SHASTRI, ANANT R.

SHASTRI, AR

Dspace at IIT Bombay

A DISCRIMINANT CRITERION FOR THE 2-DIMENSIONAL JACOBIAN PROBLEM

Osaka City University Repository

Shastri, A.R.
Osaka J. Math.
27 (1990), 217-221
A DISCRIMINANT CRITERION FOR THE TWO
DIMENSIONAL JACOBIAN PROBLEM
ANANT R. SHASTRI*
(Received April 7, 1989)
1. Let f(X, Y) and g{X, Y) be any two polynomials with complex coeffi-
cients, i.e., /, g^C[Xy Y]. The pair (/, g) is called an automorphic pair if there
exist u, v ^LC[X, Y], such that,
X=u{f{X,Y),g(X,Y))
and
Y = v(f(X,Y),g{X,Y))
The pair (/, g) is called a Jacobίan pair if the determinant of the Jacobian
matrix
IK MΛ
dx d
\dx
of (f, g) with respect to the variables X and Y, is a nonzero element of C. It is
easily seen that every automorphic pair is a Jacobian pair. The Jacobian problem
is to determine whether every Jacobian pair is an automorphic pair or not.
2. Various equivalent formluations of this problem are known. We shall
recall some of these results, relevant to our discussion, from [1].
A polynomial f^C[X, Y] is said to have r points at infinity, if its homoge-
neous component of maximal degree (i.e., the degree form) is a product of r
coprime factors. If F(X, Y, Z) is the homogenization of f(X} Y), and δ: =
{F(Xy Y, Z)=0} is the curve in P2 then, 6 intersects the line at infinity,
L:={Z=0} in precisely r distinct points. The total number of local branches
of β at all of these r points taken together is called the number of places of f at
infinity. Note that the number of points at infinity is not an automorphic
invariant, whereas, the number of places at infinity of a nonconstant polynomial
* The author is currently on leave from Tata Institute of Fundamental research, Bombay
218 A.R. SHASTRI
is an automorphic invariant. We recall a portion of the result proved in
Theorem 19.4 of [1], in the following proposition.
Proposition: Let (f,g) be a Jacobian pair. Then the following three state-
ments are equivalent:
ί) (/, g) is an automorphic pair,
it) f has only one point at infinity.
Hi) f has only one place at infinity.
REMARK: AS stated and proved, 19.4 of [1] contains only the equivalence
of *) and ίϊ). But the implications i)=Φiii)=Φii) are obvious.
3. The following notion is introduced purely for technical convenience.
A polynomial f(X, Y) is said to be in ready form with respect to X, if the
following conditions hold:
a) /(0, 0)=0
b) the degree form of/is equal to Xn for some n>2 and
c) f has no multiple factors.
Apart from the results from [1], quoted above, the only nontrivial deep
result that we use is the theory of equisingularity of plane algebroid curves,
from [2]. Given a polynomial / in the ready form with respect to X, we let
/
λ
= / + λ , λ e C ; F
λ
(X, Y, Z) be the homogenization of/
λ
 and let φ
κ
(X, Z) =
Fk(X,l,Z). LetC λ : = { Λ = 0 } , ^ : = { Λ = 0 } a n d C λ : = { F λ = 0}. Then,
^D
κ
ClC
λy and Cλ = CλU{P}, where P: = [0, 1,0] in P2. The discriminant,
A
x
(f
λ
) with respect to X, of /
λ
 is a polynomial in Y and λ. From condition a)
and b) above, it follows that, <p
κ
(0, 0)=0 and <pk is regular in X, (i.e, by defini-
tion, φ
λ
(X, 0)=tXny for some nonzero constant t; see, for instance, p. 145 of
[3]). Condition c) implies that the X-discriminant of / i s not identically zero.
Hence, the JY-discriminant of <p0 is not identically zero. This means that the X-
discriminant of φ
κ
 is not identically zero, considered as a polynomial in Z and λ.
Thus we are in a situation as described in §6 of [2]. We now state our main
result:
Theorem: Let (/, g) be a Jacobian pair with f in the ready form with respect
to X. Then each statement in the Proposition is equivalent to each of the follow-
ing two statements:
iv) £Dk:={<pλ=0} is an equisingular family of plane algebroid curves at P: =
{X=0=Z}
v) The Y-degree of A
x
(f
λ
) is independent of X.
The following two sections contain the proof of this theorem. In § 4, we
shall use a geometric argument to show the equivalence of iii) and iv). In
§ 5, we shall use the results from [2] to show the equivalence of iv) and v). We
Two DIMENSIONAL JACOBIAN PROBLEM 219
shall use the notations introduced in this section, in the rest of the paper.
4. Proof of iii) <=> iv)
iiΐ)*Φiv):
Let (/, g) be a Jacobian pair with / in the ready form with respect to X. By
iii), /
λ
 has only one place at infinity, i.e., the curve S)
κ
 has only one branch at
P. Let tn1>m2>" >mk, (mk>2), be the multiplicity sequence of the generic
member of this family of curves and let mΊ>ni2> ">m'h (tn'ι>2), be the
multiplicity sequence of some specialization. By the semicontinuity property,
it follows that, T/z^m, for each /, and hence in particular, l>k. By the equi-
valence of i) and iii), it follows that, each C
λ
 is isomorphic to C.
Note that, each C
λ
 represents the same element in the divisor class group of
JP2. Hence, they all have the same arithmetic genus: p
a
{C\)==a> s a v Since,
the only possible singular point of C
κ
> is the point at infinity P, it follows that,
2p
a
 = Σ aiffo-l) = Σ
1 ί lί = l
Since, the summands in each sum are positive, it follows that, m^m'i for each
i and k=L This implies that the family i2)
λ
 is equisingular.
Assuming iv), given a Jacobian pair (/, g)} we want to show that, / has only one
place at infinity. So, we are at liberty to change this pair by pre-composing
with an automorphic pair. Note that if/ is of degree one, then there is nothing
to prove. Hence, we assume that, deg / > 2 . By replacing Y with Y+tX2, for
a suitable tEiC, if necessary, we can assume that, / satisfies b) of §3. By sub-
stracting/(0, 0) from/, we can also assume that, /(0, 0)=0. Finally, condition
c) of § 3 is satisfied by any/ such that (/, g) is a Jacobian pair. Thus / is now in
the ready form with respect to X. Statement iv) is now applicable, i.e., the family
of curves {£)>} is equisingular.
We shall first show that, each fibre C
λ
 of the map/: C2-*C is irreducible. Note
that, C
κ
=C
λ
 U {P}. The Jacobian condition implies that, C
λ
 is smooth. Hence,
by successively blowing-up at P and its pre-images, all the £
λ
's get resolved,
simultaneously, because of the equisingularity. Indeed, there exists a morphism
π: S->P2, on a smooth surface 5, such that,
1) π is a composite of contraction of (—l)-curves,
2) if £F
λ
 is the proper trasnform of C
λ
 in S> then each £F
λ
 is smooth and
3) there exists a finite set RdS (possibly empty), such that, each £F
λ
 passes
through each point of R.
By equsingularity, it follows that, each £F
λ
 represents the same element of
the divisor class group of S. Now, by further blowing-up at points of i?, and
their pre-images, we obtain a smooth surface T and a map, ψ: T->Pι, which
220 A.R. SHASTRI
extends, the map/: C2->C. The fibres δ
λ
 oίψ are connected, and generically
smooth. Let C{ denote the proper transform of C
λ
 on T. Then, C\^6
λ
 and
generically the equality holds. Again, each Ci represents the same element
of the divisor class group of Γ. So do each <S
λ
. Hence, from the generic
equality, it follows that, Cί=ΰk for each λ. In particular, each C'λ is connected.
Being smooth it is irreducible. Hence, each C
λ
 is irreducible.
Now the arithmetic genus, p
a
(C
λ
)—a, say. By equisingularity, it follows
that, the geometric genus, g{C
λ
)=rγ, is the same for all λ and the number of
places at infinity, <r(C
λ
)=σ say, is also the same for all λ. So, it follows that
the topological Euler characteristic
is the same for all λ. Hence, we have the product formula:
X(C2) = X(C
λ
)X(C)
which implies that, X(C
λ
)=l. This in turn yields that, g(C
λ
)=0 and o (C
λ
)=l.
Thus, we have shown that, / has one place at infinity, as claimed.
5. Proof of iv)^v)
v) =#• iv)
As indicated before, we appeal to theorem 7a) of [2] to conclude that {<p
λ
=0}
defines an equisingular family of plane algebroid curves at P, if the X-discri-
minant A
x
(φ
λ
) is of the form:
A
x
(φ
λ
) = 6{Z, X)Z«
for some unit, £(Z, λ) in the power series ring C[[Z, λ]]. Writing
with 0f eC[λ], and ΘO^F0, the above condition is the same as saying that θ0 is a
nonzero constant.
Now the discriminant A
x
(F
λ
) of the homogeneous polynomial F
λ
 is a
homogeneous polynomial in Y and Z and we have,
Δ
x
(F
λ
)(l, Z): = A
x
(F
λ
(X, 1, Z)): = A
x
(φ
κ
),
and
A
x
(F
λ
)(Y, 1): = AX(FK(X, Y, 1): = Ax(fλ)
Hence, it follows that, A
x
(f
λ
) has the form:
A
x
(f
λ
) = YM(θ
ΰ
Y»+θ1Y>'-
ι+-+θk).
Two DIMENSIONAL JACOBIAN PROBLEM 221
Thus, the statement v) implies that, θ0 is a nonzero constant, which in turn, as
observed above, implies the statement iv).
Here, we need to appeal to the proof of theorem 7 b) in [2], rather than the
theorem itself. For in our context, the condition that the line Z = 0 is not
tangential to the curves i2)
λ
, is not satisfied. However, the proof of theorem
7 b) of [2] uses only Lemma 6 of [2] and a corollary to it. In this lemma, the
required condition is that the corresponding local branches of S)
x
 under the
equivalence, should have the same intersection number with the line L given
by Z = 0 . Since, we have already shown that, iv) is equivalent to iii), it follows
that, each S)
x
 has a unique branch at P. Thus, we need to verify that each S)
κ
has the same intersection number with Z = 0 . But φ
λ
(Xy 0)=Xn, and hence,
the intersection number, (iZ>
λ
, L)=n for each λ and hence, we are done.
This completes the proof of the theorem.
I am grateful to R. Narasimhan for getting me interested in the Jacobian
Problem and then keeping this interest alive by his everreadiness to discuss
this problem for hours and hours. Thanks are also due to Balwant Singh,
Mohan Kumar and Madhav Nori for many helpful discussions.
References
[1] S.S. Abhyankar: Lectures on Expansion Techniques in Algebraic geometry, T.I.
F.R. Bombay, 1977.
[2] O. Zariski: Equivalent Singularities of plane algebroid curves, Amer. Journ. Math.
87 (1968), 507-536.
[3] O. Zariski and P. Samuel: Commutative Algebra, Van Nostrand, 1960.
Department of Mathematics
Indian Institute of Technology
Bombay-400 076



A discriminant criterion for the two-dimensional Jacobian problem

Shastri, Anant R.

Osaka University Knowledge Archive

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A discriminant criterion for the two-dimensional Jacobian problem

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