UTMS 2002–10February 26, 2002

An arithmetic property of Shirosaki’s
hyperbolic projective hypersurface
byJunjiro Noguchi

TUNIVERSITY OF TOKYOGRADUATE SCHOOL OF MATHEMATICAL SCIENCES
KOMABA, TOKYO, JAPANAn Arithmetic Property of Shirosaki's Hyperbolic
Projective Hypersurface
Junjiro Noguchi
2002 January 7

1 Introduction and main result
In 1974 S. Lang [La74] conjectured that an algebraic variety V de ned over an arbitrarily xed
number eld K carries only nitely many K -rational points (here referred as the arithmetic
niteness property, while it was termed to be \Mordellic" in [La86]) if the complex space VC
with some embedding K ,! C is hyperbolic in the sense of Kobayashi [Ko70]. The conjecture
was established for curves [Fa83] and for subvarieties of Abelian varieties [Fa91]. The analogue
over function elds was proved by [No85] and [No92].
S. Kobayashi conjectured in 1970 that generic hypersurfaces of the complex projective spacePn(C) of high degree are hyperbolic ([Ko70]). In [MN96] the existence of hyperbolic projective
hypersurfaces M was proved for all n = 2 in a constructive way. Hence, it is interesting tostudy the arithmetic property of such M . In [No97] it was proved that those M satisfy theanalogue of the arithmetic niteness property over function elds, and have only nitely manyS -unit points over number elds. Moreover, it was observed that \abc    -Conjecture" ([No96a],[No96b], [Vo98]) would imply the arithmetic niteness property of M . Cf. also Sarnak and Wang
[SW95] for another arithmetic property of M . So far to the author's knowledge there had beenno example, nor existence theorem of a projective hypersurface of dimension > 1 that carries
the arithmetic niteness property.
In 1998 Shirosaki [Sh98] found a simpler method to construct hyperbolic hypersurfaces X ofdegree dn with some d > 12 by an idea using a unicity polynomial, which is di erent to that of[MN96]. The purpose of this paper is to prove the arithmetic niteness property of Shirosaki's
X.
To state the result, we take co-prime positive integers d; e 2 N such that
(1.1)

d > 2e + 8;

and set(1.2)

P (w0 ; w1 ) = w0d + w1d + w0e w1d;e :1

Following to [Sh98], we de ne inductively(1.3)

P1 (w0 ; w1 ) = P (w0 ; w1 );Pn (w0 ; w1 ; : : : ; wn ) = Pn;1 (P (w0 ; w1 ); : : : ; P (wn;1 ; wn )); n = 2; 3; : : : :Then, Pn is homogeneous and of degree dn . Set

X = fPn (w0 ; w1 ; : : : ; wn ) = 0g  PnQ :(1.4)

By Shirosaki [Sh98] XC is hyperbolic for e > 2.
Main Theorem Assume that e > 2. Then X de ned by (1.4) satis es the arithmetic
niteness property; that is, for an arbitrary number eld K , the set X (K ) of K -rational pointsof X is nite.
It is interesting to observe that there is an analogy not only in the result, but also in the wayof the proof; similar analogues were found in [No97] and [NW99]. For general references on thepresent subject, cf. [La86], [La91], [No89], [Ko98], [Vo87].

2 Lemmas
We keep the notation in x1.

Lemma 2.1 For every 0 5 k 5 n there is a number (n; k) 2 N such thatPn (0; : : : ; 0; wk ; 0; : : : ; 0) = (n; k)wkd :nProof. Note rst that for wj 2 N; 0 5 j 5 n,

P (w0 ; : : : ; wn ) 2 N:(2.2)
By de nition we havePn (0; : : : ; 0; wk ; 0; : : : ; 0) = Pn;1 (0; : : : ; 0; wkd ; wkd ; 0; : : : ; 0):
Inductively, we havePn(0; : : : ; 0; wk ; 0; : : : ; 0) = Pn;l ((l; 0)wkd ; : : : ; (l; n ; l)wkd );llwhere maxfk; n ; kg < l < n and (l; j ) 2 N. Set (n; k) = Pn;l ((l; 0); : : : ; (l; n ; l)). Thenby (2.2) we have that (n; k) 2 N andPn (0; : : : ; 0; wk ; 0; : : : ; 0) = (n; k)wkd :nQ.E.D.

Lemma 2.3 Let Fj ; 1 5 j 5 m, be holomorphic functions on C, and let dj 2 N. Assume thefollowing:2P(i) mj=1 Fj = 0;(ii) the order of every zero of Fj is at least dj ;(iii) the functions Fj ; 1 5 j 5 m ; 1, have no common zero and are linearly independent overC.
Then we have

Xm 1 = 1 :m;2j =1 dj

This is due to Cartan [Ca33] (cf., [MN96] for an application to the hyperbolicity problem).
The next lemma is a slight modi cation of [Sh98], Theorem 4.2.

Lemma 2.4 Let ; 2 C and 6= 0. Then the curve C ;  P2(C) de ned byC ; = f(w0 ; w1 ; w2 ) 2 P2 (C); P (w0 ; w1 ) = P ( w1 ; w2 )gis hyperbolic for (i) =6 0; e > 2, and for (ii) = 0; e > 3.
Proof. When = 1, the lemma was proved by [Sh98], Theorem 4.2. If 6= 0, thenP ( w1 ; w2 ) = d P (w1 ; ;1 w2 ). Thus it is reduced to the case of = 1.
Let = 0, and assume that C ;0 is not hyperbolic. Then there is a non-constant holomorphic
mapping f : C ! C ;0  P2 (C). Let f = (f0 ; f1 ; f2 ) be reduced representation of f . Then one
has(2.5)
f0d + f1d ; f2d + f0ef1d;e = 0:
If one of ffj g2j =0 vanishes identically, it follows from (2.5) that other fj must be proportional;
hence the mapping f is constant. This contradicts the hypothesis. Thus none of ffj g2j =0 vanishes
identically. If fjd ; 0 5 j 5 2, are linearly dependent, there is a non-trivial relation,(2.6)
c0 f0d + c1f1d + c2 f2d = 0; cj 2 C:
If one of fcj g2j =0 , is zero, it follows from (2.6) and (2.5) that f is constant. This is absurd, andso c0 c1 c2 6= 0. By Lemma 2.3, 3=d = 1; this contradicts d > 16. Therefore fjd ; 0 5 j 5 2, mustbe linearly independent. Then, Lemma 2.3 and (2.5) yields
3 + 1 = 1:
d e 2
It follows from (1.1) and e > 3 that
3+1;1< 3 +1;1
d e 2 2e + 8 e 2
2= ;2(e e++e4)+e 8< 0:This is again a contradiction. Q.E.D.
By Faltings' Theorem [Fa83] and Lemma 2.4 we have3Lemma 2.7 Let K be an arbitrary number eld, and let ; 2 K with 6= 0. Assume that
6 0 and e > 2, or that = 0 and e > 3. Then the set C ; (K ) is nite.
=The following is an analogue of [Sh98], Theorem 4.3.

Lemma 2.8 Let e > 2, n = 2, and let (p0 ; : : : ; pn;1) 2 Pn;1 (K ) be a point such that at least two;1 are di erent to zero. Then there are only nitely many points (w0 ; : : : ; wn ) 2 Pn (K )of fpj gnj =0such that(P (w0 ; w1 ); : : : ; P (wn;1 ; wn )) = (p0 ; : : : ; pn;1 ):Proof. We use the induction on n. If n = 2, there is a number 2 K  = K n f0g such that
P (w0 ; w1 ) = P (w1 ; w2). There are only nitely many such (w0 ; w1 ; w2) by Lemma 2.7.
Assume that the statement holds up to n ; 1. We consider the case of n > 2, and let(w0 ; : : : ; wn ) 2 Pn (K ) be such points. If P (wn;1 ; wn ) = 0, the induction hypothesis implies
the niteness of the number of points (w0 ; : : : ; wn;1 ). If wn;1 = 0, then wn = 0; if wn;1 6= 0,then wn 6= 0 and the number of ratio wn =wn;1 is at most d. Therefore the number of points(w0 ; : : : ; wn ) is nite.
Assume that P (wn;1 ; wn ) 6= 0. Then there is a number k < n such that

(2.9)

k < j 5 n ; 1;P (wj;1; wj ) = 0;
P (wk;1; wk ) 6= 0:

There is a number 2 K  such that

P (wk;1; wk ) = P (wn;1 ; wn ):If wn;1 = 0, then wj = 0 for every k 5 j 5 n ; 1. Then (2.10) yields
wkd;1 = wnd 6= 0:(2.10)

Hence the number of points(wk;1 ; : : : ; wn ) = (wk;1 ; 0; : : : ; 0; wn )is nite. If wn;1 6= 0, then wj 6= 0 for k 5 j 5 n ; 1 by (2.9). Moreover, the number of ratios
wj ;1 =wj ; k + 1 5 j 5 n ; 1, is nite. Hence there are only nitely many 2 K  such thatwn;1 = wk . It follows from (2.10) that

P (wk;1 ; wk ) = P ( wk ; wn );6= 0:
By Lemma 2.7 the number of points (wk;1 ; wk ; wn ) is nite. Hence the number of points(wk;1 ; : : : ; wn ) is nite. If there is a number 1 5 j < k with P (wj ;1 ; wj ) 6= 0, the induction
hypothesis implies the niteness of the number of points (w0 ; : : : ; wn;1 ). In all, there are only
nitely many such points (w0 ; : : : ; wn ).
Assume that P (wj ;1 ; wj ) = 0 for all 1 5 j < k. Then either wj = 0 for all 1 5 j < k, or
wj 6= 0 for all 1 5 j < k and the number of ratios wj;1=wj is at most d for every 1 5 j < k.
Henceforth there are only nitely many points (w0 ; : : : ; wn ). Q.E.D.
4

3 Proof of the main theorem
We use the induction on n = 2. The case of n = 2 is done by Lemma 2.7. Assume the case ofn ; 1 to be true. For n we recall the de nition of X :Pn(w0 ; : : : ; wn ) = Pn;1 (P (w0 ; w1 ); : : : ; P (wn;1 ; wn )) = 0:
It follows from Lemma 2.1 that either all P (wj ;1 ; wj ) = 0; 1 5 j 5 n, or at least two ofP (wj;1 ; wj ) are di erent to zero. In the rst case, the niteness of such points (w0 ; : : : ; wn)
is clear. In the latter case the induction hypothesis and Lemma 2.8 imply that there are only
nitely many those points (w0 ; : : : ; wn ). Therefore, X (K ) is a nite set.
This completes the proof of the Main Theorem.
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5

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Noguchi, Junjiro
Graduate School of Mathematical Sciences
University of Tokyo
Komaba, Meguro,Tokyo 153-8914
e-mail: noguchi@ms.u-tokyo.ac.jp6Preprint Series, Graduate School of Mathematical Sciences, The University of Tokyo
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