≥ 4. Then rk J H = 4, and there exists a
T ∈ GL6 (K), such that J T −1 H(T x) is either triangular with zeroes on the
diagonal, or of one of the following forms.


 
0
0
0
0
0
0
0
0
0
0
0
0
 0

0
0
0
0
0 
0
0
0
0
0 


  0
 x2 x1 0

  0
0
0
0
x
0
0
x
0
5
2


 
 0



0
x
0
x
0
x
−cx
−
x
x
0
−cx
−x
5
3
5
6
1
2
2 

  3
 ∗
cx6
0 x1
0
cx2 
∗
∗ x2 0 −x3   x4
x5
0
0
0
x1
0
∗
∗
∗
0 x2
0


 
0
0
0
0
0
0
0
0
0
0
0
0
 0

0
0
0
0
0 
0
0
x5
x4
0 


  0
 0



x
0
x
0
0
x
x
0
−x
−2x
−x
4
2
1
6
5
4 

  2
 x3 −x5 x1 0 −x2 0   x3 0 x1

0
x
x
6
5 

 
 x4
0
x1
0
0 
0
0 x1
0
0   x4 0
x5 0
0
0
x1
0
∗
∗
∗
∗
∗
0

where c ∈ {0, 1}. Furthermore, there exists a tame invertible map x + H̄ such
that H̄ is quadratic homogeneous and J H̄ = J H.

Proof. Suppose first that K is infinite. From [dB3, Theorem 4.4], it follows
that PE(J H, x) = 0. Since IEK(x) (J H, x) = 0 as well, it follows from [dB3,
Theorem 4.4] that there exists a T ∈ GL5 (K), such that for H̃ := T −1 H(T x),
we have


0 0 0 0 0 0
 0 0 0 0 0 0 


 0 1 0 0 0 0 

(J H̃)|x=e1 = 
 0 0 1 0 0 0 


 0 0 0 1 0 0 
0 0 0 0 1 0

Using this, one can compute all solutions, and match them against the given
classification. We did this with Maple 8, see dim6chr2qdr.pdf.
To prove that x + H̄ is tame for some quadratic homogeneous H̄ such that
J H̄ = J H, we can use the same arguments as in the proof of theorem 5.1.
26

Namely, we apply proposition 4.6 with R = Z[z1 , z2 , z3 , z4 , z5 , z6 , z7 , z8 , z9 , z10 ]
and i = 5, i = 5, i = 4, and i = 4 respectively.
Suppose next that K is finite, and let L be any infinite extension field of K.
If J H is similar over L to a triangular matrix, then by [dB4, Proposition 1.3],
J H is similar over K to a triangular matrix as well. So assume that J H is not
similar over K to a triangular matrix. We distinguish two cases.
• The columns of J H are dependent over L.
Then the columns of J H are dependent over K as well. So we may assume
that the last column of J H is zero. Then the leading principal minor N
of size 5 of J H is nilpotent as well. Just like rk J H = 6 − 2 = 4, we
deduce that rk N = 5 − 2 = 3. So we can apply theorem 4.1 on N , to infer
that we can get J H̃ of the form of the third matrix.
• The columns of J H are independent over L.
Then one can check for the solutions over L that lemma 5.3 below is
satisfied, which we did. So there exists a v ∈ K 6 , such that (J H)|4x=v 6= 0.
Hence the Jordan Normal Form of (J H)|x=v has Jordan blocks of size 1
and 5, just like that of J H.

Furthermore, IE (J H)|x=v , v = 0 = IEK(x)
 (J H, x), and it follows from
[dB3, Theorem 4.4] that PE (J H)|x=v , v = 0 = PEK(x) (J H, x). Consequently, one can verify that [dB3, Theorem 4.2] holds over K as well,
and so does this theorem.
Lemma 5.3. Let L be an extension field of K. Suppose that H ∈ K[x]n and
T ∈ GLn (L), such that for H̃ := T −1 H(T x), the ideal generated by the entries
of (J H̃)s contains a power of a polynomial f .
Then there exists a v ∈ K n , such that (J H)|sx=v 6= 0, in the following cases.
(i) #K ≥ deg f + 1;
(ii) f is homogeneous and #K ≥ deg f .
Proof. From [dB1, Lemma 5.1], it follows that there exists a v ∈ K n such that
f (T −1 v) 6= 0. Let I be the ideal over L generated by the entries of (J H̃)s . As
the radical of I contains f , the radical of I(T −1 v) contains f (T −1 v). I(T −1 v)
is generated over L by the entries of (J H̃)|sx=T −1 v and
T (J H̃)|sx=T −1 v T −1 = (J H)|sx=v
From f (T −1 v) 6= 0, it follows that I(T −1 v) 6= 0 and (J H)|sx=v 6= 0.
The cardinality of K may be too small for the computational method of
[dB3, Theorem 4.2]: the following maps H do not satisfy [dB3, Theorem 4.2]
and neither lemma 2.4:
• #K = 3 and H = (0, 21 x21 , 21 x22 , (x1 + x2 )x3 , (x1 − x2 )x4 );

27

• #K = 2 and H = (0, 0, x1 x2 , x1 x3 , x2 x4 , (x1 − x2 )x5 );
• #K = 2 and H = (0, 0, x1 x4 , x1 x3 − x2 x5 , x2 x4 , (x1 − x4 )x5 ).
Theorem 5.4. Let H be quadratic homogeneous such that J H is nilpotent and
rk J H ≤ 4.
Then the rows of J H are dependent over K. Furthermore, one of the following statements hold.
(1) Every set of 6 rows of J H is dependent over K.
(2) There exists a tame invertible map x + H̄ such that H̄ is quadratic homogeneous and J H̄ = J H.
In addition, x + H is stably tame if char K 6= 2.
Proof. The case where rk J H ≤ 3 follows from theorem 4.1, so assume that
rk J H = 4. We follow the cases of corollary 2.2.
• H is as in (1) of corollary 2.2.
Let H̃ = S −1 H(Sx). Then only the first 5 rows of J H̃ may be nonzero.
Hence every set of 6 rows of J H is dependent over K.
If n > 5, then the sixth row of J H̃ is zero. So assume that n ≤ 5. Then
it follows from (i) of lemma 5.5 below that the rows of J H are dependent
over K. From [BvdEW, Theorem 4.14] and theorem 5.1, it follows that
x + H̃ is stably tame if char K 6= 2.
• H is as in (2) of corollary 2.2.
Let H̃ = T −1 H(T x) and let N be the leading principal minor matrix of
size 3 of J H̃. Just like for the case where H is as in (2) of theorem 3.1 in
the proof of theorem 4.1, we can reduce to the case where only the first
row and the first three columns of J H̃ may be nonzero.
From proposition 3.4, it follows that we may assume that H(H̃1 |x1 =0 ) is a
product of a permutation matrix and a diagonal matrix. We distinguish
three subcases:
–

∂
∂
∂x2 H̃1 , ∂x3 H̃1

∈ K[x1 , x2 , x3 ].
∂
Since tr J H̃ = 0, we deduce that ∂x
H̃1 ∈ K[x1 , x2 , x3 ] as well.
1
We may assume that the coefficient of either x24 or x4 x5 of H̃1 is
nonzero. With techniques in the proof of corollary 3.5, x4 x5 can be
transformed to x24 − x25 , so we may assume that the coefficient of x24
of H̃1 is nonzero.
Let G := H̃(x1 , x2 , x3 , x4 , 0, 0, . . . , 0). Then J G = (J H̃)|x5 =···=xn =0 .
Consequently, the nonzero part of J G is restricted to the first four
columns. From (i) of lemma 5.5 below, it follows that the rows of
J G are dependent over K and that x + G is tame.
Since the first row of J G is independent of the other rows of J G, the
rows of J H̃ are dependent over K. From proposition 4.6, it follows
28

that x + H̃ is tame if char K = 0, which gives (2) if char K = 0.
Using techniques in the proof of theorem 5.1, the case char K 6= 0
follows as well.
∂
∂
– ∂x
H̃1 , ∂x
H̃1 ∈
/ K[x1 , x2 , x3 ].
2
3
Then we may assume that the coefficients of x2 x4 and x3 x5 of H̃1
are nonzero. Looking at the coefficients of x14 and x15 of the sum
of the determinants of the principal minors of size 2, we infer that
∂
∂
∂x1 H̃2 = 0 and ∂x1 H̃3 = 0.
From a permuted version of [dB4, Lemma 2], we deduce that Jx2 ,x3
(H̃2 , H̃3 ) is nilpotent. On account of theorem 4.1 or [dB4, Theorem
3.2], the second and the third row of J H̃ are dependent over K.
By applying [dB4, Lemma 2] twice, we see that we can replace H̃1
by 0 without affecting the nilpotency of J H̃. Now (2) follows in a
∂
∂
similar manner as in the case ∂x
H̃1 , ∂x
H̃1 ∈ K[x1 , x2 , x3 ] above.
2
3
– None of the above.
∂
∂
H̃1 ∈
/ K[x1 , x2 , x3 ] and ∂x
H̃1 ∈
Then we may assume that ∂x
2
3
K[x1 , x2 , x3 ]. Furthermore, we may assume that the coefficient of
x2 x4 of H̃1 is nonzero. Now we can apply the same arguments as in
∂
∂
H̃1 , ∂x
H̃1 ∈ K[x1 , x2 , x3 ] above.
the case ∂x
2
3
• H is as in (3) of corollary 2.2.
Let H̃ = T −1 H(T x) and let N be the leading principal minor matrix of
size 4 of J H̃. Just like for the case where H is as in (3) of theorem 3.1 in
the proof of theorem 4.1, we can reduce to the case where only the first
row and the first four columns of J H̃ may be nonzero.
From proposition 3.4, it follows that we may assume that H(H̃1 |x1 =0 ) is a
product of a permutation matrix and a diagonal matrix. We distinguish
three subcases:
∂
∂
∂
∂x2 H̃1 , ∂x3 H̃1 , ∂x4 H̃1

∈ K[x1 , x2 , x3 , x4 ].
∂
Since tr J H̃ = 0, we deduce that ∂x
H̃1 ∈ K[x1 , x2 , x3 , x4 ] as well.
1
We may assume that the coefficient of x5 x6 of H̃1 is nonzero.
Just like we reduced to the case where only the first 4 columns of
J H̃ may be nonzero above, we can reduce to the case where only
the first 6 columns of J H̃ may be nonzero. Hence the results follow
from (ii) of lemma 5.2.
∂
∂
∂
H̃1 , ∂x
H̃1 , ∂x
H̃1 ∈
/ K[x1 , x2 , x3 , x4 ].
– ∂x
2
3
4
Then we may assume that the coefficients of x2 x5 , x3 x6 and x4 x7 of
H̃1 are nonzero. Looking at the coefficients of x15 , x16 and x17 of the
sum of the determinants of the principal minors of size 2, we infer
∂
∂
∂
that ∂x
H̃2 = 0, ∂x
H̃3 = 0 and ∂x
H̃4 = 0.
1
1
1
From a permuted version of [dB4, Lemma 2], we deduce that Jx2 ,x3 ,x4
(H̃2 , H̃3 , H̃4 ) is nilpotent. On account of theorem 4.1 or [dB4, Theorem 3.2], the second, the third and the fourth row of J H̃ are dependent over K.
–

29

By applying [dB4, Lemma 2] twice, we see that we can replace H̃1 by
0 without affecting the nilpotency of J H̃. Now (2) follows in a similar
∂
∂
∂
manner as in the case ∂x
H̃1 , ∂x
H̃1 , ∂x
H̃1 ∈ K[x1 , x2 , x3 , x4 ] above.
2
3
4
– None of the above.
∂
∂
H̃1 ∈
/ K[x1 , x2 , x3 , x4 ] and ∂x
H̃1 ∈
Then we may assume that ∂x
2
4
K[x1 , x2 , x3 , x4 ]. Furthermore, we may assume that the coefficient of
x2 x5 of H̃1 is nonzero.
∂
H̃1 ∈ K[x1 , x2 , x3 , x4 ]. Then we can apply the
Suppose first that ∂x
3
∂
∂
∂
same argument as in the case ∂x
H̃1 , ∂x
H̃1 , ∂x
H̃1 ∈ K[x1 , x2 , x3 , x4 ]
2
3
4
above, to reduce to the case where only the first 5 columns of J H̃
may be nonzero, which follows from (ii) of lemma 5.5 below.
∂
Suppose next that ∂x
H̃1 ∈
/ K[x1 , x2 , x3 , x4 ]. Then we may assume
3
that the coefficient of x3 x6 of H̃1 is nonzero. Now we can apply the
∂
∂
∂
same arguments as in the case ∂x
H̃1 , ∂x
H̃1 , ∂x
H̃1 ∈ K[x1 , x2 , x3 ]
2
3
4
above, to reduce to the case where only the first 6 columns of J H̃
may be nonzero, which follows from (ii) of lemma 5.5 below.
• H is as in (4) of corollary 2.2.
Let H̃ = T −1 H(T x). Then only the first 5 columns of J H̃ may be
nonzero, so the results follow from (i) of lemma 5.5 below.
• H is as in (5) of corollary 2.2.
Let H̃ = T −1 H(T x). Then only the first 6 columns of J H̃ may be
nonzero, so the results follow from (ii) of lemma 5.5 below.
Lemma 5.5. Let H be quadratic homogeneous, such that J H is nilpotent.
(i) If char K 6= 2 and only the first 5 columns of J H may be nonzero, then
the first 5 rows of J H are dependent over K, and x + H is tame.
(ii) If char K = 2 and only the first 6 columns of J H may be nonzero, then the
first 6 rows of J H are dependent over K, and there exists a tame invertible
map x + H̄ such that H̄ is quadratic homogeneous and J H̄ = J H.
Proof. Let N be the leading principal minor matrix of size 5 of J H in case of
(i), and leading principal minor matrix of size 6 of J H in case of (ii). From
[dB4, Lemma 2], it follows that N is nilpotent.
If rk N ≤ 3, then the results follow from theorem 4.1. If rk N ≥ 4, then the
results follow from theorem 5.1 in case of (i) and from theorem 5.2 in case of
(ii).
Notice that we only used the case where only the first 4 columns of J H
may be nonzero of (i) of lemma 5.5, which does not require the calculations of
theorem 5.1 to be proved.

30

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