GROUPS OF FINITE CHARACTERISTICS
§1. Introduction.
Lrt Fq be the Galois field with q elements, and V =Fnq be the vector space. Let G be a
finite subgroup of GL(V ). G induces a linear action on the symmetric algebraFq[x1, . . . , Fn].
The invariant subring is
Fq[x1, . . . , Fn]G = {f ∈ Fq[x1, . . . , Fn] : σ(f ) = f for all σ ∈ G}
A problem on Invariant Theory is to find the invariant subrings. If G = GLn(Fq) or
SLn(Fq), the result is due to Dickson[6,1911]. If G = Sp2n(Fq), it is solved by Carlisle and
Kropholler[3;1]. When G = Un(Fq2), this problem is settled by us in [5]. If G = On(Fq), we
have found the invariants for n ≤ 4[4]. In this paper, we will find the invariant subrings of O+
n(Fq) for all n, and we also prove that the invariant subring is complete intersection.
Let q be a power of odd prime number and Fq be the Galois field with q elements. Let
V =Fq and R =Fq[x1, . . . , xn] be the associate symmetric algebra.
Consider the quadratic form
Qn = 1x21+ 2x22+ · · · + nx2n, i∈ F×q . (1-1)
and let On(Fq) be the orthogonal group associated with the o Qn. It is well-known that all
quadratic forms are equivalent to one of the following two special families: Q+n = X12− X22+ X32− · · · + (−1)nXn−12 + (−1)n+1Xn2, n ≥ 1;
Q−n = X12− X22+ X32− · · · + (−1)nXn2−1+ (−1)n+1Xn2, n≥ 1 (1-2) where is a non-square inF×q. The diof them are δ(Q+n) = (−1)bn2c; δ(Q−
n) = (−1)
n
2 if n is even, δ(Q− n) =
(−1)n−12 if n is odd.
Since we are interesting on the orthogonal groups, it is known that there are three types: O+
n(Fq), n is even or odd; On−(Fq), n is even . The orders of these groups are the following:
Proposition 1.1. [7] The order of the orthogonal groups are |O+n(Fq)| = 2(qn−1 − 1)qn−2(qn−3− 1)qn−4· · · (q2− 1)q, if n is odd; 2(qn−1− qn 2−1)(qn−2− 1)qn−3· · · (q2− 1)q, if n is even, |On−(Fq)| =2(qn−1− q n 2−1)(qn−2− 1)qn−3· · · (q2− 1)q, if n is even.
The orthogonal group induces a linear action on R. We want to find the invariant subrings Fq[x1, x2, . . . , xn]On(Fq) = {f ∈ R : σ(f) = F for all σ ∈ On(Fq)}. The answer of the
corresponding problem in function fields is
Typeset by AMS-TEX 1
Theorem 1.2. [2] The invariant subfield is Fq(x1, x2, . . . , xn)On(Fq) = Fq(Qn0, Qn1, . . . , Qn,n−1) where Qni = 1xq i+1 1 + 2xq i+1 2 + · · · + nxq i+1 n . (1-3)
Notations. (a) Let M be an m × n matrix, S be a subset of {1, 2, . . . , m} and T a subset of {1, 2, · · · , n}. Then we denote MS;T to be the submatrix of M obtained by deleting the rows
in S and the columns in T . If S = {i} and T = {j}, we always denote Mij instead of MS;T.
(b) The Legendre symbol is defined as δ q = δq−12 = 1, if δ is a square in F× q ; δ q = δq−12 = −1, if δ is a non-square in F× q . (1-4) (c) We define the weight of variables as follows.
wt xi = wt yi= 1 for all i;
wt Xi= qi+ 1, wt T = 1. (1-5)
§2. Preliminary Results and the Statement of the Main Theorem for Odd n. We first consider the infinite dimensional matrix
M(X0, X1, X2,· · · ) = X0 X1 X2 X3 X4 · · · X1 X0q X1q X2q X3q · · · X2 X1q X q2 0 X q2 1 X q2 2 · · · X3 X2q X q2 1 X q3 0 X q3 1 · · · X4 X3q X q2 2 X q3 1 X q4 0 · · · .. . ... ... ... ... ... .
The (i, j)-entry of M is X|j−i|qmin(i−1,j−1). Let M(n) be the submatrix of the first n rows and
n columns. We define four sequences fn, gn, fn0, g0n of polynomials which are essential in this
paper. In fact, we define gn0 as the quotient of two polynomials, and we will show that they are actually polynomials in Lemma 2.5.
Definition 2.1. Define f0 = 1, fn = det M(n) ∈ Fq[X0, X1, · · · , Xn−1], n ≥ 1;
f0 0= 1, fn0 = det M (n+1) 1;n+1∈ Fq[X0, X1, · · · , Xn], n ≥ 1. For example, f1 = X0, f2 = X0q+1− X12; f10 = X1, f20 = X1q+1− X q 0X2. It is easy to see that fn =−fn−2q Xn−12 + φ1(X0, X1,· · · , Xn−1), degXn−1φ1 = 1 for n ≥ 3. (2-1) fn0 = (−1)n+1fn−1q Xn+fn−20 qXnq+1−1+φ2(X0, X1, · · · , Xn−1), degXn−1φ2 = q for n ≥ 3. (2-2)
Obviously, f1 and f2 are relatively prime. Suppose that fn−1 and fn−2 are relatively
prime. Then from (2-1), it follows that fn and fn−1 are relatively prime. Moreover, from
(2-2), we see that
Definition 2.2. g+n = ( f0 n − f q+1 2 n , for n ≡ 0, 1 (mod 4) f0 n + (−fn) q+1 2 , for n≡ 2, 3 (mod 4); gn−= ( f0 n+ f q+1 2 n , for n ≡ 0, 1 (mod 4) fn0 − (−fn) q+1 2 , for n≡ 2, 3 (mod 4). Definition 2.3. g−10 ± = 1; g00+ = X0, g00−= 1; g0±n = gn±/(g0∓n−2)q for n ≥ 1. Next we show that g0±
n are actually polynomials in Lemma 2.5. We first state a Lemma.
We will use part (a) here and use part (b) in Lemma 3.8. This Lemma has also appeared in [??], now we give a direct proof.
Lemma 2.4. Let R be a ring, and M ∈ Mn(R). We have the following two identities.
(a) Let N = M1,n;1,n∈ Mn−2(R). Then
det M · det N = det Mnndet M11− det Mn1det M1n.
(b) Let N0 = Mn−1,n;1,n ∈ Mn−2(R). Then
det M · det N0 = det Mnndet Mn−1,1− det Mn1det Mn−1,n.
Proof. The two identities are easily seen to be equivalent. In fact, let
M0= M2 M3 .. . Mn−1 M1 Mn .
Apply (b) to M0 and get
det M0det N = det Mnn0 det Mn−1,10 − det Mn10 det Mn−1,n
(−1)n−2det M det N =(−1)n−2det Mnndet M11− (−1)n−2det Mn1det M1n,
(a) follows. So we only prove (b) below. Let Mi be the determinant of the matrix we obtain
by deleting the ith row, the last row, the first column and the last column from M . Multiply the (n − 1)th row of M by Mn−1 (which is det N0), then add to this row the ith row multiplied
by (−1)n−1−iM
i for all 1 ≤ i ≤ n − 2. We will find that the (n − 1)th row of M, after the
above procedure, becomes ( (−1)nM
nn, 0, 0, . . . , 0, Mn1). Expand det M along this row now,
we get the identity in (b).
Applying Lemma 2.4(a) to M = M(n) , we have
fn−2q fn = fn−1q+1− fn−10 2 = −(fn−10 + f q+1 2 n−1)(fn−10 − f q+1 2 n−1) = −gn+−1gn−1− . (2-4)
On the other hand, from (2-1) and (2-2) we have, for n ≡ 0, 1(mod 4), g±n = fn0 ∓ fq+12 n = (−1)n+1fnq−1Xn+ fn−20 q Xn−1q+1+ φ2∓ (−fn−2q Xn−12 + φ1) q+1 2 = (−1)n+1fn−1q Xn + [fn0−2∓ (−fn−2) q+1 2 ]qXq+1 n−1+ φ3(X0, X1, . . . , Xn−1) = (−1)n+1fnq−1Xn+ gn−2∓ qXn−1q+1 + φ3(X0, X1, . . . , Xn−1). (2-5)
where degXn−1φ3 = q. The last equality follows from n − 2 ≡ 2, 3(mod 4). It can be checked
that the same formula holds for n ≡ 2, 3(mod 4). (We just need to replace ± by ∓ in the first three lines.)
To prove that g0±
n are polynomials, we prove the following
Lemma 2.5. (a) fn = (−1)b
n
2cg0 + n−1gn−10 −.
(b) (gn0 ∓−2)q|g±n.
Proof. We prove by induction on n. Since g−2 = (X1q+1 − X0qX2) − (X12 − X q+1 0 )
q+1
2 , so
X0q|g20−. (a) and (b) hold for n = 1, 2 trivially. Given n ≥ 3, suppose these formulae hold for fk, k≤ n − 1. We first prove (a). From (2-4),
fn−2q fn = −gn−1+ gn−1− .
By inductive hypothesis on (a), fn−2 = (−1)bn−22 cg0 +
n−3g0 −n−3, so fn = (−1)b n 2c g + n−1 (g0 −n−3)q · gn−1− (g0 +n−3)q = (−1) bn 2cg0 + n−1g0 −n−1.
Two fractions in the above are polynomials by inductive hypotheses on (b). Now we prove (b). By (2-4) again and by inductive hypotheses on (a),
fnq−1fn+1= −gn+gn−,
fn−1q = (−1)bn−12 c(g0 +
n−2)q(gn0 −−2)q.
By (2-3) , g+n and gn− are relatively prime, so gn−20 + divides just one of gn+ and gn−. Since gn−20 + ∈ Fq[X0, X1,· · · , Xn−2] and observing the Xn−1q+1-term in (2-5), we see that g0 +n−2|g−n
and(g0 +n−2)q|g− n. Lemma 2.6. g0± n = (−1)b n 2cg0± q n−2Xn+ φ03(X0, X1, . . . , Xn−1). Proof. By (2-5), we have gn0±= gn±/(gn−20∓ )q = (−1)n+1 f q n−1 (gn−20 ∓ )qXn+ φ 0 3(X0, X1, . . . , Xn−1) = (−1)bn2cg0 ± q n−2 Xn + φ03(X0, X1, . . . , Xn−1).
The last equality follows from Lemma 2.5(a).
Definition 2.7. Given ∈F×
q. Define the polynomials
Fn,(X0, · · · , Xn−1; T ) := fn(X0+ T2, X1+ Tq+1, · · · , Xn−1+ Tq n−1+1 ), Fn,0 (X0,· · · , Xn; T ) := fn0(X0+ T2, X1+ Tq+1, · · · , Xn + Tq n+1 ), G±n,(X0, · · · , Xn; T ) := g±n(X0+ T2, X1+ Tq+1, · · · , Xn+ Tq n+1 ), G0 ±n,(X0, · · · , Xn; T ) := g0±n (X0+ T2, X1+ Tq+1, · · · , Xn+ Tq n+1 ).
We often denote Fn(T ) instead of Fn,(T ) for short, if is understood. Now we give
explicit forms for Fn(T ) and Fn0(T ).
Lemma 2.8. Let fij and fij0 be the (i, j)-minor of fn and fn0, respectively, 1 ≤ i, j ≤ n. Then
(a) Fn(T ) = fn+P1≤i,j≤n(−1)i+j fijTq
i−1+qj−1 , (b) F0 n(T ) = fn0 + P 1≤i,j≤n(−1)i+j fij0 Tq i+qj−1 . Proof. (a) Let T = (Tqi−1+qj−1)
n×n, Then Fn = det(M(n) + T). Note that rankT = 1.
Hence when we expand the determinant Fn by multilinearity on columns, the terms which two
columns of T are taken must be zero. Then the result can be deduced easily. The proof of (b) is similar.
Now we describe the weight, the T -degrees and the leading coefficients of the polynomials F ’s and G’s.
Lemma 2.9. Using the definition of weight defined in (1-4), we have that Fn,, Fn,0 Gn,, G0n,
are homogeneous polynomials. The weights are as following. (a) wt Fn, = 2(q n −1 q−1 ), wt Fn,0 = (q + 1)(q n −1 q−1 ), wt Gn,±= (q + 1)(q n −1 q−1 ). (b) wt G0± n, = q n+1−1 q−1 , if n is odd ; wt G0 + n, = (q n 2 + 1)(q n 2+1−1 q−1 ) wt G0 −n, = (qn2+1+ 1)(q n 2−1 q−1 ) if n is even.
Proof. From the definition of Fn, we see that it is homogeneous if fn is, and wtFn = wtfn.
The weight of fn can be computed from the definition of fn.
Lemma 2.10. Regarding the polynomials as polynomial in T , then (a) deg Fn, = 2qn−1, deg Fn,0 = qn−1(q + 1).
(b) deg G+1,= q − (q), deg G−1, = q + ( q); deg G+n, =deg G−n, = qn−1(q + 1), n ≥ 2. (c) deg G 0+ n, = qn+ q n 2, deg Gn,0−= qn− qn2, if n is even; ( deg G 0+ n, = qn− (q)q n−1 2 , deg G 0− n, = qn+ (q)q n−1 2 , if n is odd. Proof. (a) follows from Lemma 2.8. (b) follows from (a). (c) is proved by induction. Lemma 2.11. Let lc F denote the leading coefficient of F (T ). Then
(a) lc Fn, = fn−1, lc Fn,0 = fn−10 (b) If (q) = 1, then lc G+1, = −12X0 and lc G−1, = 2. If (q) = −1, then lc G+1,= 2 and lc G−1,= −12X0. (c) For n ≥ 2, lc G±n, = gn−1± , if n is even and (− q ) = 1; gn−1± , if n is odd and (q) = 1; gn−1∓ , otherwise.
(d) If n is odd, then lc Gn,0+ = ( (−1)n+12 1 2gn0+−1, for (q) = 1, (−1)n−12 2g 0− n−1, for (q) = −1; lc Gn,0−= ( (−1)n−12 2g 0− n−1, for (q) = 1, (−1)n+12 1 2gn−10+, for (q) = −1. If n is even, then lc Gn,0+ = ( gn−10+, for (−q ) = 1, gn−10−, for (−q ) = −1; lc G 0− n, = ( gn−10−, for (−q ) = 1, gn−10+, for (−q ) = −1. Proof. (a) follows from Lemma 2.8. (b) can be checked by direct computation. (c) If n ≡ 0(mod 4), the leading coefficient of G±
n = Fn0 ∓ F q+1 2 n is fn0−1∓ (fn−1) q+1 2 = [f0 n−1± − q (−fn−1) q+1 2 ] = ( gn±−1, for (−q ) = 1, gn−1∓ , for (−q ) = −1, since n − 1 ≡ 3(mod 4).
If n≡ 1(mod 4), then the leading coefficient of G±n = Fn0 ∓ Fq+12 n is fn−10 ∓ (fn−1) q+1 2 = [f0 n−1∓ q fq+12 n−1] = ( g±n−1, for (q) = 1, g∓n−1, for (q) = −1, since n − 1 ≡ 0(mod 4).
If n ≡ 2(mod 4), then the leading coefficient of G±
n = Fn0 ± F q+1 2 n is fn−10 ± (−fn−1)q+12 = [f0 n−1∓ − q fq+12 n−1] = ( g±n−1, for (−q ) = 1, g∓n−1, for (−q ) =−1, since n − 1 ≡ 1(mod 4).
If n ≡ 3(mod 4), then the leading coefficient of G±n = Fn0 ± F
q+1 2 n is fn0−1± (−fn−1)q+12 = [f0 n−1± q (−fn−1) q+1 2 ] = ( gn−1± , for (q) = 1, gn−1∓ , for (q) = −1, since n− 1 ≡ 2(mod 4).
(d) The cases n = 1, 2 can be checked directly. Suppose n≥ 3 is odd and (
q) = 1. Then by induction, lc Gn,0+ = lc Gn,+/(lc Gn0−−2,)q = g+n−1/((−1)n−32 2g0− n−3)q = (−1) n+1 2 1 2g 0+ n−1. lc Gn,0−= lc Gn,−/(lc Gn−2,0+ )q = gn−1− /((−1)n−12 1 2g 0+ n−3)q = (−1) n−1 2 2g0− n−1.
Consider the quadratic form Qn defined in (1-1) and the polynomials Qnidefined in (1-3). Note that x1 x2 · · · xn xq1 xq2 · · · xqn .. . ... ... xq1n−1 xq2n−1 · · · xqn−1 n 1 2 . .. n x1 xq1 · · · x qn−1 1 x2 xq2 · · · x qn−1 2 .. . ... ... xn xqn · · · xq n−1 n = Qn0 Qn1 Qn2 · · · Qn,n−1 Qn1 Qqn0 Q q n1 · · · Q q n,n−2 .. . ... ... ... Qn,n−1 Qqn,n−2 Qqn,n−32 · · · Qqn0n−1 (2-6) and xq1 xq2 · · · xq n xq12 xq22 · · · xq2 n .. . ... ... xq1n−1 xq2n−1 · · · xqn−1 n 1 2 . .. n x1 xq1 · · · x qn−1 1 x2 xq2 · · · xq n−1 2 .. . ... ... xn xqn · · · xq n−1 n = Qn1 Qqn0 Q q n1 · · · Q q n,n−2 Qn2 Qqn1 Q q2 n0 · · · Q q2 n,n−3 .. . ... ... ... Qn,n Qqn,n−1 Qq 2 n,n−2 · · · Q qn−1 n1 (2-7) Let D = det x1 x2 ··· xn xq1 xq2 ··· xqn .. . ... ... xqn−11 xqn−12 ··· xqn−1 n
and the discriminant δ(Qn) = 12· · · n. Taking the
determinants of (2-6) and (2-7),we have
D2· δ(Qn) = fn(Qn0, Qn1, · · · , Qn,n−1)
and
Dq+1· δ(Qn) = fn0(Qn0, Qn1,· · · , Qn,n).
Canceling D from above, we get
fn0(Qn0, Qn1,· · · , Qn,n) − δ(Qn) q−1 2 fn(Qn0, Qn1,· · · , Q n,n−1) q+1 2 =fn0(Qn0, Qn1, · · · , Qn,n) − δ(Qn) q fn(Qn0, Qn1, · · · , Qn,n−1) q+1 2 =0 . (2-8)
where the Legendre symbol is defined in (1-4).
Now we discuss two special families of quadratic forms Q+
n and Q−n defined in (1-2). From
Lemma 2.12. g0+
n (Q+n0, Q+n1, · · · , Q+nn) = 0,
gn0−(Q−n0, Q−n1,· · · , Q−nn) = 0 for even n. gn0+(Q−n0, Q−n1,· · · , Q−nn) = 0 for odd n. Proof. If n≡ 0, 1(mod 4), then δ(Q+
n) = 1. So (2-8) says that (fn0 − f q+1 2 n )(Q+n0, Q+n1,· · · , Q+nn) = 0. If n ≡ 2, 3(mod 4), then δ(Q+ n) = −1. So (fn0 − (−1) q−1 2 f q+1 2 n )(Q+n0, Q+n1, · · · , Q+nn) = 0.
In any case, by Definition 2.2, g+
n(Q+n0, Q+n1, · · · , Q+nn) = 0. Since gn0 −−2involves only n −1
vari-ables, and Q+n0, Q+n1, · · · , Q+n,n−1are algebraically independent, so gn0 −−2(Q+n0, Q+n1, · · · , Q+n,n−2) 6= 0. Thus g0 +
n (Q+n0, Q+n1, · · · , Q+n,n) = 0.
The case Q−
n can be proved similarly. δ(Q−n) =
(−1)n−12 n is odd; (−1)n2 n is even.. If n ≡ 0(mod4), then fn0−(q)fq+12 n = fn0+f q+1 2 n = gn−; if n≡ 2(mod4), then fn0−(−q )f q+1 2 n = fn0−(−fn) q+1 2 = g− n; if n≡ 1(mod4), then f0 n− f q+1 2 n = gn+; if n ≡ 3(mod4), then fn0 + f q+1 2 n = gn+.
Lemma 2.13. (a) For all n ≥ 1, Gn−1,0+ (Q+n0, Q+n1, · · · , Q+n,n−1; Xn) = 0, where = (−1)n;
(b) For any positive even integer n, Gn−1,0+ (Q−n0, Q−n1, · · · , Q−n,n−1; Xn) = 0;
(c) For any positive odd integer n, Gn−1,−0− (Q−n0, Q−n1, · · · , Q−n,n−1; Xn) = 0.
Proof. (a) By Lemma 2.12,
Gn−1,0+ (Q+n0, Q+n1,· · · , Q+n,n−1; Xn) =gn−10+ (Q+n0+ Xn2, Q+n1+ Xnq+1,· · · , Q+n,n−1+ Xnqn−1+1) =gn−10+ (Q+n−1,0, Q+n−1,1,· · · , Q+n−1,n−1) =0. (b) Gn−1,0+ (Q−n0, Q−n1, · · · , Q−n,n−1; Xn) =gn0+−1(Q−n0+ Xn2, Q−n1+ Xnq+1,· · · , Q−n,n−1+ Xnqn−1+1) =gn−10+ (Q−n−1,0, Q−n−1,1, · · · , Q−n−1,n−1) =0. (c) Gn−1,−0− (Q−n0, Q−n1, · · · , Q−n,n−1; Xn) =gn−10− (Q−n0− Xn2, Q−n1− Xnq+1, · · · , Q−n,n−1− Xnqn−1+1) =gn0−−1(Q−n−1,0, Q−n−1,1, · · · , Q−n−1,n−1) =0.
Lemma 2.14. (a) The degree of Xn over Fq(Q+n0, Q+n1, · · · , Q+n,n−1) is
qn−1+ qn−12 , if n is odd;
qn−1− qn
2−1, if n is even.
(b) The degree of Xn over Fq(Q−n0, Q−n1, · · · , Q−n,n−1) is qn−1+ q
n
2−1 if n is even.
Proof.
Fq(Qn0, Qn1, · · · , Qn,n−1, Xn) =Fq(Qn−1,0, Qn−1,1, · · · , Qn−1,n−1, Xn)
=Fq(Qn−1,0, Qn−1,1,· · · , Qn−1,n−2, Xn).
The last equality follows from Theorem 1.2.
[Fq(Qn0, Qn1, · · · , Qn,n−1, Xn) :Fq(Qn0, Qn1,· · · , Qn,n−1)] = [Fq(X0, X1,· · · , Xn) :Fq(Qn0, Qn1,· · · , Qn,n−1)] [Fq(X0, X1, · · · , Xn) :Fq(Qn0, Qn1, · · · , Qn,n−1, Xn)] = [Fq(X0, X1, · · · , Xn) : Fq(Qn0, Qn1, · · · , Qn,n−1)] [Fq(X0, X1, · · · , Xn−1, Xn) :Fq(Qn−1,0, Qn−1,1, · · · , Qn−1,n−2, Xn)] =|On(Fq)|/|On−1(Fq)|.
Now applying Proposition 1.1, we get the followings and the Lemma follows. If n = 2k + 1 is odd, then |On+(Fq)|/|O+n−1(Fq)| = 2(qn−1− 1)qn−2(qn−3− 1)qn−4· · · (q2− 1)q 2(qn−2− qk−1)(qn−3− 1)qn−4· · · (q2− 1)q = (q 2k− 1)q2k−1 (qk− 1)qk−1 = qk(qk+ 1) = qn−1+ qk. If n = 2k is even, then |On+(Fq)|/|O+n−1(Fq)| = 2(qn−1− qk−1)(qn−2− 1)qn−3· · · (q2− 1)q 2(qn−2− 1)qn−3(qn−4− 1)qn−5· · · (q2− 1)q = qn−1− qk−1. If n = 2k is even, then |On−(Fq)|/|O−n−1(Fq)| = 2(qn−1+ qk−1)(qn−2− 1)qn−3· · · (q2− 1)q 2(qn−2− 1)qn−3(qn−4− 1)qn−5· · · (q2− 1)q = qn−1+ qk−1.
Proposition 2.15. Concider the quadratic forms Q+
n for all n and Q−n for even n. The
minimal polynomial of Xn over Fq(Q±n0, Q±n1, · · · , Q±n,n−1) is
Gn−1,0+ (Q±n0, Q±n1, · · · , Q±n,n−1; T )/lc Gn−1,0+ where = (−1)n for “Q+n” case.
Proof. By Lemma 2.13, Xn satisfies the polynomial Gn−1,0+ (Q±n0, Q±n1, · · · , Q±n,n−1; T ), and by
Lemma 2.10, the degree of the field extension is equal to the degree of the polynomial. Corollary 2.16. Concider the quadratic forms Q+
n for all n and Q−n for even n. Gn0+−1,(Qn0±, Q±n1, · · · , Qn,n−1± ; T )/lc Gn−1,0+ ∈
Fq[x1, x2,· · · , xn]O
±
n(Fq)[T ] where = (−1)n for “Q+
n” case.
Proof. By [??,Ex. ??].
Definition 2.17. We define the polynomials R±ni(X0, · · · , Xn) in the following.
(i) If n is even, the polynomial Gn,−10+ (X0, X1, · · · , Xn; T ) has degree qn+ q
n
2. Let R+ ni be
the coefficient of Tqn−i+qn2
, i = 0, 1, · · · ,n2.
(ii) If n is odd and (q) = 1, the polynomial Gn,0+(X0, X1,· · · , Xn; T ) has degree qn−q
n−1
2 .
Let R+ni be the coefficient of Tqn−i−qn−12
, i = 0, 1, · · · ,n+12 .
(iii) If n is odd and (q) = −1, the polynomial G 0+
n,(X0, X1,· · · , Xn; T ) has degree qn + qn−12 . Let R− ni be the coefficient of Tq n−i+qn−12 , i = 0, 1,· · · ,n−1 2 ; Rn,−n+12 be the constant term.
Note that R±n−1,0is the leading coefficient of G0+n−1,(X1, . . . , Xn−1; T ), and
R±n−1,i(X0,X1,··· ,Xn−1)
R±n−1,0(X0,X1,··· ,Xn−2) ∈
Fq[x1, x2,· · · , xn]O
±
n(Fq) for all i, by Corollary 2.16.
Now we begin to state the first case of our Main Theorem. The proof is postpone to §4. The second case of Main Theorem is stated in §6.
We define a k × (k + 1) matrix A(k) as A(k) = p1,2k−1 pq1,2k−3 pq1,2k2 −5 · · · pq13k−2 rq12k−1 rq11k−1 p2,2k−1 pq2,2k−3 pq2,2k−52 · · · r24qk−2 rq23k−2 rq22k−2 .. . ... ... ... ... ... rk,2k rk,2k−1 rk,2k−2 · · · rk,k+2 rk,k+1 rkk (2-9)
where the (i, j)-entry is
(
pi,2k−2j+1qj−1 , if j ≤ k − i ri,k+1+i−jqk−i , if j > k − i.
(2-10) The entries pij’s and rij’s will be defined inductively in the proof of the following Main
Theo-rem.
Theorem Ak. Let n = 2k + 1 be a positive odd integer, Q+n = x21− x22+ x23− · · · + x2n−2−
(a) The invariant subring is Fq[x1, x2,· · · , xn]O + n(Fq) =Fq[Q+n0, Q+n1, · · · , Q+n,n−2, R+n−1,1(X0,X1,··· ,Xn−1) R+n−1,0(X0,X1,··· ,Xn−2), · · · , R+n−1,k(X0,X1,··· ,Xn−1) R+n−1,0(X0,X1,··· ,Xn−2)]. (b) We have an isomorphism Φn :Fq[X0, X1, · · · , Xn−2, Y1, Y2, · · · ,Yk]/hKn1, Kn2, · · · , Kn,k−1i −→ Fq[X0, X1, · · · , Xn−2, R+n−1,1(X0,X1,··· ,Xn−1) R+n−1,0(X0,X1,··· ,Xn−2), · · · , R+n−1,k(X0,X1,··· ,Xn−1) R+n−1,0(X0,X1,··· ,Xn−2)] defined by Xi 7→ Xi, 0≤ i ≤ n − 2 Yj 7→ R+n−1,j(X0,X1,··· ,Xn−1) R+n−1,0(X0,X1,··· ,Xn−2), 1 ≤ j ≤ k.
where the relations Knj are defined by
Kn1 Kn2 .. . Kn,k−1 = A(k)k;∅ 1 Y1 .. . Yk .
(c) The polynomials pij and rij ∈ Fq[X0, X1,· · · , Xj] are independent of n.
rk,2k = −X2k+ ψ(X0, X1, · · · , X2k−1) for some polynomial ψ. (2-11)
Moreover, they are homogeneous with respect to the weight defined in (1-5) with ( wt pij = qj+1+ q j+1 2 −i, j odd; wt rij = qj+ 1. (2-12) § 3. Lemmas.
In order to prove the Main Theorem, in this section, we first give some Lemmas. Lemma 3.1. For n≥ 3 and i ≥ 1, we have
R±ni = (−1)bn2cR ± q
n−2,i−1Xn+ φni(X0, X1, . . . , Xn−1).
Proof. Let n be even. Then by Lemma 2.6, G0+n,−1= (−1)n2G0+ q
n−2,−1(Xn− Tq
n+1
From Lemmas 2.10, 2.11, we see that the leading term of G0+n,−1 is −g0 +n−1Tqn+qn2. Also by the
definition of R+n−2,i, from (3-1) we have G0+n,−1 =(−1)n2{ n 2−1 X i=0 R+n−2,iTqn−i−2+q n 2 −1 +X j ∗ ψj(X0, X1, . . . , Xn−2)Tj}qXn− gn−10 + Tq n+qn2 + φ4(X0, X1, . . . , Xn−1; T ) = − g0 +n−1Tq n+qn2 + (−1)n2 n 2 X i=1 Rn+ q−2,i−1XnTq n−i+qn2 + (−1)n2X j ∗ ψjq(X0, X1, . . . , Xn−2)XnTjq+ φ4. (3-2) where the sum P∗ runs over all j such that j < qn−2+ qn2−1 and j is not equal to qn−3 +
qn2−1, qn−4 + qn2−1, . . . , 2qn2−1, so jq is not equal to qn−2 + qn2, qn−3 + qn2, . . . , 2qn2. The
coefficient of Tqn−i+qn2
, i≥ 1, is (−1)n2R + q
n−2,i−1Xn + φni(X0, X1, . . . , Xn−1).
Let n be odd and q= 1. Then G0+n, =(−1)n−12 G0+ q n−2,(Xn − Tq n+1 ) + φ003(X0, X1, . . . , Xn−1; T ) =(−1)n−12 { n−1 2 −1 X i=0 R+n−2,iTqn−i−2−q n−1 2 −1 +X j ∗ ψj(X0, X1, . . . , Xn−2)Tj}qXn+ (−1)n+12 1 2g 0 + n−1Tq n−1+qn−12 + φ4 =(−1)n+12 1 2g 0 + n−1Tq n−1+qn−12 + (−1)n−12 n−1 2 X i=1 Rn−2,i−1+ q XnTq n−i−qn−12 + (−1)n−12 X j ∗ ψjq(X0, X1, . . . , Xn−2)XnTjq+ φ4. (3-3)
Let n be odd and q = −1. Then G0+n,=(−1)n−12 { n−1 2 −1 X i=0 R−n−2,iTqn−i−2−q n−1 2 −1 + R− n−2,n−12 +X j ∗ ψj(X0, X1, . . . , Xn−2)Tj}qXn + (−1)n−12 2g0 − n−1Tq n−1+qn−12 + φ4 =(−1)n−12 2g0 − n−1Tq n−1+qn−12 + (−1)n−12 n−1 2 X i=1 Rn− q−2,i−1XnTq n−i−qn−12 + R − q n−2,n−12 + (−1)n−12 X j ∗ ψqj(X0, X1, . . . , Xn−2)XnTjq+ φ4. (3-4)
Lemma 3.2. Under the weight defined in (1 − 5) , we have if n is even, wt R+ni = q n+1 − 1 q− 1 − q n−i, if n is odd, wt R+ni = qn+1q−1−1 − qn−i+ qn−12 , wt R−ni = qn+1q−1−1 − qn−i− qn−12 , wt R− n,n+12 = qn+1−1 q−1 . Proof. By Lemma 2.9.
Lemma 3.3. For all n∈ N, we have (a) f2k ≡ (−1)kX 2(qk−1) q−1 k (mod X0, X1, . . . , Xk−1); (b) f2k+1≡ 2X 2(qk+1 −1) q−1 k X qk −1 q−1 k+1 + 2X 1+qk+1−1q−1 k · f(mod X0, X1, . . . , Xk−1); (c) f0 2k ≡ (−1)k+1X (q+1)(qk −1) q−1 k (mod X0, X1, . . . , Xk−1); (d) f0 2k+1≡ 2X (qk+1 −q) q−1 k X qk+1 −1 q−1 k+1 + X qk+1−1 q−1 k · f1(mod X0, X1, . . . , Xk−1)
where f and f1 are two polynomials.
Proof. (a) and (c) follow from direct computation. Using the Laplace expansion to the deter-minants defining f0
2k and f2k+10 , we can get (b) and (d).
Lemma 3.4. For n∈ N, we have (a) g2k−10+ ≡ (−1)k(k−1)2 X qk −1 q−1 k (mod X0, X1, . . . , Xk−1); (b) g2k0+≡ (−1)k(k−1)2 2X qk+1−1 q−1 k (mod X0, X1, . . . , Xk−1). (c) g2k0 −≡ (−1)kXqk −1q−1 k+1 (mod X0, X1, . . . , Xk).
Proof. Step 1. By definition 2.1
f2k+1≡ det Xk 0 Xkq ∗ .. . Xkq2 ... Xk 0 Xkqk Xk+1 Xkq Xk+1q Xkq2 ... ∗ ... ... 0 Xk+1qk−1 Xkqk
Note that the matrix defining f2k+1 is symmetric. Expanding it along the (k + 1)th column,
we see that f2k+1≡2X qk+1 −1 q−1 k det Bk(mod X0, X1, . . . , Xk−1). ≡0(mod X0, X1, . . . , Xk−1, X qk+1−1 q−1 k ) (3-5)
where Bk = Xk+1 Xkq Xk+1q Xkq2 0 ... ... * ... Xkqk−1 Xk+1qk−1 . Moreover, we have det Bk ≡ X qk −1 q−1 k+1 (mod X0, X1, . . . , Xk) (3-6) Step 2. f2k+10 ≡ det Xkq 0 Xkq2 ∗ .. . ... Xk xqkk Xk+1 Xkq ... ... ... ∗ ... xqkk 0 Xk+1qk Xkqk+1
Expanding along the (k + 2)th column, we have
f2k+10 ≡ det Dk+1· X q(qk−1) q−1 k ± X qk+1 k · φ(mod X1, X2, . . . , Xk−1) where Dk+1 = Xk+1 Xkq Xqk+1Xkq2 0 ... ... * ... Xkqk Xk+1qk Since det Dk+1= X qk+1−1 q−1 k+1 + X q k · ψ, so f2k+10 ≡(X qk+1 −1 q−1 k+1 + X q k· ψ) · X q(qk −1) q−1 k ± X qk+1 k · φ(mod X1, X2, . . . , Xk−1) ≡X qk+1−q q−1 k X qk+1 −1 q−1 k+1 (mod X0, X1, . . . , X qk+1−1 q−1 k ). (3-7) By (3.2) and (3.4), we have g2k+1± ≡ X qk+1 −q q−1 k X qk+1−1 q−1 k+1 (mod X0, X1, . . . , X qk+1−1 q−1 k ).
Step 3. Now by the definition 2.3, g2k+10 ± = g±2k+1/(g0 ∓2k−1)q and induction, the formula for g2k+10 ± is proved.
Step 4. Let Ak = Xk Xkq ∗ 0 ... xqk−1k , A0k−1 = Xkq Xkq2 ∗ 0 ... xqk−1k and Bk+1 = Xk Xkq 0 ∗ ... xqkk . Then we have f2k ≡ 0 Ak Atk 0 = (−1)kX 2(qk −1) q−1 k (mod X0, X1, . . . , Xk−1), f2k0 ≡ 0 A0k−1 Bk+1 0 = (−1)k+1X (q+1)(qk −1) q−1 k (mod X0, X1, . . . , Xk−1) By the definition 2.2 of g+ n, we have g2k+ ≡ (−1)k+12X(q+1) qk −1 q−1 k (mod X0, X1, . . . , Xk−1) (3-8)
Step 5. Now from Lemma 2.5(a), f2k+1= (−1)kg0 +2k g2k0 −. Comparing (3-1),(3-2) and (3-3), we
get the formula for g−2k
Lemma 3.5. For any j such that 1≤ j ≤ k − 1, we have g2k−10 ± ≡ ±X qk−qk−j q−1 j X qk−j −1 q−1 k+j (mod X0, X1, . . . , X 1+qk −qk−jq−1 j , . . . , Xk+j−1)
Proof. Step 1. It is not difficult to see that
f2k ≡ ... Xk+j ∗ ... ... Xk+jqk−j−1 ... Xjqk−j 0 ... ... Xjqk−1 Xjqk−j 0 0 ... ... Xkqk−1 0 Xk+j 0 0 ... ... ∗ Xk+jqk−j−1 0 ≡(−1)(k+1)(j+1)−1X2 qk−qk−j q−1 j X 2qk−j−1q−1 k+j (mod X0, X1, . . . , X 1+2(qk −qk−j )q−1 j , . . . , Xk+j−1) (3-9) Step 2. Observing the matrices defining f2k−1 and f2k−10 ,we see that the exponents of Xk+j in
f2k−1 is ≤ 2(qk−j−1q−1−1), which in f0
2k−1 is ≤ q
k−j−1
q−1 + q
qk−j−2−1
in g2k±−1 is ≤ (q + 1)(qk−j−1q−1−1). On the other hand, by induction hypothesis, the exponents of Xk+j in g2k−3± is q k−j−2−1 q−1 . So the exponents of Xk+j in g2k0 ±−1 is ≤ q k−j−1 q−1 .
Using Lemma 2.5(a) and comparing the exponent of Xk+j in (3-9) and in g0 ±2k−1, we see
that the exponents of Xk+j in g2k−10 ± is equal to q
k−j−1 q−1 . Moreover, wtg + 2k−1 = wtg−2k−1= q 2k −1 q−1 ,
so they also have the same exponents for Xj. The Lemma follows.
Lemma 3.6. (a) G0 +2k ≡ (−1)k+1Xqk+1 −1q−1 k+1 T2q k (mod X0, X1, . . . , Xk), (b) G0 −2k ≡ (−1)kXqk −1q−1 k+1 (mod X0, X1, . . . , Xk).
Proof. Step 1. Note that f2k0 ≡ 0 A (k−1) B(k+1) 0 (mod X0, X1, . . . , Xk−1, X qk −1 q−1 k ) (3-10) where A(k−1)= Xkq Xkq2 * 0 . .. Xkqk−1 and B (k+1)= Xk Xk+1 Xkq 0 Xk+1q Xkq2 . .. ... * . .. Xkqk−1 Xk+1q[k−1 0 .
Note that the (k + 1)th column in (3-5) is zero and det A(k−1)= X
qk −q q−1
k . Thus to compute all
minors f0
ij in Lemma 2.8(b) carefully, we find that the only nonzero minor is fk,k+10 . And we
get F2k0 ≡ X qk−q q−1 k X qk−1 q−1 k+1 T 2qk (mod X0, X1, . . . , Xk−1, X qk −1 q−1 k )
Step 2. Similar to Step 1. , Using f2k ≡ 0 A (k) A(k)t 0 (mod X0, X1, . . . , Xk−1 ) where A(k) = Xk Xkq2 * 0 . .. Xkqk−1
. We use Lemma 2.8 to find F2k. Most minors are zero, we have F2k ≡ X 2qk −1q−1 k + k−1 X i=0 2fk+1,i+1Tq k+qi + k−1 X i=1 2fk+2,i+1Tq k+1+qi +· · · + 2f2k,kTq 2k+qk (mod X0, X1, . . . , Xk−1)
All the minors appear in the above are multiple of X
qk −1 q−1 k , so F2k ≡ 0(mod X0, X1, Xk−1, X qk −1 q−1 k ).
Step 3. From Steps 1,2, we have G2k±≡ Xqk−qq−1 k X qk −1 q−1 k+1 T 2qk (mod X0, X1, . . . , Xk−1, X qk −1 q−1 k ) By definition of G0 ±
n and induction, the Lemma follows.
Lemma 3.7. (a) R+2k,0 ≡ X qk −1 q−1 k (mod X0, X1, . . . , Xk−1). (b) R+2k,j ≡ X qk −1 q−1 −qk−j k X qk−j k+j (mod X0, X1, . . . , bXk, . . . , bXk+j, . . . , X2k), 1 ≤ j ≤ k − 1 (c) R+2k,k ≡ (−1)k+1X qk+1−1 q−1 k+1 (mod X0, X1, . . . , Xk)
Proof. By Definition and Lemma 3.6, we have, modulohX0, X1, . . . , Xk−1i
G0 +2k = (F2k0 − F q+1 2 2k )/(G0 −2k−2)q ≡ (F2k0 − F q+1 2 2k )/(X qk−1 q−1 k )q, k even (F2k0 + (−F2k) q+1 2 )/(G0 − 2k−2)q ≡ (F2k0 + (−F2k) q+1 2 )/(−X qk −1 q−1 k )q, k odd (3-11)
Note that R+2k,j is the coefficient of Tq2k−j+qk in G0 +2k .
Step 1. We claim the term Tq2k−j+qk cannot appear in the expansion of Fq+12
2k . Let S be the
set of all exponents of T appear in F2k. By (3-6)
S =S1∪ S2∪ S3∪ {0},
S1 ={ql+ qm : 2k− j + 1 ≤ l ≤ 2k − 1, 0 ≤ m ≤ l} ∪ {q2k−j+ qm : k < m < 2k − j},
S2 ={q2k−j+ qm, 0 ≤ m ≤ k},
S3 ={ql+ qm : k ≤ l ≤ 2k − j − 2, 0 ≤ m ≤ l} ∪ {q2k−j−1+ qm : 0 ≤ m ≤ k − 1}.
The exponents of T may appears in Fq+12 2k is Pα i=1ui+ Pβ i=1vi+ Pγ i=0w3, ui∈ S1, vi ∈ S2, wi ∈ S3, α+β+γ≤q+12 . Now suppose q2k−j + qj = α X i=1 ui+ β X i=1 vi+ γ X i=0 w3. (3-12)
Since ui ≥ q2k−j + qj, so α = 0. If v1, v2 ∈ S2, then v1+ v2 > q2k−j + qk, hence β = 0, 1. If
β = 1 and v1= q2k−j+ qk, then (3-12) holds.
Suppose v1= q2k−j+ qm, m < k. then v1+ ∗ = q2k+j+ qk and qm+ ∗ = qk. ∗ is a sum
of some wi. Say
qm+ w1+ · · · + vβ = qk, β ≤
q− 1 2 .
Let ν be the valuation determined by q. The valuation of the right hand side of the above equation is k, so, in the left hand side, the terms with the minimal valuation must be cancelled. thus the only possibilities is
β = q − 1
So qm+ v
1+ · · · + vβ = qm+1, and m = k − 1. Thus we have
w1 = q2k−j + qk−1, v1 = · · · = vq−1
2 = 2q k−1.
Assume that β = 0. Suppose that there are only q−32 w0
is which are equal to 2q2k−j−1.
Then q − 3 2 (2q 2k−j−1+ w 1+ w2) = q2k−j+ qk w1+ w2 = 3q2k−j−1+ qk w1+ w2 ≤ 2(q2k−j−1+ q2k−j−2) q2k−j−1+ qk ≤ 2q2k−j−2 (q − 2)q2k−j−2+ qk ≤ 0.
A contradiction. Thus we have w1=· · · wq−1
2 = 2q
2k−j−1, wq+1 2 = q
2k−j−1+ qk.
Step 2. In F2k0 the coefficient of Tq2k−j+qj is
(−1)k+j(fk,2k0 −j+1+ f2k0 −j,k+1)(mod X0, X1, . . . , bXk, . . . , bXk+j, . . . , X2k)
. By direct computations, one can check that
fk,2k−j+10 ≡ 0(mod X0, X1, . . . , bXk, . . . , bXk+j, . . . , X2k)
f2k−j,k+10 ≡ (−1)j+1Xqk −qq−1 +qkq−1−1−qk−j
k X
qk−j
k+j (mod X0, X1, . . . , bXk, . . . , bXk+j, . . . , X2k).
Step 3. Substituting the results in Steps 1 and 2 into (3-6), the Lemma follows. Lemma 3.8. (a) Consider the ring homomorphism
Φ;Fq[X0, X1, . . . , Xn] → F(Q+n0, Q+n1, . . . , Q+nn)
Xi 7→ Q+ni.
Then ker Φ = hg0+ n i.
(b) Consider the ring homomorphism
Φ;Fq[X0, X1, . . . , Xn] → F(Q−n0, Q−n1, . . . , Q−nn)
Xi 7→ Q−ni.
Then ker Φ =hg0+
n i, if n is odd; ker Φ = hgn0−i, if n is even.
Proof. Comparing the transcendental degrees of the kernel and the image of Φ, we see that ker φ is a principle prime ideal of height one. By Lemma 2.12, g±
n ∈ ker Φ and it is irreducible,
Lemma 3.9. (a) 2γfq(q+1)2 n−2 R + q+1 n−1,1 ≡ (−1) q+1 2 g± q(q+1)
n−4 gn+(mod g0 +n−2), where γ is defined in
(3-17). (b) Define N = M1,n+1;n+1,n+2(n+2) − δq+12 n Mn;n+1(n+1)M(n+1) q+1 2 n;n . Then N ≡ g0± qn−2R+n+1,1(mod gn−10 +). Proof. fn = (−1)b n 2cg0 + n−1g0 −n−1= (−1)b n 2c g + n−1 (gn−30 − )q gn−1− (gn0 +−3)q = (−1) bn 2c f 0 2 n−1− fnq+1−1 (−1)bn−22 cf q n−2 . (3-13) Since fn−1= (−1)bn−12 cg0 + n−2gn−20 − ≡ 0(mod gn−20 + ), (3-14) so from (3-13) and (3-14), fn ≡ − f0 2 n−1 fn−2q (mod g 0 + n−2) and fq(q+12 n−2 f q+1 2 n ≡ (−1) q+1 2 f0 q+1 n (mod gn−20 + ). (3-15)
Apply Lemma 2.4 to the matrix defining f0
n and by (3-13), we get
fn−2q fn0 = fn−10 Mn−1;n(n) q − fnq−1M1,n;n,n+1(n+1) ≡ fn0−1Mn(n) q−1;n(mod gn−20 + ). (3-16) By the definition (2.2) of g0 +n−2, we have
fn−20 ≡ fq+12 n−2(mod gn−20 + ) n = 2, 3(mod 4) (−1)q−12 f q+1 2 n−2(mod g0 +n−2) n = 0, 1(mod 4) Define γ = −1, if n ≡ 0, 1(mod 4); = (−1)q+12 , if n ≡ 2, 3(mod 4). (3-17) Then f0 n−2 ≡ (−1) q+1 2 γf q+1 2 n−2(mod gn−20 + ). By (3-14) and (3-15), f q(q+1) 2 n−2 gn+= f q(q+1) 2 n−2 (fn0 + γf q+1 2 n )≡ f q(q−1) 2 n−2 fn−10 M (n) q n−1;n+ γ(−1) q+1 2 f0 q+1 n−1 = fn−10 {fq−12 n−2M (n) n−1;n+ γ(−1) q+1 2 f0 n−1}q (3-18) Moreover, Mn−1;n(n) = (−1)nfn−20 Xn−1+ ψ = (−1)n(−1)q+12 γf q+1 2 n−2Xn−1+ ψ We proceed to compute (3-16), fq−12 n−2M (n) n−1;n+ γ(−1) q+1 2 f0 n−1= (−1)n(−1) q+1 2 γfq n−2Xn−1+ ψ + (−1) q+1 2 (−1)nγfq n−2Xn−1+ ψ0 = (−1)n+q+12 {2γfq n−2Xn−1+ ξ}
From (3-18) and (2-2), f q(q+1) 2 n−2 gn+≡ (−1) q+1 2 2γfq(q+1) n−2 X q+1 n−1+ . . . . g+n = (−1)q+12 2γf q(q+1) 2 n−2 Xn−1q+1+ . . . Since R0+ q+1n−1,1 = gn−40± q(q+1)Xn−1q+1 + . . . , so 2γf q(q+1) 2 n−1 R+ q+1n−1,1 ≡ (−1) q+1 2 g0± q(q+1) n−4 g0 +n (mod g0 +n−2). (b) Define N = M1,n+1;n+1,n+2(n+2) − δq−12 M(n+1) n;n+1M (n+1) q−12 n;n = fnqXn+1+ . . . = g0+ qn−2gn−20− qXn+1+ · · · R+n+1,1= R+ qn−1,0Xn+1+ · · · = g0± qn−2Xn+1+ · · ·
Thus the result follows.
Let D1 = x1 x2 . . . xn xq1 xq2 . . . xqn .. . ... ... xq1n−1 xq2n−1 . . . xqnn−1 and D2 = x1 x2 . . . xn xq1 xq2 . . . xqn .. . ... ... c x1q n−1 c x2q n−1 . . . xcnq n−1 xq1n xq2n . . . xqn n . Then M1,n+1;n+1,n+2(n+2) = D1D2qδn Mn;n+1(n+1) = D1δnD2 Mn;n(n+1) = D12δn so N (Qn0, . . . , Qnn) = 0
Lemma 3.10. (a) Let n be even and = −1 or n be odd and = 1. (i) G0+n,−1(Q+n−1,0, . . . , Q+n−1,0; Xn) = 0
(ii) If a polynomial φ(X0, X1, Xn−1; T ) =PNk=0ak(X0, . . . , Xn−1)Tk satisfies φ(Q+n−1,0,
. . . , Q+n−1,0; Xn) = 0, then ak ∈ hg0+n−1i.
(iii) R+ni ∈ hgn−10+ i.
(b) Let n be odd and (q) = −1.
(i) G0−n,−1(Q−n−1,0, . . . , Q−n−1,0; Xn) = 0
(ii) If a polynomial φ(X0, X1, Xn−1; T ) =PNk=0ak(X0, . . . , Xn−1)Tk satisfies φ(Q−n−1,0,
(iii) R−ni ∈ hgn−10− i. Proof. (a)(i) Gn,0+(Q+n−1,0, Q+n−1,1, · · · , Q+n−1,n−1; Xn) =gn0+(Q+n−1,0+ Xn2, Q+n−1,1 + Xnq+1, · · · , Q+n−1,n−1+ Xnqn−1+1) =g0+n (Q+n,0, Q+n,1, · · · , Q+n,n−1) =0. (ii) 0 = φ(Q+n−1,0, . . . , Q+n−1,0; Xn) =PNk=0ak(Qn−1,0+ , . . . , Q+n−1,0)Tk. Since ak(Q+n−1,0, . . . , Q+n−1,0) ∈
Fq[x1, . . . , xn−1], hence ak(Q+n−1,0, . . . , Q+n−1,0) = 0 for all k. The statement follows from
Lemma 3.9(a).
(iii) Applying (ii) to Gn,0+−1, we get the result. The proofs of (b) are similar.
Lemma 3.11. All the coefficients of G0+
n belong to the Fq[X0, . . . , Xn−1]-module generated by
( R+
n0, R+n1, . . . , R+n,bn+1
2 c
, if n is enen and = −1; or n is odd and (q) = 1; R−n0, R−n1, . . . , R−
n,bn+12 c
, if n is odd and (q) = −1 .
Proof. Let Mn be the Fq[X0, . . . , Xn−1]-module generated by Rn0±, Rn1±, . . . , R±n,bn+1
2 c
. We’ll prove by induction on n. For n = 1, 2, it is easy to see check directly.
Let n > 2. First suppose n is even and = −1. Consider (3-2). In this expression ψj(X0, . . . , Xn−2) is a coefficient of G0 +n−2, so by the induction hypothesis, we have ψj ∈ Mn−2
for all j, i.e.
ψj(X0, . . . , Xn−2) = h0jRn+−2,0+ · · · + hbn−12 c,jR +
n−2,bn−12 c
for some hij ∈Fq[X0, . . . , Xn−3]. Let
ψj0(X0, . . . , Xn) := hq0jRn1+ +· · · + h q bn−1 2 c,j R+ n,bn−1 2 c+1 ∈ Mn . Then by Lemma 3.1, ψj0 − (−1)bn2cψq jXn = bXn+12 c i=1 hqi−1,j(R+n,i− (−1)bn2cR+ q n−2,i−1Xn) = bn+1 2 c X i=1 hqi−1,jφn,i ∈ Fq[X0, . . . , Xn−1]. So (−1)bn2cψq jXn = hq0jR+n1+ · · · + h q bn−12 c,j R+ n,bn−12 c+1 + bn+1 2 c X i=1 hqi−1,jφn,i. Also (−1)n2R+ q
Substituting these into (3-1) and collecting all coefficients of R+ni, we get G0 +n,−1(T ) = h0R+n0+ h1R+n1+ · · · + hbn+1 2 cR + n,bn+12 c + φ0(X0, . . . , Xn−1, T ) where hj ∈ Fq[X0, . . . , Xn−1; T ]. We write φ0(X0, . . . , Xn−1; T ) =PNk=0ak(X0, . . . , Xn−1)Tk.
Then by Lemma 3.9 ak(Q+n−1,0, . . . , Q+n−1,n−1) = 0 for all k . By Lemma 3.8, ak ∈ hg0+n−1i =
hR+n0i. Hence all coefficients of G0+n belong to Mn.
To prove the other cases, we just need to replace (3-2) by (3-3) and (3-4), and use the same arguments.
A key technical part of the proof in §4 is the use of homogeneous regular sequence. So we first develop some basic properties about regular sequence which will be needed later. See [12].
Proposition 3.12. If x1, x2, . . . , xn ∈ R is a regular sequence, then x1 is a non-zerodivisor
in R/hx2, x3, . . . , xni.
Proof. It suffices to prove for n = 2. So suppose x1, x2 is a regular sequence. We want to show
that x1r ∈ hx2i implies r ∈ hx2i. Assume that x1r = x2s. Because x2 is a non-zerodivisor in
R/hx1i, s = x1s0 for some s0 ∈ R. Then x1r = x2x1s0. Since x1 is a non-zerodivisor in R, we
may cancel it from both sides of the equation and get r = x2s0∈ hx2i.
Lemma 3.13.
(a) If x1, x2, . . . , xi, . . . , xn ∈ R is a regular sequence, and x1, x2, . . . , xi−1, x0i, xi+1, . . . , xn ∈
R is a regular sequence, then x1, x2, . . . , xi−1, xix0i, xi+1, . . . , xn is a regular sequence.
(b) If x1, x2, . . . , xn ∈ R is a regular sequence, then x1k1, xk22 , . . . , xknn is a regular sequence
for any k1, k2, . . . , kn ∈ N.
To make the proof in §4 go more smoothly, we establish several lemmas below. The first one is just a direct corollary of Lemma 3.13, but it is a situation we’ll often encounter. Lemma 3.14. Let R = F [X0, X1, . . . , Xk, Y1, Y2, . . . , Yn], the polynomial ring of k + n + 1
variables over a field F . Suppose we have a sequence S1, S2, . . . , Sm ∈ R, m ≤ n, such that
Si= n X j=1 aijYj where aij ∈ F [X0, X1, . . . , Xk] and
aij ≡ γijXi+j−2lij (mod X0, X1, . . . , Xi+j−3)
for some γij ∈ F×, lij ∈ N. (Here we require k ≥ m + n − 2 so that this condition makes sense
). Then X0, X1, . . . , Xn−2, S1, S2, . . . , Sm is a regular sequence in R.
Proof. Observe that
S1≡ γ1nXn−1l1n Yn(modhX0, X1, . . . , Xn−2i) (3-19)
S2≡ γ2nXnl2nYn(modhX0, X1, . . . , Xn−1i) (3-20)
S3≡ γ3,nXn+1l3,nYn−1(modhX0, X1, . . . , Xni) (3-22)
S3 ≡ γ3,n−1Xnl3,n−1Yn−1(modhX0, X1, . . . , Xn−1, Yni) (3-23)
S3 ≡ γ3,n−2Xn−1l3,n−2Yn−2(modhX0, X1, . . . , Xn−2, Yn−1, Yni) (3-24)
.. .
The lemma is proved by applying Lemma 3.11 over and over again. To explain the procedure, we prove the case m = 3 explicitly.
From (3-22), X1, . . . , Xn−1, Xn, S3 is a regular sequence.
From (3-23), X1, . . . , Xn−1, Yn, S3 is a regular sequence.
So by (3-20) and Lemma 3.11, X1, . . . , Xn−2, Xn−1, S2, S3 is a regular sequence.
From (3-23), X1, . . . , Xn−2, Yn, Xn−1, S3 is a regular sequence.
From (3-24), X1, . . . , Xn−2, Yn, Yn−1, S3 is a regular sequence.
So by (3-21) and Lemma 3.13, X1, . . . , Xn−2, Yn, S2, S3 is a regular sequence.
Thus by (3-19) and Lemma 3.13 again, we see that X1, . . . , Xn−2, S1, S2, S3 is a regular
sequence.
In the following Lemmas 3.15, 3.16 and 3.17, let R = R0⊕ R1⊕ R2⊕ · · · be a positively
graded ring, and for any x ∈ R, we denote its homogeneous component of the highest degree by in(x).
Lemma 3.15. Let x1, x2, . . . , xn ∈ R be homogeneous elements with positive degrees. If
x1, x2, . . . , xn is a regular sequence, then x1, x2, . . . , xn is a regular sequence in any order.
Proof. Since any permutation is a product of transpositions, it suffices to prove for n = 2. So suppose x1, x2 ∈ R are homogeneous elements with positive degrees, and x1, x2 is a regular
sequence. By Lemma 3.11, we only need to show that x2is a non-zerodivisor in R, i.e. x2y = 0
will imply y = 0. We may assume y to be homogeneous. Since x1, x2 is a regular sequence,
x2y = 0∈ hx1i implies y = x1y1 for some y1∈ R, and we may assume y1 to be homogeneous.
If y1 = 0, then we’re done; if y1 6= 0, then deg y1 < deg y because deg x1 > 0. Now x2y =
x2x1y1 = 0, and because x1 is a non-zerodivisor, x2y1 = 0. Repeat the above argument, we
have y1 = x1y2 for some homogeneous y2 ∈ R, and either y2 = 0 or deg y2 < deg y1 < deg y.
Note that since R is positively graded, this process can not be repeated indefinitely. Hence we must have y = 0.
Lemma 3.16. Let x1, x2, . . . , xn ∈ R, not necessarily homogeneous, and suppose in(x1), in(x2),
. . . , in(xn) is a regular sequence. If s ∈ hx1, x2, . . . , xni, then there exist y1, y2, . . . , yn ∈ R such
that s =Pni=1xiyi and deg xi+ deg yi ≤ deg s, 1 ≤ i ≤ n.
Proof. Induction on n. n = 1 is obvious, so suppose the statement holds for n − 1. Since s ∈ hx1, x2, . . . , xni, there exist y1, y2, . . . , yn ∈ R such that s = Pni=1xiyi. Among all the
possible choices of y1, y2, . . . , yn, choose one such that the number m := max{deg xi+ deg yi|
1 ≤ i ≤ n} is the smallest. We are going to prove by contradiction that in this case, m ≤ deg s. Suppose m > deg s. Let di= m − deg xi, and let yi,di be the homogeneous component of
degree di in yi. Because m > deg s, we have n
X
i=1
