Modular Invariants of Parabolic Subgroups of General Linear Groups

Journal of Algebra 232, 197᎐208 Ž2000. doi:10.1006rjabr.2000.8395, available online at http:rrwww.idealibrary.com on

Modular Invariants of Parabolic Subgroups of General Linear Groups Pham Anh Minh and Vo ˜ Thanh Tung ` Department of Mathematics, College of Sciences, Uni¨ ersity of Hue, Dai Hoc Khoa Hoc, Hue, Vietnam Communicated by Peter M. Neumann Received October 19, 1999

Let p be a prime number and let A be an elementary abelian p-group of rank m. The purpose of this paper is to determine a full system for the invariants of parabolic subgroups of the general linear group GLŽ m, ⺖p . in H *Ž A, ⺖p .. A relation between these invariants and Dickson ones is also obtained. 䊚 2000 Academic Press

1. INTRODUCTION Let p be a prime number and let G s Gn1 n 2 ⭈⭈⭈ n k, k G 1, n1 q ⭈⭈⭈ qn k s m, be the parabolic subgroup of GLŽ m, ⺖p . given by

¡A

Gn1 ⭈⭈⭈ n k s

~



¢

) A2 ⭈ 0

1

0 ⭈ 0

⭈⭈⭈ ⭈⭈⭈ ⭈⭈⭈ ⭈⭈⭈

) ) ⭈ Ak

0

¦

¥

A i g GL Ž n i , ⺖p . , 1 F i F k .

§

Note that GLŽ m, ⺖p . acts canonically on m

A s H * Ž ⺪rp . , ⺖p s

ž

/

½

⺖ 2 w y 1 , . . . , ym x

for p s 2

⌳ w x 1 , . . . , x m x m ⺖p w y 1 , . . . , ym x

for p ) 2,

with ⌳w x 1 , . . . , x m x Žresp. ⺖p w y 1 , . . . , ym x. the exterior Žresp. polynomial. algebra with variables x 1 , . . . , x m Žresp. y 1 , . . . , ym . over ⺖p . In this paper, we shall determine a full system for the invariants of G in A. We remind 197 0021-8693r00 $35.00 Copyright 䊚 2000 by Academic Press All rights of reproduction in any form reserved.

MINH AND TUNG `

198

the reader that, modulo nilradical, this subring of invariants was determined in w3x as a polynomial algebra of which the generators are a suitable collection of Dickson polynomials; later, in w2x, it is shown that these generators can be produced by a recursive construction, as follows. For 1 F i F m, let Vi s Vi Ž y 1 , . . . , yiy1 , yi . s

Ł Ž ␭1 y1 q ⭈⭈⭈ q␭iy1 yiy1 q yi .

␭ jg⺖ p

be the Mui invariant. For 0 F s F r, let L r s L r Ž y 1 , . . . , yr . s V1 ⭈ ⭈⭈⭈ ⭈ Vr

s

y1 y 1p ⭈ y 1p

ry 1

⭈⭈⭈ ⭈⭈⭈ ⭈⭈⭈ ⭈⭈⭈

yr yrp ⭈ yrp

ry 1

and let Q r, s s Q r, s Ž y 1 , . . . , ys . be the Dickson invariants defined recursively by Q r , 0 s L rpy 1 p Q r , s s Q ry1 , sVr py 1 q Q ry1, sy1 ,

where Q s, s s 1. It follows from w1x that Vi s Ž y1 .

iy1

iy1

Ý Ž y1.

s

s

Q iy1, s yip .

ss0

Set P s ⺖p w y 1 , . . . , ym x, m 0 s 0, m i s Ýijs1 n j for i ) 0 Žso m s m k .. For 1 F j F n iq1 , define ¨ iq1, j s Vm iq1 Ž y 1 , . . . , ym i , ym iqj .

qiq1, jy1 s Q n iq 1 , jy1 Ž ¨ iq1, 1 , . . . , ¨ iq1, n iq 1 . . so ¨ 1, j s y j , q1, jy1 s Q n1 , jy1. The following was due to Hewett. THEOREM 1 Žw2, Theorem 1.4x; compare w3, Theorem 2.2x.. P G s ⺖p w q1, 0 , . . . , q1 , n1y1 , q2 , 0 , . . . , q q , n 2y1 , . . . , qk , 0 , . . . , qk , n ky1 x .

199

MODULAR INVARIANTS

For p ) 2 and for 0 F s F r y 1, set x1 y1 ⭈⭈⭈ p Mr , s s y 1

⭈⭈⭈ ⭈⭈⭈ ⭈⭈⭈ ⭈⭈⭈

x2 y2 ⭈⭈⭈

sy 1

y 2p

sy 1

sq 1 y 1p

sq 1 y 2p

⭈⭈⭈

⭈⭈⭈

ry 1 y 1p

ry 1 y 2p

⭈⭈⭈ ⭈⭈⭈ ⭈⭈⭈

xr yr ⭈⭈⭈ yrp

sy 1

.

sq 1 yrp

⭈⭈⭈ yrp

ry 1

Let Ž s1 , . . . , s j . be a sequence of integers with 0 F s1 - ⭈⭈⭈ - s j - r. As shown in w5, Proposition I.4.5x, the product Mr, s1 ⭈ Mr, s 2 ⭈ ⭈⭈⭈ ⭈ Mr, s j has the factor L rjy1. We have then Mui invariants M r , s 1 , . . . , s j s M r , s 1 , . . . , s jŽ x 1 , . . . , x r , y 1 , . . . , y r . s Ž y1 . Note that

¡A

K s K n1 ⭈⭈⭈ n k s

~



¢

jŽ jy1 .r2

) A2 ⭈ 0

1

0 ⭈ 0

Mr , s1 ⭈ Mr , s 2 ⭈ ⭈⭈⭈ ⭈ Mr , s jrL rjy1 .

⭈⭈⭈ ⭈⭈⭈ ⭈⭈⭈ ⭈⭈⭈

) ) ⭈ Ak

0

¦

¥

A i g SL Ž n i , ⺖p . , 1 F i F k

§

is a subgroup of G. Define l i s L n iŽ ¨ i , 1 , . . . , ¨ i , n i . ni

s

Ł Ł Ž ␭1¨ i , 1 q ⭈⭈⭈ q␭ j¨ i , jy1 q ¨ i , j . js1 ␭ jg⺖ p

for 2 F i F k. We shall prove THEOREM 2 Ži. P K s ⺖p w l 1 , q1, 1 , . . . , q1, n1y1 , l 2 , q2, 1 , . . . , q2, n 2y1 , . . . , l k , qk, 1 , . . . , qk, n ky1 x. Žii. As a module o¨ er P K , A K is free and has a basis consisting of 1 and all elements of Mm i , s1 , . . . , s j N 1 F i F k, 1 F j F m i , 0 - s1 - ⭈⭈⭈ - s j - m i y 1, m iy1 F s j 4 . In other words, there exists a decomposition m

AK s P K [

[[

[

js1 m iGj 0Fs 1 - ⭈⭈⭈ -s jFm iy1 m iy1 Fs j

Mm i , s 1 , . . . , s j P K .

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Set R r, s1 , . . . , s j s Mr, s1 , . . . , s j L rpy 2 . We shall also prove THEOREM 3. As a module o¨ er P G , A G is free and has a basis consisting of 1 and all elements of R m i , s1 , . . . , s j N 1 F i F k, 1 F j F m i , 0 - s1 - ⭈⭈⭈ - s j - m i y 1, m iy1 F s j 4 . In other words, there exists a decomposition m

AG s PG [

[[

[

js1 m iGj 0Fs 1 - ⭈⭈⭈ -s jFm iy1 m iy1 Fs j

R m i , s1 , . . . , s j P G .

For 0 F s F m, it is known that Q m, s is an invariant of GLŽ m, ⺖p ., hence of G. It follows from Theorem 1 that Q m, s can be expressed in terms of the elements qi, j . Such an expression is given by THEOREM 4.

⭈⭈⭈ qs k p 3q ⭈⭈⭈ s k p k Q m, s s Ý s1q ⭈⭈⭈ qs kss q1,p s2q q2, s 2 ⭈⭈⭈ qky1, s ky 1 qk, s k . 1 s

s

s

2. DICKSON AND MUI INVARIANTS For 1 F i F m, write ⺖p w y 1 , . . . , yi x s Pi , GLŽ i, ⺖p . s GLŽ i ., and SLŽ i, ⺖p . s SL i . The following was due to Dickson w1x. THEOREM 5 ŽDickson.. ⺖p w L i , Q i, 1 , . . . , Q i, iy1 x.

PiG L i s ⺖p w Q i, 0 , . . . , Q i, iy1 x and PiS L i s

In the remainder of this section, assume that p is an odd prime number. We are now interested in the elements Mr, s1 , . . . , s j . According to w4x Žsee also w6x., these invariants can be obtained by means of Steenrod operations P i Žcomposing with the Bockstein homomorphism ␤ ., as follows. Note iy 1 that Mr, 0, . . . , ry1 s x 1 ⭈⭈⭈ x r . For 1 F i F r, define Qi s P p ⭈⭈⭈ P 1␤ . We have $

$

P ROPOSITION 1. Q 1 ⭈⭈⭈ Qs 1 y 1 ⭈⭈⭈ Qs jy 1 ⭈⭈⭈ Qry 1Ž M r , 0, . . . , ry 1 . s Ž c r, j . Mr, s1 , . . . , s j with cŽ r, j . s Žy1.w r Ž ry1.yjŽ jy1.xr2 . Proof. As

␤ Ž Mr , s . s P i Ž Mr , s . s

for s s 0 otherwise

½

Lr 0

½

Mr , sy1 0

for i s p sy 1 , otherwise,

we have, for s F r, Pp Qs Ž Mr , 0 ⭈⭈⭈Mr , ry1 . s ␤ Ž Mr , 0 . P 1 Ž Mr , 1 . ⭈⭈⭈P

sy 1

Ž Mr , s . Mr , sq1 ⭈⭈⭈Mr , ry1

s L r Mr , 0 ⭈⭈⭈ Mr , sy1 Mr , sq1 ⭈⭈⭈ Mr , ry1 .

201

MODULAR INVARIANTS

Hence

Ž y1. r

Ž ry1 .r2

$

$

Lry1 Q1 ⭈⭈⭈ Qs1y1 ⭈⭈⭈ Qs jy1 ⭈⭈⭈ Qry1 Ž Mr , 0, . . . , ry1 . r $

$

$

$

s Q1 ⭈⭈⭈ Qs1y1 ⭈⭈⭈ Qs jy1 ⭈⭈⭈ Qry1 Ž y1 .

ž

r Ž ry1 .r2

Lry1 Mr , 0, . . . , ry1 r

/

s Q1 ⭈⭈⭈ Qs1y1 ⭈⭈⭈ Qs jy1 ⭈⭈⭈ Qry1 Ž Mr , 0 ⭈⭈⭈ Mr , ry1 . s Lryj r Mr , s 1 ⭈⭈⭈ Mr , s j s Ž y1 .

jŽ jy1 .r2

jy1 Lryj Mr , s 1 , . . . , s j r Lr

s Ž y1 .

jŽ jy1 .r2

Lry1 Mr , s 1 , . . . , s k . r

$

$

Therefore Q1 ⭈⭈⭈ Qs1y1 ⭈⭈⭈ Qs jy1 ⭈⭈⭈ Qry1Ž Mr, 0, . . . , ry1 . s cŽ r, j . Mr, s1 , . . . , s j . The proposition follows. COROLLARY 1.

If 0 F s1 - ⭈⭈⭈ - s jy1 - s j s r y 1, then we ha¨ e

Mr , s1 , . . . , s jy 1 , ry1 Ž x 1 , . . . , x ry1 , x r , y 1 , . . . , yry1 , 0 . s Ž y1 .

ryj

Mry1, s1 , . . . , s jy 1 ⭈ x r .

Proof. Set f s Mr, s1 , . . . , s jy 1 , ry1Ž x 1 , . . . , x ry1 ,$ x r , y1, . $ . . , yry1 , 0.. By Proposition 1, f is obtained from cŽ r, j . Q1 ⭈⭈⭈ Qs1y1 ⭈⭈⭈ Qs jy1 ⭈⭈⭈ Qry1Ž x 0 ⭈⭈⭈ x r . by setting yr s 0. Hence f s Ž y1 .

r Ž ry1 .yjŽ jy1 .r2

$

$

Q1 ⭈⭈⭈ Qs1y1 ⭈⭈⭈ Qs jy1 ⭈⭈⭈ Qry1 Ž x 0 ⭈⭈⭈ x ry1 . ⭈ x r

s ⑀ Mry1, s1 , . . . , s jy 1 ⭈ x r , with ⑀ s Žy1. r Ž ry1.yjŽ jy1.yŽ ry1.Ž ry2.qŽ jy1.Ž jy2.r2 s Žy1. ryj. The corollary follows. We shall use the following notation: given two sequences I s

s1 , . . . , s j 4 , J s t 1 , . . . , t j 4 , then I - J if there exists an integer l such that 1 F l F j, i l - jl and i n s jn for l - n F j. Let Tm be the subgroup of GLm consisting of all triangular matrices with 1 on the main diagonal. In w5x, Mui proved PROPOSITION 2 ŽMui.. Ži.

If 0 F s1 - ⭈⭈⭈ - s j F s y 1, then Ms, s1 , . . . , s jVsq1 ⭈⭈⭈ Vn s Mn , s1 , . . . , s j q

Ý Mn , t , . . . , t g t , . . . , t , 1

j

1

j

MINH AND TUNG `

202

where the summation is taken o¨ er the sequences Ž t 1 , . . . , t j . such that Ž t 1 , . . . ,t j . ) Ž s j y j q 1, . . . , s j . and g t , . . . , t are some in¨ ariants of Tn in Pm . 1 j Žii. Let m

fs

Ý

Ý

ssj 0Fs 1- ⭈⭈⭈ -s jssy1

Ms, s1 , . . . , s j g s1 , . . . , s jŽ y 1 , . . . , ym . .

Then f s 0 if and only if all g s1 , . . . , s j s 0. Žiii. Let fs

Ý

0Fs 1- ⭈⭈⭈ s jFsy1

Ms, s1 , . . . , s j g s1 , . . . , s jŽ y 1 , . . . , ym . .

Then f s 0 if and only if all g s1 , . . . , s j s 0. The subrings of invariants of GLm , SL m , and Tm in A were also determined by Mui w5x as follows. THEOREM 6 ŽMui.. We ha¨ e the following decompositions: A S L m s ⺖p w L m , Q m , 1 , . . . , Q m , my1 x m

[[

[

js1 s 1- ⭈⭈⭈ -s jFmy1

Mm , s1 , . . . , s j ⺖p w L m , Q m , 1 , . . . , Q m , my1 x ,

A G L m s ⺖p w Q m , 0 , Q m , 1 , . . . , Q m , my1 x m

[[

[

js1 s 1- ⭈⭈⭈ -s jFmy1

R m , s1 , . . . , s j ⺖p w Q m , 0 , Q m , 1 , . . . , Q m , my1 x ,

A T m s ⺖p w V1 , . . . , Vm x m

[[

[

[

js1 iGj s 1- ⭈⭈⭈ -s jsiy1

Mi , s1 , . . . , s j ⺖p w V1 , . . . , Vm x .

3. PROOFS OF MAIN RESULTS The proof of Theorem 2 is divided into the following lemmas. LEMMA 1. P K s ⺖p w l 1 , q1, 1 , . . . , q1, n1y1 , l 2 , q2, 1 , . . . , q2, n 2y1 , . . . , l k , qk, 1 , . . . , qk, n ky1 x.

203

MODULAR INVARIANTS

Proof. For every i with 1 F i F m, let Ii be the identity element of GLi . Set

¡I

Ls

and

¡A

Ss

~



¢

1

0 ⭈ 0

~



¢

0 A2 ⭈ 0

n1

)

⭈⭈⭈

)

0

In 2

⭈⭈⭈

)

⭈ 0

⭈ 0

⭈⭈⭈ ⭈⭈⭈

⭈ In k

⭈⭈⭈ ⭈⭈⭈ ⭈⭈⭈ ⭈⭈⭈

0 0 ⭈ Ak

0

¦

0

¥

§

¦

¥

A i g SL n i , 1 F i F k .

§

It is clear that S and L are subgroups of K and P K s Ž P L . S . By Theorem 6, we have P L s ⺖p w y 1 , . . . , yn1 , ¨ 2, 1 , . . . , ¨ 2, n 2 , . . . , ¨ k , 1 , . . . , ¨ k , n k x . Hence SL SL Ž P L . s ⺖p w y 1 , . . . , yn1 x n m ⺖p w ¨ 2, 1 , . . . , ¨ 2, n 2 x n S

1

m ⭈⭈⭈ m ⺖p w ¨ k , 1 , . . . , ¨ k , n k x

S Lnk

2

.

By Theorem 5, the lemma follows. LEMMA 2. fs

Let

Ý

Ý

m iGj 0Fs 1 - ⭈⭈⭈ -s jFm iy1 m iy1 Fs j

M m i , s 1 , . . . , s j f m i , s 1 , . . . , s jŽ y 1 , . . . , y m . .

Then f s 0 if and only if all f m i , s1 , . . . , s j s 0. Proof. Assume that the elements f m i , s1 , . . . , s j are not all equal to 0. Let m l be the smallest integer such that the elements f m l , s1 , . . . , s j are not all equal to 0 and let t 1 , . . . , t j be the smallest sequence satisfying f m l , t 1 , . . . , t j / 0. Following Proposition 2 Ži., f can be decomposed as fs

Ý

s1- ⭈⭈⭈ -s jsry1 Ž t 1 , . . . , t j .F Ž s 1 , . . . , s j .

qÝ

Ý

Mr , s 1 , . . . , s j g r , s 1 , . . . , s j

r-s s 1- ⭈⭈⭈ -s jssy1

Ms, s1 , . . . , s j g s, s1 , . . . , s j

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with g r, t 1 , . . . , t j s f m l , t 1 , . . . , t jVrq1 ⭈⭈⭈ Vm l . By Proposition 2 Žii., f s 0 implies g r, t 1 , . . . , t j s 0; hence f m l , t 1 , . . . , t j s 0, a contradiction. So all the elements f m i , s1 , . . . , s j are equal to 0. The lemma follows. We now assume that f s Ý i1 - ⭈⭈⭈ - i j x i1 ⭈⭈⭈ x i j f i1 ⭈⭈⭈ i jŽ y 1 , . . . , ym . is an invariant of K in A with j G 1. Since K contains Tm as a subgroup, f is also an invariant of Tm . By Theorem 6, we also assume that f is of form fs

Ý

Ý

sGj 0Fs 1- ⭈⭈⭈ -s jssy1

Ms, s1 , . . . , s j g s, s1 , . . . , s j ,

where all g s, s1 , . . . , s j are invariants of Tm . LEMMA 3. For j F r and m i - r - m iq1 , g r, s1 , . . . , s j contains yrq1 ⭈⭈⭈ ym iq 1; hence Vrq1 ⭈ ⭈⭈⭈ ⭈ Vm iq 1 as a factor. Proof. For m i - r - t F m iq1 , let ␧ t, r be the matrix with 1 in the Ž t, r .-position and 0 elsewhere. ␻ s 1 q ␧ t, r is then an element of K. So fs␻f s

Ý

Ý

␻ Ms, s1 , . . . , s j ␻ g s, s1 , . . . , s j

Ý

Ý

Ms, s1 , . . . , s j ␻ g s, s1 , . . . , s j

sGj 0Fs 1- ⭈⭈⭈ -s jssy1

s

sGj 0Fs 1- ⭈⭈⭈ -s jssy1 ty1

q

Ý

Ý

ssr 0Fs 1- ⭈⭈⭈ -s jssy1

Ms,X s1 , . . . , s j ␻ g s, s1 , . . . , s j

Ms,X s1 , . . . , s j

with s Ms, s1 , . . . , s jŽ x 1 , . . . , x ry1 , x t , x rq1 , . . . , x s , y 1 , . . . , yry1 , yt , . yrq1 , . . . , ys . Therefore 0sfy␻f s

Ý

Ý

sGj 0Fs 1- ⭈⭈⭈ -s jssy1

Ms, s1 , . . . , s jŽ g s, s1 , . . . , s j y ␻ g s, s1 , . . . , s j .

ty1

y

Ý

Ý

ssr 0Fs 1- ⭈⭈⭈ -s jssy1

As

Ms,X s1 , . . . , s j ␻ g s, s1 , . . . , s j .

Ž 1.

g s, s 1 , . . . , s j y ␻ g s, s 1 , . . . , s j s g s, s 1 , . . . , s jŽ y 1 , . . . , y m . y g s, s 1 , . . . , s j r

Ž y 1 , . . . , yr q ˘ yt , . . . , ym ., from Ž1., setting yt s 0 yields ty1

Ý

Ý

ssr 0Fs 1- ⭈⭈⭈ -s jssy1

Ms, s1 , . . . , s jŽ x 1 , . . . , x ry1 , x t , x rq1 , . . . , x s , t

y 1 , . . . , yry1 , 0, yrq1 , . . . , ys . g r , s1 , . . . , s j y 1 , . . . , ˘ 0 , . . . , ym s 0. Ž 2 .

ž

/

205

MODULAR INVARIANTS

For t s r q 1, Ž2. becomes rq1

Ý

0Fs 1- ⭈⭈⭈ -s jsry1

MrX , s1 , . . . , s j g r , s1 , . . . , s j y 1 , . . . , ˘ 0 , . . . , ym s 0.

ž

/

By Corollary 1 and by Proposition 2 Žii., it follows that g r, s1 , . . . , s j rq1

Ž y1, . . . , ˘ 0 , . . . , ym . s 0 for 0 F s1 - ⭈⭈⭈ - s j s r y 1, which implies that g r, s1 , . . . , s j contains yrq1 as a factor. We may then suppose that, for every s satisfying r - s - m iq1 and for 0 F s1 - ⭈⭈⭈ - s j s s y 1, g s, s1 , . . . , s j contains ysq1 ⭈⭈⭈ ym iq 1 as a factor. Then Ž2. implies t

MrX , s1 , . . . , s j g r , s1 , . . . , s j y 1 , . . . , ˘ 0 , . . . , ym s 0.

ž

Ý

0Fs 1- ⭈⭈⭈ -s jsry1

/

So g r, s1 , . . . , s j contains yt as a factor. Therefore g r, s1 , . . . , s j contains yrq1 ⭈⭈⭈ ym iq 1 as a factor. Since g r, s1 , . . . , s j is an invariant of Tm , it follows from Theorem 6 that g r, s1 , . . . , s j contains Vrq1 ⭈ ⭈⭈⭈ ⭈ Vm iq 1 as a factor. The lemma follows. The following is then straightforward from Lemma 3 and Proposition 2 Ži.. LEMMA 4. There is a decomposition fs

Ý

Ý

m iGj 0Fs 1 - ⭈⭈⭈ -s jFm iy1 m iy1 Fs j

M m i , s 1 , . . . , s j h m i , s 1 , . . . , s jŽ y 1 , . . . , y m . .

Combining with Lemmas 1 and 2, the following lemma completes the proof of Theorem 2. LEMMA 5. In the decomposition gi¨ en in Lemma 4, e¨ ery h m i , s1 , . . . , s j is an in¨ ariant of K. Proof. Let ␻ be an element of K. By Theorem 6, we have 0sfy␻f s

Ý

Ý

m iGj 0Fs 1 - ⭈⭈⭈ -s jFm iy1 m iy1 Fs j

Mm i , s 1 , . . . , s j Ž h m i , s 1 , . . . , s j y ␻ h m i , s 1 , . . . , s j . .

It follows from Lemma 2 that h m i , s1 , . . . , s j y ␻ h m i , s1 , . . . , s j s 0, whenever s1 - ⭈⭈⭈ - s j . The lemma follows.

MINH AND TUNG `

206

We now prove Theorem 3. For i F m, consider GLi as a subgroup of GLm via the inclusion i: GLi ª GLm A¬

ž

A 0

0 Imy i .

/

We have LEMMA 6. Let f g P and let i F m be an integer. Assume that p is odd and, for e¨ ery ␻ g GLi , ␻ f s detŽ ␻ . py 2 f. Then f contains L ipy2 as a factor. Proof. Note that f is an invariant of Ti . Hence, by Theorem 5, we need prove that f contains y jpy 2 as a factor, for every j F i. Assume that f contains y jn as a factor, with 0 F n - p y 2. So f s y jn g. Fix ␭ g ⺖p , ␭ / 0, 1, and let ␻ s Ž ␻ r s . be the matrix given by

¡ ¢

1 ␻r s s ␭ 0

~

if r s s / j if r s s s j otherwise.

It is clear that ␻ g G, detŽ ␻ . s ␭. Therefore n

␻ f s Ž ␭ y j . g Ž y 1 , . . . , y jy1 , ␭ y j , y jq1 , . . . , ym . s ␭ n y jn g Ž y 1 , . . . , y jy1 , ␭ y j , y jq1 , . . . , ym . . So

␭ n g Ž y 1 , . . . , y jy1 , ␭ y j , y jq1 , . . . , ym . s ␭ py 2 g Ž y 1 , . . . , ym . . Setting y j s 0 then yields j

j

g y1 , . . . , ˘ 0 , . . . , ym s ␭ py 2yn g y 1 , . . . , ˘ 0 , . . . , ym .

ž

/

ž

/

As ␭ / 0, 1 and n - p y 2, the above equality shows that g Ž y 1 , . . . , j

˘0 , . . . , ym . s 0. Hence g contains yj as a factor. The lemma follows. Now, let us assume that f is an invariant of G in A. It is clear that f is also an invariant of K. Hence we also assume that f has the decomposition given in Lemmas 4 and 5. The proof of Theorem 3 then follows from LEMMA 7.

h m i , s1 , . . . , s j contains L mpyi 2 as a factor.

207

MODULAR INVARIANTS

Proof. For every ␻ g G, as

␻ Mm i , s1 , . . . , s j s det Ž ␻ . Mm i , s1 , . . . , s j , for every ␻ g G, it follows from Lemma 2 that

␻ h m i , s1 , . . . , s j s det Ž ␻ . By Lemma 6, h m i , s1 , . . . , s j contains

L mpyi 2

py 2

h m i , s1 , . . . , s j .

as a factor. The lemma follows.

Our task is now to prove Theorem 4. Let X be an indeterminate and set ¨ k , X s Vm ky 1q1 Ž y 1 , . . . , ym ky 1 , X .

V⬘ s Vn kq1 Ž ¨ k , 1 , . . . , ¨ k , n k , ¨ k , X . . We first prove LEMMA 8. V⬘ s Vmq 1Ž y 1 , . . . , ym , X .. Proof. As V⬘ s

Ł Ž ␭1¨ k , 1 q ⭈⭈⭈ q␭n ¨ k , n k

␭ ig⺖ p

m ky1

s s s

Ł

␭ ig⺖ p

Ł

␭ ig⺖ p

m Ý Ž y1.

ky 1 qs

k

q ¨k, X . s

s

Q m ky 1 , s ␭1 ymp ky 1q1 q ⭈⭈⭈ q␭ n k ymp q X p

ž

ss0

Vm ky 1q1 Ž y 1 , . . . , ym ky 1 , ␭1 ym ky 1q1 q ⭈⭈⭈ q␭ n k ym q X .

Ł Ž ␭1 y1 q ⭈⭈⭈ q␭m ym q X .

␭ ig⺖ p

s Vmq 1 Ž y 1 , . . . , ym , X . , the lemma is proved. Proof of Theorem 4. Since nk

V⬘ s

Ý Ž y1. n qi qk , i¨ kp, X i

k

is0 nk

s

Ý Ž y1.

n k qi

qk , i

is0

Ý Ž y1.

m ky 1 qj

Q m ky 1 , j X p

js0

n k m ky1

s

pi

m ky1

Ý Ý Ž y1. n qiqm k

ky 1 qj

i

qk , i Q mp ky 1 , j X p

is0 js0 n k m ky1

s

Ý Ý Ž y1. mq iqj qk , i Qmp is0 js0

i ky 1 ,

j

Xp

iq j

iq j

j

s

/

MINH AND TUNG `

208 and m

Vmq 1 s

Ý Ž y1.

mq s

s

Qm , s X p ,

ss0

it follows from Lemma 8 that Qm , s s

i

Ý

Q mp ky 1 , j qk , i .

iqjss

The theorem is then proved by induction on k.

REFERENCES 1. L. E. Dickson, A fundamental system of invariants of the general modular linear group with a solution of the form problem, Trans. Amer. Math. Soc. 12 Ž1911., 75᎐98. 2. T. J. Hewett, Modular invariant theory of parabolic subgroups of GLnŽ⺖q . and the associated Steenrod modules, Duke Math. J. 82 Ž1996., 91᎐92. Erratum, Duke Math. J. 97 Ž1999., 217. 3. N. J. Kuhn and S. A. Mitchell, The multiplicity of the Steinberg representation of GLn Fp in the symmetric algebra, Proc. Amer. Math. Soc. 96 Ž1986., 1᎐6. 4. P. A. Minh, Modular invariant theory and cohomology algebras of extra-special p-groups, Pacific J. Math. 124 Ž1986., 345᎐363. 5. H. Mui, Modular invariant theory and cohomology algebras of symmetric groups, J. Fac. Sci. Uni¨ . Tokyo 22 Ž1975., 319᎐369. 6. H. Mui, Cohomology operations derived from modular invariants, Math. Z. 193 Ž1986., 151᎐163.