Lectures on
The Fourteenth Problem of Hilbert
By
M. Nagata
Tata Institute of Fundamental Research, Bombay
1965
Lecture on
The Fourteenth Problem of Hilbert
By
M. Nagata
Notes by
M. Pavaman Murthy
No part of this book may be reproduced in any
form by print, microfilm or any other means without written permission from the Tata Institute of
Fundamental Research, Colaba, Bombay 5
Tata Institute of Fundamental Research
Bombay
1965
Contents
0
1
1 A generalisation of the original fourteenth problem
3
2 A generalization of the original fourteenth problem(contd.)
9
3 The counter example
17
4 Theorem of Weitzenböck
29
5 Zarisk’s Theorem and Rees’ counter example
35
6 Complete reducibility of rational representation....
51
iii
Chapter 0
In 1900, at the international Congress of Mathematicians in Paris, 1
Hilbert posed twenty-three problems. His complete address was published in Archiv.f. Math.U.Phys.(3),1,(1901) 44-63,213-237 (one can
also find it in Hilbert’s Gesammelte Werke).
The fourteenth problem may be formulated as follows: The Fourteenth Problems. Let K be a field and x1 , . . . , xn algebraically independents elements over K. Let L be a subfield of K(x1 , . . . , xn ) containing
T
K. Is the ring K [x1 , . . . , xn ] L finitely generated over K?
The motivation for this problem is the following special case, connected with theory of invariants.
The Original fourteenth problem. Let K be a field and G a subgroup
of the full linear group GL(n, K). Then G acts as a group of automorphisms of K[x1 , . . . , xn ]. Let IG be the ring of elements of K[x1 , . . . , xn ].
invariant under G. Is IG finitely generated over K?
Contributions to the original fourteenth problem were made in particular cases. In fact it was proved that IG is finitely generated in the
following cases.
1. K is the complex number field and G = S L(n, K) acting by means
of its tensor representations. (D. Hilbert Math. Ann. 36(1890)
473 - 534).
2. K is the complex number field and G satisfies the following condition: there exists a conjugate G∗ of G such that A ∈ G∗ ⇒ t Ā ∈
G∗ , where ‘−’ and ‘t’ indicate complex conjugation and transpose
respectively (E. Fischer Crelle Journal 140(1911) 48 - 81).
2
1
2
3. K is an arbitrary field and G a finite group (E.Noether , for characteristic K = 0, Math. Ann. 77 (1916) 89 - 92, for characteristic
K , 0 Göttinger Nachr. (1926 28-35).
4. K is the complex number field and G, a one parameter group
(R. Weitzerböck Acta Math. 58 (1932) 231 - 293).
5. K is the complex number field and G, a connected semi-simple
Lie group (H. Weyl Classical groups (1936) Princeton Univ.
Press).
3
The next significant contribution was made by 0.Zariski in 1953. He
generalized (Bull. Sci., Math.78(1954) 155-168) the fourteenth problem
in the following way.
Problem of Zariski. Let K be a field and K [a1 , . . . , an ] an affine
normal domain (i.e.a finitely generated integrally closed domain over
K). Let L be a subfield of K (a1 , . . . , an ) containing K. Is the ring K
T
[a1 , . . . , an ] L finitely generated over K?
He answered the question in the affirmative when trans. degK L ≤ 2.
Later, in 1957, D. Rees (Illinois J. of Math. 2(1958) 145 - 149) gave a
counter example to the problem of Zariski when trans. degK L = 3.
Finally, in 1958, Nagata (Proceedings of the International Congress
of Mathematicians, Edinburgh (1958) 459 - 462) gave a counter example
to the original fourteenth problem itself. This counter example was in
the case when trans. degK L= 13. Then, in 1959, Nagata (Amer. J. Math.
91, 3(1959) 766 - 772) gave another counter example in the case when
trans. degK L = 4.
The groups occurring in these examples are commutative. So in
view of Weyl’s result we seek the answer to the original fourteenth
problem in the case when G is a non- comutative, non-semi-simple Lie
group. The examples mentioned above May be made to yield one with
G non-commutative by considering what is essentially the direct product by a non-commutative group. More interesting is the case when G
is a connected Lie group such that [G, G] = G. Even in this case the
answer is in the negative as we shall see later in this course of lectures.
Chapter 1
A generalisation of the
original fourteenth problem
1. We first generalise the original fourteenth problem in the follow- 4
ing way: Generalised fourteenth problem. Let K be a field. Let R =
K[a1 , . . . , an ] be a finitely generated ring over K (R need not be an integral domain). Let G be a group of automorphism of R over K. Assume
P g
f K is a finite dimensional vector space over
that for every f ∈ R,
g∈G
K. Let IG be the ring of invariants of R under G. Is IG finitely generated
over K?
Remark 1. When a1 , . . . , an are algebraically independent and G is a
subgroup of GL(n, k) acting on R in the ‘usual way’, elements of R satisfy the finiteness condition we have imposed in the problem.
Remark 2. The omission of the assumption that a1 , . . . , an are algebraically independent is helpful. For instance let N be normal subgroup
of G such that IN is finitely generated. Then G acts on IN , as N is normal in G. Hence G/N acts on IN and (IN )G/N = IG . So, for instance
if we know that our generalized problem is true for (i) finite group (ii)
connected semi-simple Lie groups, (iii) diagonalizable groups, then we
can get immediately that: if G is an algebraic group whose radical N
is diagonalizable then the answer to generalized problem is in the affir3
4
A generalisation of the original fourteenth problem
mative. Note that IN need not be polynomial ring even if a1 , . . . , an are
algebraically independent.
5
Remark 3. We do not assume R is an integral domain, as the constant
field extension (of §3 below) of a finitely generated integral domain over
a field K need not be an integral domain.
2. Algebraic linear groups. In this section by an affine variety we
mean the set of rational points over K of a certain affine variety with
points in an universal domain. We shall give Zariski topology to an
algebraic variety. The group GL(n, k), being the complement of the hy2
persurface defined by det(xi j ) = 0 in K n , is an affine variety (in fact it
2
is isomorphic to the hypersurface z det(xi j ) = 1 in K n +1 ). The group
operation of GL(n, K) are regular i.e. GL(n, K) is an algebraic group.
In this course of lectures by an algebraic group we always mean closed
subgroups of GL(n, k). We define connectedness and irreducibility under Zariski topology. In the case of algebraic groups these two concepts
coincide. If Go is the connected component of an algebraic group G,
then Go is a normal subgroup of finite index in G. An algebraic group
contains the largest connected normal solvable subgroup of G, which
we call the radical of G. For details about algebraic groups we refer
to seminaire C. Chevalley Vol.1 (1956-1958) and A. Borol Groups
Linéaires Algébriques, Annals of Mathematics, 64 (1956). We remark
that in the generalized problem we may assume that G is a subgroup of
GL(m, K) for some m. Because of the finiteness condition on the elen P
P
g
ai K is finite dimensional vector
ments of R = K[a1 , . . . , an ], V =
i=1 g∈G
6
space. We choose a basis b1 , . . . , bm for V. Then R = K[b1 , . . . , bm ]
and G is a group of automorphisms of V. Therefore in the generalised
problem we may assume G ⊆ GL(n, k). From now on, whenever we say
that a subgroup G of GL(n, K) acts on R = K[a1 , . . . , an ] we mean that it
n
P
acts linearly on the vector space
Kai . The following theorem shows
i=1
that in the problem we may assume that G is an algebraic group
Theorem 1. Let R = K[a1 , . . . , an ] and let G ⊆ G(n, k) act as a group
of automorphisms of R over K. Let G∗ be the closure of G in GL(n, K).
A generalisation of the original fourteenth problem
5
Then G∗ acts as a group of automorphisms of R over K and IG = IG∗ .
This theorem is a consequence of the following lemma (put S = IG ).
Lemma . Let S ⊆ R and let H be the set of elements of GL(n, k) which
induce automorphisms of R over K and leave every s ∈ S invariant.
Then H is algebraic.
Proof. Let H1 denote the set of elements of GL(n, K) which induce automorphisms of R over K. Let U be the ideal of the ring of polynomials
K[x1 , . . . , xn ] such that K[a1 , . . . , an ] ≈ K[x1 , . . . , xn ]/U. Now g ∈ H1
if and only if U g = U. Let f1 , . . . , fm generate U. We may assume
that f1 , . . . , fm are linearly independent over K. Extend f1 , . . . , fm to a
linearly independent basis f1 , . . . , fm , fm+1 , . . . , fl of the vector space
V=
n
X
X
g
fi K.
Let fit
i=1 g∈GL(n,k)
=
l
X
λi j f j , i = 1, . . . , m,
j=1
where t = (trs ), trs are indeterminates. Then λi j are polynomials in trs .
The condition U g = U is equivalent to
λi j (g) = 0, i = 1, . . . , m, j = m + 1, . . . , l.
7
Hence H1 is algebraic.
Let now H = g ∈ H1 | sg = s, s ∈ S .
It is enough to prove the lemma when S consists of a single element
s. Let s, s1 , . . . , sk be a linearly independent basis of the vector space
P h
s K. Let y = (yrs ) be a generic points or may component of H1
h∈H1
over K. We have sy = λ0 s +
k
P
i=1
λi si , where λi are polynomials in yrs .
The condition sy = s is equivalent to λ◦ (s) = 1, λi (g) = 0 for i > 0.
Hence H is algebraic.
3. Constant field extension. We shall now study the behaviour of
invariants under the constant field extensions. Let R′ and R′′ be two
commutative rings containing a field K. Let G′ (respectively G′′ ) act
A generalisation of the original fourteenth problem
6
on R′ (respectively R′′ ) as a group of automorphisms over K. Then the
direct product G′ × G′′ acts on R ⊗K R′′ as a group of automorphisms;
′′
′
in fact we have only to define (a′ ⊗ a′′ )(g′ , g′′ ) = a′g ⊗ a′′g , a′ ∈ R′ ,
a′′ ∈ R′′ , (g′ , g′′ ) ∈ G′ × G′′ .
Lemma 1. Let IG′ ×G′′ denote the ring of elements of R′ ⊗K R′′ invariant
under G′ ×G′′ Let IG′ (respectively IG′′′′ ) denote the ring of elements of R′
(respectively R′′ ) invariant under G′ (respectively G′′ ). Then IG′ ×G′′ =
IG′ ⊗K IG′′′′ .
N ′′
Proof. We have only to prove that IG′ ×G′′ ⊆ IG′ ′
IG′′ . Choose a linK
8
′
′′
′′
early independent basis (i′λ′ )λ′ ∈Λ′ and (i′′
λ′′ )λ ∈Λ for the vector spaces IG ′
′′
and Ig′′ respectively over K. Extend these bases to a linearly independent bases of R′ and R′′ respectively, say
X
X
′
′
′′
′′
R′ =
i′µ′ K, R′′ =
i′′
µ′′ K with Λ ⊆ M and Λ ⊆ M .
µ′ ∈M ′
Let f =
P
(µ′ ,µ′′ )
µ′′ ∈N ′′
kµ′ µ′′ (iµ
∈M ′ ×M ′′
⊗ iµ′′ ) ∈ IG′ ×G′′ , kµ′ µ′′ ∈ K.
For every g ∈ G, we have
X
XX
′
′
′
( kµ′ µ′′ igµ ) ⊗ i′′
fg =
kµ′ µ′′ (igµ ⊗ i′′
µ′′ = f.
µ′ ) =
µ′′
(µ′ ,µ′′ )
Hence
X
µ′
′g′
kµ′ µ′′ iµ′ =
X
µ′
µ′
kµ′ µ′′ i′µ′ , g′ ∈ G′ , µ′′ ∈ M ′′ .
Hence kµ′ µ′′ = 0, for = µ′ < Λ′ . Similarly kµ′ µ′′ = 0, for µ′′ < Λ′′ .
′′ and the lemma is proved.
i.e. f ∈ IG′ ′ ⊗K LG
′′
Lemma 2. With the above notation, R′ ⊗K R′′ is finitely generated over
K if and only if R′ and R′′ are finitely generated over K.
Proof. It is clear that if R′ and R′′ are finitely generated over K, so
is R′ ⊗K R. Now let R′ ⊗K R. be finitely generated over
K, say fi =
P ′
′′
′
γi j ⊗ γi j , i = 1, . . . , l are the generators. Then R = K γi′ j and R′ =
j
K γi′′j .
A generalisation of the original fourteenth problem
7
Let R = K[a1 , . . . , an ] and let G act on R as as group of automorphisms of R over K. Let K ′ be a filed containing K. Then G acts on
K ′ ⊗K R which we denote by K ′ [a1 , . . . , an ]. Let IG′ ′ be the invariant
elements of K ′ ⊗K R under G. Then in Lemma 1 if we put R′ = K ′ ,
R′′ = R, G′ = {1}, G′′ = G, we get IG′ = K ′ ⊗K IG . As in Lemma 2 it 9
follows that:
Proposition 1. IG′ is finitely generated over K ′ if and only if IG is finitely
generated over K.
The above proposition helps us to confine ourselves to a smaller
field when K is ‘too big’. Let G′ be a subgroup of G such that G′ is
dense in the closure Ḡ of G. For instance when K is of characteristic
zero or Ḡ is a torus group we can take G′ to be finitely generated. When
K is of characteristic p , 0 we may take G′ to be countably generated.
Let K[x1 , . . . , xn ]/U ≈ K[a1 , . . . , an ] with the x1 , . . . , xn algebraically
independent over K. Let K ′ be a subfield of K such that elements of G′
are K ′ -rational and U is defined over K ′ For instance if we can choose
G′ finitely generated, K ′ can be chosen to be finitely generated over
the prime field. If G′ can be chosen countably generated, then K ′ can
be choosen to be countably generated over the prime field. Now as
the ideal U is defined over K ′ , K⊗K ′ , K ′ [a1 , . . . , an ] = K[a1 , . . . , an ].
Hence by proposition 1, IG is finitely generated over K is and only if
G′ − invariants in K ′ [a1 , . . . , an ] are finitely generated.
Of course, as we noted in §2, we can enlarge G′ to an algebraic
group.
3. Invariants of a finite group. We shall now consider the original 14th
problems (in fact the generalised problem) when G is finite.
Theorem 2 (E. Noether). Let R = K[a1 , . . . , an ] and G a finite group
acting on R as a group of automorphisms of R over K. Then IG is finitely 10
generated over K.
Proof. Let G = {1 = g1 , . . . , gh }. Let a ∈ R. Set
S1 =
h
X
i=1
agi , S 2 =
X
i< j
agi ag j , . . . , S h = ag1 . . . agn .
A generalisation of the original fourteenth problem
8
Then S i ∈ IG , i = 1, . . . , h and ah − S 1 ah−1 + · · · + (−1)h S h = 0.
Hence R is integral over IG . Now the theorem follows from the
following Lemma:
Lemma. Let R = K[a1 , . . . , an ] and let S be subring of R containing K
such that R is integral over S . Then S is finitely generated over K.
Proof. There exist Gi j ∈ S , 1 ≤ i ≤ n, 0 ≤ j ≤ mi − 1 such that
mi −1
1
am
+ . . . + Cio = 0.
i + C imi −1 ai
Set S ′ = K Ci j
rian, s is also finite
. Then R is finite S ′ module. As S ′ is noethe-
1≤i≤n
0≤ j≤ni −1
S ′ -module.
Hence S is finitely generated over K.
Corollary. Let G0 be a normal subgroup of G (G not necessarily finite)
of finite index. Let G act on R as a group of automorphisms over K. If
IG0 is finitely generated over K, then, so is IG .
Proof. The group G/G0 acts on IG0 and (IG0 )G/G0 = IG
11
Remark. Suppose that R (of the above corollary) is an integral domain
Then the converse of the above Corollary is also true. Let K0 L be the
quotient of IG0 and IG respectively. Then K0 is a finite separable algebraic extension of L. Let IG be finitely generated. As G/G0 is finite, IG0
is integral over IG . Hence IG0 is a finite IG -module and therefore finitely
generated over K.
Chapter 2
A generalization of the
original fourteenth
problem(contd.)
1. In this chapter we shall consider the generalized problems with cer- 12
tain assumption on the representation of G. In fact we shall prove the
following:
Theorem 1. Let R = K[a1 , . . . , an ] and let G ⊆ GL(n, K) act on R as a
group of automorphisms of R over K. Suppose G satisfies the following
condition: (∗) If λ is a representation of G by a finite dimensional K(G)



1 ∗... ∗
1, 0... 0
0

 0




.
module M ⊂ R and λ =  .
,
then
λ
is
equivalent
to

′ 
 :

λ
 .. λ′ 




0
0
Then IG is finitely generated. We shall prove theorem in several steps.
Lemma 1. Let G satisfy the condition (∗) and let f ∈ R. Then there
TP
P
′
exists an f ∗ ∈ IG ( f g K) such that f − f ∗ ∈ ( f g − f g )K.
g,g′
g
Proof. Let M =
P
g
f g K and N =
P
g,g′
′
( f g − f g )K. The vector spaces M
and N are G-modules. Let the dimension of M be m. We shall prove
9
10
A generalization of the original fourteenth problem(contd.)
the lemma by induction on m. Suppose that for all h ∈ R such that
P
TP
dimension of hg K is < m, there exist h∗ ∈ IG ( hg K) with h − h∗ ∈
g
g
P g
′
g
(h − h )K. The assertion is trivial when m = 0. Further if f ∈ N,
g,g′
13
we may take f ∗ = 0. Suppose f ′ < N. Then M = K f + N and f is Ginvariant module N. Hence by the condition (∗) there exists a G-Module
N ∗ of dimension 1 in M and an f ′ ∈ IG with M = K f ′ + N ∗ . If f ′ < N,
then M = K f ′ + N and the lemma is proved. Suppose f ′ ∈ N. Set
P
f = λ f ′ + h with λ ∈ K, h ∈ N ∗ . Since M1 = hg K ⊂ N ∗ , we have
g
dim K M1 6 dim N ∗ = m − 1. Hence by induction hypothesis there exists
P g
TP
(h − hg1 )K. Since
an h∗ ∈ IG ( hg K) such that h − h∗ ∈ N1 =
g,g1
g
N1 ⊆ N, the proof of Lemma 1 is complete with f ∗ = h∗ .
Remark. If R is a graded ring and f ∈ R is homogeneous, the representations of G which occur in the above proof are all given by G-modules
generated by homogeneous elements of the same degree.
Proposition 1. Let R and G be as in Theorem 1 and let R′ be a ring
containing K. Let G act on R′ as a group of automorphisms of R′ over
K. Assume that there is a surjective homomorphism ϕ of R onto R′ such
that ϕ(ag ) = ϕ(a)g for all a ∈ R. Let IG′ denote the G-invariant elements
of R′ . Then ϕ(IG ) = IG′ .
Proof. We have only to show that IG′ ⊆ ϕ(IG ). Let f ′ ∈ IG′ . Let f ∈ R
be such that ϕ( f ) = f ′ . By Lemma 1 there exists an f ∗ ∈ IG such that
P
′
f − f ∗ ∈ N = ( f g − f g )K. But as f ′ is G-invariant, N is contained in
g,g′
the kernel of ϕ. Hence ϕ( f ∗ ) = f ′ . Hence ϕ(IG ) = IG′ .
We shall now prove Theorem 1 in the case when R is a graded ring
i.e. R = K[a1 , . . . , an ] ≈ K[x1 , . . . , xn ]/U with x1 , . . . , xn algebraically
independent over K and U, a homogeneous ideal of K[x1 , . . . xn ].
14
Lemma 2. Let S =
∞
P
S i be a graded ring. Assume that the ideal
1=0
have a finite basis. Them S is finitely generated over S 0 .
P
i>0
Si
A generalization of the original fourteenth problem(contd.)
11
Proof. Let hi , i = 1, . . . , r be a set of generators for I. We may assume
that hi are all homogeneous, say hi ∈ S j(i) , j(i) > 0. Then we assert
that S = S 0 [h1 , . . . , hr ]. It is enough to prove that every homogeneous
element f of S is in S 0 [h1 , . . . , hr ]. The proof is by induction on the
degree i of f . For i = 0, there is nothing to prove. Assume i > 0.
r
P
Suppose that for all t < i, S t ⊆ S 0 [h1 , . . . , hr ]. We have f =
gk hk ,
k=1
gk ∈ S i− j (i). By induction hypothesis gk ∈ S 0 [h1 , . . . , hr ], 1 ≤ k ≤ r.
Hence f ∈ S 0 [h1 , . . . , hr ].
Lemma 3. Let R and G be as in Theorem 1. Let f 01 , . . . , fr ∈ IG . Then
r
r
P
T
P
( fi R) IG = IG fi .
i=1
i=1
Proof. The proof is by induction on r. For r = 0 there is nothing to
s
s
r
P
T
P
P
T
prove. Assume ( fi R) IG =
fi IG for s < r. Let f ∈ ( fi R) IG .
i=1
Then f =
i
P
i−1
′
hgr )X
i=1
i=1
hi fi , hi ∈ R. By Lemma 1 there exists an
such that hr + h′ ∈ IG . As, for g, g′ ∈ G,
there exist h′i ∈ R, 1 ≤ i ≤ r − 1 with
f − (hr + h′ ) fr =
r−1
P
(hi + h′i ) fi .
r−1
P
i=i
h′
∈N=
P
g,g′
g
(hr −
r P
′
P
(hgi − hgi ) fi = 0,
i=i
h′i fi + h′ fr = 0. Hence
i=1
− (hr + h′ ) fr
∈ IG and therefore by induction hypothesis, there
r−1
r−1
P
P ′′
exist h′′
(hi + h′i ) fi =
h1 fi .
i ∈ IG , 1 ≤ i ≤ r − 1 such that
But f
i=1
The proof of Lemma 3 is complete.
i=1
15
Remark. If R is graded and fi are homogeneous, it is enough to assume
the condition (*) for representations of G given by G-modules generated
by homogeneous elements of R of the same degree.
Proposition 2. Let R = K[a1 , . . . , an ] be a graded ring. Assume that
every representation λ of G given by a G-module generated by homoge-
12
A generalization of the original fourteenth problem(contd.)


1 ∗ · · · ∗

0


neous elements of R of the same degree and of the form λ =  ..

 .
λ′ 


0


1 0 . . . 0
0


. Then I is finitely generated over K.
is equivalent to  .
G

′
 ..
λ 


0
Proof. As R is graded, so is IG . Let I be ideal of IG of all elements of
positive degree.
As R is noetherian, the ideal R, I of R is finitely generated. Let
f1 , . . . , fr , ∈, I generate R, I. We may assume the fi to be homogeneous.
r
P
T
By Lemma 3 and the remark following it we have ( , R fi ) IG =
r
P
i=1
IG fi = I. Hence by Lemma 2, IG is finitely generated over K.
i=i
16
We shall now consider the non-graded case. Let R = K [a1 , . . . , an ].
Let be a transcendental element over R. Consider the homogeneous ring
R∗ = K[a1 t, . . . , an t, t]. Then G acts on R∗ if we set tg = t, for every
g ∈ G. Further the G-module homomorphism ϕ : R∗ → R defined
by ϕ ( f (a1 t, . . . an t, t) = f (a1 . . . , an , 1) induces an isomorphism of a
finite G-submodule of R∗ generated by homogeneous elements of the
same degree of R∗ onto a finite G-submodule of R. Hence the condition
(*) is satisfied by all representation λ∗ of G given by G-submodules
of R∗ generated by homogeneous elements of the same degree. Hence
∗ the ring of G-invariant elements of R∗ is finitely
by proposition 2, LG
generated over K. Further as every f ∈ R is the image of a homogeneous
in R∗ , it follows from Proposition 1 and the remark after Lemma 1 that
ϕ(IG∗ ) = IG . Hence IG is finitely generated over K.
2. We now give examples where the condition (*) is satisfied. It is obvious that if every rational representation λ of G is completely reducible
then it satisfies the condition (*). We shall later give some criteria of
complete reducibility of rational representation of an algebraic group.
A generalization of the original fourteenth problem(contd.)
13
(1) Consider a torus group G (i.e. a connected algebraic linear group
which is diagonalizable), acting on R = K [a1 , . . . , an ]. In this case
if G is diagonalized each K ai11 . . . ainn is a G-module. Thus R is the
direct sum of simple G-modules. Hence every representation λ of G
given by a G-sub-module of R is completely reducible. Hence LG is
finitely generated.
(2) Let K be a field of characteristic zero and let G ⊆ GL(n, K) be semisimple. Then it is well known that (see, Chevalley, Theorie des
groupes de Lie, III) every rational representation of G is completely
reducible. Hence in this case again IG is finitely generated.
(3) Combining (1) and (2) we have : If G is an algebraic group whose
radical is a torus group, then IG is finitely generated.
We add here a Corollary to Theorem 1:
Corollary. Assume that K is of characteristic zero and that an algebraic 17
group G ⊆ GL(n, K) acts on R = K[a1 , . . . , an ]. Let N be a normal
subgroup of G which contains the unipotent part of the radical of G. If
IN = the set of N-invariants of R is finitely generated over K, then so is
IG
Proof. G acts on IN and (IN )G = IG . The closure of N being denoted by
N̄, the action of G on IN is really an action of G/N̄, whose radical is a
torus group. Therefore by (3) above, IG is finitely generated.
As a further application of Theorem 1 we prove the following:
Theorem 2 (Generalization of Fischer’s theorem). Let K denote the
complex number field and let G ⊆ GL(n, K) act on R = K[a1 , . . . , an ] as
a group of automorphisms of R over K. Assume that for every A ∈ G,
¯ (complex conjugate of A) is in G. Then IG is finitely generated over
t−A
K.
Proof. Let Ḡ be the closure of G. Then Ḡ also satisfies the hypothesis
of the above theorem. Hence we may assume G is algebraic. Let H be
the radical of G. Let Hu be the set of unipotent elements of H (we recall
that A ∈ GL(n, K) said to be unipotent if all the eigen values of A are
14
18
A generalization of the original fourteenth problem(contd.)
1). It is well known that Hu is an algebraic group. It is enough to prove
that Hu consists of identity element only. For then H is a torus group
(A connected linear algebraic group which has no unipotent elements
other identity is a torus group if either the field is algebraically closed
or the group is solvable). It follows that IG is finitely generated by (3)
above.
It remains to prove that Hu consists of identity element only. Let
A ⊆ Hu . Now H is characteristic in G (i.e. admists all automorphisms of
t−
G). Considering the automorphism g → (t ḡ)−1 , we see that ( A)−1 ∈ H
t−
t−
t−
and hence ( A)−1 ∈ Hu since ( A)−1 is unipotent. The element AA is
t−
unipotent and hermitian. Hence AA = E, the identity matrix. Hence A
is unitary. But the only unipotent unitary matrix is the identity matrix.
Thus A = E and the theorem is proved.
3. Let R = K[a1 , . . . , an ] be an integral domain and let G ⊆ GL(n, K)
act on R and satisfy the condition (*). Let V be the affine K-variety
defined by R (the points of V lie say, in the algebraic closure K of K).
Let L be the function field of R and let LG be the field of G-invariant
elements in L. The group G acts on V in a natural way. A subset F of V
is said to be G-admissible if for every P ∈ F, Pg ∈ F, for every g ∈ G.
Theorem 3. Let F be a G-admissible closed subset of V. Let T = f f ∈
LG , f regular at every point of F . Then T is a ring of quotients of IG .
19
Proof. Let S denote the multiplicatively closed set of all s ∈ IG such
that s does not vanish at any point of F. We shall show that T = (IG )s .
We have
the ideals
only to show that T = (IG )s . Let f ∈ T . Consider
U = gg ∈ R, g f ∈ R and U(F) = hh ∈ R, h(F) = 0 . As the closed
set defined by U and F are disjoint, we have, U + U(F) = R. Hence
there exists a g ∈ U, g′ ∈ U(F) with g + g′ = 1.
It is clear that the ideals U and U(F) are G-admissible. Hence
TP
P
P
′
(gσ − gσ )K ⊆ U(F) ( gσ K) ⊂ gσ K⊆U. By Lemma 1 of
σ TP
σ
σ,σ′ ∈G
P σ
Theorem 1 there exists a g∗ ∈ IG
(g −
gσ K such that g − g∗ ∈
σ
σ,σ′
A generalization of the original fourteenth problem(contd.)
15
′
gσ )K. Hence g∗ (P) = 1 for every P ∈ F and g∗ f ∈ R. The theorem is
proved.
T G
Consider the relation ∼ in V defined by P ∼ Q if PG
Q ,φ
G
for P, Q ∈ V (where P is the orbit of P, namely the set of all Pσ with
σ ∈ G and ∞
¯ denotes the closure in V).
Theorem 4. (1) ∼ is an equivalence relation. (2) The quotient set V/ ∼
i.e. the set of all equivalence classes by ∼ acquires the structure of an
affine K-variety with coordinate ring IG such that the natural mapping
V → V/ ∼ is regular.
Proof of (1): We shall in fact give the following characterization which
proves that ∼ is an equivalence relation: P ∼ Q if and only if f (P) =
f (Q), for every f ∈ IG . Let P ∼ Q and let P′ ∈ PG ∩ QG . We have
f (P) = f (PG ) = f (P′ ) = f (QG ) = f (Q), for every f ∈ IG . To prove
the converse we remark that if F1 and F2 are two G-admissible disjoint
closed sets, then as in the proof of Theorem 3 we can find an f ∈ IG
such that f (F1 ) = 1, f (F2 ) = 0. The closed set PG , for P ∈ V is
G-admissible. If P, Q are not related by ∼ we can separate them by an
f ∈ IG . Hence (1) is proved.
Proof of (2): By Theorem 1, IG is finitely generated over K. Let
f1 (a), . . . , fl (a) generate IG . Let W be the K-affine variety defined by
IG = K[ f1 , . . . , fl ]. Consider the regular mapping ϕ : V → W defined 20
l
by ϕ(P) = P = ( f1 (P), . . . fl (P)) ∈ W ⊆ K . By
(1), ϕ(P) = ϕ(Q) if
1
and only if P ∼ Q. Hence ϕ (P) = QP ∼ Q = P, say. Let M be
the maximal ideal corresponding to a point P of W. Then by Lemma 3
T
of Theorem 1 MR IG = M. Hence ϕ surjective and we identify the
equivalence class P with the point P of W. By Theorem 3 the local ring
T
T
at P of W is (IG )P = ( ℧ x ) LG , where ℧y denotes the local ring at
x∈P
the point y of V.
Corollary 1. Let Q ∈
PG
. Then (IG )P =
!
T
℧ Q∗ T
LG .
∗
G
Q ∈ Q
16
A generalization of the original fourteenth problem(contd.)
Proof. We have only to prove
\
⊆ (IG )P .
G
∗
Q ∈ Q
Let
f ∈
℧ Q∗
\
Q∗ ∈
G
℧ Q∗ .
Q
Then by Theorem 3, f = h/g, h, g ∈ IG , g(Q∗ ) , 0, for every Q∗ ∈ QG .
But as Q ∈ PG , g(Q) = g(P′ ) for every P′ ∈ P. Hence f ∈ (IG )P .
Corollary 2. Under
the hypothesis of Corollary 1, if QG is closed, then
T
T
℧ p∗ T
LG .
℧Q LG = P∗ ∈PG T
T
℧Q∗ ) LG = ℧Q ∩ LG . As QG is closed,
Proof. For any Q ∈ V, (
∗
G
T Q ∈Q
T
T
by Corollary 1, ℧Q LG = (IG )P = (
℧P∗ ) LG .
P∗ ∈{PG }
As we noted before, our Theorem 1 implies the following
21
Theorem 1′ Let R = K[al , . . . , an ] and let G ⊆ GL(n, K) act on R as a
group of automorphisms of R. over K. If every rational representation
of G is completely reducible, then IG is finitely generated.
What we like to remark here is that when we want to prove this
Theorem 1∗ , we need not use the technique at the end of §1, and we
can prove as follows: R is a homomorphic image of a polynomial ring
K[x1 , . . . , xn ] on which G acts naturally. The polynomial ring is graded,
hence the result on the graded case and Proposition 1 prove Theorem 1∗ .
Chapter 3
The counter example
1. In this chapter we give some counter examples to the original 14th 22
problem. For this we need some results on plane curves.
Lemma 1. Let C be a curve of positive genus on the projective plane
and let P1 , . . . , Pm , Pm+1 , . . . , Pn ∈ C(m ≥ 1) and let P1 , . . . , Pm be
independent generic points of C over k(Pm+1 , . . . , Pn ), where k is a field
of definition for C. Then for any set of natural numbers αi (1 ≤ i ≤ n)
n
P
there does not exist any curve C ′ such that C ′ .C = αi Pi .
i=1
Proof. Suppose there exists a C ′ with C ′ .C =
of the same degree as C ′ with C ′′ .C =
Pl
n
P
i=1
αi .Pi , let C ′′ be a curve
β j Q j withQ j algebraic over
j=1
C ′′ .C
k. Then C ′ .C is linearly equivalent to
(notation C ′ .C ∼ C ′′ .C)
m
l
n
P
P
P
on C. Hence
αi P i ∼
β jQ j −
αt Pt . Let P be any point on
i=1
j=1
t=m+1
C. Specializing the above linear equivalence under the specialization
(P1 , . . . , Pm ) → (P, . . . , P) over k(Pm+1 , . . . , Pn ) (where k is the algem
n
P
Pl
P
braic closure of k) we get ( αi )P∼
β jQ j −
αt Pt . Thus for any
i=1 P j=1
P t=m+1
two points P, Q on C, we have ( αi )P ∼ ( βi )Q. This contradicts the
assumption that C has positive genus.
17
The counter example
18
Here after for a variety V, we denote by K(V) the smallest field of
definition for V containing k.
23
Lemma 2. Let P1 , . . . , Pt be independent generic points of the projec2
tive plane over a field k and d an integer such that a t < d +3d
2 . Let L
be the linear system of curves of degree d passing through P1 , . . . , Pt .
Let C be a generic member of L over k(P1 , . . . , Pt ). Then P1 , . . . , Pt are
independent generic points of C over k(C).
2
Proof. Set u = d +3d
2 . Let Pt+1 , . . . , Pu be independent generic points
of C over k(C) (P1 , . . . , Pt ). Then P1 , . . . , Pu are independent generic
points of the projective plane over k as u is the dimension of the linear system L′ of curves of degree d. Let R1 , . . . , Rt be independent
generic points of C over k (C, Pt+1 , . . . , Pu ). Then we have a specialization (P1 , . . . , Pu ) → (R1 , . . . , Rt , Pt+1 , . . . , Pu ) over k because P1 , . . . , Pu
are independent generic points of the projective plane. Let C specialize
to C ′ under this specialization. If C , C ′ , then R1 + · · · + Rt + Pt+1 +
· · · + Pu ⊂ C.C ′ . Since R1 , . . . , Rt , Pt+1 , . . . , Pu are independent generic
points of C over k(C) and the dimension of the trace LC′ on C of the
linear system L′ is n − 1, we get a contradiction. Hence C = C ′ . Thus
(P1 , . . . , Pt ) → (R1 , . . . , Rt ) is a specialisation over k(C) and therefore
P1 , . . . , Pt are independent generic points of C over k(C).
Proposition 1. Let P1 , . . . , Pr be independent generic points of the projective plane over the prime field.
(1) Let C be a curve is degree d passing through the Pi with multiplicity mi . Then Pdmi √1r , for r = s2 , s ≥ 4.
24
√1 , then there
r
′
Pd ′ 1 , where C ′
mi r
(2) Furthermore if r′ is a real number such that r′ >
exists a curve C ′ of degree d′ such that r′ >
passes through Pi with with multiplicity m′i .
Proof. Let C specialize to σ(C) under the specialization (P1 , . . . , Pr ) →
(Pσ(1) , . . . , Pσ(r) ) where σ runs through cyclic permutations of 1, . . . , r.
P
Considering the curve σ(C), we have only to prove (1) and (2) in the
σ
case when all the mi are equal, say mi = m. Thus we have to prove
The counter example
19
√
√
that md > r = s. If there exists a curve C with d/m ≤ r, then by
suitably raising the degree we can get actually the equality. Thus (1) is
equivalent to:
(1′ ) There does not exist a curve of degree sm passing through the Pi
with multiplicity at least m.
Similarly (2) is equivalent to:
√
(2′ ) Given r′′ > r, there exist integers d, m and a curve C ′ of degree
d passing through the Pi with multiplicity m such that r′′ > md >
√
r.
s+2
. Let CS , C2 , and CS′ ′ be
Proof of (1′ ). Case (i) s even. Set s′ =
2
independent generic curves of degree s, s′ , s′ , respectively.
Let CS′ ′ .CS ′ =
s′2
P
i=1
Pi , CS .CS ′ =
′
ss
P′
i=1
Qi and CS .CS′ =
ss
P′
Qi . As
i=1
′2
′
s′ is s +3s
>
2
the
dimension of the linear system LS of curves of degree
2s,
among the Pi there exist 2s points say P1 , . . . , P2s which are independent 25
generic points of the projective plane over k(C s ). Then by Lemma 2
P1 , P2 , . . . , P2s are independent generic points of C s′ over k(CS .CS′ ′ ),
in fact over k(CS , CS ′ , Q) as all the curves considered are independent
generic curves.
2
> 2s, there exist curves of degree s passing through
Since s +3s
2
the points P1 , . . . , P2s . Let CS∗ be the generic curve passing through
P1 , . . . , P2s .
The counter example
20
Let
CS ′ .CS∗
=
′ −2s
ssX
Q∗i
′ −2s
ssX
Q∗i +
+
i=1
CS′ .CS∗
=
Pi ,
i=1
i=1
26
′ −2s
ssX
′
2s
X
Pi .
i=1
By Lemma 2, P1 , . . . , P2s are independent generic points of CS∗ . Fur′
ther, we claim that the Q∗i and Q∗i are all distinct. For , consider the
P
P
linear system L = CS∗ .CS ′ − Pi on CS ′ . Q∗i is a generic member
′2
′
+2
of L. Since degree of L = ss′ − 2s = 4. s −3s
= 4g(CS ′ ), where
2
g(CS ′ ) denotes the genus of CS ′ , L has no base points. Hence each Q∗i is
a generic points of C s′ over k(C s′ , P1 , . . . , P2s ). In particular Q∗i do not
lie on CS′ ′ . Again the degree of the linear system L − Q∗i = ss′ − 2s − 1 =
4g(CS ′ ) − 1 ≥ 2g(CS ′ ). Hence L − Q∗i has no base point and therefore
Q∗j ( j , i) are generic points of CS ′ over k(C s′ , Q∗i ). Thus Q∗i are all dis′
tinct. Similarly considering C ′s′ , we see Q∗i , Q∗i are all distinct. We may
′
renumber Qi and Q′i such that Qi , Q′i , 1 ≤ i ≤ s −2s
2 , are specialized to
′
∗
∗
Qi , Qi under the specialization (P1 , . . . , P2s , CS ) → (P1 , . . . , P2s , CS∗ )
over k(CS , CS ).
Assume that for some m there exists a curve of degree sm passing
through r = s2 independent generic points of the projective plane with
multiplicity at least m. Then there exists a curve E of degree sm passing
2
′
through the Pi (1 ≤ i ≤ 2s), the Qi (1 ≤ i ≤ s −2s
2 ) and the Qi (1 ≤ i ≤
2
s −2s
′
2 ) with multiplicity at least m. Then C s is component of E. For
otherwise, E .C s′ is defined and contains
degree E.CS ′ is
sms′ ,
we have E.C s′ =
2s
P
mPi +
s2 −2s
2
i=1
2s
P
i=1
mPi +
P
mQi . As the
i=1
s2 −2s
2
P
mQi . Since the
i=1
genus of C s′ (= 21 s′ (s′ − 3) + 1) is positive we get a contradiction by
Lemma 1. Hence C s′ is a component of E. Let E − C s′ specialize to
E ∗ under the specialization (P1 , . . . , P2s , C s ) → (P1 , . . . , P2s , C ∗s ).E ∗ is
a curve of degree sm − s′ passing through the Pi (1 ≤ i ≤ 2s) and the
The counter example
21
2
Q∗i (1 ≤ i ≤ s −2s
2 ) with the the multiplicity at least (m-1) and through
Q′∗
with
the
multiplicity
at least m. We shall show the non-existence of
i
such a curve by induction on m. We assert that C ∗s is component of E ∗ .
For, if not, E ∗ .C ∗s is defined and contains
s2 −2s
s2 −2s
i=1
i=1
2s
2
2
X
X
X
′
(m − 1)Q∗i +
mQ∗i .
(m − 1)Pi +
i=1
Inspecting the degrees we see that
s2 −2s
s2 −2s
i=1
i=1
2s
2
2
X
X
X
′
(m − 1)Q′i +
mQ∗i .
E ∗ .C s =
(m − 1)Pi +
i=1
Hence ((m − 1)C s′ + mC ′s′ − E ∗ ).C ∗s =
2s
P
i=1
mPi . As the genus of CS∗ (=
1
2 s(s − 3) + 1) is positive and since the Pi are independent generic points
of C ∗s by lemma 2, we have a contradiction by Lemma 1.
Hence C ∗s is a component of E ∗ . If m = 1, degree of E ∗ is s − s′ .
Hence we get a contradiction. If m > 1, then E ∗ − CS∗ . is of degree
s(m − 1) − s′ and passes through the Pi , Q∗i , with multiplicity at least
m − 2, m − 2 and (m − 1) respectively. Hence by induction hypothesis
E ∗ − CS∗ . does not exist and the lemma is proved, when r is even
′
′
Case (ii) s odd. Set s′ = s+1
2 . Let C S , C S C S ′ be independent generic
curves. Let P1 , . . . , P s be s points contained in CS ′ .CS′ ′ . We take a
generic curve CS∗ of degree s passing through P1 , . . . , P s and proceed as
in (i). We omit the details.
Remark. We remark that Proposition 1 is not true for r ≤ 9. The following are the example to that effect. (1) For r = 1, 2, a line passing through
the Pi (2) r = 3, a line passing through two of the Pi (3) r = 4, 5, a conic
through the Pi (4) r = 6, a conic through 5 of the Pi (5) r = 7, a cubic
having a double point at one of the Pi and passing through all the Pi .
(6) r = 8, a curve of degree 6 having a triple point at one of the Pi and
double points at all the Pi (7) r = 9, a cubic passing through all the Pi .
27
The counter example
22
Furthermore it is not known if r ≥ 9, is sufficient to ensure the
inequality of Proposition 1.
28
Proof of (2′ ). Let Ld be the linear system of curves of degree d. Let
f (x0 , x1 , x2 ) be a homogeneous polynomial of degree d. The condition
that the curve defined by f should pass through a point p with multipliclinear conditions (not necessarily independent).
ity m imposes m(m+1)
2
Thus
r
X
d(d + 3)
m(m + 1)
dim(Ld −
mPi ) ≥
−r
≥0
2
2
i=1
if
m(m + 1)
d(d + 3)
≥r
.
2
2
r − 3( md )
i.e. if m ≥ d
. Now choose a rational number λ such that r′′ >
( m )2 − r
√
λ > r. Writing λ = md with sufficiently large m we get a curve C ′
of degree d passing through the Pi with multiplicity m such that r′′ >
√
d/m > r and (2′ ) is proved.
2. We now proceed to give the counter example where the transcendence
degree of the sub-field LG of invariants is four.
Let ai j , i = 1, 2, 3, j = 1, . . . , r be algebraically independent elements over the prime field Π . Let k be a field containing the ai j . Consider the projective plans S over k. Set Pi = (a1i , a2i , a3i ), Pi ∈ S , i =
1, 2, . . . , r. Then the Pi are independent generic points of S over the
prime field . Let x1 , . . . , xr , y1 , . . . , yr be algebraically independent elements over k. Consider the subgroup G of GL(2r, k) given by



0 

 B1





σ = 
..
 ,
σ
∈
GL(2r,
k)
G=

.






0
Br

! r
r

Y

ci ci bi X

,
Bi =
bi a ji = 0, j = 1, 2, 3,
ci = 1
.


0 ci
i=1
i=1
The group G acts on k[x1 , y1 , . . . , xr , yr ] with σ(xi ) = ci (xi + bi yi ),
σ(yi ) = ci yi , i = 1, . . . , r.
The counter example
23
Theorem 1. The ring of invariants IG is not finitely generated the k. Set 29
r
P
t = y1 , . . . , yr , ui = t/yi , vi = xi ui and w j =
a ji vi . We need some
i=1
lemmas.
Lemma 1. IG = k[x1 , y1 , . . . , xr , yr ] ∩ k(w1 , w2 , w3 , t).
Proof. It is sufficient to prove that the invariant sub-field LG = k
(w1 , w2 , w3 , t). A straight forward verification shows that w1 , w2 , w3 , t ∈
Q
LG . Hence k(w1 , w2 , w3 , t) ⊆ LG as ai j are independent over , we
have k(x1 , . . . , xr , y1 , . . . , yr ) = k(w1 , w2 , w3 , x4 , . . . , xr , y1 , . . . , yr ). Now
G operates on k[w1 , w2 , w3 , x4 , . . . , xr , y1 , . . . , yr ]. As wi , t ∈ LG , to compute LG it is enough to consider the action of G on k(w1 , w2 , w3 , t)
[x4 , . . . , xr , y1 , . . . , yr−1 ]. Consider the subgroup H of G consisting of
elements σ of G for which ci = 1, i = 1, . . . , r, bi = 0, i ≥ 5 and b4
arbitrary. Since H is infinite, k(w1 , w2 , w3 , t, x4 , . . . , xr , y1 , . . . , yr−1 ) is
a transcendental extensions of LH the fixed field of H. Now, LH ⊇
k(w1 , w2 , w3 , t, x5 , . . . , xr , y1 , . . . , yr−1 ). Therefore LH = k(w1 , w2 , w3 ,
t, x5 , . . . , xr , y1 , . . . , yr−1 ). Next we consider the action of G on
k(w1 , w2 , w3 , t)[x5 , . . . , xr , y1 , . . . , yr−1 ] and consider the subgroup H1 of
G consisting of elements σ of G with bi = 0, i ≥ 6, ci = 1, i = 1, . . . , r.
The fixed field LH1 of H1 is k(w1 , w2 , w3 , t, x6 , . . . , xr , y1 , . . . , yr−1 ). Proceeding in the same way we arrive at k(w1 , w2 , w3 , t, y1 , . . . , yr−1 ). Consider k(w1 , w2 , w3 , t)[y1 , . . . , yr−1 ]. σ ∈ G acts on k(w1 , w2 , w3 , t)
[y1 , . . . , yr−1 ] with σ(yi ) = ci yi , i = 1, . . . , r − 1, where ci are arbitrary
non-zero elements of k. Hence LG = k(w1 , w2 , w3 , t).
As w1 , w2 , w3 are algebraically independent over k, we may regard
H = k[w1 , w2 , w3 ] as the homogeneous coordinate ring of the projective 30
plans S . Let Fi = (zi , z′i ) denote the prime ideal is H corresponding to
the point Pi , where zi = a3i w1 − a1i w3 , z′i = a3i w2 − a2i w3 . Set Wn =
r
∩ Yi (n) , for n > 0 and Wn = H, for n ≤ 0.
i=1
Lemma 2. IG =
P
n
an t−n | an ∈ ⊓n .
Assuming Lemma 2 we shall first prove Theorem 1 and later prove
Lemma 2. Suppose IG is finitely generated, say IG = k[ f1 , . . . , fm ]. We
The counter example
24
may assume f j = h j t− j , h j ∈ W j , h j homogeneous, j = 1, . . . , m. Set
degree h j
rj =
, j = 1, . . . , m and r∗ = min r j . For any monomial
i≤ j≤m
j
degree (hi11 . . . himm )
≥ r∗ .
f1i1 . . . fmim = hi11 . . . himm t−i1 −2i2 −···−mim , we have
i1 + 2i2 + · · · + mim
degree an
≥ r∗ . By
Hence for any homogeneous an ∈ Wh , we have
n
√
proposition 1, r∗ > s = r, for r ≥ 4. Again by the same proposition
there exists an an ∈ Wn , for some n, such that degree an/n < r∗ . This is
a contradiction and therefore IG is not finitely generated.
It now remains to prove Lemma 2. We first prove
X
IG ⊆
an t−n | an ∈ H .
(∗)
n
As ai j are algebraically independent over k we have
k[v1 , . . . , vr ] = k[w1 , w2 , w3 , v4 , . . . , vr ].
Hence
"
#
1
1
k x1 , . . . , xr , y1 ..yr , , . . . ,
y1
yr
#
"
1
1
.
= k w1 , w2 , w3 , x4 , . . . , xr , y1 , . . . yr , , . . . ,
y1
yr
Now
"
#
1
1
k w1 , w2 , w3 , x4 , . . . , xr , y1 , . . . yr , , . . . ,
y1
yr
#
"
\
1
1
.
k(w1 , w2 , w3 , y1 , . . . , yr ) = k w1 , w2 , w3 , y1 , . . . , yr , , . . . ,
y1
yr
Now by lemma 1,
#
"
1 \
1
k(w1 , w2 , w3 , t)
IG ⊆ k w1 , w2 , w3 , y1 , . . . , yr , , . . . ,
y1
yr
The counter example
25
#
1 \
1
1
k(w1 , w2 , w3 , t)
= k w1 , w2 , w3 , t, , y2 , . . . , yr , , . . . ,
t
y2
yr
"
#
1
= k w1 , w2 , w3 , t, .
t
"
Hence
IG ⊆
(
P
n
an t−n an
31
)
∈ H . Let ℧i be the valuation ring
k[y1 , . . . , xr , y1 , . . . , yr ](yi ) , 1 ≤ i ≤ r and let Vi be the corresponding
normalized valuation. We prove
Yi m = Yi (m) = f ∈ H | Vi ( f ) ≥ m ,
∗∗
for every m. Since zi =
P
j,i
(a3i a1 j − a1i a3 j )x j u j , Vi (zi ) = 1. Simi-
larly Vi (z′i ) = 1. It is easy to check that w3 , zi /t, and z′i /t modulo the
maximal ideal Mi of ℧i are algebraically independent over k. Now
k[w1 , w2 , w3 ] = k[zi , z′i , w3 ]. If h < Yi , then h ≡ h1 (mod Mi ), with
0 , h1 ∈ k[w3 ]. Hence Vi (h) = 0. Therefore Yi (m) ⊆ h ∈ H | Vi (h) ≥
m . To prove (∗∗) we need only to prove that, for f ∈ H, Vi ( f ) ≥ m
implies f ∈ Yi m . This in turn will be proved if we prove that f ∈ Yi m ,
f < Yi m+1 implies Vi ( f ) = m. If f ∈ Yi m , f < Yi m−1 then f can be
m
P
j m− j
written as f = h j zi zi′ , with h j ∈ k[w3 ], hl , 0 mod M , for some l.
Hence f /z′m
i =
j=
m
P
j=0
h j ( zzii ) j . As zi /z′i and w3 modulo mi are algebraically
m
m
independent over k, f /z′i . 0 (mod Mi ). Hence Vi ( f ) = V(z′i ) = m
and (∗∗) is proved. Next, we prove
X
Vi ( an t−n ) = minVi (ar t−n ),
n
n
for an ∈ H. Let d = min Vi (an t−r ). For every n, Vi (an t−n ) ≥ d i.e. 32
n
Vi (an ) ≥ n + d. Hence by (∗, ∗), an ∈ Yi n+d . Let an =
Then
P
an
n+d
n t
=
PP
n
j
z′
n+d
P
j=0
j ′n+d− j
hn j zi zi
.
hn j ( zti ) j ( ti )n+d− j . If Vi (am t−m ) = d for some m, then
The counter example
26
0 , hml mod Mi ∈ k[w3 ] for some l. Since zi /t and z′i /t modulo Mi are
m
P
an
. 0 (mod Mi ).
algebraically independent over k(w3 mod Mi ), tn+d
n
P −n
Hence Vi ( an t ) = d.
n
P
Now suppose f ∈ IG . Then by (∗) f can be written as f = an t−n ,
n
an ∈ H. As IG ⊆ ℧i , 1 ≤ i ≤ r, Vi ( f ) ≥ 0. Hence
by (∗∗∗), Vi (an t−n) ≥ 0,
P
that is Vi (an ) ≥ n and an ∈ Yi n . Hence IG ⊆
an t−n | an ∈ ℧n . On
the other hand if an ∈ Wn, then Vi (an ) ≥ n, i = 1, . . . , r. Hence an is
P −n
divisible by tn . Thus IG ⊇
an t | an ∈ Wn and the proof of Lemma
2 is complete.
33
3. In the counter example we have given to the original 14th problem
in section 2, the group G is commutative. In this section we give an
example, where the group G is such that [G, G] = G.
If we make use of the structure of the ring IH of invariants of the
subgroup H of G given in the last section such that all the ci are 1 then
we can give a direct construction of the required example. But since we
did not give the explicit structure of IH , we do it in an indirect way.
Let ai j , i = 1, 2, 3, j = 1, . . . , r(r = 16) be algebraically independent
real numbers
( over the field of relational numbers. Set
P
ai j b j = 0(i = 1, 2, 3, s = an even
V = (b1 , . . . , br ) | bi real,
j
number ≥ 4,






 1 . . . . . . bil 








 .


c
B



1
1










.






.
0
(b
,
.
.
.
,
b
)
∈
V,





li
ri
.



.



Q
Go = 
B
=
,






.
i





.






=
1
c
real,
.


ci



 i


.
b


is

c
B



r
r






0
1
Let K be the real number field. Go acts on the polynomial ring
R0 = K[x11 , x12 , . . . , x1s , y1 , x21 , x22 , . . . , x2s , y2 , . . . , xr1 , x s2 , . . . , xrs , yr ].
Let I0 be the ring of Go - invariants in Ro . Then setting t =
P
(i) wi j = aiα xα j t/yα ∈ Io
α
Q
yj
The counter example
(ii) Io = Ro
27
T
K(w11 , . . . , w1s , w21 , . . . , w2s , w31 , . . . , w3s , t).
T P
Since Io /Io ( xi j R) is isomorphic to previous example IG , we see
j≥2
that Io is not finitely generated. We see also easily that every element of
P
Io can be expressed in the form ai t−i (finite sum) such that ai ∈ Y =
K[w11 , . . . , w3s ], ai t−i ∈ Io .
For each T ∈ S L(s, K), let
 ′
T

T r = 

0
i


0 
T


′
..


,
T
=

.


0
T′

0


1
(T ′ ∈ S L(s + 1, K)), T r ∈ S L(r(s + 1), K)
and G1 = T r | T ∈ S O(s, K) .
P
Then one can easily see that Io is G-admissible and the module Mi =
wi j K is also G1 -admissible and T is the transformation on Ni given by
j
T r . Therefore for a general linear transformation T ∈ S L(s, K) we have
Io/Io ∩( P xi j R) T is isomorphic to the previous example IG .
j≥2
Therefore we see that the degree ai > 4i (for i > 0) and that for any
rational number α greater than 14 . there is a homogeneous form al in wi j 34
deg a
such that rl l = α and al t−l ∈ Io .
Now consider a copy of Ro say R′o . The isomorphism of Ro onto R′o ,
we denote by “/”. Let G2 be the subgroup in S L(2r(s + 1), K) generated
by
)
!
(
B1 0
∗
| Bi ∈ Go (i.e. G∗o = Go × Go )
Go =
0 B2
(
)
!
T 0
∗
G1 =
| T ∈ G1 .
0 T
Then the ring of G∗o - invariants is Io ⊗Io′ . Let Mid be the module of homogeneous forms ai in Y of degree d such that ai t−1 is in Io . Then each Mid
which is not zero defines a representation of G1 . Hence we can choose
The counter example
28
a linearly independent basis of Mid say fidl , . . . , fidn (i, d) such that the
P
fid j fid′ j is G∗1 -invariant and
representation is orthogonal. Then fid∗ =
j
fid∗ t−i
is G2 -invariant. Therefore the same reasoning as in the previous
example IG can be applied to the ring I2 of G2 -invariant and we see that
I2 is not finitely generated.
Let C be the complex number field and let G¯2 be the closure of G2 in
S L(2r(s + 1), C). We want to show that G¯2 contains a connected normal
algebraic subgroup Ḡ such that
(i) G¯2 /Ḡ is a torus group and
(ii) [Ḡ, Ḡ] = Ḡ
If the existence of Ḡ is shown then Ḡ is the given example.
Let Goo be the subgroup of G0 defined by c1 = . . . = cr = 1. Then
we can consider the same construction as done for G∗0 , G2 and Ḡ20 .
We claim that Ḡ20 is the required Ḡ.
35
Proof. Since Ḡ20 is normal in G2 we see that Ḡ20 is normal in Ḡ2 . Obviously Ḡ2 /Ḡ20 is a torus group. It is also obvious that Ḡ20 in connected.
Therefore we have only to show that [Ḡ20 , Ḡ20 ] ⊇ Ḡ20 . Since [Ḡ20 , Ḡ20 ]
is a closed subgroup we have only to show that both G∗1 and G∗oo are in
[Ḡ20 , Ḡ20 ]. Since G∗1 is semi-simple and since G∗1 ⊆ Ḡ20 we see that
!
B O
∗
∗
G1 ⊆ [Ḡ20 , Ḡ20 ]. Consider the set {B1 } =
(E =identity in
O E
∗ ∗
S L(x(s+1), k)B ∈ G∞ ). For a fixed T ∗ ∈ G∗1 , B∗1
T ∗−1 B∗−1
1 T B1 gives
a homomorphism B∗1 and B∗1 (because B∗1 is isomorphic to a vector space
∗ ∗
over K). Since s is even, there is a T ∗ ∈ G∗1 such that T ∗−1 B∗−1
1 T B1 =
!
!
E O
E O
implies B∗1 =
. Therefore we see that image is dense
O E
O E
in B∗1 . Therefore [Ḡ20 , Ḡ20 ] contains the set {B∗1 }. Similarly [Ḡ20 , Ḡ20 ]
!
E O
contains all the
, (B ∈ Goo ). Therefore [Ḡ20 , Ḡ20 ] contain G∗oo
O B
and the proof is complete.
Chapter 4
Theorem of Weitzenböck
Theorem 1 (Weitzenböck). Let K be the complex number field and G a 36
complex one parameter Lie subgroup of GL(n, K) acting on the ring of
polynomials K[x1 , ..., xn ]. Then the ring of invariants IG of K[x1 , ..., xn ]
is finitely generated over K. (see also C.S. Seshadri; On a theorem of
Weitzenböck in invariant Theory, J. Math. Kyoto Univ. 1-3 (1962), 403409).
Let G = {etA |t ∈ K} be a one parameter Lie group, A being a constant matrix. Let A = N + S , whereN and S are nilpotent and semisimple matrices respectively with NS = S N. Let G1 = {etN |t ∈ K} and
G2 = {etS |t ∈ K}. If N , O, then the mapping t → etN of the additive group K onto G1 is an algebraic isomorphism, since N is nilpotent.
−
Since every element of G2 is semi-simple, the closure G2 of G is a torus
−
group. Further since the elements of G2 and G2 commute, the mapping
−
(g1 , g2 ) → g1 g2 of G1 × G2 onto is an algebraic homomorphism. Hence
−
G1Ḡ2 is a closed subgroup of GL(n, k). Hence the closure G of G is con−
−
−
tained in G1G2 . Since G1 ⊆ G and G2 ⊆ G, it follows that G = G1G2 .
−
Thus G is a torus group or a direct product of a torus group and the
additive group K. Hence Theorem 1 is equivalent to the following, by
virtue of Chapter 2, Section 2, Corollary to Theorem 1.
Theorem 2. Let G be a unipotent algebraic group of dimension one
29
Theorem of Weitzenböck
30
37
of GL(n, k) acting on K[x1 , . . . , xn ], where K is a field of characteristic
zero. Then IG is finitely generated over K.
Our proof is mostly due to C. S. Seshadri.
We first prove two lemmas. Unless otherwise stated K denotes a
field of characteristic zero.
Lemma 1. Let V be an affine variety and G a connected algebraic group
acting on V. Let W be a subvariety
of V andH asubgroup of G. Let W
be stable under H and suppose xh | h ∈ H = xg | g ∈ G ∩ W, for
every x ∈ W. Let f be a H-invariant regular function on W. Then there
exists a G-invariant rational function f ∗ on W G such that f ∗ is integral
−
over the local ring ℧ x in WG for every x ∈ W G ; f ∗ takes unique value at
x, for x ∈ W G , and such that f ∗ induces f on W. (Lemma of Seshadri).
−
Proof. Assume first W G is normal. The function f defines a regular
function F on W × G defined by F(x, g) = f (x), for x ∈ W, g ∈ G.
−
Let Q be the regular mapping of W ×G into W G defined by ϕ(x, y) =
′−1
′
g
x . If xg = x′g , for x, x′ ∈ W, g, g′ ∈ G then xgg = x′ = xh for some
h ∈ H, by hypothesis. Thus f (x′ ) = f (xh ) = f (x). Thus F is constant on
−
each fibre ϕ−1 (xg ), for x ∈ W, g ∈ G. Let A be the coordinate ring of W G .
Let U be the affine variety defined by the affine ring A[F]. The generic
fibre of the projection of U onto W G is reduced to a single point. As
the ground field K is of characteristic zero, U and W G are birationally
−
equivalent. Thus F induces a rational function f ∗ on W G . By definition
of F, f ∗ is G-invariant. Further f ∗ assumes the unique finite value f (x)
−
at xg , for x ∈ W, g ∈ G. As W G is normal f ∗ is regular value at xg and
−
the lemma is proved in the case W G is normal.
38
−
Suppose W G is not normal. Without loss of generality we may as−
∼
∼
∼
sume W G = V. Let V be the derived normal model of V. Let W1 , . . . , W s
∼
correspond to W, in Ṽ. How G operates on V. As W G is dense in V, so
Theorem of Weitzenböck
∼
31
∼
are WiG in V. The functions f induces H-invariant regular functions fi
∼
on Wi 1 ≤ i ≤ s. By what we have just proved, we get G-invariant ra∼
∼
tional function fi∗ , 1 ≤ i ≤ s of V which is regular on WiG , and induces
∼
∼
∼
fi . Now fi∗ , and f j∗ take the same value on WiG ∩ W Gj . As WiG contains
∼
a non-empty open-set of V, f1∗ = . . . = fs∗ = f ∗ . Thus f ∗ is regular on
∼
WiG , 1 ≤ i ≤ s. Hence f ∗ is integral over the local ring of xg in W G , for
x ∈ W, g ∈ G. Lemma 1 is completely proved.
Remark 1. If the ground field is of characteristic p , 0, then Lemma 1
is true under the following modification of the rationality condition for
∼
f ∗ : f ∗ is in a purely inseparable extension of the function field of W G .
Remark 2. It is interesting to note that W G is open. For, each fibre
ϕ−1 (xg ) is irreducible and of dimension equal to dimension of H, because of the condition on the orbits we have imposed. Thus ϕ has no
fundamental points and W G is open.
C
Corollary. If furthermore G is semi-simple and codimension of (W G )
−
in W G is at least 2, then the ring IH of H-invariants in the coordinate
ring of W is finitely generated.
−
C
Proof. As codim (W G ) ≥ 2, any function f on W G which is integral
−
over the local rings of points of W G in W G is integral over the coordinate
−
ring A of W G . Let B be the ring got by adjoining all rational functions 39
−
f ∗ on W G , as in Lemma 1. Since B is contained in the derived normal
ring of A, B is an affine ring and G acts on B. As G is semi-simple, the
ring of invariants IG of B is finitely generated. By Lemma 1, the ring IH
of H-invariants in the coordinate ring of W is the homomorphic image
of IG and hence is finitely generated.
Lemma 2. Let K be of characteristic zero and let ρ be a rational repre-
Theorem of Weitzenböck
32
!
1 λ
sentation of the additive group H =
| λ ∈ K in GL(n, k). Then
0 1
there exists a rational representation ρ∗ of S L(2, k) in GL(n, λ) such that
ρ∗ = ρ on H.
!
1 λ
Proof. We shall denote the element
of H by λ. Choose λ such
0 1
that ρ(λ) , 1(λ , 0 is enough). Let the Jordon normal form of ρ(λ) be


110 . 0 



011
0 
..0
A1




.
..
 , where Ai =  .. ..
A = 
 .





1

0
Ar


0.. . 1
Let Ai have ni rows. Let ρ∗i be the representation of S L(2, K) given by
homogeneous forms of degree ni − 1 in two variables. It is easy to check
that the Jordan normal form of ρ∗i (λ) is Ai . Hence we may assume that
!
ρ∗i 0
∗
∗
ρi (λ) = Ai . We take ρ =
.
0 ρ∗r
Remark 2. Lemma 2 is not true in the case where K is of characteristic
p , 0, because the group H has non-faithful rational representations
which are not trivial.
40
Proof of theorem
2. Let G be given by a rational representation ρ of
!
1 λ
H=
| λ ∈ K in GL(n,K). By Lemma 2 we extend this repre0 1
sentation to a rational representation ρ∗ of SL(2, K). Let
( ∗
!
)
ρ (g) 0
′
G =
| g ∈ S L(2, k)
0
g





0 !
ρ(λ)







′

1
λ
and
H =
|
λ
∈
K











0


0 1
The group G′ acts on K = [x1 , . . . , xn , xn+1 , xn+2 ]. Let V = K n × K 2
and W the subvariety defined by the equations xn+1 − 1 = 0 = xn+2 . The
Theorem of Weitzenböck
33
group G′ acts on V and W is stable under H ′ . If for a point
′
where
Q = (a1 , . . . , an , 1, 0) ∈ W, Qg ∈ W,
!
ρ∗ (g) 0
g′ =
∈ G′ , g ∈ S L(2, K).
0
g
Then (1, 0)g = (1, 0). Hence g ∈ H. That is g′ ∈ H ′ . Hence the
orbit condition of Lemma 1 is satisfied. As W G′ contains the complement of the hyperplane xn+1 = 0, W G′ is dense in V. Further (W G′ )C is
contained in the variety defined by the equations xn+1 = 0 = xn+2 . The
conditions of the corollary to Lemma 1 are satisfied. Hence the ring of
H ′ -invariants IH′ in the coordinate ring of W is finitely generated. That
is IG is finitely generated.
Chapter 5
Zarisk’s Theorem and Rees’
counter example
1. G-transform. Let R be an integral domain with quotient field L. Let 41
G be an ideal of R. The set
n
S (G; R) = f | f ∈ L, f G ⊆ R for some n
is defined to be the G- transform of R.
Remark 1. S (G; R) is an integral domain containing R.
Remark
√ is clear that if G and b are two ideals with finite basis such
√ 2. It
that G = G, then S (G; R) = s(b; R) (we
recall that for an ideal G in
√
n
R, G = x | x ∈ R, x ∈ G, for some n ).
Remark 3. If R is a noetherian normal ring and height G ≥ 2, then
S (G, R) = R.
Proof. Let f ∈ S (G, R). The f Gn ⊆ R, for some n. As G is not contained in any prime ideal of height 1, f ∈ RY , for every prime ideal Y
of height 1. Hence f ∈ R, R being normal.
Corollary . Let R be as in Remark 3 and G = G1 ∩ G2 with height
G2 ≥ 2. Then S (G, R) = s(G1 , R).
35
Zarisk’s Theorem and Rees’ counter example
36
Remark 4. If R is an affine ring and height G ≥ 2, then S (G; R) is
integral over R.
Proof. By passing to the derived normal ring we may assume that R is
normal any apply Remark 3.
For any integer n ≥ 0, set G−n = f | f ∈ L, f Gn ⊆ R , Then
S
S (G; R) = G−n . We shall abbreviate S (G; R) to S when there is no
n
confusion.
42
Remark 5. If R is noetherian, G−n is a finite R-module.
Proof. G−n ⊆ R 1a , a , 0, a ∈ G−n .
Proposition 1. Let G be an ideal in R with a finite basis a1 , . . . , an , (ai ,
0, i = 1, . . . , n). Let t1 , . . . , tn−1 be algebraically independent elements
n−1
P
over R. Set tn = (1 − ai ti ) | an ; this is an element of R[t1 , . . . , tn−1 , a1n ].
i=1
Then
(i) S = R[t] ∩ L, where t stands for (t1 , . . . , tn ).
(ii) Further if R is normal, then S = R∗ ∩ L, where R∗ is the derived
normal ring R[t].
Proof.
(i) Let c = f (t) ∈ R[t] ∩ L. Choose r greater than the degree
n
P
of f . Then since ai ti = 1, we have
i=1
ari c ∈ R[t1 , . . . , ti−1 , tˆi , ti+1 , . . . , tn ],
where b
‘’ on ti indicates that ti has been omitted. As ti , . . . , ti−1 , tˆi ,
ti+1 , . . . , tn are algebraically independent over R, ari c ∈ R, i = 1,
. . . , n. HenceGnr c ⊆ R, i.e.c ∈ S . Thus R[t] ∩ L ⊆ S . On the
other hand let c ∈ S . Then cGr ⊆ R, for some r. Let m1 , . . . , ml
be all the monomials in the ai of degree r. Raising the equation
n
P
Pl
ai ti = 1 to the rth power, we have mi fi = 1, where fi ∈ R[t].
i=1
i=1
Zarisk’s Theorem and Rees’ counter example
Since cmi ∈ R, i = 1, . . . , t, we have c =
37
Pl
i=1
(cmi ) fi ∈ R[t]. Hence
S ⊆ R[t] ∩ L and proof of (i) is complete.
(ii) To prove that S = R∗ ∩ L, we have only to show the inclusion
S ⊇ R∗ ∩ L. Let c ∈ R∗ ∩ L. Then c is integral over R[t] and
we have a monic equation cs + f1 (t)cs−1 + . . . + fs (t) = 0, fi (t) ∈
R[t], i = 1, . . . , s.
Let r ≥ max (degree of fi ). Then ari c is integral over R[t1 , . . . , ti−1 , 43
1≤i≤s
tˆi , ti+1 , . . . , tn ]. As R is normal and t1 , . . . , ti−1 , tˆi , ti+1 , . . . , tn are
algebraically independent, R[t1 , . . . , ti−1 , tˆi , ti+1 , . . . , tn ] is normal.
Hence ari c ∈ R. Therefore cGnr ⊆ R, i.e., c ∈ S . Hence R∗ ∩L ⊆ S
and (ii is proved).
Let R be an affine ring over a ground field K. Let L′ be a field with
K ⊆ L′ ⊆ L, where L is the quotient field of R. Then, is R ∩ L′ an affine
ring? Let us call this Generalized Zariski’s Problem. We recall that
this is just the restatement of Zarisk’s Problem (see Chapter 0) without
the hyper thesis of normality on R. We shall later in this chapter give a
counter example to the above problem (with trans degk L′ = 2).
Let R be an affine ring. By proposition 1 it follows that :
(i) If there exists an ideal G in R such that S (G; R) is not finitely
generated, then S (G; R) is a counter example to the Generalized
Zariski’s Problem.
(ii) If further R is normal and if there exists an ideal G in R such
that S (G; R) is not finitely generated, then S (G; R) is a counter
example to Zariski’s Problem.
2. Krull Rings and G - transforms
We shall now proceed to prove the converse of proposition 1 (see
proposition 4) in the case when R is normal. For that we need some
generalities on Krull rings. We say that an integral domain R is Krull
ring if there exists a set I of discrete valuations of the quotient field L 44
Zarisk’s Theorem and Rees’ counter example
38
of R, such that (i) R =
T
Rv , where Rv denotes the valuation ring of v.
v∈I
(ii) For a ∈ R, a , 0, v(a) = 0, for all but a finite number of v ∈ I.
Proposition 2. An integral domain R is a Krull ring if and only if the
following two conditions are satisfied:
(i) For every prime ideal Y of height 1, RY is a discrete valuation
ring.
(ii) Every principal ideal of R is the intersection of a finite number of
primary ideals of height one.
For the proof of Proposition 2, we refer to: M. Nagata “Local
Rings” Interscience Publishers, Now York, 1962).
T
Remark 1. It easily follows that R = RY , where Y runs through all
prime ideals of height 1.
T
Remark 2. Let R = Rv be a Krull ring defined by a family of discrete
v∈I
valuations {Rv }v∈I of the quotient field of R. Then for a prime ideal Y
of R of height 1, we have RY = Rv , for some v ∈ I. (For prof see M.
Nagata “Local rings”, Interscience Publishers, New York, 1962).
Proposition 3. If R is a Krull ring then the G-transform S of R is also
a Krull ring.
Proof. Let J be the set of those prime ideals Y of height 1 which do not
T
RY . Let x ∈ S . Then xGn ⊆ R,
contain G. We shall prove that S =
Y ∈J T
for some n. As Gn 1 Y , for Y ∈ J, X ∈
RY .
45
Conversely let y ∈
T
Y ∈J
RY . Now as R is a Krull ring, G is contained
Y ∈J
in a finite number of prime ideals of height 1. Therefore there exists
an n ≥ 0 such that yGn ⊆ Rg where g is any prime ideal of height 1
T
containing G. Hence yGn ⊆ RY , Y running through all prime ideals
Y
of height 1. By the remark after Proposition 2, yGn ⊆ R. Hence y ∈ S
and the proposition is proved.
Zarisk’s Theorem and Rees’ counter example
39
Proposition 4. Let R be an affine normal ring over a ground field K.
Let L′ be a field such that K ⊆ L′ ⊆ L, where L is the quotient field of R.
Set R′ = R ∩ L′ . Then there exists an affine normal ring O and an ideal
G of O such that R′ = S (G; O).
We shall prove the assertion in several steps.
(i) R′ is a Krull ring.
Proof. We have R =
T
RY , where y runs through all prime ideals
T
of height 1. Now R′ = (RY L′ ). Hence R′ is a Krull ring. YT
Y
(ii) For every prime ideal g of height 1 in R′ there exists a prime ideal
T
g of height 1 in R lying above g i.e.g R′ = g′ .
Proof. We may assume that L′ is the quotient field of R′ . We have
T
T
R′ = (RY L′ ), as in (i) By Remark 2 following Proposition 2,
Y
T
we have R′g′ = Rg L′ for some prime ideal g of height 1 of R.
This proves (ii).
(iii) There exists a normal affine ring O ′ ⊆ R′ such that for every prime
T
ideal ρ′ of height 1 of R′ , we have height (g′ O ′ ) = 1.
Proof. We may assume that L′ is the quotient field of R′ . Take a
normal affine ring R′′ ⊆ R′ such that L′ is the quotient field of R′′ .
Let Q′ be the set of prime ideals g of height 1 in R′ with height
T
(g′ R′′ ) 1. Let T be set of prime ideals Y of height 1 of R such 46
T
that heights (Y R′′ ) > 1. Let V, V ′′ be the affine varieties defined by R and R′′ respectively. For a Y ∈ T, Y ∩ R′′ defines
isolated fundamental subvariety of V ′′ with respect to V under
the morphism f : V → V ′′ defined by the inclusion R′′ ⊆ R.
Hence T is finite. Therefore by (ii), Q′ is finite. Let R1 be an
affine normal ring such that R′′ ⊆ R1 ⊆ R′ . Let Q′1 be the set of
T
prime ideals g′1 of height 1 of R′ such that height (g′1 R1 ) > 1.
Then Q′1 ⊆ Q′ . We next prove that for a g′ ∈ Q′ , there exists
Zarisk’s Theorem and Rees’ counter example
40
an affine normal ring R1 with R′′ ⊆ R1 ⊆ R′ such that g′ < Q′1 ,
where Q′1 is as above. We claim that R′g′ is a discrete valuation
ring such that trans.degR′g′ /g′ R′ g′ = trans.deg. L′ − 1. Let g
K
K
be a prime ideal of height 1 of R lying above g′ (see (ii)). Let
x1 , . . . xr ∈ R be such that (i) x1 , . . . , xr are algebraically independent over R′ (ii)x1 , . . . , xr modg from a transcendence base for R/g
over R′ /g′ . Such a choice is possible since Rg and R′g′ are valuation
rings. Since R is affine and G is of height 1, we have trans.deg.R/G
K
= trans.deg.L − 1. Also trans.deg.R/G = trans.deg.R′ /G′ + r.
K
K
K
Hence trans.deg.R′ /G′ = trans.deg.L − 1 − r ≥ trans.deg.L′ − 1.
K
K
K
Hence trans.deg.R′ /G′ = trans.degL′ − 1. Since R′′ is affine and
K
TK
′′
height (G R′′ ) > 1, we have trans. R /(R′′ T G)≤ trans.deg.L′ − 2
K
47
Let y1 , . . . , yl ∈ R′ be such that y1 , . . . , yl mod G′ from a tran′′
scendence base of R′ /G′ over R /G T R′′ . Let R′′′ be the derived
T
normal ring of R′′ [y1 , . . . , yl ]. Then height (G′ R′′ ) = 1. Since
Q′ is finite, in a finite number of steps we arrive at a normal affine
ring O ′ ⊆ R′ , with the same quotient field as R′ such that for every
T
prime ideal g′ of height 1 of R′ we have height (g′ O ′ ) ≤ 1.
But O ′ and R′ have the same quotient field. Therefore height
T
(O ′ g′ ) = 1. This proves (iii).
(iv) Let O be as in (iii). Let P be the set of prime ideals G of height 1
in G ′ for which there does not exist any prime ideal of height 1 in
R, lying over G. Let V, V ′ be the affine varieties defined by R and
G ′ and let f : V → V ′ be the morphism induced by the inclusion
G ′ ⊆ R. There are only a finite number of subvarieties of codimension 1 in V ′ to which there do not correspond any subvariety
of codimension 1 in V. Hence the set P is finite. To prove PropoT
sition 4, we take O = O ′ , G =
Y . Let F be the set of prime
Y ∈P
T
Og
ideals of height 1 which do not contain G. Then S (G; O) =
g∈F
T
(See Proposition 3). We have, R′ = R′g′ , where g′ runs through
g′
Zarisk’s Theorem and Rees’ counter example
41
prime ideals of R′ of height 1. Now it follows easily by our construction of O that R′ = S (G; O).
Let R be an integral domain and let G be an ideal of R. We say that
the G-transform of R is finite if S (G; R) = R[G−n ] for some n ≥ 0.
Theorem 1. Let R be a normal affine ring and G be an ideal of R. Then
the G-transform S of R is finite if and only if the G P- transform of P is
finite for any P = RY , where Y , is a prime ideal of R.
Proof. Clearly G−n P = (GP)−n , for every n. Hence if G′ transform of 48
R finite, then so is GP- transform of P, for every P. Conversely assume
S is not finite. We then define an increasing sequence of normal rings
by induction as follows:
Set R0 = R. Having defined
R j , 0 ≤ j ≤ i, we define Ri+1 as
the derived normal ring of Ri (G(Ri ))−1 . Then we have S ⊃ Ri ⊃
−1
Ri−1 (Gi−1 ) , where such Ri is an affine normal ring. By the defini-
tion of G-transform, S = ∪Ri . Further GRi - transform of Ri is also S .
i
We claim that height GRi = 1 for every i. For if height GRi > 1, then
the GRi −transform of Ri is integral over Ri and therefore S = Ri . This
contradicts the assumption that S is not finite. Let Gi be the intersection
of those prime ideals of height 1 of Ri which contains GRi . Then we
have Gi ⊆ Gi+1 . For let R be a prime ideals of height 1 in Ri+1 containing GRi+1 . If height (y ∩ Ri = 1), then by definition of “Gi , we have
Gi ⊆ Y ∩ Ri ”.
Otherwise let height R ∩ Ri i1. Since the G-transform of Ri is S (see
remark after the definition of G-transform), we have (GRi )−1 Gm
i ⊆ Ri ,
−1
for
Now if Gi 1 R ∩ Ri , then (GRi ) ⊆ (Ri )R∩Ri . Hence
some m.
R (GRi )−1 ⊆ (Ri )R∩Ri . Therefore, Ri+1 ⊆ (Ri )R∩Ri , since Ri is normal.
Hence (Ri+1 )R = (Ri )R∩Ri . This contradicts the assumption that height
R ∩ Ri > 1. Hence Gi ⊆ R ∩ Ri . Therefore Gi ⊆ Gi+1 , for i ≥ 0. Set
G∗ = ∪Gi . Then G∗ is a proper ideals of S = ∪Ri . Let Y ∗ be a prime 49
i
i
ideal containing G∗ , got Y = Y ∗ Ri 1 and PGRY . We now consider the
42
Zarisk’s Theorem and Rees’ counter example
GP-transfer of P. We can (Ri )i = (RY )i Where T = R − Y and (RY )i
are defined by n infection as follows:
We set (RY )o = RY and having defined (R
)i we define
Y )o , . . . , (RY
(RY )i+1 as the derived normal ring of (RY )i (G(RY )i )−1 . Since height
G(Ri )T = 1 by our construction, (Ri )T cannot be the GP-transform of P.
Hence GP-transform is not finite.
3. Geometric meaning of the G-transform.
Let V be a variety. We call an affine variety V ′ , an associated affine
variety of V if i) V ′ ⊇ V. ii) The set of divisors of V ′ coincide with that
Vi.e. the set of local rings of rank 1 of V and V ′ are the same.
Theorem 2. Let F be a proper closed set of an affine variety V and let G
be an ideal which defines F in the affine ring R of V. Then V − F has an
associated affine variety if and only if the G-transform S of R is finite;
in this case S defines an associated affine variety and S contains and is
integral over the affine ring of any associates affine variety of V − F.
Lemma 1. Let R be an integral domain and let G be an ideal of R.
Y′∩R
Set R′ = R G−n or S (G; R). Then the correspondence Y ′
establishes a 1 − 1 corresponding between the set of prime ideals of
R′ not containing G and the set of prime ideals of R not containing G.
Further, for a prime ideals Y ′ of R′ with Y ′ 2 G, we have R′Y ′ =
RY ′ ∩R .
50
Proof. Let Y be a prime ideals
not contains G. Let
of R which does
1
1
a ∈ U , a < Y . Then R′ ⊆ R a . Since Y R a is a prime ideal, so is
T
T
R′ . Further Y ′ R = Y and R′Y ′ = RY .
Y ′ = Y R a1
Conversely if Y ′ is a prime ideal of R′ which does not contain G,
T
then Y = Y ′ R does not contain G and RY = R′Y ′ .
Lemma 2. Let G be an ideal of a noetherian domain R. Let S be the
G-transform of R and R′ a subring of S containing R. Then the GR′ transfer of R′ is S .
Zarisk’s Theorem and Rees’ counter example
43
Proof. Let S ′ be the GR′ -transform of R′ . We have only to prove that
S ′ ⊆ S . Let f ∈ S ′ . Then there exists an n such that f Gn ⊆ R′ . Let
a1 , . . . , al ∈ Gn generate Gn . There exists an m such that f ai Gm ⊆ R.
Hence f Gn+m ⊆ R and the lemma is proved.
Corollary . With the above notation, if a ∈ S , then aS : GS = aS and
consequently GS is not of height 1.
Proof. Let f ∈ S , f G ⊆ S a i.e. af G ⊆ S . af ∈ S .
We now prove Theorem 2. Suppose S is finite. Then S is an affine
ring and defines an associated affine variety by lemma 1 and Corollary.
to Lemma 2. Conversely assume that V ′ is an associated affine variety
of V − F. Let R′ be the affine ring of V ′ . Let x′ ∈ R′ . Set G x′ = y |
′
y ∈ R, yx ∈ R . Since x′ ∈ RY , for Y 2 G (by the hypothesis) we have
Gx′ 1 Y , for Y 2 G. Hence Gx′ contains a power of G. Therefore
x′ ∈ S i.e., R′ ⊆ S . Since the divisors of V ′ and V − F are the same,
height (GR′ ) ≥ 2. Hence S is integral over R′ . Hence S is an affine ring
and therefore finite.
Let V be an affine variety defined by an affine ring R and let F be a 51
closed set defined by an ideal G.
Theorem 3′ The variety V − F is affine if and only if 1 ∈ GS , where S
is the G-transform of R. In this case F is pure of codimension 1 and S
is the affine ring of V − F.
Proof. Suppose V −F is affine. Then V −F is an associated affine variety
of V − f . Let R′ be the coordinated ring of V − F. Then R′ ⊆ S (by
Theorem 2). Now 1 ∈ GR′ . Hence 1 ∈ GS . Conversely
suppose that
1 ∈ GS . Then 1 ∈ GG−n for some n. Set R′ = R G−n . Since GR′ ∋ 1,
Lemma 1 of Theorem 2, proves that the affine variety defined by R′ is
V − F. R′ = S because GR′ ∋ 1 (by virtue of Lemma 2).
It now remains to prove that F is pure of codimension 1 if V − F is
affine. Suppose the contrary. Let F1 be an irreducible component of F
with codimension F1 > 1. Let f ∈ R, with f not vanishing on F1 and
Zarisk’s Theorem and Rees’ counter example
44
vanishing
on all tho other irreducible components. Then considering
1
R f we may suppose that F = F1 . But this would mean that S is
integral over R. Hence 1 < GS . Contradiction.
Theorem 3. V − F is an affine variety if and only if G(G(P)−n(P) ∋ 1 for
every local ring P of F (the integer n(P) depending on P).
52
Proof. If V − F is affine, then by Theorem 3, 1 ∈ GS i.e 1 ∈ GG−n for
some n. Hence 1 ∈ G(GP)−n for every P of F. Conversely assume that
1 ∈ G(GP)−n(P) for every P of F. Then by lemma 2 of Theorem 2, the
GP-transform of P is finite for every P of F. Hence by Theorem 2, S is
finite. Then 1 ∈ GS and V − F is an affine variety.
Corollary 1. If V is an affine variety and F a divisorial closed subset
of F such that some multiple of F is locally principle, then V − F is an
affine variety.
Corollary 2. If V is a non-singular affine variety and F a divisorial
closed subset of V, then V − F is an affine variety.
Corollary 3. If V is an affine curve and F a closed subset of V, then
V − F is again affine.
Proof. It is sufficient to prove the Corollary when F consists of a single
point P. If P is normal then by theorem 3′ , V − F is affine. If P is
not normal we consider the derived normal ring P′ of P. Let C be the
conductor of P′ with respect to P. Then Gn ⊆ C for some n, G being
an ideal which defines F. By considering R G−n we are reduced to the
case when F consists of normal points. This proves the Corollary.
4. Zariski’s theorem and some related results.
Theorem 4 (Zariski’s). Let R be an affine ring over a ground filed K
and let Ω be the quotient field of R. Let L be a subfield of Ω containing
K.
T
(1) If trans.deg .L = 1 , then R L is an affine ring
K
Zarisk’s Theorem and Rees’ counter example
(2) If R is normal and trans.degL = 2 , then R
K
45
T
L is an affine ring.
Proof of (1) is a consequence of Proposition 4, Theorem 2 and Corollary
3 to theorem 3′ .
Proof of (2) By virtue of Proposition 4, (2) is contained in the following
theorem:
Theorem 4′ Let R be an affine normal ring of dimension 2 over a ground 53
field K. Then for any ideals G if R the G-transform S (G, R) is finite.
We require the following lemma
Lemma. Let S be a Krull ring, g′ a prime ideal of height 1. Let G be an
ideal of S such that GRg = gRg . Then G : g 1 g.
Proof. Since R is a Krull ring Rg is a discrete valuation ring.Therefore
there is an a ∈ G such that aRg = gRg . Since R is a Krull ring, aR =
g ∩ g1 ∩ . . . ∩ gn , gi being primary ideals of height 1 different from g .
Then g 2 g1 ∩ . . . ∩ gn = aR : g ⊆ G : g.
Proof of Theorem 4′ . We may assume that the ideal G is pure of height
1. Choose an element b ∈ G such that VYi (b) = ni , 1 ≤ i ≤ r where UYi
is the normal valuation corresponding to Yi . Then U ∩ U 1 = bR where
U ′ is an ideal pure of height 1 such that U and U ′ do not have common
prime divisors of height 1. We have GS = bS (cf. Cor to Lemma 2,
p.50). Choose an element a a ∈ G such that a is not contained in any
of the prime divisors of G and VYi (a) > VYi (b) = ni , 1 ≤ i ≤ r. Let x
be a transcendental element over R. Extend the ground field K to K(x).
We remark that the GK(x)[R]-transform of R′ = K(x)[R] is finite if and
T T
only if the G-transfer of R is finite. Now GR′ = K1(n1 ) . . . Kr (nr )
, where K is the centre on R′ of the valuation VKi on Ω(x) define by
P
VKi ( a1 xi ) = min VYi (a j ). The element a and b do not have a common
j
prime divisor in S . The choice of the element a and the following any 54
lemma show that we may assume that bS is a prime ideal.
Lemma . Let S be a Krull ring and let a, b ∈ c with a, b not having a
common prime divisor. Then ax − b prime in S [x].
46
Zarisk’s Theorem and Rees’ counter example
Now assume that the G-transform S of R is not finite. Then as in
the proof of Theorem 1 (of this chapter), there exist normal local rings
(Pi , W ), 0 ≤ i < ∞ such that P0 = RY for suitable prime ideal Y ⊃ G
S
and (Pi , Wi ) dominates (Pi−1 , Wi−1 ). Furthermore S ∗ = Pi = S M ,
i
55
where M is a maximal ideals of S . Set W ∗ = ∪Wi . Consider W =
W ∗ ∩R. Since bS ∩R = G′ , the canonical mapping ϕ : R/V ′ → S /bS is
an injection. For s ∈ S , we have sGm ⊆ R for some m. Since Gm 1 G′ is
of dimension 1, by the theorem of Krull-Akizuki (see M. Nagata “Local
rings”, Theorem 33.2, p.115) S /bS is noetherian and S /M is of finite
length over R/W . In particular W ∗ is finitely generated. Hence exists
all such that W1 S ∗ = W ∗ and that S ∗ /W ∗ is of finite length over P1 /W1 .
We now proceed to prove that S ∗ is noetherian. Since height (W ∗ ) =
2, by virtue of a theorem of Cohen we need only prove that every prime
ideal of height 1 of S ∗ is finitely generated. (see M. Nagata, “Local
rings” Theorem 3.4, p.8).
Let g∗ be a prime ideal of height 1. Set g = g∗ ∩ R. Then S ∗ g∗ = Rg .
Hence by the lemma proved we have g′′ = S ∗ : g 1 g∗ . But gS ∗ ⊆
g′′ . We claim that height g′′ ≥ 2. For if g′′ ⊆ R, a prime ideal of
∗ = S ∗ . Hence K = g∗ a contradiction
height 1 of S ∗ , then 1. g = S K
g∗
′′
′′
to the fact that g 4 |.Hence either g′′ = S ∗ or g′′ is N W -primary.
Hence N W ∗t ⊆ g′′ for some t. Since S /N W ∗r is artinian and N W ∗r
is finitely generated we conclude that g′′ is finitely generated. Now
g′′ /gg′′ is a finitely generated over S ∗ /g∗ . But in S ∗ /g∗ the only prime
ideals are ⊞∗ /g∗ and (0) and S ∗ /g∗ is Noetherian. Hence g∗ ∩ g′′ /g∗ g′′
is finitely generated. Hence g′′ ∩ g∗ gS ∗ , being residue class module of
g′′ ∩ g∗ g∗ g′′ , is finitely generated. Hence g” ∩ g∗ is finitely generated.
Since S ∗ /g′′ and S ∗ /g∗ are noetherian, we have S ∗ g′′ ∩g∗ is noetherian.
Hence g∗ /g′′ ∩ g∗ is finitely generated. Hence g∗ is finitely generated.
Hence S ∗ is noetherian.
The ring, P1 being a geometric normal local ring, is analytically
normal. Further S ∗ and P1 have the some quotient field, N W 1 S ∗ =
N W ∗ and S ∗ /N W ∗ is of finite length over R/N W ∗ and therefore over
P1 /N W 1 P1 . Hence by Zariski’s main Theorem (see M. Megata, “Local ring” Theorem 37.4 , P.137) P1 = S ∗ . This is contradiction to the
construction of the Pi .
Zarisk’s Theorem and Rees’ counter example
47
Theorem 5. Let V be a normal affine variety of dimension 2 and lot F
be a divisorial closed subset of V. Then V − F is affine.
Proof. Let U be the ideal defining F. By Theorem 4 the U -transfer
S of R is finite. Let V ′ be the affine variety defined by S . Since height
U ≤ 2 and V is of dimension 2, V−F is isomorphic to an open subset V ′′
of V ′ such that V ′ − V ′′ consists of atmost a a finite number of points. 56
Let ×′ ∈ V ′ − V ′′ . Consider the morphism V ′ → V induced by the
inclusion R ⊆ S . Since f−1 f (×′ ) is discrete, by Zariski’s Main Theorem
f is bio-holomorphic at x′ . Let W be an irreducible component of F
passing through f (x′ ). Since x′ and f (X ′ ) are bio-holomorphic there is
subvariety of codimension 1 of V ′ passing through x′ and laying over
W. This contradicts the fact that V ′ is the associated affine variety of
V − F. Hence V − V ′′ is empty and V − F is affine.
We now proceed to give an example to show that Theorem 4(2) is
false if we do not assume that R is normal. Take R = K[X, Y, Z]/( f ).
Where
(1) f (X, Y, Z) = Y(Z + YT ) + X(u1 YZ + U2 Z 2 )
(2) f is irreducible
(3) T, U1 , ∈ U2 ∈ K[X, Y].
Let the image X, Y, Z, T, U1, U2 be denoted by x, y, z, t, u1 and u2 respectively so that R = K[x, y, z] = 0. Set Y = (x, y). Since R/Y =
K[X, Y, Z]/( f,X,Y) ≈ K[Z], the ideal Y is prime. Similarly g = (y, z) is
prime. We shall show that the Y -transfer of R is not finite. We first
prove that
(*) Y −1 = R + z1 R, z1 = (z + yt)/X.
Proof. We have (x) = Y ∩ (x, 2 + yt). For let λx + µy = ax + β(z + yt).
Then µy2 ∈ (x) and therefore µy2 t ∈ (x) i.e. µyz ∈ (x). But z is not
a zero divisor module (x). Hence | (y ∈ (x). Therefore (x) = Y ∩
(x, z + yt). Lot now g ∈ Y −1 . Then g = γx = γy i.u.γ ∈ (x) : (y).
T
Since (x) = Y (x, Z + yt), we have (x) : (y) = (x, Z + yt) : (y). 57
But ℓ ∈ K[x, y] and therefore y is not a zero divisor module (x, z +
48
Zarisk’s Theorem and Rees’ counter example
yt). Hence (x, Z + yt) : (y) = (x, Z + yt). Hence γ ∈ R + z1 F. This
proves (*). Set R1 = R[Y −1 ] = R[z1 ] = K[x, y, z1 ] K[X, Y, Z1 ]/U
(say) where Z1 = (Z + YT )/X . We have f1 (X, YZ) = X f1 (X, Y, Z1 ) where
f1 (X, YZ) = X f1 (X, Y, Z1 ) = Y(Z1 +Yt1 )+X(U1′ YZ1 +U2′ Z12 ), T 1 = U2 T 2 −
U1 T, U!′ = U1 − 2U2 T, U2′ = U2′ X. Now f (x, y, z) = X f1 (x, y, z1 ) = 0.
Hence f1 (x, y, z1 ) = 0 i.e. f1 (x, y, z1 ) = 0 ∈ U . We claim next that
the element f1 (x, y, z1 ) = 0 is prime in K[X,
Y, Z1 ]. Suppose g2 (x, y, z1 ).
Then one of the g1 say g2 is a unit in K X, Y, Z, X1 . Thus f1 (x, y, z1 ) =
X r g1 (x, y, z1 ) = 0 for some r ≥ 0. But f1 (x, Y, Z1 ) is not divisible by
in K[X, Y, Z1 ]. hence f1 (X, Y, Z1 ) is irreducible. Hence it follow that
R1 = R[Y −1 ] = K[X, Y, Z1 ]/( f1 ) . Further f1 satisfied the same condition
as f . Proceeding in the same way we see that the Y -transform S of R is
obtained by the successive adjunction of elements Z1 , Z2 , Z3 , . . . , where
n
Zn+1 zn +yt
X , tn ∈ K[x, y]. This shows that S is not finite.
5. Rees’ counter example.
Let K be a field of an arbitrary characteristic and lot C be a nonsingular plane cubic curvedefined over K. For a natural number n and
a fixed point Q of C, T n = P | np or nQ is a finite act, because C is of
58
positive genus. We choose here Q to be a point of inflexion (it is well
known that a non-singular plan cubic has 9 points of inflexion). Then
P ∈ T 3d if and only if there is a plane curve Cd of degree d such that
Cd .C = 3dP. Thus we are that the set of points P and C, such that Cd .C
is a multiple of P on suitable curve Cd of positive degree, is a countable
set. Therefore there is a point P of C such that no np (0. > 0) is linearly
equivalent to any C.Cd (on C). (Note that the system of Cd .C is complete
linear this follows from the arithmetic normality of C.) We fix such a
point P also and we enlarge K, if necessary, so that P and Q are rational
over K.
Let H = K[x, y, z] be the homogeneous coordinate ring of C, (x, y, z)
being a generic point of C over K. Let Y be the prime ideals of H
which defines P. Let t be a transcendental element over H and consider
the ring S generated by all of at−n with a ∈ Y (n) (n runs through all
natural numbers) over H[t].
Zarisk’s Theorem and Rees’ counter example
49
We want to show that:
(1) S is not finitely generated over K.
(2) There is a normal affine ring R such that S = R ∩ L,where L is the
field of quotient of S .
This S is the Roes’ counter example to Zariski problem in the case
of transcendence degree 3.
Proof of (1). Consider deg Y (n) (=minimum if degree of elements of
Y (n) ). if a homogeneous element h of degree d is in Y (n) , then h defines
Cd .C with a suitable plane curve Cd of degree d. Since h ∈ Y (n) , Cd .C
contains nP. By the choice of P, Cd .C , nP, whence 3d = degCd .C >
n and d > n/3. Let σ be one of 1, 2, 3 and such that m′ σ in divisible by 59
3. Set d = (n + σ)/3. Since C is an abelian variety, there is a point R of
C such that n(P − Q) + (R − Q) ∼ 0. Then
nP + R + (σ − 1) ∼ 3U ∼ Cd .C,
where Cd is a curve of degree d. Since C is a non-singular plane curve,
the system of all Cd .C (with fixed d) is a complete linear system, whence
there is a Cd such that nP + R + (σ − 1)Q = Cd .C. Let h be the homogeneous form of degree d in H defined by Cd . Then h Y (n). Thus we see
that
(∗)
deg Y (n) = (n + σ)/3, 1 ≤ σ ≤ 3.
This (∗) being shown, we see that S is not finitely generated over
K by the same way an in the construction of the fourteenth problem in
Chapter III.
Proof of (2). We first show that
(∗∗)
S = H[t, t−1 ] ∩ V,
where V is the valuation ring obtained as follows:
Let p be a prime element of the valuation ring HY . Then t/p is
transcendental over HY , whence we have a valuation ring HY (t/p)(=
HY [t/p] pHY [t/p] ).HY (t/p) is independent of the particular choice of p
and this valuation ring is denoted by V.
50
Zarisk’s Theorem and Rees’ counter example
It is obvious that S ≤ H[t, t−1 ] ∩ V. Let f be an arbitrary element
P
of H[t, t−1 ] ∩ V. Then f is of the form ai ti (finite sum) with ai ǫH 60
(i may be negative). Let v be a valuation defined by V. Then by the
P
construction of V, v( ai ti ) = min V(ai ti ) ≥ 0. Therefore ai ǫY (−i) if
i < U, which implies that f ǫS . Thus (∗∗) is proved.
This being settled, it remains only to prove the following lemma by
virtue of Proposition 1 (§1).
Lemma. Let R be a normal affine ring of a function field L over a field K
and let V1 , . . . , Vn be divisional valuation rings of L (i.e., Vi are discrete
valuation rings of L over K such that trans. degk L − 1 = trans. deg. of
the residue class field of Vi ). Then there is a normal affine ring V with
an ideal G such that R ∩ V1 ∩ · · · ∩ Vn = S (G; V).
The proof is substantially the same as that of Proposition 4 (§2) and
we omit the detail.
Chapter 6
Complete reducibility of
rational representation of a
matrix group
This chapter is mostly a representation of M. Nagata: Complete re- 61
ducibility of rational representation of a matric group, J. Math. Kyoto
Univ. 1-1 (1961), 87-99.
It is well known in the classical case that every rational representation of a semi-simple algebraic linear group is completely reducible.
But the same argument becomes false in the case where the universal
domain is of characteristic p , 0. For instance, when K is a universal
domain of characteristic 2, the simple group S L(2, K) has the following
rational representation ρ which is not completely reducible:


! 1 ac bd


a b
ρ
= 0 a2 b2  .


c d
0 c2 d2
(This ρ is not completely reducible because ac and bd are not linear
polynomials in a2 , b2 , c2 , d2 .) Therefore it is an interesting question to
ask conditions for an algebraic linear group G so that every rational
representation of G is completely reducible.
Now, our answer of the above question, can be stated as follows:
51
52
62
Complete reducibility of rational representation....
(1) When p , 0 : Every rational representation of G is completely
reducible if and only if there is a normal subgroup G◦ of finite
index such that (i)G◦ is a subgroup of a torus group (i.G., diagonalizable) and (ii) the index of G◦ in G is prime to p. If G is
connected, then the above condition is equivalent to the condition
that the representation of G by homogeneous forms of degree p
is completely reducible. On the other hand, if G is an algebraic
group (which may not be connected), then the complete reducibility of all rational representations of G is equivalent to the condition that every element of G is semi-simple (i.e., diagonalizable).
(2) When p = 0 : Each of the following two conditions is equivalent
to the complete reducibility of all rational representations of G.
(I) The closure of G has a faithful rational representation which is
completely reducible.
(II) The radical of the closure of G is a torus group.
We shall prove also the following interesting theorem concerning
the complete reducibility of rational representations of a connected algebraic linear group:
If G is a connected algebraic linear group, then every rational representation of G is completely reducible if (and only if) the following is
true:
!
1 τ
If ρ′ =
is a rational representation of G, then ρ′ is equivalent
0 ρ
!
1 0
to the representation
.
0 ρ
63
1. Preliminaries on connected algebraic linear groups.
Throughout this chapter, K denotes a universal domain of an arbitrary characteristic, unless the contrary is explicitly stated. Let G be a
connected algebraic linear group contained in GL(n, K). A Borel subgroup B of G is defined to be a maximal connected solvable subgroup
of G. Then as was proved by Borel, the following is true:
Lemma 1. The homogeneous variety G/B is a projective variety. On
the other hand, every element of G is in some conjugate of B.
Complete reducibility of rational representation....
53
Now we have:
Lemma 2. If u ∈ G is unipotent, G being a connected algebraic linear
group, then there is a closed connected unipotent subgroup of G which
contains u.
Proof. By the last half of Lemma 1, we see that u is in a Borel subgroup
B of G. Since B is solvable, the set U of all unipotent elements of B is a
closed connected subgroup, which proves the assertion.
On the other hand, the following was proved by Borel:
Lemma 3. If a connected algebraic linear group G consists merely of
semi-simple elements, then G is commutative, hence is a torus group.
Next we shall concern with an algebraic group which is not connected:
Lemma 4. Let G be an algebraic linear group and let G◦ be the connected component of the identity of G. Then each coset G◦ g(g ∈ G)
contains an element of finite order.
Proof. Let A be the smallest algebraic group containing g and let A◦ be
the connected component of the identity of A. Since A is commutative
and since A◦ is infinitely divisible (in the additive formulation), A◦ g 64
contains an element of finite order, which proves the assertion.
2. Preliminaries on group representations.
Let G be an abstract group, let G◦ be a normal subgroup of G and
let K be a field of characteristic p which may be zero, throughout this
section, except for in Lemmas 8 and 9.
The following lemma is well known :
Lemma 5. If a finite K-module M is a simple K − G-module, then M is
the direct sum of a finite number of K − G◦ −modules which are simple.
Corollary . If a representation ρ of G in GL(n, K) is completely reducible, then the restriction of ρ on G◦ is completely reducible.
Complete reducibility of rational representation....
54
The converse of the above Corollary is not true in general if p , 0,
but we have:
Lemma 6. Let ρ be a representation of G in GL(n, K). If the restriction
ρ◦ of ρ on G◦ is completely reducible and if the index t = [G : G◦ ] is
finite and not divisible by p, then ρ itself is completely reducible.
65
Proof. If ρ is not completely
reducible, then ρ contains a representation
!
ρ1 τ
which is not completely reducible and such that
of the form
0 ρ2
!
ρ1 τ
ρ1 , ρ2 are irreducible. Hence we may resume that ρ =
and that
0 ρ2
ρ1 , ρ2 are irreducible. Let the representation module of ρ be M ∗ .M ∗ contains the representation module M of ρ2 and M ∗ /M is the representation
module of ρ1 . Since ρ0 is completely reducible, we see that M is a direct
summand of M ∗ are a G◦ − K− module. Hence M ∗ = M ⊕ N1 ⊕ · · · ⊕ Nr ,
where Ni are simple G◦ − K− modules. For each Ni we fix a linearly
independent basis a, . . . , ai s over K; we note here that the number s is
independent of i because M ∗ /M is a simple G − K-module (remember
the well know proof of Lemma 5). For each (r, s)− matrix b = (bi j )
P
over the module M, we define N(b) = i j (ai j + bi j )K. Thus we have a
one-one correspondence between all of b and all of submodules N such
that M ∗ = M ⊕ N as a K-module. We may assume, on the other hand,
that ρ1 is given by the linearly independent basis a11 , . . . , ars modulo
M of M ∗/M. Each g ∈ G defines a linear transformation f (g) on the
module of (r, s)-matrices over M ∗ as follows: If (x11 , . . . , xrs )ρ1 (g) =
(y11 , . . . , xrs ), then (xi j ). f (g) = (xi j ). We define also an (r, s) - matrix c(g) over M by the relation N(c(g)) = N(0)g . If b and b′ are such
that N(b)g = N(b′ ), then we have (ai j + bi j )g = (ai j + b′i j ). f (g). Since
(ai j )g = (ai j + c(g)). f (g), we see that b′ = c(g) + bg . f (g)−1 . Thus:
(1)
N(b)g = N(c(g) + bg . f (g)−1 ).
If we apply this formula to the case where b = c(h) with h ∈ G, then we
have
(2)
c(hg) = c(g) + c(h)g . f (g)−1 .
Complete reducibility of rational representation....
55
P
Now, let g1 . . . , gt be such that G = G◦ gi and we set d = t−1
P
( c(gi )). We want to show that N(d)g = N(d) for every g ∈ G. Indeed,
P
P
c(g) + dg . f (g)−1 = c(g) + t−1 ( c(gi )g . f (g)−1 ) = c(g) + t−1 ( (c(gi g) −
P
c(g)) = t−1 ( c(gi g)) = d. Therefore M ∗ = M ⊕ N(d) is a representation 66
module of G, which completes the proof.
Corollary . If p = 0, then the coverer, of the Corollary to Lemma 5 is
true, provided that the index [G : G◦ ] is finite.
Lemma 7. Let H be a subgroup of finite index of G. If a representation
ρ of H in GL(n, K) is not completely reducible, then the representation
ρ∗ of G included by ρ is not completely reducible.
Proof. Let M be the representation module of ρ. M contains an K − Hmodule N which is not a direct summand of M. Let M ∗ be the repreP
sentation module of ρ∗ . Then M ∗ is of the form M ⊕ Mgi where gi
P
P
are such that G = H + Hgi (gi < H). It is obvious that Mgi is HP
admissible. M ∗ contains N ∗ = N ⊕ Ngi . If N ∗ is a direct summand of
P
P
M ∗ as a G − K-module , then we have M ⊕ Mgi = N ⊕ Ngi ⊕ N ′ as
P
an H − K − Module . Then we see that M = N ⊕ (M ∩ ( Ngi + N ′ )) as
an H − K− module, which is a contradiction. Hence N ∗ is not a direct
summand of M ∗ and ρ∗ is not completely reducible.
Corollary . If a finite group G∗ has order which is divisible by p, then
G∗ has a representation which is not completely reducible.
Proof. G∗ has an element! a whose order is p. Then the sub-group {ai }
1 i
is represented by
, and we see the assertion by Lemma 7. 0 1
Next we observe relationship between rational representations of a
matric group G and those of the closure of G.
Lemma 8. Let G be a matric group and let G∗ be the closure of G. Let 67
ρ∗ be a rational representation of G∗ and let ρ be the restriction of ρ∗ on
G. Then ρ is irreducible if and only if ρ∗ is irreducible. ρ is completely
reducible if and only if ρ∗ is completely reducible.
Proof. ρ(G) is dense in ρ∗ (G∗ ) and we see the assertions easily.
56
Complete reducibility of rational representation....
Lemma 9. Let N be a normal subgroup of a matric group G and let
ρ be an irreducible rational representation of G into GL(n, K), K being
an universal domain. If N consists only of unipotent elements, then N is
contained in the kernel of the irreducible representation ρ.
Proof. Since the set of all unipotent matrices in GL(n, K) is closed, and
since the image of unipotent element under a rational representation is
again unipotent, the closure N ∗ of ρ(N) consists only of unipotent elements. Therefore N ∗ is nilpotent, hence is solvable. Therefore we may
assume that every element (ai j ) of ρ(N) is such that ai j = 0 if i > j,
whence aii = 1 for every i. On the other hand, the Corollary to Lemma
5 says that the restriction of ρ on N is completely reducible, whence
ρ(N) must consist only of the identity, which completes the proof.
3. The main result in the case where G is connected and p , 0.
Theorem 1. Let K be a universal domain of characteristic p , 0 and
let G be a connected matric group contained in GL(n, K). Then the
following three conditions are equivalent to each other:
(I) Every rational representation of G is completely reducible.
(II) G is contained in a term group, i.e., there is an element a of
GL(n, K) such that a−1 Ga is a subgroup of the diagonal group.
68
(III) The representation of G by homogeneous forms of degree p is
completely reducible.
Proof. It is obvious by virtue of Lemma 8 that such of the above conditions for G is equivalent to that for the closure of G. Therefore we may
assume that G is a connected algebraic linear group. It is well known
that (II) implies (I) and it is obvious that (I) implies (III). Thus we have
only to show that (III) implies (II). Assume that (III) is true and that (II)
is not true and we shall lead to a contradiction. Lemma 3 shows that G
contains an element g which is not semi-simple. Then the unipotent part
gu of g is different from the identity and is contained in G (cf. Borel’s
paper “Groupes leneaires algebriques, Ann. of Math 64, No.1 (1956)
20-82), hence G contains a connected closed unipotent subgroup U , 1
Complete reducibility of rational representation....
57
by Lemma 2. The representation module F p of the representation p of G
by homogeneous forms of degree p is nothing but the module of homogeneous forms of degree p in n variables X1 , . . . , Xn on which element
g of G operates by the rule h(X1 , . . . , Xn )g = h((X1 , . . . , Xn )g). F p conP p
tains M = Xi K, which is also a representation module of G. Hence
(III) implies that M is a direct summand of F p . Thus F p = N ⊕ M.
For each monomial ni1 ...in = X1i1 . . . Xnin with i j such that i j < p and
P
i j = p, there is a uniquely determined element mi1 ...in of M such that
fi1 ...in = ni1 ...in + mi1 ...in form linearly independent basis for N. We note
that N and M are representation modules of U. Hence we have only to
show that:
The decomposition F p = N ⊕ M as a representation module of the
connected closed unipotent group Ulead us to a contradiction.
Let u = (ui j ) be a generic point of U over the universal domain K. 69
We may replace U with conjugate of U. Hence we may
first
assume
p
that ui j = 0 if i > j, where ui j = 1 for i. set K ∗ = K( ui j ), and we
choose (K, 1) that uk1 < 1∗ , ui j j∈k∗ ,ui j ∈K ∗ if i > K and such that uk j ∈ K ∗
if j > 1. for each A−1 UA (A being a triangular unipotent matrix), we
can associate such a(k, 1) and we may assume that the pair (k, 1) for U
is lexicographically smallest among those (k, 1) for A−1 UA. assume for
P
a moment that is a linear relation i αi uki ∈ K ∗ with α1 ∈ k and α1 , 0.
We may assume that α1 = 1 and that αi = 0 if uki ∈ k∗ . Hence, in
particular, α1 = . . . = αk = α1+1 = . . . αn = 0. Consider the unit matrix
1 and the matrix c1 = (c1i j ) such that (i) c1i j = 0 if j , 1, (ii) c′i1 = αi if
i , 1 and (iii) c111 = 0. Set c = 1+c′ . Then obviously c−1 = 1−c1 . Since
c−1 u ≡ u modulo k∗ , We see easily that such a (k, 1) defined for c−1 Uc
has the same K and a smaller 1 than our (k, 1), which is a contradiction.
Therefore:
X
(1)
If αi ∈ K and if α1 , 0, then
iαi uki < K ∗ .
Now, let a = (ai j ) be an arbitrary element of U. Then ua is also a
generic point of U over K. Since uk j ( j > l) is in K ∗ , the (k, j). component of ua must be in K ∗ . This shows by virtue of (1) above that al j = 0
if j > 1. since a is arbitrary, we see that ul j = 0 for every j , l.Thus X1
Complete reducibility of rational representation....
58
P
is U-invariant. Now we consider the elements fi1 . . . in (i j < p, i j = p).
We denote by g j the element fil . . . in such that i j = 1, i1 = p−1 for each
j = k, k + 1, . . . , 1 − 1, 1 + 1, . . . , n. Since we have
X
p−l
p−l
(2)
(Xk Xl )u =
uk j Z j Xl ,
70
we see that
guk =
(3)
X
uk j g j .
j,l
p
Consider the coefficient of Z1 in guk ; let it be d. (3) shows that d is
a linear combination of uk j( j , l) with coefficients in K. On the other
hand, (2) shows that d − ukl must be in K ∗ . Thus we have a contradiction
to (1) above, which completes the proof of Theorem 1.
4. The main result in the case where p , 0.
Theorem 2. Let K be a universal domain of characteristic p , 0 and
let G be a matric group contained in GL(n, k). Then the following conditions are equivalent to each other:
(I) Every rational representation of G is completely reducible.
(II) There is a normal subgroup Go of finite index such that (i) Go is
a subgroup of a torus group and (ii) the index of Go in G is not
divisible by p.
(III) The connected component Go of the identity of G is a subgroup of
a torus group and [G : Go ] is not divisible by p.
If G is an algebraic linear group, then the above conditions are
equivalent to the following condition:
(IV) Every element of G is semi- simple.
71
Proof. It is obvious that (III) implies (II) and that (II) implies (I) by
virtue of Lemma 6. Therefore, by Lemma 8, we have only to prove the
equivalence of (I), (III), (IV) in the case where G is an algebraic linear
group. Thus we assume that G is algebraic let Go be the connected
component of the identity of G. Assume first that (IV) is true. Then Go
Complete reducibility of rational representation....
59
consists merely of semi simple elements, hence Go is a torus group by
Lemma 3. If a semi-simple a has a finite order, then the order is prime to
p. Therefore Lemma 4 implies that[G : Go ] is not divisible by p. Thus
(IV) implies (III). As we have remarked above, (III) implies (I). Assume
now that (IV) is not true. Then, as we have seen in the proof of Theorem
1, there is a unipotent element u of G which is different from the identity,
If u ∈ Go , then Go has a rational representation which is not completely
reducible, hence G itself has such one by Lemma 7 or by the corollary
to Lemma 8. If u < Go , then the finite group G/G0 has a representation
which is not completely reducible, which is a rational representation of
G. Thus we see that (I) is not true. Therefore (I) implies (IV), which
completes the proof of Theorem 2.
5. The main result in the case where p = 0.
Theorem 3. Let K be a universal domain of characteristic p = 0 and
let G be a matric group contained in GL(n, k). Then the following conditions are equivalent to each other:
(I) Every rational representation of G is completely reducible.
(II) The closure of G has a faithful rational representation which is
completely reducible.
(III) The radical of the closure of G is a torus group.
Proof. It is obvious that (I) implies (II) by virtue of Lemma 8. Lemma
9 shows that (II) implies (III). In order to show that (III) implies (I), we
shall prove the following lemma:
Lemma 10. Let G be a connected algebraic linear group and let R be 72
the radical of G. If R is a torus group, then there is a closed connected
normal subgroup S such that (i) G = RS and (ii) R ∩ S is a finite group.
Furthermore, R is contained in the center of G (hence R is the connected
component of the identity of the center of G).
Proof. For the fact that R is contained in the center of G, see Borel’s
paper. Let S be the subgroup generated by all unipotent elements of
60
Complete reducibility of rational representation....
G. Then S is obviously a normal subgroup. Each unipotent element is
in a closed connected unipotent subgroup of G, hence S is generated
by closed connected subgroups, and therefore S is a closed connected
subgroup of G. Now, we may assume that R is a diagonal group and that
each g ∈ G is given by


τ12 (g) · · ·
τ1r (g)
ρ1 (g)

 0
ρ2 (g) · · ·
τ2r (g)

g = 

...................................


0
....................
ρr (g)
73
with irreducible representations ρ1 , . . . , ρr . If u ∈ G is unipotent, then
ρi (u) is unipotent, whence the determinant of ρi (u) is 1. Therefore we
see that if s ∈ S , then the determinant of ρi (s) is 1. On the other hand,
since R is in the center of G, ρi (R) is in the center of ρi (G), hence by
the famous lemma of Schur every element of ρi (R) is of the form k.ρi (1)
with k ∈ K. Therefore we see that R ∩ S is a finite group. Since S a is
closed normal subgroup. RS is a closed normal subgroup. Since G/R is
semi- simple, we see that G/RS is semi-simple, unless G = RS . If G ,
RS , then G/RS contains a non-trivial unipotent element, whence there
must be a unipotent element of G outside of RS , which is a contradiction
to our construction of S . Therefore G = RS , which completes the proof.
Now we proceed with the proof of Theorem 3. By the Corollary to
Lemma 6, we may assume that G is connected. Lemma 8 allows us to
assume that G is an algebraic linear group. Let R be the radical of G
and let S be the normal subgroup given in Lemma 10. Since R ∩ S is
a finite group and since G = RS , we see that S is semi-simple, whence
every rational representation of S is completely reducible. Let ρ be
an arbitrary rational representation of G. We may assume that ρ(R) is
a diagonal group, whence the completes reducibility of the restriction
of ρ on S implies the complete reducibility of ρ, which completes the
proof.
Complete reducibility of rational representation....
61
6. Another result.
Let G be a connected algebraic linear group with universal domain
K, throughout this section.
Theorem 4. Every rational representation of G is completely reducible
if (and only if) the !following is true:
1 0
If ρ′′ =
is a rational representation of G, then ρ′′ is equiva0 ρ′
!
1 0
.
lent to the representation
0 ρ′
!
ρ1 τ
Proof. Letρ =
be a rational representation of G. We have only
0 ρ2
!
ρ1 0
to show that ρ is equivalent to the representation
.
0 ρ2
Since quad ρ(ab) = ρ(a)ρ(b), we have
(1)
τ(ab) = ρ1 (a)τ(b) + τ(a)ρ2 (b) for any a, b − G.
Let x be a generic point of G over K consider f (x) = T (x) ρ2 (x)−1 .
f (a) is then well defined for any a ∈ G. The relation (1) implies that
f (ab) = ρ1 (a)τ(1)ρ2 (b)−1 ρ2 (a)−1 +τ(a)ρ2 (a)−1 = ρ1 (a) f (b)ρ2 (a)−1 +
f (a) for any a, b ∈ G, whence
(2)
f (xa) = ρ1 (x) f (a) fa (x)−1 + f (x) for any a ∈ G.
Let m, n be such that T is an(m,n) matrix and consider the module
L of all (m, n)-matrices over K(x). Each element C of G defines an
K-linear map φg or. L as follows:
φg (wi j (x)) = (wi j (xg)).
Thus L becomes K − G - module. Let M be the set of all ρ1 (x)cρ2
(x)−1 with (m, n)-matrices c over K. Then M is a finite K-module
contained in L. Since ρ1 (xa)cρ2 (xa)−1 = ρ1 (x)(ρ1 (a)cρ2 (a)−1 , xρ2
74
62
75
Complete reducibility of rational representation....
(x)−1 (a ∈ G), M is G-admissible. Set N = f (x)K + M. Then the
relation (2) shows that N is also a finite K −G -module. We consider
a representation ρ∗ of G by the module N. The relation (2) shows
∗
that f (x) is G-invariant modulo M, hence either f (x)
! ∈ M or ρ
1 λ
is equivalent to a representation of the form
. The former
0 ρ′
case implies that f (x) + ρ1 (x)cρ(x)−1 = 0 with some (m, n)-matrix
c over K. By our assumption, the latter case implies that there is
an element ρ1 (x)cρ2 (x)−1 of M such that f (x) + ρ1 (x)c ρ2 (x)−1 is Ginvariant. Hence, in any case, there is an (m, n)-matrix c over K
such that f (x) + ρ1 (x)cρ2 (x)−1 is G-invariant. Set τ!∗ = τ − cρ2 + ρ1 c.
ρ (1)
c
Then, transforming ρ by the matrix 1
, we see that ρ is
0
ρ2 (1)
!
ρ τ∗
. Set f ∗ (x) = τ∗ (x)ρ2 (x)−1 .
equivalent to the representation 1
0 ρ2
Then f ∗ (x) = f (x) − c + ρ1 (x)cρ2 (x)−1 , which is G-invariant by
our choice of c. Therefore f ∗ (xa) = f ∗ (x) for any a ∈ G, whence
f ∗ (x) = f ∗ (xx−1 ) = 0. This shows that τ∗ = O, which completes the
proof of Theorem 4.
We note by the way that the matrix f (x) has an interesting property
as follows:
!
ρ1 τ
Proposition . Assume that ρ =
is a rational representation of
0 ρ2
G. Set H = {h|h ∈ G, τ(h) = 0}. Then the homogeneous variety G/H
= {gH} is a quasi-affine variety, on which the coordinates of a point gH
are given by f (g).
Proof. Since τ(1) = 1, the formula (1) in the above proof
shows that
−1
−1
−1
−1
−1
τ(a ) = −ρ1 (a) τ(a)ρ2 (a) , hence τ(a b) = ρ1 (a) τ(b)ρ2 (b)−1 −
τ(a)ρ2 (a)−1 ρ2 (b). There fore f (a) = f (b) if and only if aH = bH,
which prove the assertion.
Remark . Note that the above preposition only proves that G/H is a
quasi-affine and not affine as stated in M. Nagata: Complete reducibility
of rational representations, J. Math. hyeto Univ., 1-1 (1961), 87-99.