FORMAL MODULAR SEMINVARIANTS
R. JAMES SHANK
December 13, 1996
Abstract. We construct a generating set for the ring of invariants for th*
*e four
and five dimensional indecomposable modular representations of a cyclic g*
*roup of
prime order. We then observe that for the four dimensional representation*
* the ring
of invariants is generated in degrees less than or equal to 2p - 3, and f*
*or the five
dimensional representation the ring of invariants is generated in degrees*
* less than
or equal to 2p - 2.
Introduction
Let k be a field and define a linear function oe by oe(x1) = x1 and, for i > *
*1,
oe(xi) = xi+ xi-1. Extend oe to an algebra automorphism of the polynomial algeb*
*ra
k[x1; : :;:xn]. If f 2 k[x1; : :;:xn] and oe(f) = f, then f will be called oe-*
*invariant.
Since oe is a degree preserving map, any oe-invariant polynomial is a sum of ho*
*moge-
neous oe-invariant polynomials. Thus we will restrict our attention to homogene*
*ous
polynomials. Let k[x1; : :;:xn]oedenote the ring of oe-invariant polynomials. S*
*uppose
that p is a prime number and let Fp denote the field with p elements. If k = Fp
and n p, then oe generates a group isomorphic to Z=p and we denote k[x1; : :;:*
*xn]oe
by Fp[x1; : :;:xn]Z=p. The action of Z=p induced by oe on the degree one polyno*
*mials
of Fp[x1; : :;:xn] is the indecomposable modular representation of dimension n.*
* The
study of Fp[x1; : :;:xn]Z=p has a long history going back at least to L. E. Dic*
*kson's
Madison Colloquium [5]. From Dickson's perspective the problem is an extension *
*of
classical invariant theory and the elements of Fp[x1; : :;:xn]Z=pare the formal*
* modular
semivariants of a binary (n - 1)-form [5, III]. Dickson gave a complete descrip*
*tion
of Fp[x1; : :;:xn]Z=p for n = 2 and n = 3. He gave a generating set for n = 4, *
*p = 5.
G. Almkvist, in [1], described the set of relations for n = 4, p = 5. W. L. G. *
*Williams,
in [10], constructed a generating set for n = 4, p = 7. The primary purpose of *
*this
paper is to describe a generating set for n = 4 and n = 5 for all p 5.
If the characteristic of k is zero, then oe generates a group isomorphic to Z*
*. In
this case we denote k[x1; : :;:xn]oeby k[x1; : :;:xn]Z. Let Q denote the ration*
*al num-
___________
1991 Mathematics Subject Classification. 13A50.
Key words and phrases. Invariant theory, Indecomposable modular representatio*
*n.
1
2 R. JAMES SHANK
bers. For any element f 2 Q[x1; : :;:xn]Z, a suitable scalar multiple of f li*
*es in
Z[x1; : :;:xn]Z. By reducing coefficients modulo p, an element of Z[x1; : :;:xn*
*]Z gives
rise to an element of Fp[x1; : :;:xn]Z=p. We will call elements of Fp[x1; : :;*
*:xn]Z=p
constructed in this fashion rational invariants. G. Almkvist has shown that, if
f 2 Fp[x1; : :;:xn]Z=p and the degree of f is small compared to p, then f is a *
*ra-
tional invariant [1, 2.5]. Thus characteristic zero computations will provide u*
*s with
some of our generators.
Two constructions are needed to provide us with the remaining generators. The
first of these constructions is the transfer. The transfer is a homomorphism of
Fp[x1; : :;:xn]Z=p-modules from Fp[x1; : :;:xn]to Fp[x1; : :;:xn]Z=p defined by
Xp
Tr(f) = oec(f):
c=1
The second construction is the norm. For an element, f, of Fp[x1; : :;:xn] the *
*norm of
f, denoted by N(f), is the product over the orbit of f. We shall see that, at l*
*east for
n = 4 and n = 5, Fp[x1; : :;:xn]Z=p is generated by N(xn), selected rational in*
*variants
and elements from the image of the transfer.
We compute generating sets by constructing a collection of invariants and then
using the theory of S.A.G.B.I. bases, introduced by L. Robbiano and M. Sweedler*
* in
[8], to prove that the given collection of invariants form a generating set. In*
* Section 1
we define a S.A.G.B.I. basis and discuss the properties of S.A.G.B.I. bases req*
*uired for
our purposes. Section 2 is devoted primarily to constructing rational invariant*
*s with
particular lead monomials. In this section we also discuss the ring Q[V1 ]Z fo*
*rmed
by taking the union over n of Q[x1; : :;:xn]Z. We are able to construct a vecto*
*r space
basis for Q[V1 ]Z. In Section 3 we compute the lead monomials of certain famil*
*ies
of elements in the image of the transfer. Section 4 contains the proof that a c*
*ertain
collection of invariants is a generating set for Fp[x1; : :;:x4]Z=p and Section*
* 5 contains
the analogous result for Fp[x1; : :;:x5]Z=p. Section 6 is devoted to conclusio*
*ns and
conjectures.
We recommend [9] as a good general reference for the invariant theory of fini*
*te
groups. Preliminary calculations, including the construction of a generating se*
*t for
n = 4 with p = 11, were performed using G. Kemper's Maple package INVAR ([6],[7*
*]).
I would like to thank Catherine Chambers for implementing the most recent versi*
*on
of INVAR on our computing facilities and for supervising the computations. I wo*
*uld
also like to thank Eddy Campbell, Ian Hughes and David Wehlau for their assista*
*nce
and encouragement.
1. S.A.G.B.I. Bases
Throughout the paper we use the graded reverse lexicographic monomial order w*
*ith
xm < xm+1 . We direct the reader to Chapter 2 of [4] for the appropriate defini*
*tions
and a detailed discussion of monomial orders. We use the convention that a mono*
*mial
SEMINVARIANTS 3
is a product of variables and that a term is a monomial with a coefficient. We *
*extend
the monomial order to a partial order on polynomials by comparing lead monomial*
*s.
Suppose that A is a subalgebra of k[x1; : :;:xn]. Let LT(A) denote the vector*
* space
spanned by the lead terms of elements of A. LT (A) is a subalgebra of k[x1; : :*
*;:xn].
If C is a subset of A then let LM (C) denote the set of lead monomials of eleme*
*nts of
C. If LM (C) generates the algebra LT (A) then C is called a S.A.G.B.I. basis f*
*or A.
Proposition 1.1. If C is a S.A.G.B.I. basis for A then C generates the algebra *
*A.
Proof.See [8, 1.16]. __|_ |
Suppose that M is a subspace of k[x1; : :;:xn]. Let Md denote the homogeneous
component of degree d. The Poincare series of M is given by
1X
P (M; t) = dimk(Md)td:
d=0
Proposition 1.2. If A is a subalgebra of k[x1; : :;:xn], then P (A; t) = P (LT *
*(A); t).
Proof.We will prove that Ad has a basis, B, with distinct lead monomials and he*
*nce
LM (B) is a basis for LT (A)d.
Choose a basis for Ad. Every element of this basis can be written as a linear*
* com-
bination of monomials of degree d. Using the monomial order, write the coeffici*
*ents
of each basis element as a row vector. The row vectors corresponding to the ba*
*sis
elements can be used to construct a matrix with linearly independent rows. Usi*
*ng
row operations put this matrix in echelon form. The rows of the echelon form ar*
*e the
coefficients of a new basis, say B, for Ad. Since the coefficient matrix corres*
*ponding
to B is in echelon form, the lead monomials of the elements of B are distinct. *
* __|_ |
2. Lead Monomials of Invariants
In this section we construct rational invariants with particular lead monomia*
*ls.
We also characterize those monomials which are the lead monomial of a oe-invari*
*ant.
Lemma 2.1. If fi is a monomial in k[x1; : :;:xm-1 ] then fixm and fixm-1 are *
*consec-
utive elements in the order.
Proof.Suppose that fl is a monomial with fixm-1 < fl fixm . We will prove that
fixm-1 and fixm are consecutive by showing fl = fixm .
Let bi be the exponent of xi in fi and let ei be the exponent of xi in fl. Le*
*t j be
the first position at which the exponent sequence of fixm-1 differs from the ex*
*ponent
sequence of fl. Since fixm-1 and fl have the same degree, we can assume that j *
*< m.
Thus bi = ei for i < j. If j < m - 1 then fl > fixm , contradicting our hypothe*
*ses.
Thus j = m - 1. Therefore bm-1 + 1 > em-1 bm-1 and fl = fixm . __|_ |
Theorem 2.2. If fi is a monomial in k[x1; : :;:xm-1 ] then fixm is not the lea*
*d term
of a oe-invariant.
4 R. JAMES SHANK
Proof.Suppose that f 2 k[x1; : :;:xn] for some n and that the lead term of f is*
* fixm .
Then f = fixm + h for some polynomial h with LM (h) < LM (f). We will prove that
LM O (oe - 1)(f) = fixm-1 ; hence f is not oe-invariant.
Evaluating oe(xm ) and rearranging terms gives
(oe - 1)(f) = (xm + xm-1 )(oe - 1)(fi) + oe(fi)xm-1 + (oe - 1)(h):
We extend the monomial order to a partial order on polynomials by comparing l*
*ead
monomials. Note that, for any monomial fl, (oe-1)fl < fl. Thus (oe-1)(h) < h < *
*fixm
and xm (oe - 1)(fi) < xm fi. From Lemma 2.1, fixm and fixm-1 are consecutive in*
* the
order. Thus (oe - 1)(h) < fixm-1 and LM (xm (oe - 1)(fi)) fixm-1 . Since fixm-*
*1 is in
k[x1; : :;:xm-1 ], LM (xm (oe-1)(fi)) is not equal to fixm-1 . Therefore LM O(o*
*e-1)(f) =
fixm-1 . __|_ |
Theorem 2.3. If fi is a monomial in k[x1; : :;:xm-1 ] and i 2 then fiximis th*
*e lead
monomial of a oe-invariant in k[x1; : :;:xn] for sufficiently large n.
Proof.We prove the theorem by introducing an algorithm for constructing a oe-
invariant with lead monomial fixim.
Apply oe - 1 to fiximand observe that the lead term of (oe - 1)(fixim) is ifi*
*xm-1 xi-1m.
Define f1 = fixim- ifixm-1 xi-2mxm+1 . Note that LM O (oe - 1)(f1) < fixm-1 xi*
*-1m. For
j > 1, if LM O (oe - 1)(fj-1) = 0 then fj-1 is oe-invariant, otherwise write t*
*he lead
term of (oe - 1)(fj-1) as flxkrwith fl 2 k[x1; : :;:xr-1] and k > 0 and define *
*fj =
fj-1 - flxk-1rxr+1. Observe that LM (fj) < LM (fj-1). Thus LM (fj) is a decreas*
*ing
sequence of monomials in a fixed degree.
To prove that the algorithm terminates it is sufficient to prove that the set*
* of mono-
mials in countably many variables is well ordered by the graded reverse lexicog*
*raphic
order. We will , however, take a slightly different approach; one that willQpro*
*vide us
with an upper bound on the numberPof variables required. For a monomial = sx*
*iss,
define the weight of by wt() = ssis. Note that that the monomials in (oe - 1*
*)()
all have weight less than wt(). Furthermore wt(flxk-1rxr+1) = wt(flxkr) - 1. He*
*nce
any monomial appearing in fj has weight less than or equal to wt(fixim). Since *
*there
are only a finite number of monomials in a given degree with a given weight, we*
* see
that there are only finitely many fj. In fact, if we let d denote the degree of*
* fiximand
define ` = wt(fixim) - d + 1, then xd-11x` is the smallest monomial of degree d*
* and
weight wt(fixim), and fj 2 k[x1; : :;:x`]. __|_ |
Note that all monomials except those of the form xi1satisfy the hypotheses of*
* either
Theorem 2.2 or Theorem 2.3. We will call a monomial admissible if it satisfies*
* the
hypotheses of Theorem 2.3 or if the monomial is of the form xi1.
Corollary 2.4. The set of admissible monomials equals LM (Q[V1 ]Z).
If fl is a monomial satisfying the hypotheses of Theorem 2.3, then let inv(fl*
*) be the
invariant produced by the algorithm. For convenience we define inv(xi1) = xi1.
SEMINVARIANTS 5
Remark 2.5. Reviewing the algorithm, we observe that fl is the only admissible
monomial appearing in inv(fl). Furthermore, if ` = wt(fl) - degree(fl) + 1, th*
*en
inv(fl) is in k[x1; : :;:x`]oe.
Let B denote the set {inv(fl) | fl is admissable}.
Theorem 2.6. B is a basis for the vector space Q[V1 ]Z.
Proof.Since the elements of B have distinct lead monomials the set is linearly *
*inde-
pendent. To see that B is a spanning set, consider a polynomial f 2 Q[V1 ]Z. Let
be the set of admissible monomials appearing in f and for fl 2 let cfldenote t*
*he
coefficient of fl in f. Let X
fe= cflinv(fl):
fl2
The lead monomial of the invariant f - efis not admissible and, therefore, from
Corollary 2.4, f - ef= 0. Thus B spans Q[V1 ]Z as required. __|_ |
Using the algorithm we see that inv(x22) = x22- x1(x2 + 2x3) ,
inv(x23) = x23- x2(x3 + 2x4) + x1(x3 + 3x4 + 2x5);
and inv(x32) = x32+ x21(3x4- x2) - 3x1x2x3: Explicit calculation shows that, al*
*though
inv(x33) involves x1 through x7, if we define
___ 3 3 2 2
inv(x3) = 2 inv(x3) - 3 inv(x2x3) + 9x1inv(x4);
we get an element of k[x1; : :;:x5]oewith lead monomial x33. Similarly define
___ 2 2 2 2 i 2 3 j 2i 2 *
*2 j
inv(x2x3) = 3 inv(x2x3) + x1 6 inv(x2x3) - 8 inv(x3)- x1 9 inv(x4) + 8 inv(x*
*3);
___ ___ ___
and inv(x22x23x24) = 4 inv(x23)3_- inv(x33)2: Clearly LM (inv(x22x23))_= x22x23*
*. Careful
computation_shows that inv(x22x23) 2 k[x1; : :;:x4]oe, LM (inv(x22x23x24)) = x2*
*2x23x24and
inv(x22x23x24) 2 k[x1; : :;:x5]oe.
3. Lead Monomials of Transfers
In this section we compute the lead monomial for various elements in the imag*
*e of
the transfer. We will use LM to denote the operation which takes a polynomial *
*and
produces its lead monomial. Observe that
! ! !
c c c
(3.1) oec(xm ) = xm + xm-1 + xm-2 + . .+. x1:
1 2 m - 1
Therefore
0 ! 1j
i j X m-2X c! !i m-1X c
(3.2) Tr xim-1xjm = xm-1-k @ xm-j A :
c2Fp k=0 k j=0 j
6 R. JAMES SHANK
i j
Note that ciis a polynomial in c of degree i. The following lemma is well-kn*
*own.
Lemma 3.1. Suppose that ` is a positive integer. Then
(
X -1; if p - 1 divides `;
c` =
c2Fp 0; if p - 1 does not divide `.
Proof.See, for example, [3, 9.4]. __|_ |
Theorem 3.2. If (p - 1)=2 i p - 1 then
i j 2i-(p-1)
LM O Tr xim = xp-1-im-2xm-1 :
Proof.Referring to Equation 3.1, we see that the coefficient of xp-1-im-2x2i-(p*
*-1)m-1in
i jp-1-iij2i-(p-1)i j
oec(xim) is c2 c1 p-i1-i. Using Lemma 3.1, we see that
! p-1-i !2i-(p-1) ! !
X c c i 1 p-1-i i
= - __
c2Fp 2 1 p - 1 - i 2 p - 1 - i
and, since i p - 1, this is non-zero. All of the monomials appearing in oec(x*
*im)
which are greater than xp-1-im-2x2i-(p-1)m-1have coefficients which, as polynom*
*ials in c,
have degree less than p-1 and hence, by Lemma 3.1, these monomials do not appear
in Tr(xim). __|_ |
Theorem 3.3. If 1 i p - 1 then
i j
LM O Tr xim-1xp-1m= xi+p-1m-1:
Proof.Using Equation 3.2 we see that the coefficient of xi+p-1m-1in oec(xim-1xp*
*-1m) is cp-1.
Thus, using Lemma 3.1, the coefficient of xi+p-1m-1in Tr(xim-1xp-1m) is -1. All*
* of the
monomials appearing in oec(xim-1xp-1m) which are greater than xi+p-1m-1have coe*
*fficients
which, as polynomials in c, have degree less than p - 1 and hence, by Lemma 3.1,
these monomials do not appear in Tr(xim-1xp-1m). __|_ |
Theorem 3.4. If 2 i p - 1 then
i j
LM O Tr xim-1xp-2m= xm-2 xi+p-3m-1:
Proof.UsingiEquationj3.2 we see that the coefficient of xm-2 xi+p-3m-1in oec(xi*
*m-1xp-2m) is
icp-1+(p-2) c2cp-3. Thus, using Lemma 3.1, the coefficient of xi+p-1m-1in Tr(xi*
*m-1xp-1m)
is -i + 1. As long as i 6= 1, this coefficient is non-zero. All of the monomi*
*als ap-
pearing in oec(xim-1xp-2m) which are greater than xm-2 xi+p-3m-1have coefficien*
*ts which,
as polynomials in c, have degree less than p - 1 and hence, by Lemma 3.1, these
monomials do not appear in Tr(xim-1xp-2m). __|_ |
SEMINVARIANTS 7
Theorem 3.5. If (p - 1)=2 - 1 i p - 1 then
i j
LM O Tr x2m-1xim = xp-1-im-2x2i-p+3m-1:
Proof.We use Equation 3.2 to compute the coefficient of xp-1-im-2x2i-p+3m-1in o*
*ec(x2m-1xim).
oec(x2m-1) contributes x2m-1, 2cxm-1 xm-2 or c2x2m-2with the rest of the term c*
*oming
from oec(xim). Thus the coefficient of xp-1-im-2x2i-p+3m-1in oec(x2m-1xim) is
!p-1-i ! !p-2-i ! !p-3-i !
c i 2i-p+3 c i 2i-p+5 c i
c2i-p+1 +2c +c *
* :
2 p - 1 - i 2 p - 2 - i 2 p - 3 - i
Thus, using Lemma 3.1, the coefficient of xp-1-im-2x2i-p+3m-1in Tr(x2m-1xim) is
1 p-1-i i ! 1 p-2-i i ! 1 p-3-i i !
- __ - 2 __ - __ :
2 p - 1 - i 2 p - 2 - i 2 p - 3 - i
We need to show that this coefficient is non-zero. If i = p - 1 or i = p - 2 th*
*en the
coefficient is -1. If i < p - 2 then, after factoring, simplifying and reducing*
* modulo
p, the coefficient is !
__-1___ 1_ p-3-i i
:
2(i + 2) 2 p - 3 - i
Thus the coefficient is non-zero. All of the monomials appearing in oec(x2m-1x*
*im)
which are greater than xp-1-im-2x2i-p+3m-1have coefficients which, as polynomia*
*ls in c,
have degree less than p-1 and hence, by Lemma 3.1, these monomials do not appear
in Tr(x2m-1xim). __|_ |
Theorem 3.6. Suppose that LM O Tr(fi) 2 Fp[xm ; : :;:xn]. Then
LM O Tr(xm fi) = xm LM O Tr(fi):
Proof.Using Equation 3.1, we see that
X
Tr(xm fi)= oec(xm )oec(fi)
c2Fp
0 ! 1
X m-1X c
= @ xm-j Aoec(fi)
c2Fp j=0 j
0 ! 1
X m-1X c
= xm Tr(fi) + @ xm-j Aoec(fi):
c2Fp j=1 j
Thus Tr(xm fi) is congruent, modulo the ideal generated by x1 through xm-1 , to
xm Tr(fi). Since LM O Tr(fi) is in Fp[xm ; : :;:xn], the lead monomial of Tr(xm*
* fi) comes
from xm Tr(fi). Therefore LM O Tr(xm fi) = xm LM O Tr(fi). __|_ |
8 R. JAMES SHANK
4.The Four Dimensional Representation
In this section we construct a generating set for Fp[x1; : :;:x4]Z=p. If p 1*
* (mod 3)
then define ` = (p-1)=3 and q = 2`+1. If p -1 (mod 3) then define ` = (p+1)=3
and q = 2` - 1. If i is a integer, define "(i) to be 0 if i is even and 1 if i *
*is odd.
___
Theorem 4.1. Fp[x1; : :;:x4]Z=pis generated by x1, inv(x22), inv(x32), inv(x22*
*x23), N(x4)
and the following families:
(i) Tr(xi3xp-14) for 0 i p - 2,
(ii) Tr(xi3xp-24) for 3 i p - 2,
(iii) Tr(xj4) for q j p - 2 and
(iv) Tr(x23xj4) for 2` - 1 j p - 2.
This rest of this section is devoted to the proof of Theorem 4.1. Let C denot*
*e the
collection of invariants given in the statement of the preceding theorem. We p*
*rove
the theorem by showing that C is a S.A.G.B.I. basis for Fp[x1; : :;:x4]Z=p, i.e*
*., the lead
monomials of the elements of C generate the lead term algebra of Fp[x1; : :;:x4*
*]Z=p.
We do this by computing the Poincare series of the algebra generated by the lead
monomials of C and comparing the result with the Poincare series of Fp[x1; : :;*
*:x4]Z=p
as computed by G. Almkvist and R. Fossum [2, 3.1]. We observe that the two seri*
*es
are equal and, using Proposition 1.2, we conclude that the lead monomials of C
generate the lead term algebra of Fp[x1; : :;:x4]Z=p. Therefore, by Proposition*
* 1.1, C
generates Fp[x1; : :;:x4]Z=p.
Let A denote the algebra generated by the lead monomials of C. We wish to
compute the Poincare series of A. Using Theorem 3.2 and the fact that the lead
monomial of N (xn) is xpn, we have
LM {x1; inv(x22); Tr(xp-14); N(x4)} = {x1; x22; xp-13; xp4}:
Note that this set is algebraically independent. Let R denote the ring generate*
*d by
{x1; x22; xp-13; xp4}. Using elementary properties of Poincare series we see th*
*at
1
P (R; t) = _____________________________
(1 - t)(1 - t2)(1 - tp-1)(1 - tp)
We will use the R-module structure of A to compute its Poincare series. In orde*
*r to
understand the R-module structure we need to find module generators for A. Let
___ 2 2 i+1 3 ___ 2 2 i
D = {Tr(x3xp-14) . Tr(xp-24)} [ {inv(x2x3) ; inv(x2) . inv(x2x3) | 1 i `=2 *
*- 1}
and let M be the R-module generated by 1, LM (C) and LM (D). Observe that M
is an R-submodule of A. We will now compute the Poincare series of M. We shall
see that the Poincare series of M is equal to the Poincare series of Fp[x1; : :*
*;:x4]Z=p
and thus M = A = LT(Fp[x1; : :;:x4]Z=p).
We start by imposing a Z=2 x Z=(p - 1)-grading on Fp[x1; : :;:x4]. A monomial
xi11xi22xi33xi44will be assigned the multidegree (i2; i3) 2 Z=2 x Z=(p - 1). Ob*
*serve that
SEMINVARIANTS 9
the action of R preserves the multidegree. Since A is generated by monomials, A*
* is
a Z=2 x Z=(p - 1)-graded R-module. In particular all generators and relations c*
*an
be chosen to be homogeneous with respect to the Z=2 x Z=(p - 1)-grading.
If fi and fl are monomials in Fp[x1; : :;:x4] with the same multidegree, then*
* the
intersection of Rfi with Rfl is the free R-module generated by the least common
multiple of fi and fl. In particular, an R-module generated by two monomials wi*
*th
the same multidegree has a single free relation.
Using the results of Section 3, we see that the R-module M is generated by the
following families of monomials:
(i) 1 and x2x2p-33,
(ii) (x22x23)i, x32(x22x23)i-1for 1 i `=2,
(iii) xp-1+i3for 1 i p - 2,
(iv) x2xp-3+i3for 3 i p - 2,
(v) xp-1-j2x2j-p+13for q j p - 2 and
(vi) xp-1-j2x2j-p+33for 2` - 1 j p - 2.
This list of monomials contributes either one or two elements to each multide*
*gree.
When there is one element in a given multidegree then the homogeneous component
of M in that multidegree is a free R-module of rank one. If there are two eleme*
*nts
in a given multidegree then the homogeneous component has two generators and a
single free relation. Therefore we can write the Poincare series of M as
g(t) - r(t)
P (M; t) = _____________________________
(1 - t)(1 - t2)(1 - tp-1)(1 - tp)
where g(t) is the Poincare series for the generators and r(t) is the Poincare s*
*eries for
the relations. Referring to our list of generators we see that
`=2Xi j p-2X
g(t) = 1 + t2p-2+ t4i+ t4i-1+ tp-1+i
i=1 i=1
p-2X p-2X p-2X
+ tp-2+i+ tj + tj+2:
i=3 j=q j=2`-1
We rewrite our expression for g(t) by combining the second sum and the third sum
and making minor adjustments to the range of the indices giving
`=2Xi j
g(t) = 1 - t2p-3- tp - tp-1+ t4i+ t4i-1
i=1
p-1Xi j p-2X p-2X
+ tp-1+i+ tp-2+i+ tj + tj+2:
i=1 j=q j=2`-1
10 R. JAMES SHANK
To compute r(t) we need to identify the multidegrees containing two generators
and compute the degree of the least common multiple of the two generators.
Sorting our generators into homogeneous components leads to the following rel*
*a-
tions:
(i) lcm((x2x3)2i; x2i+p-13) = x2i2x2i+p-13for 1 i `=2,
(ii) lcm(x2i+12x2i-23; x2x2i+p-33) = x2i+12x2i+p-33for 1 i `=2,
(iii) lcm(xp-1-j2x2j-p+13; x"(j)2x2j3) = xp-1-j2x2j3for q j p - 3, and
(iv) lcm(xp-1-j2x2j-p+33; x"(j)2x2j+23) = xp-1-j2x2j+23for 2` - 1 j p - 3.
Thus
`=2Xi j p-3X p-3X
r(t) = t4i+p-1+ t4i+p-2+ tp-1+j+ tp+1+j:
i=1 j=q j=2`-1
Now form the polynomial g(t) - r(t) collecting terms when possible to get
`=2Xii j i jj
g(t) - r(t) = 1 - t2p-3- tp - tp-1+ t4i+ t4i-1- t4i+p-1+ t4i+p-2
i=1
p-3Xi j p-3Xi j
+tp-2+ tj - tp-1+j+ tp + tj+2- tp+1+j
j=q j=2`-1
p-1Xi j
+ tp-1+i+ tp-2+i:
i=1
Factoring gives
i ji j`=2X
g(t) - r(t)= 1 - t2p-2- tp-1+ tp-2+ 1 - tp-1 t3 + t4 t4i-4
i=1
i jp-3X i j p-3X
+ 1 - tp-1 tj + 1 - tp-1 tj+2
j=q j=2`-1
p-1X
+tp-1(1 + t) ti-1:
i=1
Evaluate the various geometric series to get
SEMINVARIANTS 11
i ji j 1 - t2`!
g(t) - r(t)= (1 - tp-1)(1 + tp-2) + 1 - tp-1 t3 + t4 ______
1 - t4
i j tq - tp-2! i j t2`+1- tp!
+ 1 - tp-1 ________ + 1 - tp-1 _________
1 - t 1 - t
!
1 - tp-1
+tp-1(1 + t) ________ :
1 - t
Factoring gives
!
1 - tp-1 h 4 p-2 3 4 2`
g(t) - r(t)= ________ (1 - t )(1 + t ) + (t + t )(1 - t )
1 - t4
i i j
+(1 + t + t2 + t3) tq - tp-2+ t2`+1- tp + tp-1(1 +:t)
Expand to get
!
1 - tp-1 i 3 q q+1 q+2 q+3 2`+1 2`+2j
g(t) - r(t)= ________ 1 + t + t + t + t + t + t + t :
1 - t4
If p 1 (mod 3), then 2` = q - 1 and
1 + t3 + 2tq + 2tq+1+ tq+2+ tq+3
P (M; t) = ______________________________:
(1 - t)(1 - t2)(1 - t4)(1 - tp)
If p -1 (mod 3), then 2` = q + 1 and
1 + t3 + tq + tq+1+ 2tq+2+ 2tq+3
P (M; t) = ______________________________:
(1 - t)(1 - t2)(1 - t4)(1 - tp)
Comparing with [2, 3.1] we see that P (M; t) = P (Fp[x1; : :;:x4]Z=p; t) as req*
*uired.
Corollary 4.2. Fp[x1; : :;:x4]Z=p is generated in degrees less than or equal to*
* 2p - 3.
5. The Five Dimensional Representation
*
* ___
Theorem 5.1. Fp[x1; : :;:x5]Z=pis generated by x1, inv(x22), inv(x23), inv(x32*
*), inv(x33),
___ (p-1)=2
inv(x22x23x24), N (x5), Tr(x2x3x5 ) and the following families:
(i) Tr(xi4xp-15) and Tr(x2xi4xp-15) for 0 i p - 2,
(ii) Tr(xi4xp-25) and Tr(x2xi4xp-25) for 3 i p - 2,
(iii) Tr(x24xj5) and Tr(x2x24xj5) for (p - 1)=2 j p - 2.
12 R. JAMES SHANK
(iv) Tr(xj5) for (p + 1)=2 j p - 1, and
(v) Tr(x2xj5) for (p - 1)=2 j p - 2.
This section is devoted the proof of Theorem 5.1. The methods used are similar
to those used in Section 4.
Let C denote the collection of invariants given in the statement of the prece*
*ding
theorem. Let A denote the algebra generated by the lead monomials of C. Using
Theorem 3.2, we see that
LM {x1; inv(x22); inv(x23); Tr(xp-15); N(x5)} = {x1; x22; x23; xp-14; *
*xp5}:
This is an algebraically independent subset of A. Let R denote the ring generat*
*ed
by {x1; x22; x23; xp-14; xp5}. As in Section 4, if p 1 (mod 3) then define ` *
*= (p - 1)=3
and q = 2` + 1, if p -1 (mod 3) then define ` = (p + 1)=3 and q = 2` - 1, and*
* if i
is a integer, define "(i) to be 0 if i is even and 1 if i is odd. Let
___ 3 j ___ 2 2 2 k
D0= {inv(x32)i. inv(x3) . inv(x2x3x4) | i; j 2 {0; 1}; 0 k `=2 - 1 - j};
and let
D = {Tr(x4xp-15) . Tr(xp-25); Tr(x2x4xp-15) . Tr(xp-25)} [ D0:
Let M be the R-module generated by LM (C) and LM (D). Note that M is a subset
of A. Impose a Z=2 x Z=2 x Z=(p - 1)-grading on Fp[x1; : :;:x5]Z=p. A monomial
xi11xi22xi33xi44xi55will be assigned the multidegree
(i2; i3; i4) 2 Z=2 x Z=2 x Z=(p - 1):
Observe that the action of R preserves the multidegree. Since M is generated by
monomials, all generators and relations can be chosen to be homogeneous with re*
*spect
to the Z=2 x Z=2 x Z=(p - 1)-grading.
Using the results of Section 3 we see that the R-module M is generated by the
following families of monomials:
(i) 1, x3x2p-34and x2x3x2p-34,
(ii) (x2x3x4)2i, x2i+13(x2x4)2(i-1)and (x2x3)2i+1x2(i-1)4for 1 i `=2 - 1,
x2i+12(x3x4)2(i-1)for 1 i `=2,
(iii) xp-1+i4for 1 i p - 2, x2xp-1+i4for 0 i p - 2,
x3xp-3+i4and x2x3xp-3+i4for 3 i p - 2, and
(iv) x2xp-1-i3x2i-p+14and xp-1-i3x2i-p+34for (p - 1)=2 i p - 2,
x2xp-1-i3x2i-p+34for (p - 1)=2 - 1 i p - 2,
xp-1-i3x2i-p+14for (p - 1)=2 + 1 i p - 2.
SEMINVARIANTS 13
Thus the Poincare series of the generators is given by
`=2-1Xi j
g(t) = 1 + (1 + t)t2p-2+ t3`-3+ 2t6i+ 2t6i-3
i=1
p-2X p-2X p-2X
+ tp-1+i+ tp+i+ (1 + t) tp-2+i
i=1 i=0 i=3
p-2X p-2X p-2X
+ tj + tj(t + t2) + tj+3:
j=(p+1)=2 j=(p-1)=2 j=(p-3)=2
Evaluating the geometric series gives
!
1 - t3`-6
g(t) = 1 + (1 + t)t2p-2+ t3`-3+ 2t3(1 + t3) ________
1 - t6
1 h
+ _____ tp(1 - tp-2) + tp(1 - tp-1) + (1 + t)tp-2(t3 - tp-1)
1 - t i
+t(p+1)=2- tp-1+ (t + t2)(tp-1)=2- tp-1) + t3(t(p-1)=2-1- tp-1)
Simplifying and collecting terms gives
!
1 + t3 3`-3 p-1 1 + t h (p+1)=2 2 2p-3i
g(t) = ______ (1 - t ) - t + _____ 2t - (1 + t )t :
1 - t3 1 - t
Observe that each homogeneous component contains one, two, or three generator*
*s.
If the component contains one generator then the component is a free module of *
*rank
one. If the component contains two generators then there is a single free rela*
*tion
generated by the least common multiple. If the component contains three generat*
*ors
then there are three relations given by the pairwise least common multiples and*
* a
single syzygy given by the least common multiple of all three generators. Thus *
*the
Poincare series can be written as
g(t) - r1(t) + s(t) - r2(t)
P (M; t) = ______________________________
(1 - t)(1 - t2)2(1 - tp-1)(1 - tp)
where r1(t) is the Poincare series for the free relations, s(t) is the Poincare*
* series for
the syzygies and r2(t) is the Poincare series for the relations associated the *
*homoge-
neous components with three generators.
The free relations are given by:
(i) lcm(x33; xp-14) = x33xp-14,
14 R. JAMES SHANK
(ii) lcm(xp-1-j3x2j-p+14; x"(j)3x2j4) = xp-1-j3x2j4and
lcm(x2xp-1-j3x2j-p+14; x2x"(j)3x2j4) = x2xp-1-j3x2j4for 2` - 1 j p - 3,
(iii) lcm(xp-1-j3x2j-p+34; x"(j)3x2j+24) = xp-1-j3x2j+24and
lcm(x2xp-1-j3x2j-p+34; x2x"(j)3x2j+24) = x2xp-1-j3x2j+24for q - 2 j p - 3.
Thus 0 1
p-3X p-3X
r1(t) = tp+2+ (1 + t)tp-1@ tj + tj+2A:
j=2`-1 j=q-2
Evaluate the geometric series to get
!
t2`-1- tp-2+ tq - tp
r1(t) = tp+2+ (1 + t)tp-1 ___________________ :
1 - t
We now describe the syzygies and the relations associated the homogeneous com-
ponents with three generators. In the first line we list the three generators *
*and in
the second line we give the pairwise least common multiples followed by the lea*
*st
common multiple of the three monomials.
(i) xp-1-j3x2j-p+14; x"(j)3x2j4; x3"(j)3(x2x3x4)2j-p+1
xp-1-j3x2j4; xp-1-j3(x2x3)2j-p+1; x2j-p+12x2j-p+1+3"(j)3x2j4; x2j-p+12xp-1-j3*
*x2j4
for (p + 1)=2 j 2` - 2.
(ii) x2xp-1-j3x2j-p+14; x2x"(j)3x2j4; x32x3"(j)3(x2x3x4)2j-p+1
x2xp-1-j3x2j4; x32xp-1-j3(x2x3)2j-p+1; x2j-p+42x2j-p+1+3"(j)3x2j4; x2j-p+42x*
*p-1-j3x2j4
for (p - 1)=2 j 2` - 2.
(iii) xp-1-j3x2j-p+34; x"(j)3x2j+24; x3"(j)3(x2x3x4)2j-p+3
xp-1-j3x2j+24; xp-1-j3(x2x3)2j-p+3; x2j-p+32x2j-p+3+3"(j)3x2j+24; x2j-p+32x*
*p-1-j3x2j+24
for (p - 1)=2 j q - 3.
(iv) x2xp-1-j3x2j-p+34; x2x"(j)3x2j+24; x32x3"(j)3(x2x3x4)2j-p+3
x2xp-1-j3x2j+24; x32xp-1-j3(x2x3)2j-p+3; x2j-p+62x2j-p+3+3"(j)3x2j+24; x2j-p*
*+62xp-1-j3x2j+24
for (p - 1)=2 - 1 j q - 3.
Thus
2`-2X 2`-2X q-3X q-3X
s(t) = t3j+ t3j+3+ t3j+4+ t3j+7:
j=(p+1)=2 j=(p-1)=2 j=(p-1)=2 j=(p-3)=2
Evaluate the geometric series to get
1 h i
s(t) = ______ 2t3(p+1)=2- t6`-3- t6`+ 2t3(p+1)=2+1- t3q-2- t3q+1:
1 - t3
Simplifying gives
SEMINVARIANTS 15
1 h i
s(t) = ______ 2t3(p+1)=2(1 + t) - (1 + t3)(t6`-3+ t3q-2):
1 - t3
The Poincare series for the relations associated the homogeneous components w*
*ith
three generators is given by
2`-2X 2`-2X q-3X q-3X
r2(t) = tj+p-1+ tj+p+ tj+p+1+ tj+p+2
j=(p+1)=2 j=(p-1)=2 j=(p-1)=2 j=(p-3)=2
2`-2X 2`-2X q-3X q-3X
+ t3j-p+1+ t3j-p+4+ t3j-p+5+ t3j-p+8
j=(p+1)=2 j=(p-1)=2 j=(p-1)=2 j=(p-3)=2
2`-2X 2`-2X
+ t6j-2(p-1)+3"(j)+ t6j-2(p-1)+3"(j)+3
j=(p+1)=2 j=(p-1)=2
q-3X q-3X
+ t6j-2(p-1)+3"(j)+6+ t6j-2(p-1)+3"(j)+9:
j=(p-1)=2 j=(p-3)=2
Evaluating the geometric series in the first two lines and reindexing and reorg*
*anizing
the sums in the third and fourth lines gives
1 h i
r2(t) = _____ 2t3(p-1)=2+1- t2`+p-2- t2`+p-1+ 2t3(p-1)=2+2- tq+p-1- tq+p)
1 - t
1 h (p-1)=2+3 6`-p-2 6`-p+1 (p-1)=2+4 3q-p-1 3q-p+2i
+ ______ 2t - t - t + 2t - t - t
1 - t3
`=2-1X `=2-1X `=2-2X `=2-2X
+ t6j+(p-1)+ t6j+(p-1)+3+ t6j+(p-1)+3+ t6j+(p-1)+6:
i=1 i=0 i=1 j=0
Evaluate the geometric series in the last line and simplify to get
1 h i
r2(t) = _____ 2t3(p-1)=2+1(1 + t) - (1 + t)(t2`+p-2+ tq+p-1)
1 - t
1 h (p-1)=2+3 3 6`-p-2 3q-p-1 i
+ ______ 2t (1 + t) - (1 + t )(t + t )
1 - t3
1 p-1h 3 6 9 3`-3 3 6 i
+ ______ t t + 2t + t - t (1 + 2t +:t )
1 - t6
Further simplification gives
16 R. JAMES SHANK
1 h i
r2(t) = _____ 2t3(p-1)=2+1(1 + t) - (1 + t)(t2`+p-2+ tq+p-1)
1 - t
1 h (p-1)=2+3 3 6`-p-2 3q-p-1 i
+ ______ 2t (1 + t) - (1 + t )(t + t )
1 - t3!
1 + t3 p-1i 3 3`-3j
+ ______ t t - t :
1 - t3
Let n(t) = g(t) - r1(t) + s(t) - r2(t). Combining the previous expressions gi*
*ves
( ! )
1 + t3 3`-3 p-1 1 + t h (p+1)=2 2 2p-3i
n(t) = ______ (1 - t ) - t + _____ 2t - (1 + t )t
1 - t3 1 - t
( !)
t2`-1- tp-2+ tq - tp
- tp+2+ (1 + t)tp-1 ___________________
1 - t
ae 1 i oe j
+ ______ 2t3(p+1)=2(1 + t) - (1 + t3)(t6`-3+ t3q-2)
ae1 - t3
1 i 3(p-1)=2+1 2`+p-2 q+p-1j
- _____ 2t (1 + t) - (1 + t)(t + t
1 - t
1 i (p-1)=2+3 3 6`-p-2 3q-p-1 j
+ ______ 2t (1 + t) - (1 + t )(t + t )
1 - t3! )
1 + t3 p-1i 3 3`-3j
+ ______ t t - t :
1 - t3
Collecting terms with common factors gives
!
tp-1- 1 h (p-1)=2+3 3 6`-p-2 3q-p-1 i
n(t) = ________ 2t (1 + t) - (1 + t )(t + t )
1 - t3
!
1 + t3 h 3`-3 p-1 3 p-1 3 3`-3 i
+ ______ 1 - t - t (1 - t ) - t (t - t )
1 - t3
1 + t h
+ _____ 2t(p+1)=2- (1 + t2)t2p-3- tp-1(t2`-1- tp-2+ tq - tp)
1 - t i
+t2`+p-2+ tq+p-1- 2t3(p-1)=2+1:
Simplify to get
SEMINVARIANTS 17
!
tp-1- 1 h (p+1)=2+2 3 6`-p-2 3q-p-1 i
n(t) = ________ 2t (1 + t) - (1 + t )(t + t )
1 - t3
!
1 + t3 p-1 3`-3 1 + t (p+1)=2 p-1
+ ______ (1 - t )(1 - t ) + _____ 2t (1 - t ):
1 - t3 1 - t
Using the fact that 1=(1 - t) = (1 + t + t2)=(1 - t3) to simplify the expressio*
*n gives
!
1 - tp-1 h 3 3`-3 6`-p-2 3q-p-1
n(t) = ________ (1 + t )(1 - t + t + t )
1 - t3
i ij
+(1 + t)2t(p+1)=2(1 + t + t2) - t2:
Further simplification gives
!
1 - tp-1 h 3 3`-3 6`-p-2 3q-p-1 (p+1)=2 2i
n(t) = ________ (1 + t )(1 - t + t + t ) + 2t (1 +:t)
1 - t3
Note that, for any prime p, t3q-p-1+ t6`-p-2- t3`-3= tp. Thus
(1 + t3)(1 + tp) + 2t(p+1)=2(1 + t)2
P (M; t) = _______________________________:
(1 - t)(1 - t2)2(1 - t3)(1 - tp)
Comparing with [2, 3.1] we see that P (M; t) = P (Fp[x1; : :;:x5]Z=p; t) as req*
*uired.
This completes the proof of Theorem 5.1
Corollary 5.2. Fp[x1; : :;:x5]Z=p is generated in degrees less than or equal to*
* 2p - 2.
6. Concluding Remarks
We believe that, in principle, the methods used here could be extended to n >*
* 5
but that the computations required will become increasingly more complicated. In
particular the problem of computing the rational invariants becomes significant*
*ly
more difficult as n increases. Instead we suggest a more conceptual approach al*
*ong
the lines of the following conjecture.
Conjecture 6.1. Fp[x1; : :;:xn]Z=p is generated by rational invariants, the ima*
*ge of
the transfer and N (xn).
18 R. JAMES SHANK
A proof of this conjecture would reduce the problem of finding an upper bound*
* on
the degrees of the generators to the relatively accessible problem of computing*
* the
image of the transfer. As philosophical evidence for the conjecture we conclude*
* with
the following theorem.
Theorem 6.2. Fp[x1; : :;:xn]Z=p is an integral extension of the subalgebra gen*
*erated
by N (xn) and the image of the transfer.
Proof.It is sufficient to find a homogeneous system of parameters for Fp[x1; : *
*:;:xn]
inside Fp[x1; : :;:xn]Z=p. Consider the set
C = {Tr(xp-12); Tr(xp-13); : :;:Tr(xp-1n); N(xn)}
Using Theorem 3.2, LM (C) = {xp-11; xp-12; : :;:xp-1n-1; xpn}: Since LM (C) is *
*a homoge-
neous system of parameters and we are using the graded reverse lexicographic or*
*der,
C is a homogeneous system of parameters . __|_ |
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Trans. of the A.M.S. 22 (1921) 56-79.
Department of Mathematics and Statistics, Queen's University,
Kingston, Ontario, Canada, K7L 3N6
E-mail address: shank@mast.queensu.ca