459
O N THE F O U R T E E N T H P R O B L E M OF H I L B E R T
By M A S A Y O S H I
NAGATA
The purpose of the present paper is to show that the answer to the
14th problem of HilbertC1] is negative, even in the following restricted
case, which may be called the original 14th problem of Hilbert :
Let ffbea subgroup of the full linear group of the polynomial ring
in indeterminates x1,...,xn over a field h, and let o be the set of elements
of k[xl9..., xn] which are invariant under 0. Is o finitely generated ?
Our construction of a counter-example is independent of the characteristic of h, and 1c can be the field of complex numbers.
1. The construction of a counter-example
Let {atj} (i = 1,2,3; j = 1,2,..., 16) be algebraically independent
elements over the prime field n of arbitrary characteristic, and let h be
a field containing the a^. Let V be the vector space of dimension 16
over h and let F* be the set of vectors in V which are orthogonal to the
vectors (au,ai2, ...,am) (i = 1,2,3). (F* is a subspace of dimension 13.)
Let xv...,xle9
tv...,t16 be algebraically independent elements over
h and let 0 be the set of linear transformations cr such that (i) er(^) = ti
for any i and (ii) cr(a^) = xi + biti with (bx,..., 616) € V*. Then:
The set o of elements of k[xx,...,x16, t%,..., t1Q] which are invariant
under 0 is notfinitelygenerated.
2. A lemma on plane curves
In order to prove the example, we need the following lemma on plane
curves:
Fundamental lemma. Let Pv..., P16 be independent generic points of the
projective plane S over the prime field n. For any curve G of degree d, the
sum of the multiplicities of PionO is less than éd.
Proof. Assume that there exists a curve C of degree d such that
Sm^ ^ éd, where mi is the multiplicity of Pi on G. Since the Pt are
independent generic points, the i^ can be specialized to any permutation
of the Pi and therefore we see that there exists a curve of degree d' such
that the multiplicity of the Pi is equal to m for every i and d' < 4m.
Therefore it is sufficient to prove the following lemma (which is equivalent to the fundamental lemma) :
460
MASAYOSHI NAGATA
Lemma. For any natural number m, there is no curve of degree Am which
goes through every Pi with multiplicity at least m.
Proof. Let C4, Cs and (73 be independent generic curves of degree 4, 3
and 3 respectively in S. Let Qx, ...,Q8 be 8 points among C3.C'Z. Then
the Qi are independent generic points of S over TT((74). Let Q\, ...,Q*
be independent generic points of C4 over n(Gé), let C* and C*' be general
curves going through the Qf and let B\,..., jßf be such that
of.o 4 = s<3*+s}2?5, of.c4 = SQ?+sf IP$.
We consider a speciahzation
(Ql5 • • • 5 $8> ^3> ^3> Q ) "*" ( ö l > • • • > 6 8> C*> ^ 3 ? Q )
over Tr. We take i?1? ...,iî 8 so that (1) SfE^ s C3.Cé, 2 | ^ . c <74.C?i, and
(2) the i?^ are speciahzed to the i2f by the speciahzation considered
above.
Assume that for some m there exists a curve of degree 4m which goes
through every i^ with multiplicity at least m. Then we see that there
exists a curve E of degree 4m which goes through the Qi and the B^ with
multiplicity at least m. We shall show that E must contain 0 3 as a component. Assume the contrary. Then CZ.E contains T,mQi + yLfmBj.
Since degCg.i? = 12m, we see that GS.E = SmÇ^ + SfmJS^, which gives
a contradiction because Ql9...9QS9 Bl9 ...,Bé are independent generic
points of Cg, and C3 is of positive genus. Thus E must contain C3 as a
component. Similarly, E must contain (73 as a component. Then
U' = E — C3 — C3 is a curve of degree 4(m — 1 ) — 2 which goes through the
Bj with multiplicity at least m — 1 and goes through the Qi with multiplicity at least m — 2. Specializing the Qi and Bj to the Qf and jß*., we
see that there exists a curve U* of degree U — 2 (t = m— 1) which goes
through the Q* with multiplicity at least t — 1 and the B% with multiplicity at least t. Assume that JE7* does not contain C4 as a component.
Since C^.E* contains H(t-1)Qf+ IltBf and since deg04.jE7* = 16«-8,
we see that <74.E* = S(*-1) Qf + 2Ü?*.
Since
we have
C4. (OS + Of) = 22Q* + 21?*.,
Cé.(tC* + tC¥-E*)
= H(t+1)Q%.
Since the Q* are independent generic points of (74 and since C4 is of
positive genus, we have a contradiction. Thus E* must contain (74 as
a component. Then E* — Cé is in the same situation with t—1 instead
of t. Therefore, by induction, we have a contradiction (observe that if
t = 1, then degU* = 2 < deg<74 and E* cannot contain 04). Thus the
lemma is proved.
THE FOURTEENTH PROBLEM OF HILBERT
461
3. The proof of the example
Setu = tx ...£ 16 ,^ = u\ti9wi = v^y^ = 2 a ^ (i = 1,..., 16;j = 1,2,3).
Then yl9y2,yz, tv..., t16 are invariant under G. Since
we nave
k[wl9..., wlß] = k[yv y2, y3, w 4 ,..., w1Q],
fcfóì,...,x1Q, tl9...,r16) = k(yl9y2,y3, # 4 ,...,x 16 , tl9...,t16),
and G is the set of linear transformations o* of
ku/iy y%i y 39 #4? • • • ? # i e > H, • • • ? HQ\
such that <r(2^) = yi9 a(tj) =fyand cr(#z) = xl + bltl with arbitrary elements 6 4 ,..., 616 of &. Therefore we see that the invariant subfield of G
is k(yx,y2,y3, t1}...,tu). Thus
(1) o = k[xv ...,x16, tx, ...,£16] n k(yl9y29yd9 tl9 ...,£16).
Since yl9 y29 yd are algebraically independent over k9 we can regard
$ = 1c[yl9 y29 y3] as a homogeneous co-ordinate ring of the projective
plane S. Let Pi be the point on S with co-ordinates (ali9 a2i9 a3i). Then
the Pi are independent generic points of S over the prime field n. Let
pi be the homogeneous prime ideal of Pi and set (x(nl9 ..., w16) = fi* #f*>
w^ = max (0, %) (for arbitrary integers nt). We shall show that
(2) o is the set of elements of the form 2a(%,..., n16) tïni... Êj^ie with
a(nl9 ...,% 6 ) € a(nl9 ...,% 6 ) (finite sum).
Proof. Since k[wx,..., w16] = k[yvy2,y3, w 4 ,..., w1Q], we have
# L # 1 J • • • » #16' ^1? • • • J ^16» 1/^1? • • • > l/^ieJ
=
k[yi> y2, y3)
x&,...,
x1Q, tl9...,
c 16 , ljtl9...,
l/^igj.
The intersection of this last ring with k(yt,y2,yz, tl9 ...,£16) is equal to
kb/vy&ysitv-'hv
Ifiv-'llhelSi n c e this last ring contains o (by
virtue of (1) above), it follows that for any element c of o there exists
an integer s such that cus is in k[yl9 y2, y3, tl9..., t16]. Therefore
(i) Any element of o can be expressed in the form
Sa(w 1 ,...,n l e )«f n i...^ n ii
(finite sum, n^ may be negative) with a(nv ...,n1Q) e ip.
Let t)€ be the valuation ring k\xx, ...,# 16 , tl9 ...5£i6](y and let vi be the
normalized valuation defined by fy. o is contained in every fy. I t is
obvious that
(ii) / / an element f of § has value not less than m ( > 0) under vi9 then
f is divisible by tf in k[xl9...,x16, tl9..., t16\.
pt is generated by zi = a3iy1-aliyz
and z\ = aziy2-a2iyz.
Obviously,
v^Zi) = vfoi) = 1. Furthermore, yi9 zj^ and «J/fy are algebraically
462
MASAYOSHI NAGATA
independent modulo the maximal ideal tn^ offyover k. Therefore 2J2J is
transcendental over k(y3 modulo m$). Hence
$[*il*ihi (^ = Ht* n #fa/zj])
is a valuation ring dominated byfy.Therefore we have
(iii) An element f of $Q has value not less than m ( ^ 0) under vi if and only
Since ys,Zilti9Zilti9tl9 ...9ti__1,ti+l9 ...,tw modulo mt are algebraically
independent over k9 we see that
(iv) If a finite number of a(nl9..., nu) are elements of <p, then
^(2a(%,...,n 16 ) tïni... ££^16) = min(v(a(nv ...,n16) t^ni... *£*«)).
Now, (ii) and (iii) show that if a(nx, ...,% 6 ) are in a(nl9 ...9n1%)9 then
Tia(nl9 ...9n16)tïni...tïQnM (finite sum) is in 0 (by virtue of (1)). Conversely, (i) and (iv) show that if c is an element of 0 then e is of the form
stated in (2). Thus (2) is proved completely.
Now we shall prove that 0 is not finitely generated. Assume the contrary. Then there are a finite number of homogeneous forms h(nl9..., n16)
(in §) such that h(nl9 ...,% 6 ) e a(nl9 ...,% 6 ) a n ^ ° *s generated by the
h(nl9 ...9n16)tîni...tÏQni* (each % ^ 0 and Hni>0) over
^[tl9...9tle1.
Let r be the minimum of degA(%, ...,% 6 )/S^. By the fundamental
lemma, we have r > \. Let / b e a n y rational number such that r > r' > J.
Let m be a sufficiently large natural number such that 16r'm is an integer. Since r' > £, there exists a curve of degree 16r'm which goes
through every Pt with multiplicity at least m, i.e. there is a homogeneous
form h of degree 16mr' which is in a(m9..., m). Then htj~m... t^71 is in 0
and therefore ht^m...t^1
can be expressed in a polynomial in the
h(nx, ...,n16)tïni ...tÏQnie and therefore degA/16m = r' cannot be less
than r, which is a contradiction. Therefore 0 is not finitely generated.
Bemark. k[xl9..., x16, tl9 ..., t1Q] n k(yx, y2, ys, u) is not finitely generated,
which gives another counter-example to the 14th problem of Hilbert.
REFERENCES
[1] Hilbert, D. Mathematische Probleme. Archiv Math. Phys. (3), 1, 44-63,
213-237 (1901).
[2] Zariski, O. Interprétations algébrico-géometriques du quatorzième problème de Hubert. Bull. Sci. Math. (2) 78, 155-168 (1954).
[3] Nagata, M. A treatise on the 14th problem of Hilbert. Mem. Coll. Sci.
Univ. Kyoto, A, 30,. 57-70 (1956-57). Addition and correction to it,
pp. 197-200.
[4] Rees, D. On a problem of Zariski. Illinois J. Math. 2, 145-149 (1958).