BABA, KIYOSHI

NAKAI, YOSHIKAZU

Nakai, Yoshikazu

Baba, Kiyoshi

Osaka University Knowledge Archive

A generalization of Magnus' theorem

Osaka City University Repository

Nakai, Y. and Baba, K.
Osaka J. Math.
14 (1977), 403-409
A GENERALIZATION OF MAGNUS' THEOREM
YOSHIKAZU NAKAI* AND KIYOSHI BAB A
(Received May 4, 1976)
Let f(x, y) and g(x, y) be polynomials in two variables with integral
coefficients. O.H. Keller raised the problem in [1]: If the functional deter-
minant 3(/, g)/d(x> y) is equal to 1, then is it possible to represent x and y as
polynomials of / and g with integral coefficients ? This problem drew many
mathematicians' attension and several attempts have been made by enlarging
the coefficient domain to the complex number field C. But no success has
been reported yet. On the other hand A. Magnus studied the volume preser-
ving transformation of complex planes and obtained a result which is relevant
to Keller's problem ([2]). From his results it is immediately deduced that
Keller's problem is answered affirmativiely provided one of f(x, y) and g(x, y)
has prime degree. For the proof Maguns used recursive formulas. But
these formulas are complicated and not easy to handle. In this paper we shall
give a simple proof of his theorem based on the notion of quasi-homogeneity
for generalized polynomials. Moreover we shall go one step further than he did.
Our results ensure that Keller's problem is valid provided one of f(x> y) and
g(x, y) has degree 4 or larger degree is of the form 2p with an odd prime p.
Since a complete solution of Keller's problem is not found yet our paper will
be of some interest and worth-while publication.
1. Quasi-homogeneous generalized polynomials
Let x and y be two indeterminates. We shall set Λ= Σ C xiyi where C
is the complex number field and Z\s the ring of rational integers. A is a graded
ring and the polynomial ring C[x, y] is a graded subring. Hereafter we shall
call an element f(x, y) of A a generalized polynomial or simply a ^-polynomial.
We shall denote by S(f) the set of lattice points (i,j) in the real two space R2
such that the monomial x* yj appears in f(x> y) with a non-zero coefficient.
S(f) will be called the supoprt of f(x, y). A ^-polynomial f(x, y) is called a
homogeneous ^-polynomial or a £-form if S(f) lies in the straight line of the
form X-\-Y=m where w G Z a n d is called the degree of the £-form /(x, 3;).
*) Supported by Takeda Science Foundation.
404 Y. NAKAI AND K. BABA
We shall use the symbol S[f] to denote the set of monomials x* yj such that
the lattice point (iyj) is in S(f).
Proposition 1. Let f(x, y) and g(xy y) be non-constant g-forms of degrees
m and n respectively such that the functional deterimίnant 3(/, g)/d(xy y) is
equal to zero. We shall define an integer d by the rule', (a) d is equal to the
GCD of ImI and \n\ if one of m and n is postive, (b) d is equal to the negative
of GCD (\m\y \n\) if both of m aud n are negative. We shall set m/d=^m/ and
n/d=n'. Then we have the following:
(i) If one of m and n is zero, so is the other and f(xy y) and g(xy y) are
g-polynomials in one variable (y/x).
(ii) If mn<0, then both of f(x, y) and g(xy y) are monomials and there exist
a monomial h(xy y) of degree d such that f=c1hm\ and g=c2hn' where ci(ί=\, 2)
are constants.
(iii) If mn>0, there exists a g-form h(xy y) of degree d such that f=cjιm
and g=c2h
n\
Proof. Assume first m=0 and nφO. It follows from d(fy g)jd(xy y)=0 that
we have dfldx=dfldy=O. This is against the assumption. Since a g-form of
degree zero is necessarily of the form 2 ai(y/xY w e g e t the assertion (i). To
itΞZ
prove (ii) we assume m>0 and n<0 and let f1=f~n and g1 = gm. Then
i> gi)ld(xi> J i ) = 0 Since the degrees of f
λ
 a n d ^ differ only in sign we see
immediately that f
λ
-^ + g
λ
~^ = 0 , or equivalent^, d(f1g1)/dx=0. Similarly
ox dx
we have d(f1g1)/dy=0. Hence fλgx must be a constant. But such a case can
occur only when fly hence /, is a monomial because gλ is a ^-polynomial.
The rest follows easily from this. The proof of (iii) will be carried out by a
similar device and the detailed proof will be omitted.
DEFINITION. A ^-polynomial f(xy y) is called a quasi homogeneous g-
polynomial (or simply a quasi £-form) if the support S(f) of f(x>y) is contained
in the straight line. When the equation of that straight line has the form
Y-\-aX~\. We shall say that the quasi ^-form f(x, y) is (α)-homogeneous of
degree λ.
It should be noticed that if a is an irrational number, monomials only
can be (α)-homogeneous ^-forms.
Proposition 2. Let f(xy y) and g(xy y) be (a)-homogeneous g-forms of
positive degrees λ and μ respectively such that 3(/, g)ld{x, y)=0. Assume that
a is a rational nnmber q/p with comprime integers p ( > 0 ) and q. Let d =
GCD(/>λ, pμ). Then there exists an (a)-homogeneous g-form h(x, y) of degree d/p
such that f=c1hm' and g=c2hn' where m'=pXjdy n'=pμ\d and Ci(i=ly 2) are
constants.
GENERALIZATION OF MAGNUS' THEOREM 405
Proof. Let u, v be new indeterminates and let x=up and y=vq. Then
F(u, v)=f(up, vq) and G(w, v)=g(up, vq) are £-forms of degrees pX and pμ
respectively. The rest follows easily from Proposition 1.
Let γ be an arbitrary real number. Then we can define a grading on A in
the following way. Let λ be a real number and let A
λ
 be the vector space over
C generated by the set of ^-monomials x'yj such t h a t / + γ ί = λ . Then we have
A= φA
λ
 where the sum is extended over all real numbers contained in the
λ
additive subgroup of R generated by 1 and γ. In case γ = l we have the
standard grading and its degree function is the ordinary function. The term
''homogeneous'' is reserved for this standard grading.
Proposition 3. Letf(x, y) and g(x, y) be g-polynomials in x and y such
that d(f,g)ld(x,y)<^C. Let a be any real number and fet / = θ /
λ
 and g=
be the direct sum decomposition by the (a)-gradίng. Then we have
2 Q
 m
\+u=s d(x. γ)
The proof is immediate and will be mitted.
2. Magnus' Theorem
For future reference we shall give Magnus* Theorem in a slightly different
formulation from Magnus' original one.
Theorem 1. Let f(x, y) and g(x> y) be polynomials iu two ariables x and
y with complex coefficients and let m and n be the degrees of f(x, y) and g(x, y)
respectkely. Assume that the functional determinant 9(/, g)/d{Xy y) is a nonzero
constant. If Min (m> ή)> 1, then we have GCD(τra, n)> 1.
Proof. Assume that GCD(m, ή)=\. Let/W and gn be the degree forms of
/ and g respectively. From proposition 1, there is a linear form, say h, such
that f
m
=S
ι
hm and g=£2h
n
. Without loss of generalities we can assume that
h=x and £f = l . We shall pick up a point P=(pv p2) in S(f) in the following
way. Let L be the line defined by the equation X=m and let L rotate around the
point M=(m> 0) counterclockwise until L meets a point in S(f) other than M.
Let / be the line thus obtained. The point in S(f)f]l with the smallest X-
coordinate is the desired point P. Pick up a point Q=(qly q2) in S(g) in a
similar way.
Now ssume we have either (m>) p2>0 or (n>) q2>0. Then we easily
verify that one of the following situation takes place.
(1) The lines MP and NQ are not parallel where N=(n, 0).
(2) The three points P, Q and the origin are not collinear.
If the case (a) occurs let
406 Y. NAKAI AND K. BABA
Y+aX = am, Y+bx = bn
be the equations of the lines MP and NQ respectively. Then we have
If a>b let γ be a real number such that α > 7 > £ . If we choose γ near enough
to α, then χpiyp2 will have the highest (7)-degree in S[f] and xn will have the
highest (γ)-degree in S[g]. Hence by Proposition 3, d(xpiyp*, xn)/d(xy y)=
np2x
n+pi~ι yp2-1=0. But this is impossible. Similarly we have a ontradicition
if a<b.
Now assume the lines MP and NQ are parallel, i.e., a=b then we have the
case (2), i.e., />2?iΦ?i/>2 Let y=a—S with £ <0. If we choose 6 small enough,
then χpiyp2 will have the highest (γ)-degree in S[f] and xqiyq* will have the
highest (γ)-degree in S[g]. But this contradicts Proposition 3 because we have
9lp2^Q2pl'
Thus we have seen that p2=q2=0y i.e., f(x, y) and g(x, y) are polynomials
in x alone. But this is impossible because 9(/, g)ld(x, y) is a non-zero constant,
and the proof of Theorem 1 is complete.
For the sake of reference we shall call the method adopted in this proof
"the method of rotation of lines around the points M and N"'.
3. A generalization of Magnus' Theorem
Theorem 2. Under the same notations and assumptions as Theorem 1, we
hove the following: //Min(/w, w)>2, then we have GCD(m, ή)>2.
Proof. Assume that Min(m, n)>2 and GCD(τw, n)=2 and we shall draw
a contradiction. Let f
m
 and g
n
 be degree forms of/ and g respectively. From
Proposition 2 it follows that there exists a quadratic form h(x, y) such t h a t /
w
=
ahm' and g
r
=bhn\ where m=2m' and n=2n'. There are two possibilites.
(I) A is a product of two independent linear forms. In this case we can
assume without loss of generalities that f
m
=(xy)m' and g
n
—(xy)n'. Apply the
method of rotation of lines around the points M1=(m /, m') and Nί=(n\ n').
Then we can easily see that any point (i,j) in S(f) satisfies the condition j<m\
and any point (s, t) in S(g) satisfies the condition t<n\
Now consider the (O)-grading in A. The degree forms of / and g are
respectively of the forms
fW = y*ϊa
o
+a
ι
x+ +aM,-ιxr'-
1+*"')
From Propositions 2 and 3 there is a linear form c-\-x such that
/<?) = y>»Xc+xy' and g$> = y''(c+x)*'
If we set x1=c
Ji-x and consider / and g as polynomials in new variables x1 and
GENERALIZATION OF MAGNUS* THEOREM 407
y
ι
=y>then the support «SΊ(/) have no point (i, j) with j>m' except the point
(m\ mf). Similarly S
x
{g) have no point (s, t) with t>n'. Apply again the
method of rotation of lines around the points M
γ
 and N
v
 Then we can see
finally that no point (i,j) with i<j is in S(f) and no point (s, t) with s<t is in
S(g). This means that f(x, y) and g(x, y) lack the terms ys(s>l). This is
impossible because of the assumption d(f> g)/d(x, y) is an element of C*.
(II) h is a power of a linear form: In this case we can assume as before
that the degree forms are of the forms f
m
=
z
x
m
 and g
n
=x
n
 respectively. Then
we can see, following the method of rotations of lines around the point Λf=(m,
0) and N(n, 0), that S(f) is contained in the region defined by the inequality
Y + — X < — and (g) is in the region Y+ — X ^ —. Consider (l/2)-grading
Δι L* At Li
and apply Propositions 2 and 3. Then we see that degree forms of / and g by
this grading are
(ay+x2)™' and {ay-\-x2)m'
respectively. If a=0 we can proceed further and we see that no point (i, j)
with j>0 is in S(f) and no point (ί, t) with ΐ>0 is in S(g). This is a
contradiction. Hence we must have αΦO. Then apply de Jonquiere trans-
formation
Y\ = ay-\-x2y x1 = x .
Since we have
/(*, y) = (ay+χY+
 jΈ<matjχ
iyj
and
We easily see that
Σ ,*'iMyί
and
gι(^yι) = yΐ+ Σ b'uxiyi
j + i/2<n'
where
/i(*i> yι) =Ax» a~\yι-xl)) and g
λ
(xly y,) = ^(x, α" 1 ^!—*?)).
By the method of (clockwise) rotation of lines around the points (0, m') and
(0, n') applied to the pair of polynomials f
x
(xly yλ) and gx{xιy ji), we see that
408 Y. NAKAI AND K. BABA
S(fj) is in the half plane X+ Y <m! and £(&) is in the half plane X+Y<n'.
This means that f
λ
(x
ιy yλ) is of degree rri and gλ(xly yλ) is of degree n\
Moreover ^L/ίijgi)
 =
 g-i9(/>^) is in C*. Since Min(m, n)>2, we have Min
9(*i, Jι) d(χy y)
(m\nf)>\. Moreover GCD(wι', w')=l. This is the situation negated in
Theorem 1.
4. Application to Keller's problem
Theorem 3. Let f(x, y) and g(x, y) be polynomials of degrees m and n
respectively with complex coefficients and assume that the functional determinant
®(f>g)/Q(x>y) *s a non-zero constant. Then we have C[x, y] = C[f, y), g(x, y)]
in the following three cases:
(1) m or n is a prime number \
(2) m or n is 4;
(3) nι=2p^tn where p is an odd prime.
Proof. In any case it follows from Theorems 1 and 2 that smaller de-
gree, say n, divides larger degree m. Then from Proposition 2 and 3 the degree
forms f
m
 and g
n
 are related like this, f
m
=Sgn/n. Then
fι=f-{εn/mg)m/n
has lower degree than / and 9(/i, g)l®(x, y)=®(f> g)l®(x> y) ιs> a non-zero
constant. Thus we can use induction on the sum m-\-n of degrees to get the
conclusion. q.e.d.
Keller's Original problem is also settled in these three cases cited in The-
orem 3 because of the following
Proposition 4. Let f(x> y) and g(x, y) be the polynomials in x and y with
integer coefficients such that the functional determinant is equal to 1 and C [/, g]
= C [x, y]. Then we hove necessarily Z [/, g] = Z [x, y].
Proof. It suffices to prove that x and y are in Z[f, g]. By assumption
we have
If we set
f(x,y)=flox+foiy+
then the assumption implies that/ 1 0 £ 0 1 —/ 0 1 £ 1 0 = 1. Apply the unimodular
transformation of variables
GENERALIZATION OF MAGNUS' THEOREM 409
Then Z[x, y] = Z[x', y'] and f=χ'-\-(higher degree terms) and g =y'+ (higher
degree terms). Hence to prove the assertion we can assume without loss
of generalities that linear parts of / and g are x and y respectively. We shall
define a linear order in the set (/, j) of lattice points in R2 by the way: (i,j)>
(i\ / ) if (i) i-\-j>i/Jrj' or (ii) i-\-j=i'+f and i> /'. We shall show that every
Cij is in Z by induction on the linear order just defined. Assume every £ fγ
with (i\jf)<{}>]) is integer. Then the coefficients of the polynomial
cIJfigJ+ci+1J-1fi+1gi-1+ "+coi+J+1gi+J+1+ "
are integers. In this polynomial xiyi appears once with the coefficient ci}.
Hence cij must be an integer. Similarly y is in Z[f, g] and the assertion in
proved completely.
OSAKA UNIVERSITY
References
[1] O.H. Keller: Ganze Cremona-Transformationen, Monatsh. Math, und Phys. 47
(1939), 299-306.
[2] A. Magnus: On polynomial solutions of a differential equation, Math. Scand. 3
(1955), 255-260.



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A generalization of Magnus' theorem

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