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Finitely Generated Algebras Associated with Rational Vector Fields
198, 481]498 Ž1997. JA977154
JOURNAL OF ALGEBRA ARTICLE NO.
Finitely Generated Algebras Associated with Rational Vector Fields Hisayo Aoki and Masayoshi Miyanishi Department of Mathematics, Graduate School of Science, Osaka Uni¨ ersity, Toyonaka, Osaka, 560, Japan Communicated by Craig Huneke Received December 9, 1996
INTRODUCTION Let k be an algebraically closed field of characteristic zero and R a polynomial ring in n variables over k. A normal k-subalgebra S of R which is finitely generated over k is called cofinite if R is integral over S. It is one of the central problems in affine algebraic geometry to consider the structures of cofinite normal k-subalgebras of R. If n s 2, it is known Žcf. w2x. that Spec S is isomorphic to the quotient of the affine plane A2k modulo a finite subgroup G of GLŽ2, k . acting linearly on A2k . In the case n G 3, nothing essential is known. In the present article, we propose one concrete method to construct a k-subalgebra of R. For R s k w x 1 , . . . , x n x, we consider a rational vector field
ds
1
f1 x1
q ??? q
1
fn x n
,
where f 1 , . . . , f n are homogeneous polynomials of degree m1 , . . . , m n , respectively, such that the set f 1 , . . . , f n4 is a system of parameters of the maximal ideal Ž x 1 , . . . , x n . of R. If n s 2 this is equivalent to saying that gcdŽ f 1 , f 2 . s 1. Let A be the k-subalgebra of R which is generated by all elements w of R such that d Ž w . g R, that is, A s k w g R dŽw. g R . In the present paper, we consider when A is finitely generated over k. Since most results will be stated in the case n s 2, we denote, for the sake 481 0021-8693r97 $25.00 Copyright Q 1997 by Academic Press All rights of reproduction in any form reserved.
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AOKI AND MIYANISHI
of simplifying the notations, two variables x 1 , x 2 by x, y, f 1 , f 2 by f, g and m1 , m 2 by m, n, respectively. Our main results are the following: THEOREM 1.10. Let A˜ be the integral closure of A in its quotient field. Then the following conditions are equi¨ alent: Ž1. Ž2. Ž3.
A is a finitely generated algebra o¨ er k. ˜ f, g g A. A˜ s k w x, y x.
THEOREM 3.1. Let A 0 s k w f i g j < i G 2, j G 2x. Assume that A l k w f, g x p A 0 and that deg f ¦ deg g and deg g ¦ deg f. If A is finitely generated o¨ er k then f M g A or g N g A for some positi¨ e integers M, N G 2. For polynomials w , c of R, we write w ; c if c s c w with a nonzero element c of k. In order to indicate that a polynomial w is the zero polynomial we denote w ' 0 and distinguish it from the equation w s 0. For relevant results on regular vector fields on A2 s Spec k w x, y x, we refer to w3x. 1. THE INTEGRAL CLOSURE A˜ OF A Note that the hypothesis that the set f 1 , . . . , f n4 is a system of parameters of Ž x 1 , . . . , x n . implies that any two of the f i have no common irreducible factors. We shall begin with the following result. LEMMA 1.1. For an element w of R, w belongs to A if and only if f i < w x i for e¨ ery 1 F i F n, where w x i s wr x i . Proof. If w g A, we write Ž1rf 1 . w x 1 q ??? qŽ1rf n . w x n s h with h g R. Then we have n
f 2 ??? f n w x 1 s f 1 f 2 ??? f n h y
ž
Ý f2
??? fˇi ??? f n w x i .
is2
/
Since gcdŽ f 1 , f 2 , . . . , f n . s 1, we obtain f 1 < w x 1. Similarly, f i < w x i for 2 F i F n. It is clear that w belongs to A if f i < w x i for every 1 F i F n. Q.E.D. LEMMA 1.2. Ž1. A is a graded ring. Ž2. If n s 2 the quotient field QŽ A. of A is k Ž x, y .. s Proof. Ž1. For w g A, we write w as w s Ý rs1 wr , where wr is the r th homogeneous part of w . Then f i < w x i if and only if f i N Ž wr . x i for any i,
483
FINITELY GENERATED ALGEBRAS
where 1 F i F n. Hence w g A if and only if wr g A for any r. So A is a graded ring. Ž2. Write f s l a f 1 , g s l9 b g 1 , where l, l9 are linear polynomials, Ž gcd l, f 1 . s 1, and gcdŽ l9, g 1 . s 1. Since gcdŽ f, g . s 1, it follows that kl q kl9 s kx q ky as k-vector spaces. Let
w i j s l i f 12 l9 j g 12
with i G a q 1 and j G b q 1.
Then it is straightforward to see that w i j is an element of A. Since l s w iq1, jrw i j
and
lX s w i , jq1rw i j ,
it follows that QŽ A. s k Ž x, y ..
Q.E.D.
The following result gives a sufficient condition that A be a finitely generated k-algebra. LEMMA 1.3. The following assertions hold: Ž1. There are some positi¨ e integers si G 2 Ž1 F i F n. with f is i g A if and only if, after a change of indices 1, 2, . . . , n4 which allows us to assume m1 F m 2 F ??? F m n , the f i are written in the forms f 1 ; x 1m 1 f 2 ; x 1m 1q1s 21 Ž x 1 , x 2 . q c 2 x 2m 2 f 3 ; x 1m 1q1s 31 Ž x 1 , x 2 , x 3 . q x 2m 2q1s 32 Ž x 2 , x 3 . q c 3 x 3m 3 ??? fn ;
x 1m 1q1sn1
Ž x 1 , . . . , x n . q x 2m 2q1sn2 Ž x 2 , . . . , x n .
m ny 1 q1 q ??? qx ny sn , ny1 Ž x ny1 , x n . q c n x nm n , 1
where c i g k* and the si j Ž x j , . . . , x i . are the homogeneous polynomials of degree m i y Ž m j q 1. satisfying the conditions deri¨ ed from the conditions f j Ž fi . x j
for all pairs Ž i , j . with 1 F j - i , 2 F i F n.
Ž2. If the f i are written in the abo¨ e forms in Ž1., then A is a finitely generated k-algebra. Proof. Ž1. Suppose f is i g A with si G 2 for all 1 F i F n. Then, by Lemma 1.1, f j N Ž f i . x j whenever i / j because gcdŽ f i , f j . s 1. This implies that m j F m i y 1 provided Ž f i . x j k 0. If the indices 1, . . . , n4 are changed so that m1 F m 2 F ??? F m n , this gives rise to the condition that Ž f i . x j ' 0 whenever j ) i. Namely, f i is a polynomial in x 1 , . . . , x i . So, f 1 ; x 1m 1 .
484
AOKI AND MIYANISHI
Since f 1 N Ž f 2 . x 1, we have f 2 s bx 1m 1r Ž x 1 , x 2 . dx 1 q cx 2m 2
H
; x 1m 1q1s 21 Ž x 1 , x 2 . q c 2 x 2m 2 . Suppose f iy1 is written as iy2
f iy1 ;
m y1 . Ý x jm q1siy1, j Ž x j , . . . , x iy1 . q c iy1 x iy1 j
i
js1
Since f iy1 N Ž f i . x iy 1, we can write iy1
fi ;
Ý x jm q1si j Ž x 1 , . . . , x i . q si i Ž x 1 , . . . , x iy2 , x i . , j
js1
where we may assume that the following condition is satisfied:
si j does not contain monomial terms whose x k-exponent is Ž ). greater than or equal to m k q 1 for any k with 1 F k - j. We show by induction on j that si j Ž x 1 , . . . , x i . is in fact a polynomial in x j , . . . , x i . It is easy to see that the condition x 1m 1 N Ž f i . x 1 implies iy1
x 1m 1
žÝ
js2
x jm jq1 Ž si j . x 1 q Ž sii . x 1 .
/
Since si j Ž x 1 , . . . , x i . Ž2 F j - i . contains no monomial terms whose x 1-exponent is greater than or equal to m1 q 1, it follows that Ž si j . x 1 ' 0 for 2 F j F i. Suppose si k , . . . , sii are polynomials in x k , . . . , x i for 1 F k - j. Write m jy 2 q1 Ž f i . x jy 1 ; x 1m 1q1 Ž si1 . x jy 1 q ??? qx jy2 Ž si , jy2 . x jy 1 m jy 1 q x jy1
žŽm
jy1
q 1 . si , jy1 q x j Ž si , jy1 . x jy 1
/
iy1
q Ý x km kq1 Ž si k . x jy 1 q Ž sii . x jy 1 . ksj
Note that f jy 1 N Ž f i . x j-1. Write Ž f i . x jy 1 s f jy1 h with h g R. In both sides of this equation, we equate, by making use of the condition Ž)., the terms divisible by x 1m 1q1 first, the terms divisible by x 2m 2q1 next in the remaining m jy 2 q1 terms, and continue this way until the terms divisible by x jy2 . Looking
485
FINITELY GENERATED ALGEBRAS
at the remaining terms, we then know that iy1
m jy 1 x jy1
žÝ
x km kq1 Ž si k . x jy 1 q Ž sii . x j
ksj
y1
/
.
It then follows by virtue of the condition Ž). that si j , . . . , sii are polynomials in x j , . . . , x i . We show that c i g k*. Indeed, if c i s 0 then Ž f 1 , . . . , f i . ; Ž x 1 , . . . , x iy1 ., which is impossible because f 1 , . . . , f n4 is a system of parameters of the ideal Ž x 1 , . . . , x n .. Hence c i g k*. Conversely, if f 1 , . . . , f n are written in the above forms and satisfy the condition that f j N Ž f i . x j for 1 F j - i and 2 F i F n, one can readily show that f 12 , . . . , f n2 g A. Ž2. With the expressions of f 1 , . . . , f n in Ž1. above, define a k-subalgebra B of A by B s k f 12 , f 22 , . . . , f n2 . Let B˜ be the integral closure of B in k Ž x 1 , . . . , x n . and let Bˆ s k w x 1 , f 2 , . . . , f n x. Then Bˆ > B and Bˆ is integral over B. Furthermore, it is easy to show that x 1 , . . . , x n are integral over Bˆ because c i g k* for 2 F i F n. Hence x 1 , . . . , x n are integral over B and B˜ s R. Note that B˜ is a finitely generated B-module and B˜ contains A as a B-submodule. Since B is a Noetherian ring, A is a finitely generated B-module. Hence A is a finitely generated k-algebra. Q.E.D. We define a k-subalgebra A 0 of A by A 0 s k f 1i1 f 2i 2 ??? f ni n i j G 2, 1 F j F n . In connection with Lemma 1.3, we wish to show that f is i g A for some positive integer si G 2 Ž1 F i F n. provided A contains A 0 and A is a finitely generated k-algebra. Modifying the settings a bit, we consider the following problem. Problem 1.4. Let p1 , . . . , pn be homogeneous polynomials of R of respective degrees m1 , . . . , m n such that p1 , . . . , pn4 is a system of parameters of the ideal Ž x 1 , . . . , x n .. Let C be a finitely generated graded k-subalgebra of R which contains the subalgebra C0 s k p1i1 p 2i 2 ??? pni n i j G 1, 1 F j F n . Then, does it follow that pis i g C for some positive integers si , 1 F i F n?
486
AOKI AND MIYANISHI
With the notations of Lemma 1.3, if we take C s A and pi s f i2 Ž1 F i F n., then A > k w p1i1 ??? pni n < i j G 1, 1 F j F n x. If we can answer Problem 1.4 affirmatively, we obtain f is i g A and the converse of the assertion Ž2. of Lemma 1.3 will hold. Problem 1.4 has, however, a counterexample which will be given in Proposition 1.11. Furthermore, we note that tr.deg k k Ž f 1 , . . . , f n . s n since f 1 , . . . , f n4 is a system of parameters of Ž x 1 , . . . , x n .. THEOREM 1.5. Let C˜ be the integral closure of C in its quotient field QŽ C .. Then we ha¨ e p1 , . . . , pn g C˜ pro¨ ided C˜ l k Ž p1 , . . . , pn . is finitely generated o¨ er k. Note that the condition in Theorem 1.5 is satisfied when n s 2 Žcf. w4x.. In order to prove Theorem 1.5, we may and shall replace, if necessary, p1 , . . . , pn by their powers p1u1 , . . . , pnu n and assume m1 s ??? s m n . Our proof consists of several lemmas. The following result is well known. ŽSee, for example, Gurjar w1, Lemma 3x for a geometric proof in the case n s 2.. LEMMA 1.6.
R s k w x 1 , . . . , x n x is integral o¨ er k w p1 , . . . , pn x.
Proof. Set S s k w p1 , . . . , pn x. By the hypothesis that p1 , . . . , pn4 is a system of parameters of Ž x 1 , . . . , x n ., RrŽ p1 , . . . , pn . R is a finite-dimensional k-vector space. Let w 1 , . . . , w N 4 be a set of elements of R such that the set w 1 , . . . , wn4 of the residue classes of w 1 , . . . , wn modulo Ž p1 , . . . , pn . R is a k-basis of RrŽ p1 , . . . , pn . R. We shall show that R is generated by w 1 , . . . , wn as an S-module. Let c be a homogeneous polynomial of degree, say r, in R. Then one can write N
cs
n
Ý a i w i q Ý pj c j , is1
js1
where a i g k Ž1 F i F N . and the c j Ž1 F j F n. are homogeneous polynomials of degree less than r s deg c . By induction on r, every c j is written as N
cj s
Ý h jl w l ,
where h jl g S.
ls1
Then we have N
cs
Ý ls1
½
n
al q
Ý pj h jl js1
5
wl .
Hence R is a finite S-module generated by w 1 , . . . , wn . So, R is integral over S. Q.E.D.
FINITELY GENERATED ALGEBRAS
487
LEMMA 1.7. Let w , c be homogeneous polynomials of R such that gcdŽ w , c . s 1 in R. Then gcdŽ w Ž p1 , . . . , pn ., c Ž p1 , . . . , pn .. s 1 in R. Proof. Write w Ž p . s w Ž p1 , . . . , pn . and c Ž p . s c Ž p1 , . . . , pn .. Suppose to the contrary that gcdŽ w Ž p ., c Ž p .. / 1. Then there exists a prime ideal P of height 1 of R such that w Ž p . ; P and c Ž p . ; P. Let p s P l S, where S s k w p1 , . . . , pn x. Since R is integral over S by Lemma 1.6, the going-down theorem implies that p is a prime ideal of height 1. Furthermore, p contains w 9 and c 9, which are the copies of w and c , respectively, and considered as elements of S. This contradicts the hypothesis that gcdŽ w , c . s 1. Q.E.D. LEMMA 1.8. The following assertions hold: Ž1. Ž2.
C˜ l k Ž p1 , . . . , pn . s C˜ l k w p1 , . . . , pn x. C l k Ž p1 , . . . , pn . s C l k w p1 , . . . , pn x.
Proof. We shall prove only the assertion Ž2.. The assertion Ž1. can be proved in a similar fashion. Note first that C l k Ž p1 , . . . , pn . is a graded ring since C is a graded ring. In fact, let j be any element of C l k Ž p1 , . . . , pn . and let
j s j n q j nq1 q ??? qj l
with j i g C
be the homogeneous decomposition. Then we have j s c Ž p .rw Ž p ., where w Ž p ., c Ž p . g k w p1 , . . . , pn x. Let the homogeneous decompositions of w Ž p . and c Ž p . be
c Ž p . s cc Ž p . q ccq1 Ž p . q ??? qcd Ž p . w Ž p . s we Ž p . q weq1 Ž p . q ??? qw f Ž p . . Since
Ž j n q j nq1 q ??? qj l . Ž we q weq1 q ??? qw f . s cc q ccq1 q ??? qcd , we obtain the relations
j n we s cc , j nq1 we q j n weq1 s ccq1 , . . . , j l w f s cd . Then it follows that j n s ccrwe , j nq1 s Ž ccq1 we y cc weq1 .rwe2 , . . . , j l s cdrw f . They are elements of C l k Ž p1 , . . . , pn .. Hence C l k Ž p1 , . . . , pn . is a graded ring. We shall show the assertion Ž2.. Let j s c Ž p .rw Ž p . be any homogeneous element of C l k Ž p1 , . . . , pn ., where w , c are homogeneous polynomials in k w p1 , . . . , pn x such that gcdŽ w , c . s 1. Since j is an element of k w x 1 , . . . , x n x and gcdŽ w Ž p ., c Ž p .. s 1 in k w x 1 , . . . , x n x by Lemma 1.7, w Ž p . must be a constant. Hence j is an element of C l k w p1 , . . . , pn x. Q.E.D.
488
AOKI AND MIYANISHI
Note that tr.deg k k Ž p1 , . . . , pn . s n. Now we assume that C˜ l k Ž p1 , . . . , pn . is a finitely generated k-algebra. If n s 2, a theorem of Zariski w4x says that C˜ l k Ž p1 , p 2 . is finitely generated over k. We replace C˜ by C˜ l k Ž p1 , . . . , pn . and assume that k w p1 , . . . , pn x > C˜ > C0 [ k p1i1 ??? pni n i j ) 0, 1 F j F n . On the other hand, it is clear that Q Ž k w p1 , . . . , pn x . s Q Ž C0 . s k Ž p1 , . . . , pn . . Then, in order to prove Theorem 1.5, it suffices to show the following result. LEMMA 1.9. C˜ s k w p1 , . . . , pn x. Proof. Since C˜ is finitely generated over k, let w 1 , . . . , wr be homogeneous polynomials of degree d i which generate C˜ over k. We need the following auxiliary result. CLAIM.
'Ž w Ž p . , . . . , w Ž p . . R s Ž x , . . . , x .. r
1
n
1
Proof. Let P be a prime divisor of Ž w 1Ž p ., . . . , wr Ž p .. R and let p s P l S, where S s k w p1 , . . . , pn x. Then w 1Ž p ., . . . , wr Ž p . g p. We have only to show that p1 , . . . , pn g p. In fact, p has then height n and P has therefore height n because R is integral over S. Let Ž O , m . be a discrete valuation ring of the quotient field k Ž p1 , . . . , pn . such that Ž O , m . dominates Ž Sp , p Sp ., and let ¨ be the associated valuation. Then ¨ Ž w 1Ž p .. ) 0, . . . , ¨ Ž wr Ž p .. ) 0. Since p1N p 2 ??? pn is an element of C0 , we can write p1N p 2 ??? pn s FN Ž w 1 Ž p . , . . . , wr Ž p . . s
Ý
a 1 d 1q ??? q a r d rsNqny1
ca w 1 Ž p .
a1
??? wr Ž p .
ar
,
where FN is a weighted homogeneous polynomial. If N tends to infinity, at least one of a 1 , . . . , a r should tend to infinity. Hence ¨ Ž p1N p 2 ??? pn . tends to infinity as N tends to infinity. This implies that ¨ Ž p1 . ) 0. Hence p1 g p. Similarly, p 2 , . . . , pn g p. Q.E.D. Now let us return to the proof of Lemma 1.9. By the above claim, we know that RrŽ w 1Ž p ., . . . , wr Ž p .. R is a finite-dimensional k-algebra. Since C˜ is a graded subalgebra of k w x 1 , . . . , x n x, we conclude by the same argument as in the proof of Lemma 1.6 that k w x 1 , . . . , x n x is a finitely ˜ kw x1, . . . , x n x generated C˜ module. Since k w x 1 , . . . , x n x > k w p1 , . . . , pn x > C, ˜ contains k w p1 , . . . , pn x as a C-submodule. Since C˜ is a Noetherian ring,
FINITELY GENERATED ALGEBRAS
k w p1 , . . . , pn x is a finitely ˜ Since integral over C. k w p1 , . . . , pn x. This implies proved. Now we shall state and paper.
489
˜ generated C-module. Hence k w p1 , . . . , pn x is QŽ k w p1 , . . . , pn x. s QŽ C˜., we have C˜ s ˜ and Theorem 1.5 is thus that p1 , . . . , pn g C, prove one of the main results of the present
THEOREM 1.10. Suppose n s 2. With the same notations and assumptions as in the Introduction, let A˜ be the integral closure of A in its quotient field. Then the following conditions are equi¨ alent: Ž1. Ž2. Ž3.
A is finitely generated o¨ er k. ˜ f, g g A. A˜ s k w x, y x.
Proof. The condition Ž1. implies the condition Ž2. by Theorem 1.5. ˜ Note that the condition gcdŽ p1 , p 2 . s 1 is the only Suppose that f, g g A. condition which is necessary in the proof of Lemma 1.6. Hence the hypothesis gcdŽ f, g . s 1 implies that k w x, y x is integral over k w f, g x. Hence k w x, y x is integral over A. In view of Lemma 1.2, this implies that A˜ s k w x, ˜ Thus the conditions y x. Conversely, if A˜ s k w x, y x it is clear that f, g g A. Ž2. and Ž3. are equivalent. We shall show that the condition Ž3. implies the condition Ž1.. Let F Ž x . s x M q c1 x My 1 q ??? qc My1 x q c M s 0 C Ž y . s y N q d1 y Ny 1 q ??? qd Ny1 y q d N s 0 be the integral relations of x and y over A, where c1 , . . . , c M , d1 , . . . , d N g A. Let C s k w c1 , . . . , c M , d1 , . . . , d N x. Then k w x, y x is a finitely generated C-module and A is its C-submodule. Hence A is a finitely generated k-algebra. Q.E.D. Problem 1.4 itself has the following counter-example in dimension 2. PROPOSITION 1.11. Let C be a subalgebra of k w x, y x defined as C s k w x q y, x i y j < ; i ) 0,;j ) 0x. Then C has the following properties, Ž1. Ž2. Ž3.
C > C0 [ k w x i y j < i ) 0, j ) 0x. C is finitely generated o¨ er k. C does not contain x s, y t for e¨ ery s ) 0 and e¨ ery t ) 0.
Proof. Ž1. It is clear. Ž2. Define a subalgebra C1 of C as C1 s k w x q y, xy, x 2 y x. We shall show that C s C1. First we prove x n y g C1 by induction on n. If n s 1, 2,
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AOKI AND MIYANISHI
this holds clearly. Suppose that this holds for x i y with i F n y 1, that is, x ny 1 y, x ny2 y, . . . , xy g C1. Since x n y s x ny1 y Ž x q y . y xy ? x ny 2 y, we then have x n y g C1 , where n G 3. Suppose n G r G 2. Since n
Ž xy . , x y s Ž xy . Ž x ny 1 y ry1 . , n
½
r
nsr n)r
we have x n y r g C1 by induction on r. Finally, when n F r, we suppose that x n y r 9 g C1 for r 9 with n F r 9 - r. Since x n y r s x n y ry1 Ž x q y . y x nq 1 y ry1 g C1 , we have x n y r g C1 by induction on r. Hence C s C1. Ž3. Note that C is a graded subring of k w x, y x. If x s g C, we may write s
x s s a s00 Ž x q y . q
a
Ý
aq2 b q3g ss b )0 or g )0
aa bg Ž x q y . Ž xy .
b
Ž x2 y.
b
Ž x2 y.
g
,
where a s00 / 0. Then we have with r g k w x, y x .
a s00 y s s x r
This is a contradiction. Similarly, if y t g C, write t
y t s a t 00 Ž x q y . q
Ý
a
aq2 b q3g st b )0 or g )0
aa bg Ž x q y . Ž xy .
g
,
where a t 00 / 0. Then we have a t 00 x t s yu
with u g k w x, y x .
This is a contradiction as well.
Q.E.D.
2. CRITERIA FOR THE FINITE GENERATION OF A In this section, we treat the case n s 2 and consider whether or not A is finitely generated over k for some specific pairs of f, g. These results will be used in the next section. Here we write A as AŽ f, g . whenever it is necessary to recall the vector field d s 1rf r x q 1rg r y. We shall begin with the following result. LEMMA 2.1. Ž1.
For positi¨ e integers m, n, we ha¨ e
A Ž x m , y n . s k x mqi , y nqj 1 F i F m q 1, 1 F j F n q 1 . Hence AŽ x m , y n . is finitely generated o¨ er k.
FINITELY GENERATED ALGEBRAS
491
Ž2. We ha¨ e A Ž y, x . s k x i y j i G 1, j G 1 . Then A is not finitely generated o¨ er k. Proof. Ž1. A homogeneous polynomial w belongs to A if and only if x m < w x and y n < w y . So, w g A if and only if w is a sum of monomials ax s y t with a g k such that s s 0 or s G m q 1 as well as t s 0 or t G n q 1. Then it is easy to see that A is generated by the x mq i and the y nq j with 1 F i F m q 1 and 1 F j F n q 1. Ž2. It is straightforward to show that A is generated by the elements i j x y with i G 1 and j G 1. Then A is not finitely generated because A contains x i y with i 4 0. Q.E.D. In the following two lemmas, we shall consider the case where f and g are linear homogeneous polynomials. LEMMA 2.2. The following assertions hold true: Ž1.
Let f s y and g s y y a x with a g k*. Then we ha¨ e
A Ž y, y y a x . s k u i ¨ j Ž i G 1, j G 2 . , u n y nu ny1 ¨ Ž n G 2 . , where u s y and ¨ s y y a x. The subalgebra A is not finitely generated o¨ er k. Ž2. Let f s x and g s y y a x with a g k*. Then we ha¨ e A Ž x, y y a x . s k u i ¨ j Ž i G 2, j G 2 . , ¨ n qn a u¨ ny 1 Ž n G 3 . , u n Ž n G 2 . , where u s x and ¨ s y y a x. The subalgebra A is then finitely generated o¨ er k. Ž3. Let f s y y a x and g s x. Then we ha¨ e A Ž y y a x, x . s k u i ¨ j Ž i G 2, j G 1 . , ¨ n y nu¨ ny1 Ž n G 2 . , where u s y y a x and ¨ s x. The subalgebra A is not finitely generated o¨ er k. Ž4. Let f s y y a x and g s y with a g k*. Then we ha¨ e A Ž y y a x, y . s k u i ¨ j Ž i G 2, j G 2 . , u n ynu ny 1 ¨ Ž n G 3 . , ¨ n Ž n G 2 . , where u s y y a x and ¨ s y. The subalgebra A is finitely generated o¨ er k.
492
AOKI AND MIYANISHI
Proof. We prove the assertions Ž1. and Ž2.. The assertions Ž3. and Ž4. are verified in the same fashion as for Ž1. and Ž2., respectively. Ž1. By the change of variables u s y, ¨ s y y a x, the vector field d is written as
ds
1 ¨ u
q
u y a¨ u¨
¨
.
Let w g R [ k w x, y x s k w u, ¨ x be a homogeneous polynomial of degree n and write it as
ws
ai j u i ¨ j .
Ý iqjsn
Then we have
dŽw. s
1 u¨
Ž nan0 q any1, 1 . u n q u¨ c y n a a0 n¨ n 4 ,
where c is a homogeneous polynomial of degree n y 2. Hence d Ž w . g R if and only if na0 n q a ny1, 1 s 0 and a0 n s 0. Namely, w s a20 Ž u 2 y 2 u¨ . if n s 2 and
w s a n0 Ž u n y nu ny 1 ¨ . q
a i j u i ¨ j q a1, ny1 u¨ ny 1
Ý iqjsn i , jG2
if n G 3. Hence A is generated by the elements as in the statement. Since A contains u¨ j Ž j 4 0., the subalgebra A is not finitely generated over k. Ž2. By the change of variables u s x and ¨ s y y a x, the vector field d is written as
ds
1 u u
q
u y a¨ u¨
¨
.
So, if w s Ý iqjsn a i j u i ¨ j is a homogeneous polynomial of degree n, d Ž w . is computed as
dŽw. s
1 u¨
any 1, 1 u n q u¨ c q Ž a1, ny1 y n a a0 n . ¨ n 4 ,
493
FINITELY GENERATED ALGEBRAS
where c is a homogeneous polynomial of degree n y 2. Hence d Ž w . g R if and only if a ny 1, 1 s 0 and a1, ny1 y n a a0 n s 0. The last condition implies that the subalgebra A is generated by the elements as given in the statement. In order to prove that the subalgebra A is finitely generated over k, we first show that u, ¨ are integral over A. It is clear that u is integral over A. The element ¨ is integral over A since there is the following integral relation with coefficients in A: ¨ 7 y Ž ¨ 3 q 3 a u¨ 2 . ¨ 4 y Ž ¨ 4 q 4a u¨ 3 . ¨ 3 q Ž ¨ 7 q 7a u¨ 6 . s 0.
Then R is a finite B-module, where B s k w u 2 , ¨ 3 q 3 a u¨ 2 , ¨ 4 q 4a u¨ 3 , ¨ 7 q 7a u¨ 6 x which is a subalgebra of A. Then A is a finite B-module, and A is therefore finitely generated over k. Q.E.D. The following result will complement Lemma 2.2. Let f s y y a x and g s y y b x with a , b g k* and a / b .
LEMMA 2.3. Then we ha¨ e
A Ž y y a x, y y b x . s k u 2 y 2 u¨ q
a b
b
¨ 2 , u n y nu ny1 ¨ , ¨ n y
a
n¨ ny1 u Ž n G 3 . ,
u i ¨ j Ž i G 2, j G 2 . , where u s y y a x and ¨ s y y b x. The subalgebra A is then finitely generated o¨ er k. Proof. By the change of variables, the vector field d is written as
ds
ž
u y a¨ u¨
q
/ ž u
u y b¨ u¨
/
¨
.
Let w s Ý iqjsn a i j u i ¨ j be a homogeneous polynomial of degree n G 2 in R [ k w x, y x s k w u, ¨ x. We then compute
dŽw. s
1 u¨
Ž nan0 q any1 , 1 . u n q u¨ c y Ž a a1, ny1 q n b a0 n . ¨ n 4 ,
where c is a homogeneous polynomial of degree n y 2. Hence d Ž w . g R if and only if nan0 q a ny1, 1 s 0 and a a1, ny1 q n b a0 n s 0. The last condition implies that the subalgebra A is generated by the elements as given in the statement. The element ¨ is integral over A since there is a monic relation in ¨ with coefficients in A: ¨ 11 y ¨ 5 y
ž
5b
a
¨ 4u ¨ 6 y ¨6 y
/ ž
6b
a
¨ 5 u ¨ 5 q ¨ 11 y
/ ž
11 b
a
¨ 10 u s 0.
/
494
AOKI AND MIYANISHI
Then the element u is integral over Aw ¨ x, hence over A because there is a monic relation
a
u2 y 2 ¨ u q
b
a
¨ 2 y u 2 y 2 u¨ q
ž
b
¨ 2 s 0.
/
Then it follows that A is finitely generated over k.
Q.E.D.
Lemma 2.3 implies that the polynomial ring contains a one-parameter family of two-dimensional subalgebras A t 4t g k such that A t is finitely generated over k for each value of t / 0, b and A 0 is not finitely generated over k, where b is some fixed element of k. In fact, we have only to take A t to be AŽ y y tx, y y b x .. Then the assertion will follow from Lemmas 2.2 and 2.3. In Section 3, we need the following result which is a variant of Lemma 2.2. LEMMA 2.4. The algebra A s AŽ f, g . is finitely generated o¨ er k in each of the following cases, where s G 2 and t G 2 are integers: Ž1. Ž2. Ž3.
f s x s and g s Ž y y a x . t with a g k*. f s Ž y y a x . s and g s y t with a g k*. f s Ž y y a x . s and g s Ž y y b x . t with a , b g k* and a / b .
Proof. We prove only the case Ž3., as the other cases can be treated similarly. Let u s y y a x and ¨ s y y b x. The vector field d is then written as
ds
ž
us y a ¨ t u ¨ s
t
/
q
u
ž
us y b ¨ t u ¨ s
t
/
¨
.
A computation as in the proof of Lemma 2.2 shows that the following elements are contained in A, t
w Ž m . s u myt u t ym C1 u ty1 ¨ qm C2 u ty2 ¨ 2 q ??? q Ž y1 . m Ct¨ t 4
½
c Ž n . s ¨ ny s ¨ s yn C1
b
ž / a
¨ sy1 u qn C2
b
ž / a
s
q ??? q Ž y1 . n C s
2
¨ sy2 u 2
b
s
ž / 5 a
us ,
where m Ci and n C j are the binomial coefficients and m, n G s q t q 2. In order to show that A is finitely generated over k, it suffices to find two elements w Ž m. and c Ž n. which are prime to each other. Since w Ž m. and
495
FINITELY GENERATED ALGEBRAS
c Ž n. are homogeneous in u, ¨ , it suffices to show that there are no elements j in k satisfying the equations t
j t ym C1 j ty1 qm C2 j ty2 q ??? q Ž y1 . m Ct s 0 1 yn C1
b
ž / a
j qn C2
b
ž / a
2 s
j q ??? q Ž y1 . n C s 2
b
ž / a
s
jss0
provided m and n are sufficiently large and mutually independent. This assertion can be easily verified. Q.E.D. Suppose that f s f 1 f 2 with deg f 1 ) 0, deg f 2 ) 0, and gcdŽ f 1 , f 2 . s 1. Then it is easy to verify that AŽ f 1 , g . l AŽ f 2 , g . s AŽ f, g .. Similarly, AŽ f, g 1 g 2 . s AŽ f, g 1 . l AŽ f, g 2 . if g s g 1 g 2 with deg g 1 ) 0, deg g 2 ) 0, and gcdŽ g 1 , g 2 . s 1. This easy remark entails the following result. LEMMA 2.5. Suppose that either y < f or x < g. Then A s AŽ f, g . is not finitely generated o¨ er k. Proof. Suppose, to the contrary, that A is finitely generated over k. Then the integral closure A˜ of A in its quotient field is finitely generated and contains f, g by Theorem 1.10. Hence k w x, y x is integral over A since k w x, y x is integral over k w f, g x by Lemma 1.6. Suppose y < f. Then we have A ; A Ž y, y y a x . ; k w x, y x , where y y a x is a linear factor of g and hence a g k*. Hence AŽ y, y y a x . is finitely generated over k. This is, however, a contradiction because AŽ y, y y a x . is not finitely generated over k by Lemma 2.2Ž1.. The case x < g is proved in a similar way. Q.E.D. This lemma has immediate applications. COROLLARY 2.6. true:
For positi¨ e integers m, n, the following assertions hold
Ž1. AŽ y m , x n . is not finitely generated o¨ er k. Ž2. AŽ y m , y n q ax n . is not finitely generated o¨ er k, where a g k*. Ž3. AŽ y m q bx m , x n . is not finitely generated o¨ er k, where b g k*. In connection with these obser¨ ations, we pose the following problem. Problem 2.7. Let f, g be as above. Suppose f s f 1 f 2 with deg f 1 ) 0 and deg f 2 ) 0. Suppose, furthermore, that AŽ f 1 , g . and AŽ f 2 , g . are finitely generated over k. Is AŽ f, g . then finitely generated over k?
496
AOKI AND MIYANISHI
3. A SUFFICIENT CONDITION FOR POWERS OF f AND g ARE CONTAINED IN A Let f, g be as above and let A 0 be the k-subalgebra defined by all elements f i g j with i G 2 and j G 2. Namely, A 0 s k f i g j < i G 2, j G 2 . Our objective is to prove the following result. THEOREM 3.1. Let A 0 s k w f i g j < i G 2, j G 2x. Assume that A l k w f, g x p A 0 and that deg f ¦ deg g and deg g ¦ deg f. Then if A is finitely generated o¨ er k then f M g A or g N g A for some positi¨ e integers M, N G 2. The proof will be given in this section. In what follows, we assume that A is finitely generated, A l k w f, g x p A 0 , and m ¦ n and n ¦ m, where m s deg f and n s deg g. Furthermore, we assume that f M f A and g N f A for all M G 2 and all N G 2. We shall show that this assumption leads to a contradiction. Write m s m9d and n s n9d with d s gcdŽ m, n.. Then the hypothesis A l k w f, g x p A 0 implies that A contains a nonzero homogeneous element F of the form F s af a q bf b g q cfg g q dg d ,
Ž 1.
where a, b, c, d g k and a , b , g , d are positive integers. The following two lemmas will finish the proof of Theorem 3.1. LEMMA 3.2. With the abo¨ e notations and assumptions, F is not of the form F ; f a , F ; f b g, F ; fg g , or F ; g d. Proof. By the hypothesis, it is clear that F is not of the form F ; f a or F ; g d. Assume that F ; f b g. Then we have Fx ; b f by1 f x g q f b g x
and
Fy ; b f by1 f y g q f b g y
and g < Fy implies g < g y . Hence g y ' 0. So g ; x n. Then A is not finitely generated by Lemma 2.5. Similarly, the case F ; fg g is excluded. Q.E.D. LEMMA 3.3. With the abo¨ e notations and assumptions, F does not contain two or more terms among f a , f b g, fg g , and g g. Proof. Our proof proceeds by the comparison of the degrees of these terms, which are given as deg f a s m a s dm9a ,
deg f b g s m b q n s d Ž m9b q n9 . ,
deg fg g s m q ng s d Ž m9 q n9g . ,
deg g d s n d s dn9d .
Ž 2.
FINITELY GENERATED ALGEBRAS
497
In view of the assumption that m ¦ n and n ¦ m, it follows that m9 ) 1 and n9 ) 1. First we assume that a / 0. Ži. Suppose ab / 0. Note that F is a homogeneous polynomial. By comparison of degrees in Ž2., we have m9a s m9b q n9, which yields m9Ž a y b . s n9. Since gcdŽ m9, n9. s 1, this is a contradiction. Hence b s 0 if a / 0. By exchanging the roles of a and d, we conclude that cd s 0, where we do not have to assume a / 0. Žii. Suppose ac / 0. By comparison of degrees, we have m9a s m9 q n9g . Hence a s n9l q 1 and g s m9l for some integer l G 0. Then we can write F ; f n9lq1 q cfg m9l
with c g k*.
If l s 0 then F ; f and f g A. This case is excluded by the hypothesis. So l ) 0. Then we have Fx ; Ž n9l q 1 . f n9l f x q cf x g m9l q cm9lfg m9ly1 g x . Since f < Fx Žcf. Lemma 1.1., we have f < f x . So f x s 0 and f ; y m . This is impossible by Lemma 2.5. Žiii. Suppose ad / 0. By comparison of degrees, we have m9a s n9d . Hence we may write a s n9l and d s m9l for some positive integer l. Then F is written as F ; f n9l q dg m9l
with d g k*
Ž 3.
which gives rise to Fx ; n9lf n9ly1 f x q dm9lg m9ly1 g x , Fy ; n9lf n9ly1 f y q dm9lg m9ly1 g y , where f < Fx implies f < g x , and g < Fy implies g < f y . If g x k 0 and f y k 0, the comparison of degrees gives m F n y 1 and n F m y 1. This is a contradiction. So, f y ' 0 or g x ' 0. If f y ' 0 then f ; x m and f 2 g A, which contradicts the assumption. Similarly, g 2 g A if g x ' 0, which is a contradiction. By Ži. ] Žiii. above, we conclude that a s 0. By exchanging the roles of f and g, we conclude also that d s 0. It then remains to consider the following case. Živ. Suppose bc / 0. The comparison of degrees implies m9b q n9 s m9 q n9g , which yields m9Ž b y 1. s n9Žg y 1.. Since gcdŽ m9, n9. s 1, we can write b s n9l q 1 and g s m9l q 1 for some integer l G 0. Then we have with c g k*, F ; f n9lq1 g q cfg m9lq1 Fx ; Ž n9l q 1 . f n9l f x g q f n9lq1 g x q cf x g m9lq1 q c Ž m9l q 1 . fg m9l g x .
498
AOKI AND MIYANISHI
If l s 0 then F ; fg, and this is impossible by Lemma 3.2. So, l ) 0. Then f < Fx implies f < f x . Hence f x ' 0 and f ; y m . This is impossible by Lemma 2.5. Q.E.D. Thus we have proved Theorem 3.1. Remark 3.4. Ž1. The assumption that deg f ¦ deg g and deg g ¦ deg f cannot be dropped in the assertion Ž1. of Theorem 3.1. Indeed, let f s y y a x and g s y y b x with a , b g k* and a / b as in Lemma 2.3. Then A l k w f, g x p A 0 , and f M f A and g N f A for any M, N G 2. Ž2. Notwithstanding the above remark, the arguments in the proof of Lemma 3.3 work also in the cases where Ža. m9 ) 1 and n9 s 1, Žb. m9 s 1 and n9 ) 1, and Žc. m9 s n9 s 1 except for the following point. Namely, one cannot conclude that the k-subalgebra A is not finitely generated if f ; y m q ag m r n or g ; x n q bf n r m with a, b g k*. We do not know, however, whether or not A is finitely generated if f ; y m q ag m r n or g ; x n q bf n r m with a, b g k*.
REFERENCES 1. R. V. Gurjar, Graded subrings of Cw X, Y x, Proc. Indian Acad. Sci. Math. Sci. 99 Ž1989., 209]215. 2. M. Miyanishi, Normal affine subalgebras of a polynomial ring, in ‘‘Algebraic and Topological Theories}To the Memory of Dr. Takehiko Miyata, Kinokuniya, Tokyo, 1985,’’ pp. 37]51. 3. M. Miyanishi, Vector fields on factorial schemes, J. Algebra 173 Ž1995., 144]165. 4. M. Nagata, On the fourteenth problem of Hilbert, in Lecture Notes at Tata Institute of Fundamental Research, Bombay, 1965.
JOURNAL OF ALGEBRA ARTICLE NO.
Finitely Generated Algebras Associated with Rational Vector Fields Hisayo Aoki and Masayoshi Miyanishi Department of Mathematics, Graduate School of Science, Osaka Uni¨ ersity, Toyonaka, Osaka, 560, Japan Communicated by Craig Huneke Received December 9, 1996
INTRODUCTION Let k be an algebraically closed field of characteristic zero and R a polynomial ring in n variables over k. A normal k-subalgebra S of R which is finitely generated over k is called cofinite if R is integral over S. It is one of the central problems in affine algebraic geometry to consider the structures of cofinite normal k-subalgebras of R. If n s 2, it is known Žcf. w2x. that Spec S is isomorphic to the quotient of the affine plane A2k modulo a finite subgroup G of GLŽ2, k . acting linearly on A2k . In the case n G 3, nothing essential is known. In the present article, we propose one concrete method to construct a k-subalgebra of R. For R s k w x 1 , . . . , x n x, we consider a rational vector field
ds
1
f1 x1
q ??? q
1
fn x n
,
where f 1 , . . . , f n are homogeneous polynomials of degree m1 , . . . , m n , respectively, such that the set f 1 , . . . , f n4 is a system of parameters of the maximal ideal Ž x 1 , . . . , x n . of R. If n s 2 this is equivalent to saying that gcdŽ f 1 , f 2 . s 1. Let A be the k-subalgebra of R which is generated by all elements w of R such that d Ž w . g R, that is, A s k w g R dŽw. g R . In the present paper, we consider when A is finitely generated over k. Since most results will be stated in the case n s 2, we denote, for the sake 481 0021-8693r97 $25.00 Copyright Q 1997 by Academic Press All rights of reproduction in any form reserved.
482
AOKI AND MIYANISHI
of simplifying the notations, two variables x 1 , x 2 by x, y, f 1 , f 2 by f, g and m1 , m 2 by m, n, respectively. Our main results are the following: THEOREM 1.10. Let A˜ be the integral closure of A in its quotient field. Then the following conditions are equi¨ alent: Ž1. Ž2. Ž3.
A is a finitely generated algebra o¨ er k. ˜ f, g g A. A˜ s k w x, y x.
THEOREM 3.1. Let A 0 s k w f i g j < i G 2, j G 2x. Assume that A l k w f, g x p A 0 and that deg f ¦ deg g and deg g ¦ deg f. If A is finitely generated o¨ er k then f M g A or g N g A for some positi¨ e integers M, N G 2. For polynomials w , c of R, we write w ; c if c s c w with a nonzero element c of k. In order to indicate that a polynomial w is the zero polynomial we denote w ' 0 and distinguish it from the equation w s 0. For relevant results on regular vector fields on A2 s Spec k w x, y x, we refer to w3x. 1. THE INTEGRAL CLOSURE A˜ OF A Note that the hypothesis that the set f 1 , . . . , f n4 is a system of parameters of Ž x 1 , . . . , x n . implies that any two of the f i have no common irreducible factors. We shall begin with the following result. LEMMA 1.1. For an element w of R, w belongs to A if and only if f i < w x i for e¨ ery 1 F i F n, where w x i s wr x i . Proof. If w g A, we write Ž1rf 1 . w x 1 q ??? qŽ1rf n . w x n s h with h g R. Then we have n
f 2 ??? f n w x 1 s f 1 f 2 ??? f n h y
ž
Ý f2
??? fˇi ??? f n w x i .
is2
/
Since gcdŽ f 1 , f 2 , . . . , f n . s 1, we obtain f 1 < w x 1. Similarly, f i < w x i for 2 F i F n. It is clear that w belongs to A if f i < w x i for every 1 F i F n. Q.E.D. LEMMA 1.2. Ž1. A is a graded ring. Ž2. If n s 2 the quotient field QŽ A. of A is k Ž x, y .. s Proof. Ž1. For w g A, we write w as w s Ý rs1 wr , where wr is the r th homogeneous part of w . Then f i < w x i if and only if f i N Ž wr . x i for any i,
483
FINITELY GENERATED ALGEBRAS
where 1 F i F n. Hence w g A if and only if wr g A for any r. So A is a graded ring. Ž2. Write f s l a f 1 , g s l9 b g 1 , where l, l9 are linear polynomials, Ž gcd l, f 1 . s 1, and gcdŽ l9, g 1 . s 1. Since gcdŽ f, g . s 1, it follows that kl q kl9 s kx q ky as k-vector spaces. Let
w i j s l i f 12 l9 j g 12
with i G a q 1 and j G b q 1.
Then it is straightforward to see that w i j is an element of A. Since l s w iq1, jrw i j
and
lX s w i , jq1rw i j ,
it follows that QŽ A. s k Ž x, y ..
Q.E.D.
The following result gives a sufficient condition that A be a finitely generated k-algebra. LEMMA 1.3. The following assertions hold: Ž1. There are some positi¨ e integers si G 2 Ž1 F i F n. with f is i g A if and only if, after a change of indices 1, 2, . . . , n4 which allows us to assume m1 F m 2 F ??? F m n , the f i are written in the forms f 1 ; x 1m 1 f 2 ; x 1m 1q1s 21 Ž x 1 , x 2 . q c 2 x 2m 2 f 3 ; x 1m 1q1s 31 Ž x 1 , x 2 , x 3 . q x 2m 2q1s 32 Ž x 2 , x 3 . q c 3 x 3m 3 ??? fn ;
x 1m 1q1sn1
Ž x 1 , . . . , x n . q x 2m 2q1sn2 Ž x 2 , . . . , x n .
m ny 1 q1 q ??? qx ny sn , ny1 Ž x ny1 , x n . q c n x nm n , 1
where c i g k* and the si j Ž x j , . . . , x i . are the homogeneous polynomials of degree m i y Ž m j q 1. satisfying the conditions deri¨ ed from the conditions f j Ž fi . x j
for all pairs Ž i , j . with 1 F j - i , 2 F i F n.
Ž2. If the f i are written in the abo¨ e forms in Ž1., then A is a finitely generated k-algebra. Proof. Ž1. Suppose f is i g A with si G 2 for all 1 F i F n. Then, by Lemma 1.1, f j N Ž f i . x j whenever i / j because gcdŽ f i , f j . s 1. This implies that m j F m i y 1 provided Ž f i . x j k 0. If the indices 1, . . . , n4 are changed so that m1 F m 2 F ??? F m n , this gives rise to the condition that Ž f i . x j ' 0 whenever j ) i. Namely, f i is a polynomial in x 1 , . . . , x i . So, f 1 ; x 1m 1 .
484
AOKI AND MIYANISHI
Since f 1 N Ž f 2 . x 1, we have f 2 s bx 1m 1r Ž x 1 , x 2 . dx 1 q cx 2m 2
H
; x 1m 1q1s 21 Ž x 1 , x 2 . q c 2 x 2m 2 . Suppose f iy1 is written as iy2
f iy1 ;
m y1 . Ý x jm q1siy1, j Ž x j , . . . , x iy1 . q c iy1 x iy1 j
i
js1
Since f iy1 N Ž f i . x iy 1, we can write iy1
fi ;
Ý x jm q1si j Ž x 1 , . . . , x i . q si i Ž x 1 , . . . , x iy2 , x i . , j
js1
where we may assume that the following condition is satisfied:
si j does not contain monomial terms whose x k-exponent is Ž ). greater than or equal to m k q 1 for any k with 1 F k - j. We show by induction on j that si j Ž x 1 , . . . , x i . is in fact a polynomial in x j , . . . , x i . It is easy to see that the condition x 1m 1 N Ž f i . x 1 implies iy1
x 1m 1
žÝ
js2
x jm jq1 Ž si j . x 1 q Ž sii . x 1 .
/
Since si j Ž x 1 , . . . , x i . Ž2 F j - i . contains no monomial terms whose x 1-exponent is greater than or equal to m1 q 1, it follows that Ž si j . x 1 ' 0 for 2 F j F i. Suppose si k , . . . , sii are polynomials in x k , . . . , x i for 1 F k - j. Write m jy 2 q1 Ž f i . x jy 1 ; x 1m 1q1 Ž si1 . x jy 1 q ??? qx jy2 Ž si , jy2 . x jy 1 m jy 1 q x jy1
žŽm
jy1
q 1 . si , jy1 q x j Ž si , jy1 . x jy 1
/
iy1
q Ý x km kq1 Ž si k . x jy 1 q Ž sii . x jy 1 . ksj
Note that f jy 1 N Ž f i . x j-1. Write Ž f i . x jy 1 s f jy1 h with h g R. In both sides of this equation, we equate, by making use of the condition Ž)., the terms divisible by x 1m 1q1 first, the terms divisible by x 2m 2q1 next in the remaining m jy 2 q1 terms, and continue this way until the terms divisible by x jy2 . Looking
485
FINITELY GENERATED ALGEBRAS
at the remaining terms, we then know that iy1
m jy 1 x jy1
žÝ
x km kq1 Ž si k . x jy 1 q Ž sii . x j
ksj
y1
/
.
It then follows by virtue of the condition Ž). that si j , . . . , sii are polynomials in x j , . . . , x i . We show that c i g k*. Indeed, if c i s 0 then Ž f 1 , . . . , f i . ; Ž x 1 , . . . , x iy1 ., which is impossible because f 1 , . . . , f n4 is a system of parameters of the ideal Ž x 1 , . . . , x n .. Hence c i g k*. Conversely, if f 1 , . . . , f n are written in the above forms and satisfy the condition that f j N Ž f i . x j for 1 F j - i and 2 F i F n, one can readily show that f 12 , . . . , f n2 g A. Ž2. With the expressions of f 1 , . . . , f n in Ž1. above, define a k-subalgebra B of A by B s k f 12 , f 22 , . . . , f n2 . Let B˜ be the integral closure of B in k Ž x 1 , . . . , x n . and let Bˆ s k w x 1 , f 2 , . . . , f n x. Then Bˆ > B and Bˆ is integral over B. Furthermore, it is easy to show that x 1 , . . . , x n are integral over Bˆ because c i g k* for 2 F i F n. Hence x 1 , . . . , x n are integral over B and B˜ s R. Note that B˜ is a finitely generated B-module and B˜ contains A as a B-submodule. Since B is a Noetherian ring, A is a finitely generated B-module. Hence A is a finitely generated k-algebra. Q.E.D. We define a k-subalgebra A 0 of A by A 0 s k f 1i1 f 2i 2 ??? f ni n i j G 2, 1 F j F n . In connection with Lemma 1.3, we wish to show that f is i g A for some positive integer si G 2 Ž1 F i F n. provided A contains A 0 and A is a finitely generated k-algebra. Modifying the settings a bit, we consider the following problem. Problem 1.4. Let p1 , . . . , pn be homogeneous polynomials of R of respective degrees m1 , . . . , m n such that p1 , . . . , pn4 is a system of parameters of the ideal Ž x 1 , . . . , x n .. Let C be a finitely generated graded k-subalgebra of R which contains the subalgebra C0 s k p1i1 p 2i 2 ??? pni n i j G 1, 1 F j F n . Then, does it follow that pis i g C for some positive integers si , 1 F i F n?
486
AOKI AND MIYANISHI
With the notations of Lemma 1.3, if we take C s A and pi s f i2 Ž1 F i F n., then A > k w p1i1 ??? pni n < i j G 1, 1 F j F n x. If we can answer Problem 1.4 affirmatively, we obtain f is i g A and the converse of the assertion Ž2. of Lemma 1.3 will hold. Problem 1.4 has, however, a counterexample which will be given in Proposition 1.11. Furthermore, we note that tr.deg k k Ž f 1 , . . . , f n . s n since f 1 , . . . , f n4 is a system of parameters of Ž x 1 , . . . , x n .. THEOREM 1.5. Let C˜ be the integral closure of C in its quotient field QŽ C .. Then we ha¨ e p1 , . . . , pn g C˜ pro¨ ided C˜ l k Ž p1 , . . . , pn . is finitely generated o¨ er k. Note that the condition in Theorem 1.5 is satisfied when n s 2 Žcf. w4x.. In order to prove Theorem 1.5, we may and shall replace, if necessary, p1 , . . . , pn by their powers p1u1 , . . . , pnu n and assume m1 s ??? s m n . Our proof consists of several lemmas. The following result is well known. ŽSee, for example, Gurjar w1, Lemma 3x for a geometric proof in the case n s 2.. LEMMA 1.6.
R s k w x 1 , . . . , x n x is integral o¨ er k w p1 , . . . , pn x.
Proof. Set S s k w p1 , . . . , pn x. By the hypothesis that p1 , . . . , pn4 is a system of parameters of Ž x 1 , . . . , x n ., RrŽ p1 , . . . , pn . R is a finite-dimensional k-vector space. Let w 1 , . . . , w N 4 be a set of elements of R such that the set w 1 , . . . , wn4 of the residue classes of w 1 , . . . , wn modulo Ž p1 , . . . , pn . R is a k-basis of RrŽ p1 , . . . , pn . R. We shall show that R is generated by w 1 , . . . , wn as an S-module. Let c be a homogeneous polynomial of degree, say r, in R. Then one can write N
cs
n
Ý a i w i q Ý pj c j , is1
js1
where a i g k Ž1 F i F N . and the c j Ž1 F j F n. are homogeneous polynomials of degree less than r s deg c . By induction on r, every c j is written as N
cj s
Ý h jl w l ,
where h jl g S.
ls1
Then we have N
cs
Ý ls1
½
n
al q
Ý pj h jl js1
5
wl .
Hence R is a finite S-module generated by w 1 , . . . , wn . So, R is integral over S. Q.E.D.
FINITELY GENERATED ALGEBRAS
487
LEMMA 1.7. Let w , c be homogeneous polynomials of R such that gcdŽ w , c . s 1 in R. Then gcdŽ w Ž p1 , . . . , pn ., c Ž p1 , . . . , pn .. s 1 in R. Proof. Write w Ž p . s w Ž p1 , . . . , pn . and c Ž p . s c Ž p1 , . . . , pn .. Suppose to the contrary that gcdŽ w Ž p ., c Ž p .. / 1. Then there exists a prime ideal P of height 1 of R such that w Ž p . ; P and c Ž p . ; P. Let p s P l S, where S s k w p1 , . . . , pn x. Since R is integral over S by Lemma 1.6, the going-down theorem implies that p is a prime ideal of height 1. Furthermore, p contains w 9 and c 9, which are the copies of w and c , respectively, and considered as elements of S. This contradicts the hypothesis that gcdŽ w , c . s 1. Q.E.D. LEMMA 1.8. The following assertions hold: Ž1. Ž2.
C˜ l k Ž p1 , . . . , pn . s C˜ l k w p1 , . . . , pn x. C l k Ž p1 , . . . , pn . s C l k w p1 , . . . , pn x.
Proof. We shall prove only the assertion Ž2.. The assertion Ž1. can be proved in a similar fashion. Note first that C l k Ž p1 , . . . , pn . is a graded ring since C is a graded ring. In fact, let j be any element of C l k Ž p1 , . . . , pn . and let
j s j n q j nq1 q ??? qj l
with j i g C
be the homogeneous decomposition. Then we have j s c Ž p .rw Ž p ., where w Ž p ., c Ž p . g k w p1 , . . . , pn x. Let the homogeneous decompositions of w Ž p . and c Ž p . be
c Ž p . s cc Ž p . q ccq1 Ž p . q ??? qcd Ž p . w Ž p . s we Ž p . q weq1 Ž p . q ??? qw f Ž p . . Since
Ž j n q j nq1 q ??? qj l . Ž we q weq1 q ??? qw f . s cc q ccq1 q ??? qcd , we obtain the relations
j n we s cc , j nq1 we q j n weq1 s ccq1 , . . . , j l w f s cd . Then it follows that j n s ccrwe , j nq1 s Ž ccq1 we y cc weq1 .rwe2 , . . . , j l s cdrw f . They are elements of C l k Ž p1 , . . . , pn .. Hence C l k Ž p1 , . . . , pn . is a graded ring. We shall show the assertion Ž2.. Let j s c Ž p .rw Ž p . be any homogeneous element of C l k Ž p1 , . . . , pn ., where w , c are homogeneous polynomials in k w p1 , . . . , pn x such that gcdŽ w , c . s 1. Since j is an element of k w x 1 , . . . , x n x and gcdŽ w Ž p ., c Ž p .. s 1 in k w x 1 , . . . , x n x by Lemma 1.7, w Ž p . must be a constant. Hence j is an element of C l k w p1 , . . . , pn x. Q.E.D.
488
AOKI AND MIYANISHI
Note that tr.deg k k Ž p1 , . . . , pn . s n. Now we assume that C˜ l k Ž p1 , . . . , pn . is a finitely generated k-algebra. If n s 2, a theorem of Zariski w4x says that C˜ l k Ž p1 , p 2 . is finitely generated over k. We replace C˜ by C˜ l k Ž p1 , . . . , pn . and assume that k w p1 , . . . , pn x > C˜ > C0 [ k p1i1 ??? pni n i j ) 0, 1 F j F n . On the other hand, it is clear that Q Ž k w p1 , . . . , pn x . s Q Ž C0 . s k Ž p1 , . . . , pn . . Then, in order to prove Theorem 1.5, it suffices to show the following result. LEMMA 1.9. C˜ s k w p1 , . . . , pn x. Proof. Since C˜ is finitely generated over k, let w 1 , . . . , wr be homogeneous polynomials of degree d i which generate C˜ over k. We need the following auxiliary result. CLAIM.
'Ž w Ž p . , . . . , w Ž p . . R s Ž x , . . . , x .. r
1
n
1
Proof. Let P be a prime divisor of Ž w 1Ž p ., . . . , wr Ž p .. R and let p s P l S, where S s k w p1 , . . . , pn x. Then w 1Ž p ., . . . , wr Ž p . g p. We have only to show that p1 , . . . , pn g p. In fact, p has then height n and P has therefore height n because R is integral over S. Let Ž O , m . be a discrete valuation ring of the quotient field k Ž p1 , . . . , pn . such that Ž O , m . dominates Ž Sp , p Sp ., and let ¨ be the associated valuation. Then ¨ Ž w 1Ž p .. ) 0, . . . , ¨ Ž wr Ž p .. ) 0. Since p1N p 2 ??? pn is an element of C0 , we can write p1N p 2 ??? pn s FN Ž w 1 Ž p . , . . . , wr Ž p . . s
Ý
a 1 d 1q ??? q a r d rsNqny1
ca w 1 Ž p .
a1
??? wr Ž p .
ar
,
where FN is a weighted homogeneous polynomial. If N tends to infinity, at least one of a 1 , . . . , a r should tend to infinity. Hence ¨ Ž p1N p 2 ??? pn . tends to infinity as N tends to infinity. This implies that ¨ Ž p1 . ) 0. Hence p1 g p. Similarly, p 2 , . . . , pn g p. Q.E.D. Now let us return to the proof of Lemma 1.9. By the above claim, we know that RrŽ w 1Ž p ., . . . , wr Ž p .. R is a finite-dimensional k-algebra. Since C˜ is a graded subalgebra of k w x 1 , . . . , x n x, we conclude by the same argument as in the proof of Lemma 1.6 that k w x 1 , . . . , x n x is a finitely ˜ kw x1, . . . , x n x generated C˜ module. Since k w x 1 , . . . , x n x > k w p1 , . . . , pn x > C, ˜ contains k w p1 , . . . , pn x as a C-submodule. Since C˜ is a Noetherian ring,
FINITELY GENERATED ALGEBRAS
k w p1 , . . . , pn x is a finitely ˜ Since integral over C. k w p1 , . . . , pn x. This implies proved. Now we shall state and paper.
489
˜ generated C-module. Hence k w p1 , . . . , pn x is QŽ k w p1 , . . . , pn x. s QŽ C˜., we have C˜ s ˜ and Theorem 1.5 is thus that p1 , . . . , pn g C, prove one of the main results of the present
THEOREM 1.10. Suppose n s 2. With the same notations and assumptions as in the Introduction, let A˜ be the integral closure of A in its quotient field. Then the following conditions are equi¨ alent: Ž1. Ž2. Ž3.
A is finitely generated o¨ er k. ˜ f, g g A. A˜ s k w x, y x.
Proof. The condition Ž1. implies the condition Ž2. by Theorem 1.5. ˜ Note that the condition gcdŽ p1 , p 2 . s 1 is the only Suppose that f, g g A. condition which is necessary in the proof of Lemma 1.6. Hence the hypothesis gcdŽ f, g . s 1 implies that k w x, y x is integral over k w f, g x. Hence k w x, y x is integral over A. In view of Lemma 1.2, this implies that A˜ s k w x, ˜ Thus the conditions y x. Conversely, if A˜ s k w x, y x it is clear that f, g g A. Ž2. and Ž3. are equivalent. We shall show that the condition Ž3. implies the condition Ž1.. Let F Ž x . s x M q c1 x My 1 q ??? qc My1 x q c M s 0 C Ž y . s y N q d1 y Ny 1 q ??? qd Ny1 y q d N s 0 be the integral relations of x and y over A, where c1 , . . . , c M , d1 , . . . , d N g A. Let C s k w c1 , . . . , c M , d1 , . . . , d N x. Then k w x, y x is a finitely generated C-module and A is its C-submodule. Hence A is a finitely generated k-algebra. Q.E.D. Problem 1.4 itself has the following counter-example in dimension 2. PROPOSITION 1.11. Let C be a subalgebra of k w x, y x defined as C s k w x q y, x i y j < ; i ) 0,;j ) 0x. Then C has the following properties, Ž1. Ž2. Ž3.
C > C0 [ k w x i y j < i ) 0, j ) 0x. C is finitely generated o¨ er k. C does not contain x s, y t for e¨ ery s ) 0 and e¨ ery t ) 0.
Proof. Ž1. It is clear. Ž2. Define a subalgebra C1 of C as C1 s k w x q y, xy, x 2 y x. We shall show that C s C1. First we prove x n y g C1 by induction on n. If n s 1, 2,
490
AOKI AND MIYANISHI
this holds clearly. Suppose that this holds for x i y with i F n y 1, that is, x ny 1 y, x ny2 y, . . . , xy g C1. Since x n y s x ny1 y Ž x q y . y xy ? x ny 2 y, we then have x n y g C1 , where n G 3. Suppose n G r G 2. Since n
Ž xy . , x y s Ž xy . Ž x ny 1 y ry1 . , n
½
r
nsr n)r
we have x n y r g C1 by induction on r. Finally, when n F r, we suppose that x n y r 9 g C1 for r 9 with n F r 9 - r. Since x n y r s x n y ry1 Ž x q y . y x nq 1 y ry1 g C1 , we have x n y r g C1 by induction on r. Hence C s C1. Ž3. Note that C is a graded subring of k w x, y x. If x s g C, we may write s
x s s a s00 Ž x q y . q
a
Ý
aq2 b q3g ss b )0 or g )0
aa bg Ž x q y . Ž xy .
b
Ž x2 y.
b
Ž x2 y.
g
,
where a s00 / 0. Then we have with r g k w x, y x .
a s00 y s s x r
This is a contradiction. Similarly, if y t g C, write t
y t s a t 00 Ž x q y . q
Ý
a
aq2 b q3g st b )0 or g )0
aa bg Ž x q y . Ž xy .
g
,
where a t 00 / 0. Then we have a t 00 x t s yu
with u g k w x, y x .
This is a contradiction as well.
Q.E.D.
2. CRITERIA FOR THE FINITE GENERATION OF A In this section, we treat the case n s 2 and consider whether or not A is finitely generated over k for some specific pairs of f, g. These results will be used in the next section. Here we write A as AŽ f, g . whenever it is necessary to recall the vector field d s 1rf r x q 1rg r y. We shall begin with the following result. LEMMA 2.1. Ž1.
For positi¨ e integers m, n, we ha¨ e
A Ž x m , y n . s k x mqi , y nqj 1 F i F m q 1, 1 F j F n q 1 . Hence AŽ x m , y n . is finitely generated o¨ er k.
FINITELY GENERATED ALGEBRAS
491
Ž2. We ha¨ e A Ž y, x . s k x i y j i G 1, j G 1 . Then A is not finitely generated o¨ er k. Proof. Ž1. A homogeneous polynomial w belongs to A if and only if x m < w x and y n < w y . So, w g A if and only if w is a sum of monomials ax s y t with a g k such that s s 0 or s G m q 1 as well as t s 0 or t G n q 1. Then it is easy to see that A is generated by the x mq i and the y nq j with 1 F i F m q 1 and 1 F j F n q 1. Ž2. It is straightforward to show that A is generated by the elements i j x y with i G 1 and j G 1. Then A is not finitely generated because A contains x i y with i 4 0. Q.E.D. In the following two lemmas, we shall consider the case where f and g are linear homogeneous polynomials. LEMMA 2.2. The following assertions hold true: Ž1.
Let f s y and g s y y a x with a g k*. Then we ha¨ e
A Ž y, y y a x . s k u i ¨ j Ž i G 1, j G 2 . , u n y nu ny1 ¨ Ž n G 2 . , where u s y and ¨ s y y a x. The subalgebra A is not finitely generated o¨ er k. Ž2. Let f s x and g s y y a x with a g k*. Then we ha¨ e A Ž x, y y a x . s k u i ¨ j Ž i G 2, j G 2 . , ¨ n qn a u¨ ny 1 Ž n G 3 . , u n Ž n G 2 . , where u s x and ¨ s y y a x. The subalgebra A is then finitely generated o¨ er k. Ž3. Let f s y y a x and g s x. Then we ha¨ e A Ž y y a x, x . s k u i ¨ j Ž i G 2, j G 1 . , ¨ n y nu¨ ny1 Ž n G 2 . , where u s y y a x and ¨ s x. The subalgebra A is not finitely generated o¨ er k. Ž4. Let f s y y a x and g s y with a g k*. Then we ha¨ e A Ž y y a x, y . s k u i ¨ j Ž i G 2, j G 2 . , u n ynu ny 1 ¨ Ž n G 3 . , ¨ n Ž n G 2 . , where u s y y a x and ¨ s y. The subalgebra A is finitely generated o¨ er k.
492
AOKI AND MIYANISHI
Proof. We prove the assertions Ž1. and Ž2.. The assertions Ž3. and Ž4. are verified in the same fashion as for Ž1. and Ž2., respectively. Ž1. By the change of variables u s y, ¨ s y y a x, the vector field d is written as
ds
1 ¨ u
q
u y a¨ u¨
¨
.
Let w g R [ k w x, y x s k w u, ¨ x be a homogeneous polynomial of degree n and write it as
ws
ai j u i ¨ j .
Ý iqjsn
Then we have
dŽw. s
1 u¨
Ž nan0 q any1, 1 . u n q u¨ c y n a a0 n¨ n 4 ,
where c is a homogeneous polynomial of degree n y 2. Hence d Ž w . g R if and only if na0 n q a ny1, 1 s 0 and a0 n s 0. Namely, w s a20 Ž u 2 y 2 u¨ . if n s 2 and
w s a n0 Ž u n y nu ny 1 ¨ . q
a i j u i ¨ j q a1, ny1 u¨ ny 1
Ý iqjsn i , jG2
if n G 3. Hence A is generated by the elements as in the statement. Since A contains u¨ j Ž j 4 0., the subalgebra A is not finitely generated over k. Ž2. By the change of variables u s x and ¨ s y y a x, the vector field d is written as
ds
1 u u
q
u y a¨ u¨
¨
.
So, if w s Ý iqjsn a i j u i ¨ j is a homogeneous polynomial of degree n, d Ž w . is computed as
dŽw. s
1 u¨
any 1, 1 u n q u¨ c q Ž a1, ny1 y n a a0 n . ¨ n 4 ,
493
FINITELY GENERATED ALGEBRAS
where c is a homogeneous polynomial of degree n y 2. Hence d Ž w . g R if and only if a ny 1, 1 s 0 and a1, ny1 y n a a0 n s 0. The last condition implies that the subalgebra A is generated by the elements as given in the statement. In order to prove that the subalgebra A is finitely generated over k, we first show that u, ¨ are integral over A. It is clear that u is integral over A. The element ¨ is integral over A since there is the following integral relation with coefficients in A: ¨ 7 y Ž ¨ 3 q 3 a u¨ 2 . ¨ 4 y Ž ¨ 4 q 4a u¨ 3 . ¨ 3 q Ž ¨ 7 q 7a u¨ 6 . s 0.
Then R is a finite B-module, where B s k w u 2 , ¨ 3 q 3 a u¨ 2 , ¨ 4 q 4a u¨ 3 , ¨ 7 q 7a u¨ 6 x which is a subalgebra of A. Then A is a finite B-module, and A is therefore finitely generated over k. Q.E.D. The following result will complement Lemma 2.2. Let f s y y a x and g s y y b x with a , b g k* and a / b .
LEMMA 2.3. Then we ha¨ e
A Ž y y a x, y y b x . s k u 2 y 2 u¨ q
a b
b
¨ 2 , u n y nu ny1 ¨ , ¨ n y
a
n¨ ny1 u Ž n G 3 . ,
u i ¨ j Ž i G 2, j G 2 . , where u s y y a x and ¨ s y y b x. The subalgebra A is then finitely generated o¨ er k. Proof. By the change of variables, the vector field d is written as
ds
ž
u y a¨ u¨
q
/ ž u
u y b¨ u¨
/
¨
.
Let w s Ý iqjsn a i j u i ¨ j be a homogeneous polynomial of degree n G 2 in R [ k w x, y x s k w u, ¨ x. We then compute
dŽw. s
1 u¨
Ž nan0 q any1 , 1 . u n q u¨ c y Ž a a1, ny1 q n b a0 n . ¨ n 4 ,
where c is a homogeneous polynomial of degree n y 2. Hence d Ž w . g R if and only if nan0 q a ny1, 1 s 0 and a a1, ny1 q n b a0 n s 0. The last condition implies that the subalgebra A is generated by the elements as given in the statement. The element ¨ is integral over A since there is a monic relation in ¨ with coefficients in A: ¨ 11 y ¨ 5 y
ž
5b
a
¨ 4u ¨ 6 y ¨6 y
/ ž
6b
a
¨ 5 u ¨ 5 q ¨ 11 y
/ ž
11 b
a
¨ 10 u s 0.
/
494
AOKI AND MIYANISHI
Then the element u is integral over Aw ¨ x, hence over A because there is a monic relation
a
u2 y 2 ¨ u q
b
a
¨ 2 y u 2 y 2 u¨ q
ž
b
¨ 2 s 0.
/
Then it follows that A is finitely generated over k.
Q.E.D.
Lemma 2.3 implies that the polynomial ring contains a one-parameter family of two-dimensional subalgebras A t 4t g k such that A t is finitely generated over k for each value of t / 0, b and A 0 is not finitely generated over k, where b is some fixed element of k. In fact, we have only to take A t to be AŽ y y tx, y y b x .. Then the assertion will follow from Lemmas 2.2 and 2.3. In Section 3, we need the following result which is a variant of Lemma 2.2. LEMMA 2.4. The algebra A s AŽ f, g . is finitely generated o¨ er k in each of the following cases, where s G 2 and t G 2 are integers: Ž1. Ž2. Ž3.
f s x s and g s Ž y y a x . t with a g k*. f s Ž y y a x . s and g s y t with a g k*. f s Ž y y a x . s and g s Ž y y b x . t with a , b g k* and a / b .
Proof. We prove only the case Ž3., as the other cases can be treated similarly. Let u s y y a x and ¨ s y y b x. The vector field d is then written as
ds
ž
us y a ¨ t u ¨ s
t
/
q
u
ž
us y b ¨ t u ¨ s
t
/
¨
.
A computation as in the proof of Lemma 2.2 shows that the following elements are contained in A, t
w Ž m . s u myt u t ym C1 u ty1 ¨ qm C2 u ty2 ¨ 2 q ??? q Ž y1 . m Ct¨ t 4
½
c Ž n . s ¨ ny s ¨ s yn C1
b
ž / a
¨ sy1 u qn C2
b
ž / a
s
q ??? q Ž y1 . n C s
2
¨ sy2 u 2
b
s
ž / 5 a
us ,
where m Ci and n C j are the binomial coefficients and m, n G s q t q 2. In order to show that A is finitely generated over k, it suffices to find two elements w Ž m. and c Ž n. which are prime to each other. Since w Ž m. and
495
FINITELY GENERATED ALGEBRAS
c Ž n. are homogeneous in u, ¨ , it suffices to show that there are no elements j in k satisfying the equations t
j t ym C1 j ty1 qm C2 j ty2 q ??? q Ž y1 . m Ct s 0 1 yn C1
b
ž / a
j qn C2
b
ž / a
2 s
j q ??? q Ž y1 . n C s 2
b
ž / a
s
jss0
provided m and n are sufficiently large and mutually independent. This assertion can be easily verified. Q.E.D. Suppose that f s f 1 f 2 with deg f 1 ) 0, deg f 2 ) 0, and gcdŽ f 1 , f 2 . s 1. Then it is easy to verify that AŽ f 1 , g . l AŽ f 2 , g . s AŽ f, g .. Similarly, AŽ f, g 1 g 2 . s AŽ f, g 1 . l AŽ f, g 2 . if g s g 1 g 2 with deg g 1 ) 0, deg g 2 ) 0, and gcdŽ g 1 , g 2 . s 1. This easy remark entails the following result. LEMMA 2.5. Suppose that either y < f or x < g. Then A s AŽ f, g . is not finitely generated o¨ er k. Proof. Suppose, to the contrary, that A is finitely generated over k. Then the integral closure A˜ of A in its quotient field is finitely generated and contains f, g by Theorem 1.10. Hence k w x, y x is integral over A since k w x, y x is integral over k w f, g x by Lemma 1.6. Suppose y < f. Then we have A ; A Ž y, y y a x . ; k w x, y x , where y y a x is a linear factor of g and hence a g k*. Hence AŽ y, y y a x . is finitely generated over k. This is, however, a contradiction because AŽ y, y y a x . is not finitely generated over k by Lemma 2.2Ž1.. The case x < g is proved in a similar way. Q.E.D. This lemma has immediate applications. COROLLARY 2.6. true:
For positi¨ e integers m, n, the following assertions hold
Ž1. AŽ y m , x n . is not finitely generated o¨ er k. Ž2. AŽ y m , y n q ax n . is not finitely generated o¨ er k, where a g k*. Ž3. AŽ y m q bx m , x n . is not finitely generated o¨ er k, where b g k*. In connection with these obser¨ ations, we pose the following problem. Problem 2.7. Let f, g be as above. Suppose f s f 1 f 2 with deg f 1 ) 0 and deg f 2 ) 0. Suppose, furthermore, that AŽ f 1 , g . and AŽ f 2 , g . are finitely generated over k. Is AŽ f, g . then finitely generated over k?
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AOKI AND MIYANISHI
3. A SUFFICIENT CONDITION FOR POWERS OF f AND g ARE CONTAINED IN A Let f, g be as above and let A 0 be the k-subalgebra defined by all elements f i g j with i G 2 and j G 2. Namely, A 0 s k f i g j < i G 2, j G 2 . Our objective is to prove the following result. THEOREM 3.1. Let A 0 s k w f i g j < i G 2, j G 2x. Assume that A l k w f, g x p A 0 and that deg f ¦ deg g and deg g ¦ deg f. Then if A is finitely generated o¨ er k then f M g A or g N g A for some positi¨ e integers M, N G 2. The proof will be given in this section. In what follows, we assume that A is finitely generated, A l k w f, g x p A 0 , and m ¦ n and n ¦ m, where m s deg f and n s deg g. Furthermore, we assume that f M f A and g N f A for all M G 2 and all N G 2. We shall show that this assumption leads to a contradiction. Write m s m9d and n s n9d with d s gcdŽ m, n.. Then the hypothesis A l k w f, g x p A 0 implies that A contains a nonzero homogeneous element F of the form F s af a q bf b g q cfg g q dg d ,
Ž 1.
where a, b, c, d g k and a , b , g , d are positive integers. The following two lemmas will finish the proof of Theorem 3.1. LEMMA 3.2. With the abo¨ e notations and assumptions, F is not of the form F ; f a , F ; f b g, F ; fg g , or F ; g d. Proof. By the hypothesis, it is clear that F is not of the form F ; f a or F ; g d. Assume that F ; f b g. Then we have Fx ; b f by1 f x g q f b g x
and
Fy ; b f by1 f y g q f b g y
and g < Fy implies g < g y . Hence g y ' 0. So g ; x n. Then A is not finitely generated by Lemma 2.5. Similarly, the case F ; fg g is excluded. Q.E.D. LEMMA 3.3. With the abo¨ e notations and assumptions, F does not contain two or more terms among f a , f b g, fg g , and g g. Proof. Our proof proceeds by the comparison of the degrees of these terms, which are given as deg f a s m a s dm9a ,
deg f b g s m b q n s d Ž m9b q n9 . ,
deg fg g s m q ng s d Ž m9 q n9g . ,
deg g d s n d s dn9d .
Ž 2.
FINITELY GENERATED ALGEBRAS
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In view of the assumption that m ¦ n and n ¦ m, it follows that m9 ) 1 and n9 ) 1. First we assume that a / 0. Ži. Suppose ab / 0. Note that F is a homogeneous polynomial. By comparison of degrees in Ž2., we have m9a s m9b q n9, which yields m9Ž a y b . s n9. Since gcdŽ m9, n9. s 1, this is a contradiction. Hence b s 0 if a / 0. By exchanging the roles of a and d, we conclude that cd s 0, where we do not have to assume a / 0. Žii. Suppose ac / 0. By comparison of degrees, we have m9a s m9 q n9g . Hence a s n9l q 1 and g s m9l for some integer l G 0. Then we can write F ; f n9lq1 q cfg m9l
with c g k*.
If l s 0 then F ; f and f g A. This case is excluded by the hypothesis. So l ) 0. Then we have Fx ; Ž n9l q 1 . f n9l f x q cf x g m9l q cm9lfg m9ly1 g x . Since f < Fx Žcf. Lemma 1.1., we have f < f x . So f x s 0 and f ; y m . This is impossible by Lemma 2.5. Žiii. Suppose ad / 0. By comparison of degrees, we have m9a s n9d . Hence we may write a s n9l and d s m9l for some positive integer l. Then F is written as F ; f n9l q dg m9l
with d g k*
Ž 3.
which gives rise to Fx ; n9lf n9ly1 f x q dm9lg m9ly1 g x , Fy ; n9lf n9ly1 f y q dm9lg m9ly1 g y , where f < Fx implies f < g x , and g < Fy implies g < f y . If g x k 0 and f y k 0, the comparison of degrees gives m F n y 1 and n F m y 1. This is a contradiction. So, f y ' 0 or g x ' 0. If f y ' 0 then f ; x m and f 2 g A, which contradicts the assumption. Similarly, g 2 g A if g x ' 0, which is a contradiction. By Ži. ] Žiii. above, we conclude that a s 0. By exchanging the roles of f and g, we conclude also that d s 0. It then remains to consider the following case. Živ. Suppose bc / 0. The comparison of degrees implies m9b q n9 s m9 q n9g , which yields m9Ž b y 1. s n9Žg y 1.. Since gcdŽ m9, n9. s 1, we can write b s n9l q 1 and g s m9l q 1 for some integer l G 0. Then we have with c g k*, F ; f n9lq1 g q cfg m9lq1 Fx ; Ž n9l q 1 . f n9l f x g q f n9lq1 g x q cf x g m9lq1 q c Ž m9l q 1 . fg m9l g x .
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If l s 0 then F ; fg, and this is impossible by Lemma 3.2. So, l ) 0. Then f < Fx implies f < f x . Hence f x ' 0 and f ; y m . This is impossible by Lemma 2.5. Q.E.D. Thus we have proved Theorem 3.1. Remark 3.4. Ž1. The assumption that deg f ¦ deg g and deg g ¦ deg f cannot be dropped in the assertion Ž1. of Theorem 3.1. Indeed, let f s y y a x and g s y y b x with a , b g k* and a / b as in Lemma 2.3. Then A l k w f, g x p A 0 , and f M f A and g N f A for any M, N G 2. Ž2. Notwithstanding the above remark, the arguments in the proof of Lemma 3.3 work also in the cases where Ža. m9 ) 1 and n9 s 1, Žb. m9 s 1 and n9 ) 1, and Žc. m9 s n9 s 1 except for the following point. Namely, one cannot conclude that the k-subalgebra A is not finitely generated if f ; y m q ag m r n or g ; x n q bf n r m with a, b g k*. We do not know, however, whether or not A is finitely generated if f ; y m q ag m r n or g ; x n q bf n r m with a, b g k*.
REFERENCES 1. R. V. Gurjar, Graded subrings of Cw X, Y x, Proc. Indian Acad. Sci. Math. Sci. 99 Ž1989., 209]215. 2. M. Miyanishi, Normal affine subalgebras of a polynomial ring, in ‘‘Algebraic and Topological Theories}To the Memory of Dr. Takehiko Miyata, Kinokuniya, Tokyo, 1985,’’ pp. 37]51. 3. M. Miyanishi, Vector fields on factorial schemes, J. Algebra 173 Ž1995., 144]165. 4. M. Nagata, On the fourteenth problem of Hilbert, in Lecture Notes at Tata Institute of Fundamental Research, Bombay, 1965.