BASES FOR RINGS OF COINVARIANTS

H E A CAMPBELL, I P HUGHES, R J SHANK, AND D L WEHLAU

November 20, 1996

Abstract. We study the multiplicative structure of rings of coinvariants for finite groups. We develop methods which give rise to natural monomial bases for such rings over their ground fields and explicitly determine precisely which monomials are zero in the ring of coinvariants. We apply our methods to the Dickson, upper triangular and symmetric coinvariants. Along the way, we recover theorems of Steinberg [17] and E. Artin [1]. Using these monomial bases we prove that the image of the transfer for a general linear group over a finite field is a principal ideal in the ring of invariants.

Introduction

We let V be a vector space of dimension n over a field k. We take P_{n} to be the symmetric algebra of V over k, in other words P_{n} = k[x_{1}, …, x_{n}] for a choice of basis {x_{1}, …, x_{n}}. We observe that elements of GL(V) act as degree-preserving algebra automorphisms of P_{n}. If G is any subgroup of GL(V) then we denote by P_{n}^{G} the ring of G-invariant polynomials. The ring of coinvariants, which we denote by (P_{n})_{G}, is the quotient of P_{n} by the ideal generated by the homogeneous elements in P_{n}^{G} with positive degree.

We describe sufficient conditions for the existence of a block basis for (P_{n})_{G}. A block basis is a particularly nice basis consisting of the monomial factors of a single monomial. Such bases simplify computations in two ways. Generally, the combinatorics involved in writing a polynomial in terms of such a basis is less complicated since we are dealing with monomials. As well, the product of two elements in the block basis is frequently also an element of the block basis. Moreover, we can sometimes identify exactly when the product of two elements in the block basis is zero in the quotient. We do this by studying certain critical monomials associated to the block basis. As an application, a block basis can be used to determine the image of the transfer over the group G. The transfer is a P_{n}^{G}-module homomorphism from P_{n} to P_{n}^{G} and, therefore, the image of the transfer is an ideal in P_{n}^{G}. Evaluating the transfer on the elements of a basis for P_{n} over P_{n}^{G} gives a set of generators for the image. If the order of G is a unit in k then the transfer is surjective. In the modular case the image is a proper ideal which, in general, is not well understood.

1991 Mathematics Subject Classification. 13A50.

Key words and phrases. Invariant theory, Dickson invariants, free module, coinvariants, transfer.

This research is supported in part by the Natural Sciences and Engineering Research Council of Canada.

We apply our methods to three interesting rings of coinvariants, namely those associated to the symmetric group, Σ_{n}, the general linear group, GL_{n}(F_{q}), and the p-Sylow subgroup of GL_{n}(F_{q}), U_{n}(F_{q}). Here F_{q} is the finite field of order q = p^{s}. In all three cases we first construct block bases. In the first two cases the block bases are known — for the symmetric coinvariants see page 40 of [1] and for the Dickson coinvariants see [17]. The Schubert calculus gives a basis for the coinvariants of any Weyl group, including the symmetric group (see [8]). These bases are not monomial bases and consequently many computations are more complicated.

Our block bases combined with our description of the set of trivial monomials gives a partial description of the multiplicative structure. We are able to give a complete description in the case of the upper triangular coinvariants. Here the ring of coinvariants is isomorphic to a tensor product of truncated polynomial algebras — see Corollary 7.2. In the case of the symmetric coinvariants we provide a relation, Corollary 8.5, which, in conjunction with our basis and the set of trivial monomials, gives an algorithm for computing products. The case of the Dickson coinvariants is especially interesting and more complicated than the other two examples. Although we cannot determine the complete multiplication table in this case, we do determine a Gröbner basis for the ideal generated by the Dickson invariants. This yields an algorithm for computing products in the Dickson coinvariants.

Using block bases, we study the image of the transfer for GL_{n}(F_{q}), U_{n}(F_{q}) and Σ_{n}. For GL_{n}(F_{q}) and U_{n}(F_{q}) we prove that the image is a principal ideal and, in each case, we produce a generator. The image of the transfer for GL_{n}(F_{2}) has been computed independantly by Clarence Wilkerson. For Σ_{n} the ideal is in general not principal; however we are able to construct a relatively small set of generators.

We observe that our rings of coinvariants are examples of Poincaré duality algebras. A non-zero element in top degree will be called a top class. It is well-known, again thanks to Steinberg [16], that if a_{1}, …, a_{n} is a regular sequence then, over a field of characteristic zero, the Jacobian det[\frac{∂a_{i}}{∂x_{j}}] is a top class for the algebra P_{n}/(a_{1}, …, a_{n}). We show here that for two of our examples, GL_{n}(F_{q}) and U_{n}(F_{q}), the Jacobian is zero in the ring of coinvariants. We further recall a result due to Munkholm and Smith, see [14], that det[b_{ij}] is a top class for the algebra P_{n}/(a_{1}, …, a_{n}) over a field of arbitrary characteristic provided the {b_{ij}} satisfy a_{i} = ∑_{j} b_{ij}x_{j}. One view of our results is that they give alternate constructions of the top class in our examples which also go some way towards describing the multiplicative structure.

We thank the referee of an earlier version of this paper for suggesting the use of block bases to study the image of the transfer. We note that the referee was aware of Theorem 9.17.

1. Preliminaries

We first recall some standard results of commutative algebra as it applies to invariant theory. These results can be found in any standard textbook of commutative algebra, for example [11]. For modern treatments explicitly tailored to the invariant theory of finite groups, we refer the reader to [2] or [15].

A homogeneous system of parameters for P_{n} is any collection of homogeneous elements {a_{1}, …, a_{n}} with the property that they generate a polynomial subalgebra A = k[a_{1}, …, a_{n}] over which P_{n} is finitely generated as a module. P_{n} is a polynomial algebra and hence Cohen-Macaulay, that is to say, P_{n} must be a free module over any subalgebra generated by a homogeneous system. Considering the quotient field of P_{n} as a field extension of the quotient field of A, it follows that the degree of this field extension must be ∏_{i=1}^{n} deg(a_{i}). Therefore P_{n} is a free module over A on ∏_{i=1}^{n} deg(a_{i}) basis elements.

We note that the Hilbert series for P_{n} is (1 − t)^{−n} and that of A is given by ∏_{i=1}^{n}(1−t^{deg(a_{i})})^{−1}. It follows that the quotient algebra P_{n}/(A^{+}) is finite dimensional with Hilbert polynomial ∏_{i=1}^{n}(1 − t^{deg(a_{i})})/(1 − t)). Here (A^{+}) = (a_{1}, …, a_{n}) denotes the ideal of P_{n} generated by the homogeneous elements of positive degree in A. Moreover, a basis for P_{n} as a free module over A is any set of elements of P_{n} that projects to a k-basis for B = P_{n}/(A^{+}).

We observe that the Hilbert polynomial factors as ∏_{i=1}^{n}(1 + t + t^{2} + ⋯ + t^{deg(a_{i})−1}). This suggests that a basis {x_{1}, …, x_{n}} for V might be found with the property that {x_{1}^{i_{1}} ⋯ x_{n}^{i_{n}} | 0 ≤ i_{1} < deg(a_{1}), …, 0 ≤ i_{n} < deg(a_{n})} is a basis for P_{n} as a module over A. This is not necessarily the case, see [3] for a discussion. Such a basis, when it exists, is called a block basis.

Definition 1.1. A block basis is a basis for P_{n}, as a free A-module, consisting of the monomial factors of a single monomial.

We observe that P_{n} is integral and therefore finitely generated over P_{n}^{G}. To see this, we take any f ∈ P_{n} and consider the monic polynomial ∏_{g∈G}(t − g(f)) ∈ P_{n}^{G}[t]. In the event that P_{n}^{G} is a polynomial algebra the ring of coinvariants (P_{n})_{G} is a Poincaré duality algebra, since it is a complete intersection as the quotient of one polynomial algebra by the ideal generated by another. In particular, (P_{n})_{G} is a finite dimensional algebra. It is well-known that if P_{n}^{G} = k[a_{1}, …, a_{n}] is a polynomial algebra then the product of the degrees of the generators is the order of the G. Hence P_{n} is free on |G| elements as a module over P_{n}^{G}.

The Dickson invariants D_{n} are the polynomials left fixed by all of GL_{n}(F_{q}), that is D_{n} = P_{n}^{GL_{n}(F_{q})}. We know from [6] that D_{n} is the polynomial algebra F_{q}[d_{1,n}, …, d_{n,n}] where the degree of d_{i,n} is q^{n} − q^{n−i}. One way to construct these generating Dickson invariants is via the polynomial F_{n}(t) = ∏_{v∈V} (t − v) = ∑_{i=0}^{n} d_{i,n}t^{q^{n−i}}. Note that d_{0,n} = 1 and, by convention, d_{i,j} = 0 if i > j. For a lovely modern treatment see [18]. Putting all these observations together, we see that P_{n} is a free module over the Dickson invariants D_{n} of rank |GL_{n}(F_{q})| = (q^{n} − 1)(q^{n} − q) ⋯ (q^{n} − q^{n−1}).

We let U_{n}(F_{q}) denote the subgroup of GL_{n}(F_{q}) which consists of upper triangular matrices with 1’s along the diagonal. We will denote by H_{n} the ring P_{n}^{U_{n}(F_{q})}. We know from [12] that H_{n} = F_{q}[h_{1}, …, h_{n}] where deg(h_{i}) = q^{i−1}. It is easy to describe the polynomials h_{i}. We write V_{i} for the span of {x_{1}, …, x_{i}} and F_{i}(t) = ∏_{v∈V_{i}}(t − v). Then h_{i} = F_{i−1}(x_{i}) is the product of all elements in the orbit of x_{i} under the group U_{n}(F_{q}).

Proposition 1.1. (i) F_{n}(t) = F_{n−1}^{q}(t) − F_{n−1}^{q−1}(x_{n})F_{n−1}(t), (ii) d_{i,n} = d_{i,n−1}^{q} − d_{i−1,n−1}F_{n−1}^{q−1}(x_{n}).

Proof. See Proposition 1.3(b) of [18].

Observe that, since h_{n} = F_{n−1}(x_{n}), Proposition 1.1(ii) is equivalent to

d_{i,n} = d_{i,n−1}^{q} − d_{i−1,n−1}h_{n}^{q−1}.

Proposition 1.2. −d_{1,n} = (h_{1}^{q^{n−1}})^{q−1} + ⋯ + (h_{n})^{q−1} and d_{n,n} = (−1)^{n}(h_{1} ⋯ h_{n})^{q−1}.

Proof. See Theorem 4.3 of [3].

It will be useful to introduce the algebra homomorphism θ: P_{n} → P_{n−1} defined by θ(x_{i}) = x_{i} for 1 ≤ i ≤ n − 1 and θ(x_{n}) = 0. It is clear that ker(θ) is the ideal generated by x_{n}.

Lemma 1.3. (i) θ(d_{n,n}) = 0 and θ(d_{i,n}) = d_{i,n−1}^{q} if i < n. (ii) θ(h_{n}) = 0 and θ(h_{i}) = h_{i} if i < n.

Proof. It follows from the definition of d_{n,n} that x_{n} divides d_{n,n} and, therefore, θ(d_{n,n}) = 0. If i < n then from Proposition 1.1 we have d_{i,n} = d_{i,n−1}^{q}−d_{i−1,n−1}F_{n−1}^{q−1}(x_{n}). Since x_{n} divides F_{n−1}(x_{n}) and θ acts as the identity on P_{n−1} ⊆ P_{n}, it follows that θ(d_{i,n}) = d_{i,n−1}^{q}. To prove (ii) observe that h_{n} = F_{n−1}(x_{n}) and, if i < n, h_{i} ∈ P_{n−1}.

As a consequence of Lemma 1.3, θ restricts to an algebra homomorphism from D_{n} to D_{n−1} whose kernel is generated by d_{n,n}. Similarly θ restricts to an algebra homomorphism from H_{n} to H_{n−1} whose kernel is generated by h_{n}.

2. P_{n} as a module over H_{n}

Here we prove that there is a block basis for P_{n} as a free H_{n}-module. Let

Ψ_{n} = {x_{1}^{i_{1}} ⋯ x_{n}^{i_{n}} | ∀k(0 ≤ i_{k} < q^{k−1})} .

We will prove that Ψ_{n} is a basis for P_{n} over H_{n}. Note that the cardinality of Ψ_{n} equals the order of U_{n}(F_{q}).

Lemma 2.1. If α ∈ Ψ_{n} then x_{n}α is in the H_{n}-module spanned by Ψ_{n}.

Proof. Suppose α = x_{1}^{i_{1}} ⋯ x_{n}^{i_{n}}. If i_{n} < q^{n−1} − 1 then x_{n}α ∈ Ψ_{n} and, therefore, is in the span of Ψ_{n}. If i_{n} = q^{n−1} − 1 then x_{n}α = x_{1}^{i_{1}} ⋯ x_{n−1}^{i_{n−1}}x_{n}^{q^{n−1}}. Recall that

h_{n} = F_{n−1}(x_{n}) = x_{n}^{q^{n−1}} + ∑_{i=1}^{n−1} d_{i,n−1}x_{n}^{q^{n−1−i}}.

Therefore

x_{n}^{q^{n−1}} = h_{n} − ∑_{i=1}^{n−1} d_{i,n−1}x_{n}^{q^{n−1−i}},

and

x_{n}α = h_{n}x_{1}^{i_{1}} ⋯ x_{n−1}^{i_{n−1}} − ∑_{i=1}^{n−1} d_{i,n−1}x_{1}^{i_{1}} ⋯ x_{n−1}^{i_{n−1}}x_{n}^{q^{n−1−i}}.

Since d_{i,n−1} ∈ H_{n−1} ⊆ H_{n}, we see that x_{n}α is in the H_{n}-span of Ψ_{n}.

Theorem 2.2. Ψ_{n} is a basis for P_{n} as a free module over H_{n}.

Proof. Since the cardinality of Ψ_{n} equals the rank of P_{n} over H_{n}, it is sufficient to prove that Ψ_{n} is a spanning set. We use induction on n. For n = 1 the result is trivial. Suppose n > 1. We will use induction on degree to prove that Ψ_{n} spans P_{n}. The result is clear for polynomials of degree zero. Suppose f ∈ P_{n} and deg(f) > 0. θ(f) ∈ P_{n−1} and so by the primary induction hypothesis θ(f) = ∑_{β∈Ψ_{n−1}} c_{β}β for some c_{β} ∈ H_{n−1}. The kernel of θ is the ideal generated by x_{n} and therefore f = θ(f)+x_{n}f′ for some f′ ∈ P_{n}. Note that the degree of f′ is less than the degree of f and thus we may apply the secondary induction hypothesis to f′. This gives f′ = ∑_{α∈Ψ_{n}} b_{α}α for some b_{α} ∈ H_{n}. Therefore

f = θ(f) + x_{n}f′ = ∑_{β∈Ψ_{n−1}} c_{β}β + ∑_{α∈Ψ_{n}} b_{α}x_{n}α.

From Lemma 2.1 we know that x_{n}α is in the span of Ψ_{n}. Since Ψ_{n−1} ⊆ Ψ_{n} and H_{n−1} ⊆ H_{n}, it follows that f is in the H_{n}-span of Ψ_{n}.

3. P_{n} as a module over D_{n}

In this section we prove that there is a block basis for P_{n} as a free module over D_{n}, giving a different proof of a result of Steinberg [17, Theorem B]. In fact we prove a slightly stronger result. For ℓ ≥ 0 let D_{n}(ℓ) denote the polynomial algebra F_{q}[d_{1,n}^{q^{ℓ}}, …, d_{n,n}^{q^{ℓ}}] and define

Ω_{n}(ℓ) = {x_{1}^{i_{1}} ⋯ x_{n}^{i_{n}} | ∀k (0 ≤ i_{k} < q^{ℓ}(q^{n} − q^{n−k}))} .

We will prove that Ω_{n}(ℓ) is a basis for P_{n} as a free module over D_{n}(ℓ).

It follows from the definition that if {a_{1}, …, a_{n}} is a homogeneous system of parameters then so also is {a_{1}^{i_{1}}, …, a_{n}^{i_{n}}}, for any collection of positive integers i_{j}. In particular, {d_{1,n}^{q^{ℓ}}, …, d_{n,n}^{q^{ℓ}}} is a homogeneous system of parameters for P_{n}. Thus P_{n} is free as a module over D_{n}(ℓ) on q^{ℓ}(q^{n} − 1)q^{ℓ}(q^{n} − q) ⋯ q^{ℓ}(q^{n} − q^{n−1}) many basis elements.

Lemma 3.1. If α ∈ Ω_{n}(ℓ) then x_{n}α is in the D_{n}(ℓ)-module generated by Ω_{n}(ℓ).

Proof. Suppose α = x_{1}^{i_{1}} ⋯ x_{n}^{i_{n}}. If i_{n} < q^{ℓ}(q^{n} −1)−1 then x_{n}α ∈ Ω_{n}(ℓ) and, therefore, is in the span of Ω_{n}(ℓ). If i_{n} = q^{ℓ}(q^{n} − 1) − 1 then x_{n}α = x_{1}^{i_{1}} ⋯ x_{n−1}^{i_{n−1}}x_{n}^{q^{ℓ}(q^{n}−1)}. Recall that

0 = F_{n}(x_{n}) = ∑_{i=0}^{n} d_{i,n}x_{n}^{q^{n−i}}.

Therefore

x_{n}^{q^{n}−1} = − ∑_{i=1}^{n} d_{i,n}x_{n}^{q^{n−i}−1},

and

x_{n}α = − ∑_{i=1}^{n} d_{i,n}^{q^{ℓ}}x_{1}^{i_{1}} ⋯ x_{n−1}^{i_{n−1}}x_{n}^{q^{ℓ}(q^{n−i}−1)}.

Hence x_{n}α is in the D_{n}(ℓ)-span of Ω_{n}(ℓ) as required.

Theorem 3.2. Ω_{n}(ℓ) is a basis for P_{n} as a free module over D_{n}(ℓ).

Proof. We proceed by induction on n. We show that Ω_{n}(ℓ) is a spanning set for P_{n} over D_{n}(ℓ). Then since Ω_{n}(ℓ) has cardinality equal to the product of the degrees of the generators of D_{n}(ℓ) it follows that Ω_{n}(ℓ) is a basis.

If n = 1 the theorem is trivial, so we may assume that Ω_{n−1}(ℓ) is a basis for P_{n−1} over D_{n−1}(ℓ) for all ℓ. Next we induct on degree in P_{n} and we note there is nothing to be proved when the degree is 0. So we take f to be any element of P_{n} with the property that if f′ ∈ P_{n} has deg(f′) < deg(f) then f′ is in the submodule of P_{n} spanned by Ω_{n}(ℓ) over D_{n}. Note that θ(f) ∈ P_{n−1} and by induction on n, θ(f) = ∑_{β∈Ω_{n−1}(ℓ+1)} c_{β}β for some c_{β} ∈ D_{n−1}(ℓ + 1).

We want to lift θ(f) back into P_{n}. To do this we start by defining an algebra homomorphism, ϕ, from D_{n−1}(1) to D_{n}(0) by ϕ(d_{i,n−1}^{q}) = d_{i,n}. Observe that ϕ restricts to an algebra homomorphism from D_{n−1}(ℓ + 1) to D_{n}(ℓ) and that θ ∘ ϕ = 1. Define \overline{θ(f)} = ∑_{β∈Ω_{n−1}(ℓ+1)} ϕ(c_{β})β. Observe that \overline{θ(f)} ∈ P_{n} and that f − \overline{θ(f)} is in the kernel of θ. \overline{θ(f)} is the required lift of θ(f). Since the kernel of θ is the ideal generated by x_{n} we have f = \overline{θ(f)}+x_{n}f′ for some f′ ∈ P_{n}. Note that the degree of f′ is less than the degree of f and so by induction on degree we have f′ = ∑_{α∈Ω_{n}(ℓ)} b_{α}α for some b_{α} ∈ D_{n}(ℓ). Therefore

f = \overline{θ(f)} + x_{n}f′ = ∑_{β∈Ω_{n−1}(ℓ+1)} ϕ(c_{β})β + ∑_{α∈Ω_{n}(ℓ)} b_{α}x_{n}α.

From Lemma 3.1, x_{n}α is in the D_{n}(ℓ)-span of Ω_{n}(ℓ). Furthermore, Ω_{n−1}(ℓ + 1) ⊆ Ω_{n}(ℓ). Therefore f is in the D_{n−1}-span of P_{n}.

4. Critical Monomials

Suppose A = k[a_{1} … a_{n}] where {a_{1}, … a_{n}} is a regular sequence in P_{n}. Let I be the ideal generated by {a_{1}, … a_{n}} and let d_{i} denote the degree of a_{i}. In this section we introduce sufficient conditions for the existence of a block basis for P_{n}/I. We remind the reader that P_{n}/I has a block basis if and only if there is a monomial, say α = x_{1}^{t_{1}} … x_{n}^{t_{n}}, such that the monomial factors of α form a basis for P_{n}/I. In this case we will say that α generates a block basis for P_{n}/I. Note that the symmetric group, Σ_{n}, acts on P_{n} by permuting the variables. Let Σ(A) be the subgroup of permutations fixing A pointwise. We shall say that a ∈ P_{n} is trivial if a ∈ I and we will write a ≡ b if a − b ∈ I.

Theorem 4.1. Suppose α generates a block basis. Then the sequence t_{i} is a permutation of the sequence d_{i} − 1 and the degree of α is (∑_{i=1}^{n} d_{i}) − n. Furthermore, if g ∈ Σ(A) then g(α) also generates a block basis.

Proof. Let J be the ideal generated by {x_{1}^{t_{1}+1}, …, x_{n}^{t_{n}+1}}. Then P_{n}/J also has a block basis generated by α and thus P_{n}/I and P_{n}/J are isomorphic as graded vector spaces. Therefore, the Hilbert series of P_{n}/I and P_{n}/J are the same. Thus

∏_{i=1}^{n} \frac{1 − t^{d_{i}}}{1 − t} = ∏_{i=1}^{n} \frac{1 − t^{t_{i}+1}}{1 − t} .

This proves the first assertion. The second assertion follows easily from the fact that g carries the block basis generated by α to a block basis generated by g(α).

To find a block basis for P_{n}/I, start by considering the monomials with acceptable exponent sequences and choose one candidate from each Σ(A)-orbit. If all of the candidates are trivial in P_{n}/I then there is no block basis. For example, take k = F_{p} with p odd and consider (P_{2})_{SL_{2}(F_{p})}. We may take a_{1} to be the first Dickson invariant, d_{1,2}, and a_{2} to be the (p − 1)^{st} root of d_{2,2} (see [3, Section 5]). Since Σ(A) = Σ_{2}, we need only examine one candidate, say x_{1}^{p}x_{2}^{p^{2}−p−1}. This element is in the ideal generated by a_{1} and a_{2}. Therefore (P_{2})_{SL_{2}(F_{p})} does not have a block basis. One could, however, extend the field and consider the representation of SL_{2}(F_{p}) over F_{q}. In particular, the coinvariants of the representation of SL_{2}(F_{3}) over F_{9} does have a block basis.

If we can find a non-trivial monomial with an acceptable exponent sequence, then the next step is to determine whether the monomial factors are linearly independent modulo I. In an attempt to answer this question we make the following definition.

Definition 4.1. We will say that w = x_{1}^{m_{1}} … x_{n}^{m_{n}} is a critical monomial associated to α = x_{1}^{t_{1}} ⋯ x_{n}^{t_{n}} if:

1. For every g ∈ Σ(A) there is an i such that m_{i} > t_{g(i)},

2. No proper factor of w satisfies the first condition.

For example, if Σ(A) = {1} then the critical monomials are x_{i}^{t_{i}+1}. On the other hand, if Σ(A) = Σ_{n} and the t_{i} are distinct, then we may choose α so that the t_{i} form a decreasing sequence. In this case, the critical monomials are (x_{1} ⋯ x_{i})^{t_{i}+1}.

Theorem 4.2. Suppose α and β are monomials and that α has trivial critical monomials. Further suppose that for all g ∈ Σ(A), β does not divide g(α). Then β ≡ 0. In particular, if β and α have the same degree and β is not in the Σ(A)-orbit of α, then β ≡ 0.

Proof. Suppose β = x_{1}^{r_{1}} ⋯ x_{n}^{r_{n}}. For every g ∈ Σ(A), β does not divide g(α). Therefore there is an i such that r_{i} > t_{g(i)}. Thus β is divisible by a critical monomial. Since the critical monomials are trivial, β ≡ 0.

Theorem 4.3. Suppose α is non-trivial and all of the associated critical monomials are trivial. Then the monomial factors of α are linearly independent.

Proof. Without loss of generality we may assume that α is the largest element, with respect to the reverse lexicographic ordering, in its Σ(A)-orbit. Observe that if β is a monomial with β > α, then β is divisible by a critical monomial and is therefore congruent to zero.

We suppose ∑_{i=1}^{m} γ_{i}β_{i} ≡ 0 where γ_{i} ∈ k − {0}, β_{i} is a monomial factor of α and β_{i} > β_{i−1}. Multiply this relation by the complement of β_{1}, β_{1}^{c} = α/β_{1}, to get γ_{1}α + ∑_{i=2}^{m} γ_{i}β_{1}^{c}β_{i} ≡ 0. Since β_{i} > β_{1}, we have β_{1}^{c}β_{i} > β_{1}^{c}β_{1} = α and therefore β_{1}^{c}β_{i} ≡ 0.Thus γ_{1}α ≡ 0. However γ_{1} ≠ 0, and so α ≡ 0. This contradicts our hypothesis and therefore the monomial factors of α are linearly independent.

Theorem 4.4. Suppose α has trivial critical monomials and the sequence t_{i} is a permutation of d_{i} − 1. Then α generates a block basis for P_{n}/I.

Proof. Since the exponent sequence of α is a permutation of d_{i} − 1, we have the correct number of monomial factors in the correct degrees to give a basis. If α is non-trivial then, from Theorem 4.3, these factors are linearly independent and we have the required result. Thus it is sufficient to prove that α ≢ 0.

We know that P_{n}/I has dimension one in degree (∑_{i=1}^{n} d_{i}) − n. Hence there is at least one non-trivial monomial in this degree. From Theorem 4.2 we see that any monomial outside of the Σ(A)-orbit of α is trivial. Therefore at least one monomial in the Σ(A)-orbit of α is non-trivial. Thus α is non-trivial

This theorem provides a sufficient but not a necessary condition for the existence of a block basis. For example, take n = 2 and let a_{1} = x_{1}^{2} + x_{1}x_{2} and a_{2} = x_{2}^{2}. Observe that α = x_{1}x_{2} has an acceptable exponent sequence and {1, x_{1}, x_{2}, x_{1}x_{2}} is a block basis for P_{n}/I. Here Σ(k[a_{1}, a_{2}]) = {1} and thus x_{1}^{2} is a critical monomial. However, x_{1}^{2} is clearly non-trivial.

Corollary 4.5. Suppose α generates a block basis and that all of the associated critical monomials are trivial. Then a monomial, β, is non-trivial if and only if β divides g(α) for some g ∈ Σ(A).

Proof. If β divides g(α) then β is an element of the block basis generated by g(α). Therefore β ≢ 0. If β does not divide g(α) for all g ∈ Σ(A) then, using Theorem 4.2, β ≡ 0.

Corollary 4.6. Suppose α generates a block basis and that all of the associated critical monomials are trivial. Then a monomial, β, is trivial if and only if there is a critical monomial w that divides β.

Proof. If w is a critical monomial then by hypothesis w is trivial and if w divides β then β is trivial.

On the other hand, suppose that β is trivial. By Corollary 4.5, β does not divide g(α) for any g ∈ Σ(A). Thus β satisfies the first condition in Definition 4.1. Therefore, either β is a critical monomial or a proper factor of β is a critical monomial.

Theorem 4.7. Suppose that Σ(A) = 1. Then the critical monomials associated to α are x_{i}^{d_{i}}. If these critical monomials are trivial then P_{n}/I is isomorphic, as an algebra, to

⨂_{i=1}^{n} k[x_{i}]/(x_{i}^{d_{i}}).

Proof. It follows immediately from Definition 4.1 that the critical monomials are x_{i}^{d_{i}}. As in the proof of Theorem 4.1, we take J to be the ideal generated by {x_{1}^{d_{1}}, …, x_{n}^{d_{n}}}. Since x_{i}^{d_{i}} ∈ I, J ⊆ I. Therefore there is an algebra homomorphism from P_{n}/J to P_{n}/I. This map induces a bijection between the block bases for P_{n}/J and P_{n}/I.

5. Characterizing the monomials in (D_{n}^{+})

When A = D_{n} we may choose α by taking t_{i} = q^{n}−q^{i−1}−1. In this case Σ(A) = Σ_{n} and the corresponding critical monomials are w_{j} = (x_{1} … x_{j})^{q^{n}−q^{j−1}}. In this section we will show that w_{j} ≡ 0. Using Theorem 4.4 this gives an alternate proof that x_{1}^{q^{n}−2}x_{2}^{q^{n}−q−1} ⋯ x_{n}^{q^{n}−q^{n−1}−1} generates a block basis for (P_{n})_{GL_{n}(F_{q})}.

Theorem 5.1. (x_{1} … x_{j})^{q^{n}−q^{j−1}} ≡ 0.

Applying Corollary 4.5 gives us a characterization of the non-trivial monomials.

Corollary 5.2. A monomial is non-trivial if and only if it is a permutation of an element in the block basis generated by α.

Most of this section will be devoted to the proof of Theorem 5.1. Our methods make considerable use of polynomials in P_{n}[t]. Let I_{n} denote the ideal generated by D_{n}^{+}. If f(t) and g(t) are elements of P_{n}[t], we will write f(t) ≡ g(t) if the coefficients of f(t) − g(t) are elements of I_{n}. The following lemma allows us to divide, in (P_{n}/I_{n})[t], by a monic polynomial.

Lemma 5.3. Suppose f(t) ≡ g(t) and f(t) = u(t)q_{1}(t) with u(t) monic. Write g(t) = u(t)q_{2}(t) + r(t) with deg(r) < deg(u). Then r(t) ≡ 0 and q_{1}(t) ≡ q_{2}(t).

Proof. Since r(t) ≡ u(t)(q_{1}(t) − q_{2}(t)), it is sufficient to prove that q_{1}(t) ≡ q_{2}(t). We will use proof by contradiction.

Suppose q_{1}(t) − q_{2}(t) ≢ 0. Let i be the degree of q_{1}(t) − q_{2}(t) in (P_{n}/I_{n})[t]. Then there exits a polynomial, say s(t), with deg(s) < i and a non-zero c ∈ P_{n}/I_{n} such that q_{1}(t) − q_{2}(t) ≡ ct^{i} + s(t). Let k be the degree of u(t). Since u(t) is monic there exists a polynomial, say u_{1}(t), with deg(u_{1}) < k, satisfying u(t) = t^{k} + u_{1}(t). Since r(t) ≡ u(t)(q_{1}(t) − q_{2}(t)), we have

r(t) ≡ ct^{i+k} + s(t)t^{k} + u_{1}(t)ct^{i} + u_{1}(t)s(t).

Solving for ct^{i+k} gives

ct^{i+k} ≡ r(t) − (s(t)t^{k} + u_{1}(t)ct^{i} + u_{1}(t)s(t)) .

The polynomial on the left has degree k + i while the polynomial on the right has degree less than k + i. This contradiction proves the lemma.

We start our proof of Theorem 5.1 by attempting to compute F_{n}(t)/F_{k}(t). Following Steinberg ([17]), we observe that the coefficients of the quotient are polynomials in x_{1}, …, x_{k} and d_{1,n}, …, d_{n,n}. As a consequence of this, there is a monic polynomial in (F_{q}[x_{1}, …, x_{k}])[t] congruent to F_{n}(t)/F_{k}(t). We now construct such a polynomial. Applying the division algorithm gives t^{q^{n}} = F_{k}(t)f_{k}(t) + r_{k}(t). By definition F_{k}(t) ∈ (F_{q}[x_{1}, …, x_{k}])[t] and therefore f_{k}(t) ∈ (F_{q}[x_{1}, …, x_{k}])[t]. Furthermore, since F_{n}(t) ≡ t^{q^{n}} and F_{k}(t) is a monic polynomial dividing F_{n}(t), applying Lemma 5.3 gives f_{k}(t) ≡ F_{n}(t)/F_{k}(t). Hence f_{k}(t) is a monic polynomial with the required properties. Since x_{k+1} is a root of F_{n}(t) but not a root of F_{k}(t), f_{k}(x_{k+1}) lies in I_{n}. In fact, in Section 6 we prove that {f_{0}(x_{1}), …, f_{n−1}(x_{n})} is a Gröbner basis for I_{n} associated to the lexicographic monomial order. In principle this Gröbner basis should allow us to determine whether or not (x_{1} … x_{j})^{q^{n}−q^{j−1}} is in I_{n}. Unfortunately we do not have a sufficiently explicit description of f_{k}(t) for an arbitrary k. Instead we introduce a family of polynomials which, in a certain sense, approximate F_{n}(t)/F_{k}(t). With this in mind, we define g_{k,ℓ}(t) in P_{n}[t]. Let g_{k,0}(t) = t^{q^{k}−1} and, for n ≥ k ≥ ℓ > 0, let

g_{k,ℓ}(t) = ∑_{i=0}^{k−ℓ} g_{ℓ−1+i,ℓ−1}^{q^{k−ℓ−i+1}}(x_{ℓ})F_{ℓ}^{q^{k−ℓ−i}−1}(t).

We also define G_{k}(t) = g_{k,1}(t)F_{1}(t). We now prove a series of lemmas describing properties of g_{k,ℓ}(t) and G_{k}(t).

Lemma 5.4. g_{k,k}(t) = 1.

Proof. The proof is by induction on k. Since g_{k,0}(t) = t^{q^{k}−1}, when k = 0 we have g_{0,0}(t) = 1. For k > 0, g_{k,k}(t) = g_{k−1,k−1}^{q}(x_{k}). Using the induction hypothesis gives g_{k,k}(t) = 1.

Lemma 5.5. G_{k}(t) = t^{q^{k}} − x_{1}^{q^{k}−1}t.

Proof. Using the definitions of G_{k}(t) and g_{k,1}(t), and the fact that F_{1}(t) = t^{q} − x_{1}^{q−1}t, gives

G_{k}(t) = ∑_{i=0}^{k−1} x_{1}^{q^{k}−q^{k−i}} (t^{q^{k−i}} − x_{1}^{q^{k−i}−q^{k−i−1}}t^{q^{k−i−1}}) .

Simplifying this gives G_{k}(t) = t^{q^{k}} − x_{1}^{q^{k}−1}t.

Lemma 5.6. G_{k}(t) = G_{k−1}^{q}(t) + x_{1}^{q^{k}−q}F_{1}(t).

Proof. Using Lemma 5.5 we observe that G_{k−1}^{q}(t) = t^{q^{k}} − x_{1}^{q^{k}−q}t^{q}. Therefore

G_{k−1}^{q}(t) + x_{1}^{q^{k}−q}F_{1}(t) = t^{q^{k}} − x_{1}^{q^{k}−1}t = G_{k}(t).

Lemma 5.7. For ℓ > 0, g_{k,ℓ}(t) = F_{ℓ}^{q−1}(t)g_{k−1,ℓ}^{q}(t) + g_{k−1,l−1}^{q}(x_{ℓ}).

Proof. We start from the definition of g_{k,ℓ}(t) and pull out a common factor from the first k − ℓ terms to get

g_{k,ℓ}(t) = F_{ℓ}^{q−1}(t) ( ∑_{i=0}^{k−ℓ−1} g_{ℓ−1+i,ℓ−1}^{q^{k−ℓ−i+1}}(x_{ℓ})F_{ℓ}^{q^{k−ℓ−i}−q}(t)) + g_{k−1,ℓ−1}^{q}(x_{ℓ}).

Observe that

∑_{i=0}^{k−ℓ−1} g_{ℓ−1+i,ℓ−1}^{q^{k−ℓ−i+1}}(x_{ℓ})F_{ℓ}^{q^{k−ℓ−i}−q}(t) = ( ∑_{i=0}^{k−ℓ−1} g_{ℓ−1+i,ℓ−1}^{q^{k−ℓ−i}}(x_{ℓ})F_{ℓ}^{q^{k−ℓ−i−1}−1}(t))^{q} = g_{k−1,ℓ}^{q}(t).

Substituting the second equation into the first gives the desired result.

Lemma 5.8. G_{k}(t) = ∑_{i=1}^{k} g_{k,i}(x_{i+1})F_{i}(t).

Proof. The proof is by induction. For k = 1 the result follows from Lemma 5.4 and the definition of G_{1}(t).

Suppose k > 1. From Lemma 5.6 we have G_{k}(t) = G_{k−1}^{q}(t) + x_{1}^{q^{k}−q}F_{1}(t). Using the induction hypothesis gives

G_{k}(t) = (∑_{i=1}^{k−1} g_{k−1,i}(x_{i+1})F_{i}(t))^{q} + x_{1}^{q^{k}−q}F_{1}(t).

It follows from Proposition 1.1 that F_{i}^{q}(t) = F_{i+1}(t)+F_{i}^{q−1}(x_{i+1})F_{i}(t). Using this and the fact that taking the q^{th} power is linear gives

G_{k}(t) = ∑_{i=1}^{k−1} g_{k−1,i}^{q}(x_{i+1}) (F_{i+1}(t) + F_{i}^{q−1}(x_{i+1})F_{i}(t)) + x_{1}^{q^{k}−q}F_{1}(t).

Split the sum into two pieces, pull the last term out of the first piece and pull the first term out of the second piece to get

G_{k}(t) = g_{k−1,k−1}^{q}(x_{k})F_{k}(t) + ∑_{i=1}^{k−2} g_{k−1,i}^{q}(x_{i+1})F_{i+1}(t)

+ ∑_{i=2}^{k−1} g_{k−1,i}^{q}(x_{i+1})F_{i}^{q−1}(x_{i+1})F_{i}(t)
+ g_{k−1,1}^{q}(x_{2})F_{1}^{q−1}(x_{2})F_{1}(t) + x_{1}^{q^{k}−q}F_{1}(t).

Re-indexing the first sum and combining with the second gives

G_{k}(t) = g_{k−1,k−1}^{q}(x_{k})F_{k}(t) + ∑_{i=2}^{k−1} (g_{k−1,i−1}^{q}(x_{i})F_{i}(t) + g_{k−1,i}^{q}(x_{i+1})F_{i}^{q−1}(x_{i+1})F_{i}(t))
+ g_{k−1,1}^{q}(x_{2})F_{1}^{q−1}(x_{2})F_{1}(t) + x_{1}^{q^{k}−q}F_{1}(t).

Pull out the common factor and use the fact that x_{1}^{q^{k−1}−1} = g_{k−1,0}(x_{1}) to get

G_{k}(t) = g_{k−1,k−1}^{q}(x_{k})F_{k}(t) + ∑_{i=1}^{k−1} (g_{k−1,i}^{q}(x_{i+1})F_{i}^{q−1}(x_{i+1}) + g_{k−1,i−1}^{q}(x_{i})) F_{i}(t).

Setting t = x_{i+1} and ℓ = i in Lemma 5.7 gives

g_{k,i}(x_{i+1}) = g_{k−1,i}^{q}(x_{i+1})F_{i}^{q−1}(x_{i+1}) + g_{k−1,i−1}^{q}(x_{i}). (5.1)

Therefore

G_{k}(t) = g_{k−1,k−1}^{q}(x_{k})F_{k}(t) + ∑_{i=1}^{k−1} g_{k,i}(x_{i+1})F_{i}(t).

From Lemma 5.4, g_{k−1,k−1}^{q}(x_{k}) = 1 = g_{k,k}(x_{k+1}) and the result follows.

Lemma 5.9. G_{k}(t) = g_{k,ℓ}(t)F_{ℓ}(t) + ∑_{i=1}^{ℓ−1} g_{k,i}(x_{i+1})F_{i}(t).

Proof. The proof is by induction on k. For k = ℓ this is Lemma 5.8.

Suppose k > ℓ. From Lemma 5.6 we have G_{k}(t) = G_{k−1}^{q}(t) + x_{1}^{q^{k}−q}F_{1}(t). The induction hypothesis gives G_{k−1}(t) = g_{k−1,ℓ}(t)F_{ℓ}(t) + ∑_{i=1}^{ℓ−1} g_{k−1,i}(x_{i+1})F_{i}(t). Using this and the fact that taking a q^{th} power is linear gives

G_{k}(t) = g_{k−1,ℓ}^{q}(t)F_{ℓ}^{q}(t) + ∑_{i=1}^{ℓ−1} g_{k−1,i}^{q}(x_{i+1})F_{i}^{q}(t) + x_{1}^{q^{k}−q}F_{1}(t).

Applying Lemma 5.7 to the first term and Proposition 1.1 to the sum gives

G_{k}(t) = (g_{k,ℓ}(t) − g_{k−1,ℓ−1}^{q}(x_{ℓ})) F_{ℓ}(t) + ∑_{i=1}^{ℓ−1} g_{k−1,i}^{q}(x_{i+1}) (F_{i+1}(t) + F_{i}^{q−1}(x_{i+1})F_{i}(t))
+ x_{1}^{q^{k}−q}F_{1}(t).

As in the proof of Lemma 5.8, we split the sum into two pieces and re-index one of the pieces to get

G_{k}(t) = g_{k,ℓ}(t)F_{ℓ}(t) + ∑_{i=1}^{ℓ−1} (g_{k−1,i}^{q}(x_{i+1})F_{i}^{q−1}(x_{i+1}) + g_{k−1,i−1}^{q}(x_{i})) F_{i}(t).

Using Equation 5.1 gives

G_{k}(t) = g_{k,ℓ}(t)F_{ℓ}(t) + ∑_{i=1}^{ℓ−1} g_{k,i}(x_{i+1})F_{i}(t)

completing the proof.

To this point we have been working in P_{n}[t]. We now turn our attention to P_{n}/I_{n}.

Lemma 5.10. x_{1}^{q^{n}−1} ≡ 0.

Proof. Since x_{1} is a root of F_{n}(t) we have

F_{n}(x_{1}) = ∑_{i=0}^{n} d_{i,n}x_{1}^{q^{n−i}} = 0.

Therefore x_{1}^{q^{n}−1} = − ∑_{i=1}^{n} d_{i,n}x_{1}^{q^{n−i}−1}, so x_{1}^{q^{n}−1} ∈ I_{n}.

Lemma 5.11. x_{1}^{q^{n}−q^{k}}G_{k}(t) ≡ x_{1}^{q^{n}−q^{k}}t^{q^{k}}.

Proof. From Lemma 5.5 we have G_{k}(t) = t^{q^{k}} − x_{1}^{q^{k}−1}t. Multiplying by x_{1}^{q^{n}−q^{k}} gives x_{1}^{q^{n}−q^{k}}G_{k}(t) = x_{1}^{q^{n}−q^{k}}(t^{q^{k}} − x_{1}^{q^{k}−1}t). Apply Lemma 5.10.

In light of Lemma 5.11 we can think of G_{k}(t) as a sort of approximation or substitute for t^{q^{k}}. We shall see that in the same sense t^{q^{n}−q^{k}}g_{k,ℓ}(t) is an approximation to F_{n}(t)/F_{ℓ}(t).

Theorem 5.12. (x_{1} … x_{ℓ+1})^{q^{n}−q^{k}}g_{k,ℓ}(x_{ℓ+1}) ≡ 0 Proof. The proof is by induction on ℓ. For ℓ = 0 this is Lemma 5.10.

Suppose ℓ > 0. Observe that F_{n}(t) ≡ t^{q^{n}}. Using Lemma 5.11 gives

x_{1}^{q^{n}−q^{k}}F_{n}(t) ≡ (x_{1}t)^{q^{n}−q^{k}}G_{k}(t).

Therefore, using Lemma 5.9,

(x_{1} … x_{ℓ})^{q^{n}−q^{k}}F_{n}(t) ≡ (x_{1} … x_{ℓ}t)^{q^{n}−q^{k}} (g_{k,ℓ}(t)F_{ℓ}(t) + ∑_{i=1}^{ℓ−1} g_{k,i}(x_{i+1})F_{i}(t)) .

Applying the induction hypothesis gives

(x_{1} … x_{ℓ})^{q^{n}−q^{k}}F_{n}(t) ≡ (x_{1} … x_{ℓ}t)^{q^{n}−q^{k}}g_{k,ℓ}(t)F_{ℓ}(t).

Apply Lemma 5.3 to get

(x_{1} … x_{ℓ})^{q^{n}−q^{k}} (F_{n}(t)/F_{ℓ}(t)) ≡ (x_{1} … x_{ℓ}t)^{q^{n}−q^{k}}g_{k,ℓ}(t).

Since x_{ℓ+1} is a root for F_{n}(t)/F_{ℓ}(t), setting t = x_{ℓ+1} gives the required result.

Setting k = ℓ = j − 1 in Theorem 5.12 gives (x_{1} … x_{j})^{q^{n}−q^{j−1}} ≡ 0. This proves Theorem 5.1.

The approximation of F_{n}(t)/F_{ℓ}(t) referred to above is given by the following corollary.

Corollary 5.13. (x_{1} … x_{ℓ})^{q^{n}−q^{k}} (F_{n}(t)/F_{ℓ}(t)) ≡ (x_{1} … x_{ℓ})^{q^{n}−q^{k}} (t^{q^{n}−q^{k}}g_{k,ℓ}(t)) .

We conclude by briefly examining the case A = D_{n}(ℓ). Here α is defined by taking t_{i} = q^{ℓ}(q^{n} − q^{i−1}) − 1. The corresponding critical monomials are w_{j}(ℓ) = (x_{1} ⋯ x_{j})^{q^{ℓ}(q^{n}−q^{j−1})}. These monomials are the q^{ℓ} power of the critical monomials for D_{n}. Since taking the q^{ℓ} power is linear, we can use a description of w_{i} as an element of I_{n} to get a description of w_{i}(ℓ) as an element of I_{n}(ℓ). This explicitly determines the non-trivial monomials in D_{n}(ℓ) and gives an alternate proof that D_{n}(ℓ) admits a block basis.

6. A Gröbner Basis for (D_{n}^{+})

We continue to use the notation of Section 5. Recall that f_{k}(t) is a monic polynomial of degree q^{n} − q^{k} in (F_{q}[x_{1}, …, x_{k}])[t] with f_{k}(t) ≡ F_{n}(t)/F_{k}(t). Furthermore f_{k}(x_{k+1}) ∈ I_{n}. Let g_{i} = f_{i−1}(x_{i}). We use the lexicographic monomial order with x_{i} > x_{j} if i > j. We will show that {g_{1}, …, g_{n}} is a Gröbner basis for I_{n} with respect to this order. We refer the reader to Chapter 2 of [4] for the appropriate definitions.

Theorem 6.1. The set {g_{1}, …, g_{n}} is the reduced Gröbner basis for I_{n}.

Proof. Let LT (I_{n}) denote the ideal in P_{n} generated by the lead terms from I_{n} and let J be the ideal in P_{n} generated by {LT (g_{1}), …, LT (g_{n})}. Clearly J ⊆ LT (I_{n}). By definition, if these ideals are equal then {g_{1}, …, g_{n}} is a Gröbner basis for I_{n}.

Observe that LT (g_{i}) = x_{i}^{q^{n}−q^{i−1}} and, therefore, {LT (g_{1}), …, LT (g_{n})} is a regular sequence in P_{n}. The degrees of {LT (g_{1}), …, LT (g_{n})} are a permutation of the degrees of {d_{1,n}, …, d_{n,n}} and both sequences are regular. Thus the Hilbert series of I_{n} is equal to the Hilbert series of J. Note that the Hilbert series of I_{n} equals the Hilbert series of LT (I_{n}), see, for example, Proposition 9 on page 433 of [4]. Thus the Hilbert series of J equals the Hilbert series of LT (I_{n}). Since J ⊆ LT (I_{n}), it follows that J = LT (I_{n}) and {g_{1}, …, g_{n}} is a Gröbner basis for I_{n}.

To see that this basis is reduced observe that the lead term of g_{i} is x_{i}^{q^{n}−q^{i−1}}. For reasons of degree this lead term can not divide any monomial in {g_{i+1}, …, g_{n}}. Since g_{1}, …, g_{i−1} only involve the variables x_{1}, …, x_{i−1}, none of their monomials is divisible by x_{i}^{q^{n}−q^{i−1}}.

We note that we have a recursive algorithm for computing products in the ring of Dickson coinvariants since the relations g_{1}, …, g_{n} allow us to write x_{i}^{q^{n}−q^{i−1}} in terms of our block basis.

We have three methods of computing g_{i}. The first is by performing the division used in their definition. The second is Buchberger’s algorithm, see page 89 of [4]. Finally, we can compute them recursively: we may substitute x_{i}^{q^{n}−q^{i−1}} −g_{i} for x_{i}^{q^{n}−q^{i−1}} for all i > j in d_{j,n} to obtain g_{j}. We believe the first algorithm is the most efficient.

7. The Upper Triangular Coinvariants

In this section we will show that when A = H_{n}, the hypotheses of Theorem 4.4 are satisfied by the appropriate monomial. This gives another proof of the existence of a block basis for P_{n} as a module over H_{n}. In this case Σ(A) = {1} and we choose α by taking t_{i} = q^{i−1} − 1. The critical monomials are then w_{j} = x_{j}^{q^{j−1}}.

Theorem 7.1. x_{j}^{q^{j−1}} ≡ 0.

Proof. Recall that h_{j} = F_{j−1}(x_{j}) = x_{j}^{q^{j−1}} + ∑_{i=1}^{j} d_{i,j−1}x_{j}^{q^{j−1−i}}. Furthermore d_{i,j−1} is an element of D_{j−1} and D_{j−1} ⊂ H_{j−1} ⊂ H_{j}. Therefore d_{i,j−1} ∈ H_{j} and x_{j}^{q^{j−1}} ≡ 0.

Since Σ(A) = {1} we may apply Theorem 4.7 to get the following description of the algebra structure of (P_{n})_{U_{n}(F_{q})}.

Corollary 7.2.

(P_{n})_{U_{n}(F_{q})} ≅ ⨂_{j=1}^{n} k[x_{j}]/(x_{j}^{q^{j−1}}).

Note that this corollary also follows easily from Mùi’s description of the generators of P_{n}^{U_{n}(F_{q})}([12]).

8. The Symmetric Coinvariants

In [3] the first two authors constructed a block basis for the symmetric coinvariants. We will start this section by using our method to provide an alternate proof.

Let σ_{m,ℓ} denote the elementary symmetric function of degree m in ℓ variables. We will use σ_{m} to denote σ_{m,n}. For this example a_{m} = σ_{m} and we choose α by taking t_{i} = i − 1.

Clearly Σ(A) = Σ_{n} and the critical monomials are w_{i} = (x_{n} … x_{n+1−i})^{n+1−i}. We use the following lemma to prove that the critical monomials are congruent to zero.

Lemma 8.1. (x_{n} … x_{n+1−i})^{m}σ_{n+1−m−i,n−i} ≡ 0.

Proof. The proof is by induction on i. When i = 1 the polynomial in question is x_{n}^{m}σ_{n−m,n−1}. We will use induction on m to prove that this polynomial is in the ideal. When m = 1 we have x_{n}σ_{n−1,n−1} = σ_{n} ∈ I. Suppose m > 1. Observe that

σ_{ℓ,j} = σ_{ℓ,j−1} + x_{j}σ_{ℓ−1,j−1} (8.1)

Therefore x_{n}^{m−1}σ_{n+1−m} = x_{n}^{m−1}σ_{n+1−m,n−1} + x_{n}^{m}σ_{n−m,n−1}. The term on the left is clearly in the ideal. The first term on the right is in the ideal by the induction hypothesis. Thus x_{n}^{m}σ_{n−m,n−1} ≡ 0.

Suppose i > 1. Again we proceed by induction on m. When m = 1 we have x_{n} … x_{n+1−i}σ_{n−i,n−i} = σ_{n} ∈ I. Suppose m > 1. Applying Equation 8.1 gives

(x_{n} … x_{n+2−i})^{m} x_{n+1−i}^{i−1}σ_{n+2−m−i,n+1−i} = (x_{n} … x_{n+2−i})^{m} x_{n+1−i}^{i−1}σ_{n+2−m−i,n−i}
+ (x_{n} … x_{n+1−i})^{m} σ_{n+1−i−m,n−i}.

Using the primary induction hypothesis, the term on the left is in the ideal. Using the secondary induction hypothesis, the first term on the right hand side is in the ideal. Thus (x_{n} … x_{n+1−i})^{m} σ_{n+1−i−m,n−i} ≡ 0.

Setting i = n + 1 − m in the preceding lemma gives the following.

Corollary 8.2. The critical monomials w_{i} = (x_{n} … x_{n+1−i})^{n+1−i} are trivial.

Since the exponent sequence is acceptable, we obtain

Corollary 8.3. The monomial x_{2}^{1}x_{3}^{2} ⋯ x_{n}^{n−1} generates a block basis for the symmetric coinvariants.

At this point it will be convenient to work with the block basis given by the monomial x_{1}^{n−1}x_{2}^{n−2} … x_{n−1}. To describe the product in (P_{n})_{Σ_{n}} we need to be able to write x_{n+1−m}^{m} in terms of our block basis. Let s_{m,ℓ} be the sum of all of the monomials of degree m in the first ℓ variables. These polynomials are the complete symmetric polynomials — see [10, I.2].

Theorem 8.4. If m + ℓ ≥ n + 1 then s_{m,n+1−ℓ} ≡ 0.

Proof. The proof is by induction on m + ℓ. We begin by reducing the general step to the initial step. Assume m + ℓ > n + 1. We observe from the definition that s_{m,ℓ} = x_{ℓ}s_{m−1,ℓ} + s_{m,ℓ−1} and both of the terms on the right are zero by induction. For the initial step, we must show that s_{m,n+1−m} ≡ 0 for 1 ≤ m ≤ n and we do this by induction on m. The case m = 1 is easy. Assume m > 1.

By repeated applications of the identity just used, we observe that

s_{m,n} = x_{n}s_{m−1,n} + s_{m,n−1}
= x_{n}s_{m−1,n} + x_{n−1}s_{m−1,n−1} + s_{m,n−2}
⋮
= x_{n}s_{m−1,n} + x_{n−1}s_{m−1,n−1} + ⋯ + x_{n+2−m}s_{m−1,n+2−m} + s_{m,n+1−m}.

Each of the terms on the right-hand side of this equation, except for the last, is trivial by induction on m. The left-hand side is trivial since s_{m,n} is a symmetric function in the variables x_{1}, …, x_{n}. Thus s_{m,n+1−m} ≡ 0.

Let S_{m} be the set of all monomials of degree m in the first n+1−m variables except x_{n+1−m}^{m}. Note that S_{m} is a subset of our block basis. As an immediate consequence of Theorem 8.4 we have

Corollary 8.5. x_{n+1−m}^{m} ≡ ∑_{β∈S_{m}} −β.

This provides a recursive algorithm for computing products in the ring of symmetric coinvariants, however such algorithms are well-known.

Finally, define f_{i} = s_{n+1−i,i} so that deg(f_{i}) = n + 1 − i and f_{i} is a polynomial in x_{1}, …, x_{i}. Using the lexicographic monomial order the leading term of f_{i} is x_{i}^{n+1−i}. Using the arguments found in the proof of Theorem 6.1 we obtain the following.

Corollary 8.6. The set {f_{1}, …, f_{n}} is the reduced Gröbner basis for the ideal generated by the elementary symmetric functions.

9. The Modular Transfer

For a finite group G, the transfer, T r_{G}, is the function from P_{n} to P_{n}^{G} defined by

T r_{G}(f) = ∑_{g∈G} g(f).

The transfer is map of P_{n}^{G}-modules and, therefore, the image of the transfer is an ideal in P_{n}^{G}. If the characteristic of the field does not divide the order of the group, then the transfer is surjective. However, for a modular representation, the image of the transfer is a proper ideal in P_{n}^{G}. For a discussion of the modular transfer see [15, 11.5]. In this section we use block bases to study T r_{GL_{n}(F_{q})} and T r_{U_{n}(F_{q})}. We will prove that, in both cases, the image of the transfer is a principal ideal. We conclude the section with a few remarks on the modular transfer for Σ_{n}.

When P_{n}^{G} is a polynomial algebra, P_{n} is a free P_{n}^{G}-module and a basis for (P_{n})_{G} over k lifts to a basis for P_{n} over P_{n}^{G}. The image of the transfer is thus generated by the image of a basis for (P_{n})_{G}. Our goal is to compute the T r_{GL_{n}(F_{q})} and T r_{U_{n}(F_{q})} on the respective block bases. We start with a few definitions and preliminary lemmas.

Definition 9.1. Suppose m is a non-negative integer and the q-adic expansion of m is m = c_{0} + c_{1}q + … + c_{j}q^{j} where q = p^{s} and each c_{i} is an integer greater than or equal to 0 and less than q. Then we define

α_{q}(m) = c_{0} + c_{1} + … + c_{j}.

Note that q −1 divides m if and only if q −1 divides α_{q}(m), since q^{i} ≡ 1 (mod (q − 1)). We use this fact frequently in this section. Also, α_{q}(ℓ + m) ≤ α_{q}(ℓ) + α_{q}(m) with equality if and only if the addition ℓ + m can be performed base q without carrying.

Definition 9.2. For a sequence of non-negative integers I = (i_{1}, …, i_{k}), with m = i_{1} + … + i_{k}, the multinomial coefficient, \binom{m}{I}, is defined by

\binom{m}{I} = \binom{m}{i_{1}, …, i_{k}} = \frac{m!}{i_{1}! … i_{k}!}.

If m ≠ i_{1} + … + i_{k} then we define \binom{m}{I} = 0.

Lemma 9.1. If m = c_{0} + c_{1}q + … + c_{j}q^{j} is the q-adic expansion of m and i_{r} = i_{r,0} + i_{r,1}q + … + i_{r,j}q^{j} is the q-adic expansion of i_{r} then

\binom{m}{I} = \binom{m}{i_{1}, …, i_{k}} ≡ \binom{c_{0}}{i_{1,0}, …, i_{k,0}} ⋯ \binom{c_{s}}{i_{1,j}, …, i_{k,j}} (mod p).

Proof. Consider the expression (x_{1} +⋯+x_{r})^{m} as a polynomial with cofficients in F_{p}. Expanding gives an expression involving the multinomial coefficients \binom{m}{I}. However, using the fact that q is a power of p,

(x_{1} + ⋯ + x_{j})^{m} = ∏_{i=0}^{j}(x_{1}^{q^{i}} + ⋯ + x_{r}^{q^{i}})^{c_{i}}.

Expanding the right hand side of the above equation and comparing coefficients with the original expansion gives the required result.

Remark 9.2. The preceeding lemma is essentially due to Dickson — see [5]. There he also proves that the highest power of p dividing \binom{m}{I} is p^{e} where e is the total amount carried when the addition i_{1} + ⋯ + i_{k} = m is performed base p.

Remark 9.3. We will need to understand when p divides \binom{m}{I}. It follows from either

Lemma 9.1 or the preceeding remark that p does not divide \binom{m}{I} if and only if in performing the addition i_{1} + ⋯ + i_{k} = m base p there are no carries. This last condition is equivalent to α_{p}(i_{1}) + ⋯ + α_{p}(i_{k}) = α_{p}(m). Furthermore, if there are no carries base p, there are no carries base q (but not conversely) and therefore, if p does not divide \binom{m}{I} then α_{q}(i_{1}) + ⋯ + α_{q}(i_{k}) = α_{q}(m).

The following lemma is well-known, but we include a proof for completeness.

Lemma 9.4. Suppose that ℓ is a positive integer. Then

∑_{c∈F_{q}} c^{ℓ} = \{ \begin{matrix}−1, & if q − 1 divides ℓ; \\ 0, & if q − 1 does not divide ℓ. \end{matrix}

Proof. It is sufficient to sum over the non-zero elements of F_{q}. Let η be a generator for the group of units of F_{q}. If c ∈ F_{q} and c ≠ 0 then c = η^{i} and, therefore,

∑_{c∈F_{q}} c^{ℓ} = ∑_{i=0}^{q−2}(η^{i})^{ℓ} = ∑_{i=0}^{q−2}(η^{ℓ})^{i}.

Every unit in F_{q} is a root of x^{q} − 1 = (x − 1)(1 + x + … + x^{q−2}). If q − 1 divides ℓ then η^{ℓ} = 1 and the sum is q − 1. If q − 1 does not divide ℓ then η^{ℓ} is a root of 1 + x + … + x^{q−2} and the sum is zero.

The following proposition plays a critical role in our computations of both T r_{U_{n}(F_{q})} and T r_{GL_{n}(F_{q})}. We remind the reader that V_{i} denotes the subspace of V spanned by {x_{1}, …, x_{i}}.

Proposition 9.5. (i) If ∑_{u∈V_{i}} u^{ℓ} ≠ 0 then q − 1 divides ℓ and α_{q}(ℓ) ≥ i(q − 1), (ii) d_{i,i} divides ∑_{u∈V_{i}} u^{ℓ} and ∑_{u∈V_{i}} u^{q^{i}−1} = d_{i,i}.

Proof. Note that when q = p, the second part of (ii) is contained in Lemma 11.1 of [3]. We adopt the convention that 0^{0} = 1 and so the proposition is true when ℓ = 0. In the following we assume ℓ > 0.

The proof is by induction on i. When i = 1 we have

∑_{u∈V_{1}} u^{ℓ} = ∑_{c∈F_{q}}(cx_{1})^{ℓ} = x_{1}^{ℓ} ∑_{c∈F_{q}} c^{ℓ}.

It follows from Lemma 9.4 that

∑_{u∈V_{1}} u^{ℓ} = \{ \begin{matrix}−x_{1}^{ℓ}, & if q − 1 divides ℓ; \\ 0, & if q − 1 does not divide ℓ. \end{matrix}

Note that if q − 1 divides ℓ, q − 1 also divides α_{q}(ℓ) and α_{q}(ℓ) ≥ q − 1. This proves (i) when i = 1. To prove (ii), observe that d_{1,1} = −x_{1}^{q−1}.

Suppose i > 1. We start by expanding the sum:

∑_{u∈V_{i}} u^{ℓ} = ∑_{u∈V_{i−1}} ∑_{c∈F_{q}}(cx_{i} + u)^{ℓ}

= ∑_{u∈V_{i−1}} ∑_{c∈F_{q}} ∑_{k=0}^{ℓ} \binom{ℓ}{k}(cx_{i})^{ℓ−k}u^{k}

= ∑_{k=1}^{ℓ−1} \binom{ℓ}{k}x_{i}^{ℓ−k} ∑_{c∈F_{q}} c^{ℓ−k} ∑_{u∈V_{i−1}} u^{k}.

Thus ∑_{u∈V_{i}} u^{ℓ} is non-zero if and only if there is an integer k with 1 ≤ k ≤ ℓ − 1 such that \binom{ℓ}{k} ∑_{c∈F_{q}} c^{ℓ−k} ∑_{u∈V_{i−1}} u^{k} is non-zero. By induction, ∑_{u∈V_{i−1}} u^{k} ≠ 0 only when q − 1 divides k. Furthermore, ∑_{c∈F_{q}} c^{ℓ−k} ≠ 0 if and only if q − 1 divides ℓ − k. Therefore we can restrict our attention to ℓ and k which are divisible by q − 1.

Suppose \binom{ℓ}{k} is non-zero and consider the q-adic expansions ℓ = ℓ_{0} + ℓ_{1}q + ⋯ + ℓ_{r}q^{r}

and k = k_{0} + ⋯ + k_{r}q^{r}. Using Lemma 9.1, we see that, since \binom{ℓ}{k} ≠ 0, k_{j} ≤ ℓ_{j}. Furthermore, k < ℓ so for some j, k_{j} < ℓ_{j}. Thus α_{q}(ℓ) > α_{q}(k). By induction α_{q}(k) ≥ (i − 1)(q − 1) and thus α_{q}(ℓ) > (i − 1)(q − 1). Since α_{q}(ℓ) is divisible by q − 1, we have α_{q}(ℓ) ≥ i(q − 1).

To prove (ii) we start by describing d_{i,i}. Let R_{i} be any subset of V_{i} consisting of exactly one non-zero element from each line through the origin in V_{i}. It follows from the definition of the Dickson invariants in Section 1 that d_{i,i} = (−1)^{n} ∏_{ϕ∈R_{i}} ϕ^{q−1} Note that x_{i}^{q−1} divides ∑_{u∈V_{i}} u^{ℓ} and ∑_{u∈V_{i}} u^{ℓ} is a GL_{n}(F_{q}) invariant. Therefore ϕ^{q−1} divides ∑_{u∈V_{i}} u^{ℓ} for each ϕ in R_{i}. Since distinct elements of R_{i} have no common factors, d_{i,i} divides ∑_{u∈V_{i}} u^{ℓ}.

Finally, to show that ∑_{u∈V_{i}} u^{q^{i}−1} = d_{i,i} we view these two expressions as polynomials in x_{i} and compare the coefficients of the leading term. We know that d_{i,i} divides ∑_{u∈V_{i}} u^{q^{i}−1} so if the leading terms are equal the expressions are equal. The smallest k such that α(k) ≥ (i − 1)(q − 1) is q^{i−1} − 1. Thus the lead term of ∑_{u∈V_{i}} u^{q^{i}−1} is (−1)\binom{q^{i}−1}{q^{i−1}−1}x_{i}^{q^{i}−q^{i−1}} ∑_{u∈V_{i−1}} u^{q^{i−1}−1}. Simplifying and using induction

gives (−1)x_{i}^{q^{i}−q^{i−1}}d_{i−1,i−1}. On the other hand, using Proposition 1.2, the leading term of d_{i,i} is (−1)^{i}(h_{1} ⋯ h_{i−1})^{q−1}x_{i}^{q^{i}−q^{i−1}}. Since (−1)^{i−1}(h_{1} ⋯ h_{i−1})^{q−1} = d_{i−1,i−1}, the leading terms are equal.

When q = p, a minor modification of the preceding argument shows that ∑_{u∈V_{i}} u^{ℓ} ≠ 0 if and only if p − 1 divides ℓ and α_{p}(ℓ) ≥ i(p − 1). However, if p = 2 and q = 4, then Proposition 9.5 shows that the ∑_{u∈V_{2}} u^{21} = 0 since α_{4}(21) = 3. Thus ∑_{u∈V_{2}} u^{42} = (∑_{u∈V_{2}} u^{21})^{2} = 0 despite that fact that α_{4}(42) = 6 = 2(q − 1).

We now use Proposition 9.5 to evaluate T r_{U_{n}(F_{q})} on the block basis monomials for P_{n} over H_{n}. We remind the reader that β = x_{2}^{ℓ_{2}} ⋯ x_{n}^{ℓ_{n}} is an element of the block basis if and only if, for all i, ℓ_{i} ≤ q^{i−1} − 1.

Theorem 9.6. Let β be an element of the block basis for P_{n} over H_{n}. Then

T r_{U_{n}(F_{q})}(β) = \{ \begin{matrix}(−1)^{n(n−1)/2}(h_{1}^{n−1}h_{2}^{n−2} ⋯ h_{n−1})^{q−1}, & if β = x_{2}^{q−1} ⋯ x_{n}^{q^{n−1}−1}; \\ 0, & otherwise. \end{matrix}

Proof. Observe that

T r_{U_{n}(F_{q})}(x_{2}^{ℓ_{2}} ⋯ x_{n}^{ℓ_{n}}) = ∏_{i=2}^{n} ∑_{u∈V_{i−1}}(x_{i} + u)^{ℓ_{i}}.

Furthermore

∑_{u∈V_{i−1}}(x_{i} + u)^{ℓ_{i}} = ∑_{u∈V_{i−1}} ∑_{k=0}^{ℓ_{i}} \binom{ℓ_{i}}{k}x_{i}^{k}u^{ℓ_{i}−k}

= ∑_{k=0}^{ℓ_{i}} \binom{ℓ_{i}}{k}x_{i}^{k} ∑_{u∈V_{i−1}} u^{ℓ_{i}−k}.

However, using Proposition 9.5, ∑_{u∈V_{i−1}} u^{ℓ_{i}−k} = 0 unless q − 1 divides ℓ_{i} − k and α_{q}(ℓ_{i} − k) ≥ (i − 1)(q − 1). For a basis monomial, ℓ_{i} ≤ q^{i−1} − 1. Therefore ∑_{u∈V_{i−1}} u^{ℓ_{i}−k} = 0 unless ℓ_{i} = q^{i−1} − 1 and k = 0. In particular, if for some i, ℓ_{i} < q^{i−1} − 1 then T r_{U_{n}(F_{q})}(β) = 0. If ℓ_{i} = q^{i−1} − 1 and k = 0 we obtain ∑_{u∈V_{i−1}}(x_{i} + u)^{ℓ_{i}} = ∑_{u∈V_{i−1}} u^{q^{i−1}−1}. Again by Proposition 9.5 we obtain ∑_{u∈V_{i−1}} u^{q^{i−1}−1} = d_{i−1,i−1}. Thus

T r_{U_{n}(F_{q})}(x_{2}^{q−1} ⋯ x_{n}^{q^{n−1}−1}) = ∏_{i=2}^{n} d_{i−1,i−1} = ∏_{i=1}^{n−1} d_{i,i}.

From Proposition 1.2, d_{i,i} = (−1)^{i}(h_{1} ⋯ h_{i})^{q−1}.

Since the image of T r_{U_{n}(F_{q})} is generated as an ideal by the image of a basis for P_{n} over H_{n}, we have the following corollary.

Corollary 9.7. The image of T r_{U_{n}(F_{q})} is the principal ideal in H_{n} generated by (h_{1}^{n−1}h_{2}^{n−2} ⋯ h_{n−1})^{q−1}.

We begin our computations involving T r_{GL_{n}(F_{q})} with the following lemma.

Lemma 9.8. If T r_{GL_{n}(F_{q})}(x_{1}^{ℓ_{1}} ⋯ x_{n}^{ℓ_{n}}) ≠ 0 then, for each i, ℓ_{i} > 0.

Proof. If ℓ_{i} = 0 then the isotropy subgroup of the monomial contains a subgroup of order q. Since the transfer is the product of the orbit sum and the order of the isotropy group, the transfer of the monomial is zero.

Let M_{n} be the set of all n × n matrices with entries in F_{q}. For M ∈ M_{n}, let m_{i} denote the i^{th} column of M. For k < n, define G_{k} to be the set of matrices M = [m_{1} … m_{n}] such that {m_{1}, …, m_{k}} is linearly independant and {m_{1}, …, m_{k+1}}is linearly dependant. Let G_{n} be the set of non-singular matrices. Note that if m_{1} is non-zero, there is a unique k such that M ∈ G_{k}.

Lemma 9.9. Suppose β = x_{1}^{ℓ_{1}} ⋯ x_{n}^{ℓ_{n}} and ℓ_{1} ≠ 0. Then

T r_{GL_{n}(F_{q})}(β) = ∑_{M∈M_{n}} M(β) − ∑_{k=1}^{n−1} ∑_{M∈G_{k}} M(β).

Proof. We first sum over all matrices. Since ℓ_{1} ≠ 0, matrices with m_{1} = 0 do not contribute. We then then subtract the sum over the singular matrices (with m_{1} ≠ 0). The result is the sum over the non-singular matrices, as required.

Lemma 9.10. Suppose β = x_{1}^{ℓ_{1}} ⋯ x_{n}^{ℓ_{n}} and, for all i, 0 < ℓ_{i} ≤ q^{n} − q^{i−1} − 1. Then

∑_{M∈M_{n}} M(β) = 0.

Proof. Observe that

∑_{M∈M_{n}} M(β) = ∑_{u_{1},... ,u_{n}∈V} u_{1}^{ℓ_{1}} ⋯ u_{n}^{ℓ_{n}}

= ∏_{i=1}^{n} ∑_{u∈V} u^{ℓ_{i}}.

Since ℓ_{i} ≤ q^{n} − q^{i−1} − 1, we have α_{q}(ℓ_{i}) < n(q − 1). Applying Proposition 9.5 gives ∑_{u∈V} u^{ℓ_{i}} = 0 and the result follows.

Thus to compute the transfer on a basis element, β, it is sufficient to compute ∑_{M∈G_{k}} M(β) for 1 ≤ k ≤ n − 1.

Lemma 9.11. Suppose β = x_{1}^{ℓ_{1}} ⋯ x_{n}^{ℓ_{n}} and k is fixed. Then

∑_{M∈G_{k}} M(β) = ∑_{u_{1},... ,u_{n−1}∈V} ∑_{c_{1},... ,c_{k}∈F_{q}} u_{1}^{ℓ_{1}} ⋯ u_{k}^{ℓ_{k}}(c_{1}u_{1} + ⋯ + c_{k}u_{k})^{ℓ_{k+1}}u_{k+1}^{ℓ_{k+2}} ⋯ u_{n−1}^{ℓ_{n}}.

Proof. Suppose M = [u_{1} … u_{k} (c_{1}u_{1} + ⋯ + c_{k}u_{k}) u_{k+1} … u_{n−1}]. If M ∈ G_{k} then the sequence c_{1}, …, c_{k} is unique and M(β) appears exactly once on the right hand side. If M ∈ G_{i} with i < k then {u_{1}, …, u_{k}} is linearly dependant and the coordinates (c_{1}, …, c_{k}) satisfying c_{1}u_{1} + ⋯ + c_{k}u_{k} = m_{k+1} form a vector space with dimension at least one. Therefore M(β) appears q^{r} times (with r > 0) on the right hand side of the equation and makes no contribution to the sum.

Let J_{k}(ℓ) be the set of sequences I = (i_{1}, … i_{k}) of non-negative integers satisfying i_{1} + ⋯ + i_{k} = ℓ. Then

(c_{1}u_{1} + ⋯ + c_{k}u_{k})^{ℓ} = ∑_{I∈J_{k}(ℓ)} \binom{ℓ}{I}(c_{1}u_{1})^{i_{1}} ⋯ (c_{k}u_{k})^{i_{k}}.

Lemma 9.12. Suppose β = x_{1}^{ℓ_{1}} ⋯ x_{n}^{ℓ_{n}} and 0 < ℓ_{i} ≤ q^{n} − q^{i−1} − 1. Then if k < n − 1

∑_{M∈G_{k}} M(β) = 0

and

∑_{M∈G_{n−1}} M(β) = ∑_{I∈J_{n−1}(ℓ_{n})} \binom{ℓ_{n}}{I} ∏_{j=1}^{n−1} ( ∑_{c∈F_{q}} c^{i_{j}} ) ( ∑_{u∈V} u^{ℓ_{j}+i_{j}} )

Proof. Expanding the right hand side of the equation in Lemma 9.11 gives

∑_{M∈G_{k}} M(β) = ∑_{u_{1},... ,u_{n−1}∈V} ∑_{c_{1},... ,c_{k}∈F_{q}} ∑_{I∈J_{k}(ℓ_{k+1})} \binom{ℓ_{k+1}}{I}c_{1}^{i_{1}}u_{1}^{ℓ_{1}+i_{1}} ⋯ c_{k}^{i_{k}}u_{k}^{ℓ_{k}+i_{k}}u_{k+1}^{ℓ_{k+2}} ⋯ u_{n−1}^{ℓ_{n}}

= ( ∑_{I∈J_{k}(ℓ_{k+1})} \binom{ℓ_{k+1}}{I} ∏_{j=1}^{k} ( ∑_{c∈F_{q}} c^{i_{j}} ) ( ∑_{u∈V} u^{ℓ_{j}+i_{j}} )) ∏_{j=k+2}^{n} ( ∑_{u∈V} u^{ℓ_{j}} ) .

Since ℓ_{n} < q^{n} − 1, α_{q}(ℓ_{n}) < n(q − 1). Applying Proposition 9.5, ∑_{u∈V} u^{ℓ_{n}} = 0. Therefore, unless k = n − 1, ∑_{M∈G_{k}} M(β) = 0.

Theorem 9.13. Suppose β = x_{1}^{ℓ_{1}} ⋯ x_{n}^{ℓ_{n}} and 0 ≤ ℓ_{i} ≤ q^{n} − q^{i−1} − 1. Then

T r_{GL_{n}(F_{q})}(β) = \{ \begin{matrix}(−1)^{n}\binom{ℓ_{n}}{q^{n}−1−ℓ_{1},... ,q^{n}−1−ℓ_{n−1}}d_{n,n}^{n−1}, & if α_{q}(ℓ_{i}) = (n − 1)(q − 1) for i = 1 … n \\ & and deg(β) = (n − 1)(q^{n} − 1); \\ 0, & otherwise. \end{matrix}

Proof. Referring to Lemma 9.12 we see that it is sufficient to compute

∑_{M∈G_{n−1}} M(β) = ∑_{I∈J_{n−1}(ℓ_{n})} \binom{ℓ_{n}}{I} ∏_{j=1}^{n−1} ( ∑_{c∈F_{q}} c^{i_{j}} ) ( ∑_{u∈V} u^{ℓ_{j}+i_{j}} ) .

From Lemma 9.4, ∑_{c∈F_{q}} c^{i_{j}} is non-zero if and only if q−1 divides i_{j}. From Proposition 9.5, ∑_{u∈V} u^{ℓ_{j}+i_{j}} is non-zero only when q−1 divides ℓ_{j}+i_{j} and α(ℓ_{j}+i_{j}) ≥ n(q−1). Therefore we need only consider monomials for which q − 1 divides each ℓ_{j}. Recall that when q − 1 divides ℓ_{j}, q − 1 divides α_{q}(ℓ_{j}). Since ℓ_{j} ≤ q^{n} − q^{j−1} − 1, we have α_{q}(ℓ_{j}) < n(q − 1). Since α_{q}(ℓ_{j}) is divisible by q − 1, we have α_{q}(ℓ_{j}) ≤ (n − 1)(q − 1).

Furthermore, we require \binom{ℓ_{n}}{I} to be non-zero and, hence, using Remark 9.3, α_{q}(ℓ_{n}) = α_{q}(i_{1}) + ⋯ + α_{q}(ℓ_{i_{n−1}}). Therefore

n(n − 1)(q − 1) ≤ ∑_{j=1}^{n−1} α_{q}(ℓ_{j} + i_{j})

≤ ∑_{j=1}^{n−1} α_{q}(ℓ_{j}) + α_{q}(i_{j})
≤ (n − 1)(n − 1)(q − 1) + α_{q}(ℓ_{n})
≤ n(n − 1)(q − 1).

Thus each inequality must be an equality. In particular α_{q}(ℓ_{j}) = (n − 1)(q − 1) for all j = 1, …, n. We also have n(q − 1) = α_{q}(ℓ_{j} + i_{j}). Therefore ℓ_{j} + i_{j} ≥ q^{n} − 1. Using α_{q}(ℓ_{j} + i_{j}) = α_{q}(ℓ_{j}) + α_{q}(i_{j}) and ℓ_{j}, i_{j} ≤ q^{n} − 1 we obtain ℓ_{j} + i_{j} ≤ q^{n} − 1. Hence ℓ_{j} + i_{j} = q^{n} − 1 for j = 1, …, n − 1. Finally

deg(β) = ℓ_{1} + ⋯ ℓ_{n} = ∑_{j=1}^{n−1}(ℓ_{j} + i_{j}) = (n − 1)(q^{n} − 1).

Therefore T r_{GL_{n}(F_{q})}(β) = 0 unless deg(β) = (n − 1)(q^{n} − 1) and, for each i, α_{q}(ℓ_{i}) = (n − 1)(q − 1). When these conditions are satisfied

T r_{GL_{n}(F_{q})}(β) = (−1)\binom{ℓ_{n}}{q^{n} − 1 − ℓ_{1}, …, q^{n} − 1 − ℓ_{n−1}} ∏_{j=1}^{n−1} (( ∑_{c∈F_{q}} c^{i_{j}} ) ( ∑_{u∈V} u^{q^{n}−1}))

= (−1)^{n}\binom{ℓ_{n}}{q^{n} − 1 − ℓ_{1}, …, q^{n} − 1 − ℓ_{n−1}}d_{n,n}^{n−1}

as required.

Applying this theorem when β = ∏_{i=1}^{n} x_{i}^{q^{n}−1−(q−1)q^{i−1}} and using Lemma 9.1 to evaluate the multinomial coefficient gives T r_{GL_{n}(F_{q})}(β) = (−1)^{n}d_{n,n}^{n−1}.

Corollary 9.14. The image of T r_{GL_{n}(F_{q})} is the principal ideal generated by d_{n,n}^{n−1}.

Remark 9.15. In [13], Stewart Priddy and Clarence Wilkerson prove that d_{n,n}^{n−1} is in the image of T r_{GL_{n}(F_{q})}. It was also known that the radical of the image of T r_{GL_{n}(F_{q})} is generated by d_{n,n} — see, for example, Theorem 11.5.8 of [15].

Remark 9.16. We let H be a subgroup of a group G. Then T r_{G} factors as T r_{G/H} ∘ T r_{H} where T r_{G/H}(f) = ∑_{i=1}^{s} g_{i}(f) where {g_{1}, …, g_{s}} is a set of left coset representatives for H in G and f is invariant under H. Conversely we also have the factorization T r_{G} = T r_{H} ∘ T r_{G\H} where T r_{G\H}(f) = ∑_{i=1}^{s} g_{i}(f) where {g_{1}, …, g_{s}} is a fixed set of right coset representatives for H in G. Let H be any subgroup of GL_{n}(F_{q}). The first factorization shows that if T r_{GL_{n}(F_{q})}(f) ≠ 0 then T r_{H}(f) ≠ 0. The second factorization shows that d_{n,n}^{n−1} lies in the image of T r_{H}.

We turn our attention to the modular transfer for the usual permutation representation of Σ_{n}. First observe that for any permutation representation over a field of characteristic p, the transfer of a monomial is zero if and only if the order of the isotropy subgroup is divisible by p. For a monomial β, let k be its largest exponent and for i = 0, …, k, let b_{i} be the number of exponents equal to i. Note that b_{0} + ⋯ + b_{k} = n and the isotropy subgroup of β is isomorphic to Σ_{b_{0}} × ⋯ × Σ_{b_{k}}. If, for some i, b_{i} ≥ p, then the order of the isotropy subgroup of β is divisible by p and T r_{Σ_{n}}(β) = 0. Therefore it suffices to consider monomials in the block basis for which no p of the exponents are equal. We will use the block basis generated by x_{1}^{n−1}x_{2}^{n−2} ⋯ x_{n−1}^{1}. Since two monomials in the same orbit have the same transfer, we can restrict out attention to monomials with ℓ_{1} ≥ ℓ_{2} ≥ ⋯ ≥ ℓ_{n}. Note that

x_{1}^{ℓ_{1}} ⋯ x_{n}^{ℓ_{n}} = (x_{1} ⋯ x_{b_{k}})^{k}(x_{b_{k}+1} ⋯ x_{b_{k}+b_{k−1}})^{k−1} ⋯ (x_{b_{k}+···+b_{1}+1} ⋯ x_{n})^{0}.

Theorem 9.17. When p = 2 and n ≥ 2, then the image of T r_{Σ_{n}} is the principal ideal in P_{n}^{Σ_{n}} generated by the discriminant.

Proof. The only element of the block basis with distinct exponents is the top class, x_{1}^{n−1}x_{2}^{n−2} ⋯ x_{n−1}^{1}. It is well known that, in characteristic 2, the orbit sum of this monomial is equal to ∏_{i<j}(x_{i} − x_{j}) — the discriminant.

In the general case we have the following description of a generating set for the image of T r_{Σ_{n}}.

Theorem 9.18. The image of T r_{Σ_{n}} is generated by T r_{Σ_{n}}(β) where β = x_{1}^{ℓ_{1}} ⋯ x_{n}^{ℓ_{n}} is a basis monomial with ℓ_{1} ≥ ℓ_{2} ≥ ⋯ ≥ ℓ_{n} satisfying:

(i) b_{i} < p for i = 0, …, k, and (ii) b_{i} + b_{i+1} ≥ p for i = 0, …, k − 1.

Proof. If β fails (i), then, for some i, b_{i} > p and β has at least p equal exponents. Thus T r_{Σ_{n}}(β) = 0 and contributes nothing to the image of the transfer.

Consider (ii). Let I be the image of the transfer. Note that I is generated by the transfer of those basis monomials with ℓ_{1} ≥ ℓ_{2} ≥ ⋯ ≥ ℓ_{n}. Let J be the ideal in P_{n}^{Σ_{n}} generated by the transfer of the basis monomials which also satisfy (ii). We will show that I = J. By way of contradiction assume that β is the minimal basis monomial (with respect to the graded lexicographic ordering) such that ℓ_{1} ≥ ℓ_{2} ≥ ⋯ ≥ ℓ_{n} and T r_{Σ_{n}}(β) is not in J. The monomial β must fail (ii). Let j be the largest index, less than k, such that b_{j} + b_{j+1} < p. Let m = b_{k} + ⋯ + b_{j+1} and let β̄ = β/(x_{1} ⋯ x_{m}).

Let b̄_{i} be the number of exponents of β̄ equal to i. Thus

b̄_{i} = \{ \begin{matrix}b_{i+1}, & if j < i < k; \\ b_{j} + b_{j+1}, & if i = j; \\ b_{i}, & if i < j. \end{matrix}

Note that the integers b̄_{i} determine the exponents for β̄ and that β̄ is still a basis monomial with a non-increasing sequence of exponents. Furthermore β̄ is less than β and so, since β is a minimal counter-example, T r_{Σ_{n}}(β̄) ∈ J. Observe that the leading term of σ_{m}β̄ is β. Take each monomial in σ_{m}β̄ and permute the variables so that the exponents form a non-increasing sequence. Each of the resulting monomials is less than or equal to β. Furthermore β occurs \binom{b_{j+1}+b_{j}}{b_{j+1}} times. To see this, let τ be one of the summands of σ_{m}, i.e., τ is a square-free monomial of degree m, and consider the product τβ̄. Permute the variables in τβ̄ and call the result γ. The monomial γ is maximized only when x_{1} ⋯ x_{m−b_{j+1}} divides τ and the remaining variables dividing τ are chosen from {x_{m−b_{j+1}}, …, x_{m+b_{j}}}. In this case γ is in fact β. All other choices of τ give γ’s which, even after permuting the variables, are less than β.

Consider the set of monomials formed by taking the summands of σ_{m}β̄ and permuting the variables so that the sequence of exponents is non-increasing. Discard β from this set and call the resulting set Γ. It follows from the preceeding paragraph that, for some choice of c_{γ} ∈ F_{p},

T r_{Σ_{n}}(σ_{m}β̄) = \binom{b_{j+1} + b_{j}}{b_{j+1}}T r_{Σ_{n}}(β) + ∑_{γ∈Γ} c_{γ}T r_{Σ_{n}}(γ).

Since \binom{b_{j+1}+b_{j}}{b_{j+1}} < p, T r_{Σ_{n}}(β) is in the ideal generated by T r_{Σ_{n}}(β̄) and T r_{Σ_{n}}(γ) for γ ∈ Γ. Thus to show that T r_{Σ_{n}}(β) ∈ J, it is sufficent to show that, for each γ ∈ Γ, T r_{Σ_{n}}(γ) ∈ J. If γ is basis monomial then, since γ < β and β is a minimal counter-example, T r_{Σ_{n}}(γ) ∈ J. If γ is not a basis monomial, then for some i, x_{i}^{n−i+1} divides γ. Since γ has a non-increasing sequence of exponents, the critical monomial (x_{1} ⋯ x_{i})^{n−i+1} divides γ and therefore γ is in the ideal generated by (P_{n}^{Σ_{n}})^{+}. Thus when γ is written as a linear combination of basis monomials (with coefficients in P_{n}^{Σ_{n}}) all of the non-zero coefficients have positive degree. Therefore the basis monomials appearing are all less than β. Hence T r_{Σ_{n}}(γ) ∈ J. Thus T r_{Σ_{n}}(β) ∈ J, giving a contradiction.

A simple example is provided by the case n = p. Here the monomials satisfying the hypotheses of Theorem 9.18 are x_{1}, x_{1}x_{2}, …, x_{1} ⋯ x_{n−1}. By taking the transfer of these monomials we see that the image of T r_{Σ_{p}} is generated by σ_{1}, …, σ_{p−1}. In this example Theorem 9.18 provides us with a minimal set of generators. This is not true in general.

Remark 9.19. In the proof of the preceeding theorem we are essentially ordering the orbits of the monomials. This order was used by Göbel in [7]. We observe that this is just the lexicographic order on the sequences (b_{k}, …, b_{0}).

10. Modular Jacobians

Suppose A = k[x_{1}, …, x_{n}]^{G} = k[a_{1}, …, a_{n}] is a polynomial algebra. Then, as discussed above, the finite dimensional quotient algebra B = k[x_{1}, …, x_{n}]/(A^{+}) is a Poincaré duality algebra. In particular, B has dimension one in the top degree, ∑(deg(a_{i}) − 1). In [16], Steinberg proves that over a field of characteristic zero the Jacobian, J_{n}(a, x) = det [\frac{∂a_{i}}{∂x_{j}} ] , is a top class for B. For any characteristic the Jacobian is a relative invariant associated to the one dimensional representation det(g) (see [2, Section 5.4]). However, in positive characteristic J_{n}(a, x) may be an element of the ideal generated by A^{+}. In this section we compute J_{n}(d, x) = det [\frac{∂d_{i,n}}{∂x_{j}} ],

J_{n}(h, x) = det [\frac{∂h_{i}}{∂x_{j}} ], and J_{n}(d, h) = det [\frac{∂d_{i,n}}{∂h_{j}} ]. Observe that since U_{n}(F_{q}) is a subgroup of SL_{n}(F_{q}), J_{n}(h, x) is U_{n}(F_{q})-invariant. We shall see that although J_{n}(d, x) is not invariant (except for p = 2), J_{n}(d, x) is an element of the ideal (D_{n}^{+}).

We recall here a result due to Munkholm and Smith, see [14], that det[b_{ij}] gives a top class for the algebra P_{n}/(a_{1}, …, a_{n}) over a field of arbitrary characteristic for any choice of {b_{ij}} satisfying a_{i} = ∑_{j} b_{ij}x_{j}.

The chain rule gives us that J_{n}(d, x) = J_{n}(h, x)J_{n}(d, h). We shall compute J_{n}(h, x) and J_{n}(d, h) directly and then take the product to get J_{n}(d, x).

Theorem 10.1. J_{n}(h, x) = (−1)^{n(n−1)/2}(h_{1}^{n−1}h_{2}^{n−2} ⋯ h_{n−1})^{q−1}.

Proof. Since h_{i} ∈ P_{i}, [\frac{∂h_{i}}{∂x_{j}} ] is a lower triangular matrix. Therefore

J_{n}(h, x) = det [\frac{∂h_{i}}{∂x_{j}} ] = ∏_{i=1}^{n} \frac{∂h_{i}}{∂x_{i}} .

Furthermore, since

h_{i} = F_{i−1}(x_{i}) = ∑_{j=0}^{i−1} d_{j,i−1}x_{i}^{q^{(i−1)−j}},

\frac{∂h_{i}}{∂x_{i}} = d_{i−1,i−1}. From Proposition 1.2, d_{i−1,i−1} = (−1)^{i−1}(h_{1} ⋯ h_{i−1})^{q−1}. Therefore J_{n}(h, x) = (−1)^{n(n−1)/2}(h_{1}^{n−1}h_{2}^{n−2} ⋯ h_{n−1})^{q−1}.

Comparing Theorem 10.1 and Corollary 9.7, we see that the image of T r_{U_{n}(F_{q})} is the principal ideal in H_{n} generated by J_{n}(h, x).

We now start our computation of J_{n}(d, h). From Proposition 1.1,

d_{i,n} = d_{i,n−1}^{q} − d_{i−1,n−1}h_{n}^{q−1}.

28 H E A CAMPBELL, I P HUGHES, R J SHANK, AND D L WEHLAU

Therefore

\frac{∂d_{i,n}}{∂h_{k}} = −\frac{∂d_{i−1,n−1}}{∂h_{k}} h_{n}^{q−1} − δ_{k,n}(q − 1)d_{i−1,n−1}h_{n}^{q−2} (10.1)

where δ_{k,n} is one if k = n and zero otherwise. Lemma 10.2.

\frac{∂d_{i,n}}{∂h_{k}} = 0 if k ≤ n − i.

Proof. The proof is by induction on i. For i = 1 and k ≤ n − 1, using Equation 10.1 and the fact that d_{0,n−1} = 1, we get \frac{∂d_{1,n}}{∂h_{k}} = 0.

Suppose i > 1. If k ≤ n − i then, from Equation 10.1, \frac{∂d_{i,n}}{∂h_{k}} = −\frac{∂d_{i−1,n−1}}{∂h_{k}} h_{n}^{q−1}. However, if k ≤ n − i then k ≤ (n − 1) − (i − 1) and therefore we may apply the induction hypothesis to get \frac{∂d_{i−1,n−1}}{∂h_{k}} = 0.

As a consequence of Lemma 10.2, \frac{∂d_{i,n}}{∂h_{k}} is only non-zero in the lower right corner. This leads to the following result. Corollary 10.3.

J_{n}(d, h) = (−1)^{n(n−1)/2} ∏_{i=1}^{n} \frac{∂d_{i,n}}{∂h_{n−i+1}} .

Thus to compute J_{n}(d, h) we need only compute \frac{∂d_{i,n}}{∂h_{n−i+1}} . Lemma 10.4.

\frac{∂d_{i,n}}{∂h_{n+1−i}} = (−1)^{i+1}h_{n−i+1}^{q−2} (h_{n+2−i}h_{n+3−i} ⋯ h_{n})^{q−1} .

Proof. The proof is by induction on i. For i = 1, using Equation 10.1,

\frac{∂d_{1,n}}{∂h_{n}} = −(q − 1)h_{n}^{q−2} = h_{n}^{q−2}.

Suppose i > 1. From Equation 10.1 and Lemma 10.2 it follows that

\frac{∂d_{i,n}}{∂h_{n−i+1}} = −\frac{∂d_{i−1,n−1}}{∂h_{n−i+1}} h_{n}^{q−1}.

Using the induction hypothesis to describe \frac{∂d_{i−1,n−1}}{∂h_{n−i+1}} gives the required result. Theorem 10.5. J_{n}(d, h) = (h_{1}h_{2} ⋯ h_{n})^{q−2}(h_{2}h_{3}^{2} ⋯ h_{n}^{n−1})^{q−1}. Proof. The theorem follows from Lemma 10.4 and Corollary 10.3.

We now compute J_{n}(d, x) by taking the product of J_{n}(d, h) and J_{n}(h, x).

Corollary 10.6.

J_{n}(d, x) = (−1)^{n(n−1)/2}(h_{1} ⋯ h_{n})^{n(q−1)−1} = (−1)^{n(n−1)/2}(h_{1} ⋯ h_{n})^{q−2}d_{n,n}^{n−1}.

Observe that J_{n}(d, x) is not an element of D_{n} (except for p = 2) but it is an element of I_{n}. Therefore J_{n}(d, x) does not represent the top class for P_{n}/I_{n}.

Remark 10.7. Kemper has recently proved that if f_{1}, …, f_{n} ∈ P_{n}^{G} have non-zero Jacobian and the product of their degrees is equal to the order of G then P_{n}^{G} is the polynomial algebra generated by f_{1}, …, f_{n} (See [9, Proposition 16]). Combining this result with Theorem 10.1 gives another proof of Mùi’s result that P_{n}^{U_{n}(F_{q})} is the polynomial algebra generated by h_{1}, …, h_{n}. Similarly, Corollary 10.6 allows us to recover Dickson’s result that P_{n}^{GL_{n}(F_{q})} is the polynomial algebra generated by d_{1,n}, …, d_{n,n}.

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Department of Mathematics and Statistics, Queen’s University, Kingston, Ontario, Canada, K7L 3N6 E-mail address: eddy@mast.queensu.ca, hughesi@queensu.ca, shank@mast.queensu.ca, wehlau@mast.queensu.ca

Department of Mathematics and Computer Science, Royal Military College, Kingston, Ontario, Canada K7K 5L0 E-mail address: wehlau@rmc.ca