PACIFIC JOURNAL OF MATHEMATICS Vol. 195, No. 1, 2000

GROUP ACTIONS ON POLYNOMIAL AND POWER SERIES RINGS

Peter Symonds

When a finite group G acts faithfully on a graded integral domain S which is an algebra over a field k, such as a polynomial ring, we consider S as a kG-module. We show that S is asymptotically mostly projective in each degree, and also that it is in fact mostly free in an appropriate sense. Similar results also hold for filtered algebras, such as power series rings.

1. Introduction.

Let S = ⨁_{n=0}^{∞} S_{n} be a graded algebra over a field k. We suppose that S is finitely generated over k as a k-algebra and that the homogeneous components S_{n} are finite dimensional vector spaces over k. Let G be a finite group of grading preserving automorphisms of S (so G acts faithfully). We are concerned with the structure of S as a kG-module.

The classical theory of Hilbert and Serre asserts that for large n, dim_{k} S_{n} is given by a function

ϕ_{S}(n) = c_{d−1}(n)n^{d−1} + c_{d−2}(n)n^{d−2} + ⋯ + c_{0}(n),

where the c_{i}(n) are rational valued functions periodic in n, i.e., ϕ_{i}(n + p) = ϕ_{i}(n) for some integer p (see Section 2). If c_{d−1} is assumed not to be identically zero then d is equal to the dimension of the ring in various senses. If S is a polynomial ring then d is equal to the number of variables.

From now on, we assume that S is an integral domain. Let P_{n} denote the maximal projective summand of S_{n} (defined up to isomorphism).

Theorem 1.1. dim_{k}(S_{n}/P_{n}) is bounded by a polynomial in n of degree d−2.

Thus S_{n} is mostly projective, and if S is a polynomial ring then the non-projective part grows like a polynomial ring in one fewer variables.

In fact S is mostly free, although the individual S_{n} do not have to contain a free module at all; the different projectives can occur in different degrees. To explain this let R = S^{G}, the ring of invariants.

Theorem 1.2. S contains a free kG-submodule F of rank 1, a sum of homogeneous pieces, such that the product map R ⊗_{k} F → S is injective. Denote its image by RF = ⨁_{n}(RF)_{n}. Then RF is a free summand of S and dim_{k}(S_{n}/(RF)_{n}) is bounded by a polynomial of degree d − 2.

Of course, the first theorem is a corollary to the second. Versions of these theorems were proven by Howe [4] in characteristic 0 and by Bryant [2, 3] for polynomial rings.

Section 2 contains the main proof, except for some technical details which appear in Section 3. Section 4 proves similar results for filtered algebras.

2. Main Proof.

Proof. We can assume that k is a splitting field for G, since if a kG-module contains a free or projective summand after extension of scalars then it did so before. Let Q_{S} (resp. Q_{R}) denote the fields of fractions of S (resp. R), so Q_{S} ≅ Q_{R} ⊗_{R} S. By the Normal Basis Theorem, Q_{S} is a free module of rank 1 over Q_{R}G; let e be a generator. Then, over kG, e generates a free submodule E of rank 1 such that Q_{S} ≅ Q_{R} ⊗_{k} E. Now there is an r ∈ R such that re ∈ S. Let F be the kG-module generated by re, so F ⊂ S and F = rE ≅ kG. Also the product map R ⊗_{k} F → RF ⊂ S is injective.

We claim that F can be assumed to be a sum of homogeneous pieces, F = ⨁_{i} F_{n_{i}}. The proof of this plausible statement is somewhat delicate, and we postpone it to the next section.

Let x_{1}, …, x_{s} be homogeneous generators for S as a k-algebra. Then x_{i} = ∑_{j} \frac{a_{ij}}{b_{ij}} e_{j}, where a_{i,j}, b_{i,j} ∈ R and the e_{j} form a homogeneous k-basis for F. By writing b_{i,j}x_{i} = ∑_{j} a_{i,j}e_{j} and taking the homogeneous component of this equation in some degree where b_{i,j}x_{i} is non-zero, we see that we may assume that the b_{i,j} are homogeneous. Let α ∈ R_{a} be the product of all the b_{i,j}. Then each x_{i} ∈ α^{−1}RF, so S ⊂ α^{−1}RF.

Thus

(RF)_{n} ⊂ S_{n} ⊂ α^{−1}(RF)_{n+a},

and so, identifying RF with R ⊗_{k} F, we have

⨁_{i} R_{n−n_{i}} ⊗ F_{n_{i}} ⊂ S_{n} ⊂ α^{−1} ⨁_{i} R_{n+a−n_{i}} ⊗ F_{n_{i}}.

In particular, the dimension of S_{n}/(RF)_{n} is bounded by the difference in the dimensions of the two sides, i.e., by

∑_{i}(ϕ_{R}(n + a − n_{i}) − ϕ_{R}(n − n_{i})) dim_{k} F_{n_{i}}.

But

ϕ_{R}(n + a − n_{i}) − ϕ_{R}(n − n_{i}) = c_{d−1}(n + a − n_{i})(n + a − n_{i})^{d−1}

− c_{d−1}(n − n_{i})(n − n_{i})^{d−1} + lower degree terms,

and c_{d−1} is periodic, with period dividing a (see 3.1), so the n^{d−1} term cancels and we are done. □

3. Technical Details.

The form of ϕ_{S}(n) given above is not quite the standard one, although it is quoted in [4]. The usual references deal with a module over a polynomial ring which has all the variables in degree 1, and then all the coefficients of ϕ are constants. To deduce the version given in the introduction, note that if S is generated by x_{1}, …, x_{s} then it is a finitely generated module over k[x_{1}, …, x_{s}]. By taking suitable powers y_{i} of the x_{i} we can get all the y_{i} in the same degree b, and S will still be finitely generated over k[y_{1}, …, y_{s}]. For 0 ≤ j ≤ b − 1, set T_{j} = ⨁_{l=0}^{∞} S_{j+lb}. Then R ≅ ⨁_{j} T_{j} as a k[y_{1}, …, y_{s}]- module, and after regrading each T_{j} so that each y_{i} can have degree 1, we can apply the usual theory ([1] 11.2, [5] VII Theorem 41) to each T_{j} and sum the results. It is the summation that leads to the periodic coefficients.

Lemma 3.1 ([4]). If R is an integral domain (as it always is for us), c = gcd{r ∈ ℤ|R_{r} ≠ 0} and ϕ_{R}(n) = c_{d−1}(n)n^{d−1} + ⋯ + c_{0}(n), then there is a constant b such that

c_{d−1}(n) = \{ \begin{matrix}b, & if n|c, \\ 0, & otherwise. \end{matrix}

Proof. If 0 ≠ α ∈ R_{a}, then multiplication by α embeds R_{n} in R_{n+a}, so for large n, ϕ_{R}(n) ≤ ϕ_{R}(n + a). Now consider the limit of ϕ_{R}(n)/n^{d−1} as n → ∞ through elements of the same residue class modulo the period of c_{d−1} to see that c_{d−1}(n) ≤ c_{d−1}(n + a). This, together with the periodicity, implies the result. □

Now we prove the claim made in Section 2.

Proposition 3.2. The free module F ⊂ S can be assumed to be a sum of homogeneous pieces in such a way that the product map Q_{R} ⊗_{k} F → Q_{S} is still an isomorphism.

Proof. For each simple kG-module V, let T_{V} = Hom_{kG}(V, S), a graded Rmodule. Now soc F is a direct sum of simples. Let soc_{V} (F) denote the sum of those isomorphic to V, so soc_{V} (F) = V^{1}⊕⋯⊕V^{s}, where V^{i} ≅ V, and let P_{V^{i}} be a projective summand of F with soc(P_{V^{i}}) = V^{i}. The inclusions of the V^{i} in S give us s homomorphisms f^{i} ∈ T_{V}, which are linearly independent over R.

Lemma 3.3. Let f^{1}, …, f^{s} be elements of a graded R-module T which are linearly independent over R. Write each f^{j} as a sum of its homogeneous components; f^{j} = ∑_{k} f_{k}^{j}, f_{k}^{j} ∈ T_{k}. Then for each j there is an integer k_{j} such that f_{k_{1}}^{1}, …, f_{k_{s}}^{s} are linearly independent over R.

Proof. For each 0 ≤ t ≤ s, let P_{t} be the claim that there exist integers k_{1}, …, k_{t} such that f_{k_{1}}^{1}, …, f_{k_{t}}^{t}, f^{t+1}, …, f^{s} are linearly independent over R. P_{0} is true by hypothesis and we want P_{s}. We give a proof by induction on t, so assume P_{t}.

If P_{t+1} is false, then for each k ∈ ℤ we can find u_{k}, r_{k}^{i} ∈ R, u_{k} ≠ 0, such that

u_{k}f_{k}^{t+1} = r_{k}^{1}f_{k_{1}}^{1} + ⋯ + r_{k}^{t}f_{k_{t}}^{t} + r_{k}^{t+2}f^{t+2} + ⋯ + r_{k}^{s}f^{s}.

Let u be the product of the u_{k} for which f_{k}^{t+1} ≠ 0. Then uf^{t+1} = ∑_{k}(\frac{u}{u_{k}})u_{k}f_{k}^{t+1}, contradicting P_{t}. □

Applying this to the {f^{i}} ⊂ T_{V} we obtain homogeneous {f̄^{i}} ⊂ T_{V} , f̄^{i} ∈ T_{a_{i}}, say, linearly independent over R.

Lemma 3.4. The evaluation map ev : T_{V} ⊗_{k} V → S is injective.

Proof. In fact ev : Hom_{kG}(V, M) ⊗_{k} V → M is injective for any kG-module M. This is because it factors through soc_{V} (M), which is a direct sum of V ’s, so we are reduced to proving the case M = V. But then ev is an isomorphism, since Hom_{kG}(V, V) ≅ k, by the assumption that k is a splitting field. □

Corollary. The product map R ⊗_{k} (⨁_{i} f̄^{i}(V)) → S is injective.

Now let P̄_{V^{i}} be the image of the projection of P_{V^{i}} to S_{a_{i}}. The projection map is injective on soc(P_{V}^{i}), by the construction of a_{i}, so P̄_{V^{i}} ≅ P_{V^{i}} and soc(P̄_{V^{i}}) = f̄^{i}(V). Let P̄_{V} = ⨁_{i} P̄_{V^{i}} and consider the product map R ⊗_{k} P̄_{V} → S. Since soc(R ⊗_{k} P̄_{V}) = R ⊗ ⨁_{i} f̄^{i}(V), it is injective on the socle, so is injective.

Finally, we sum the P̄_{V} over the simples V to obtain F̄, a free kG-module of rank 1, which is a sum of homogeneous pieces, as required. □

Remark. If G is a p-group, where p is the characteristic of k, then the proof is much simpler because soc(F) ≅ k. Under at least one of the projections of F onto its homogeneous components the image of soc(F) must be nonzero. Let F̄ be the image of F under this projection. Then F̄ ≅ kG and Q_{R} ⊗_{k} F → Q_{S} is an isomorphism because it is injective on the socle, and both sides have the same dimension over Q_{R}.

This is enough to prove 1.1 for general G. For if P = Syl_{p}(G) then S is a direct summand of Ind_{P}^{G} Res_{P}^{G} S.

Remark. It is not hard to see that, given any degree m, the summands of F̄ can be moved by multiplication by a scalar to lie in T_{m+lc} = S_{m+lc} ⊕⋯⊕ S_{m+(l+1)c−1}, for some l. The argument of the proof of 1.2 now shows that the non-free part of T_{n} has dimension bounded by a polynomial of degree d − 2 (cf. [2, 3]).

4. Filtered Rings.

The case of filtered rings is slightly different. Consider the power series ring k[[x]] in characteristic 2 and let the group of order 2 act by x ↦ x/(x + 1) = x + x^{2} + x^{3} + · · ·. The action on the associated graded ring is trivial, yet the action on k[[x]] certainly contains free summands (the only alternative is trivial).

We consider finitely generated k-algebras S which are integral domains and have a filtration S = I_{0} ⊂ I_{1} ⊂ I_{2} ⊂ · · · . Each S/I_{n} is assumed to be finite dimensional over k, and ∩I_{n} = {0}. There is a finite group G of automorphisms of S, which preserves the filtration. The invariants are R = S^{G} with the induced filtration J_{n} = R ∩ I_{n}. Again there is a function

χ_{S}(n) = c_{d}(n)n^{d} + c_{d−1}n^{d−1} + ⋯ c_{0}(n),

where the c_{i} are periodic, such that dim_{k}(S/I_{n}) = χ_{S}(n) for large n. If S is a power series ring, then d is equal to the number of variables.

As before there is a free kG-module of rank 1 in S, and the product map R ⊗_{k} F → S is injective. Since F is finite dimensional there is some integer f such that F ∩ I_{f} = 0, so F injects into S/I_{f}.

For each n, let K_{n} be a vector space complement to J_{n} in R. Then the product map K_{n} ⊗ F → S/I_{f+n} is injective, so its image, K_{n}F is a free summand of S/I_{f+n}.

Proceeding in the same way as before we can prove:

Theorem 4.1. dim_{k}((S/I_{f+n})/K_{n}F) is bounded by a polynomial of degree d − 1.

So S is mostly free. Again, for a power series ring, the non-free part grows like a power series ring in one fewer variables.

References

[1] M.F. Atiyah and I.G. Macdonald, Introduction to Commutative Algebra, Addison-Wesley, Reading, MA, 1969.

[2] R.M. Bryant, Symmetric powers of representation of finite groups, Jour. Algebra, 154 (1993), 416-436.

[3] , Groups acting on polynomial algebras, Finite and locally finite groups. NATO Advanced Science Institutes Series C: Mathematical and Physical Sciences, 471, 327-346, Kluwer Acad. Publ. Dordrecht, 1995.

[4] R. Howe, Asymptotics of dimensions of invariants for finite groups, Jour. Algebra, 122 (1989), 374-379.

[5] O. Zariski and P. Samuel, Commutative Algebra, Vol. II, Van Nostrand, Princeton, 1960.

Received October 13, 1998.

U.M.I.S.T. Manchester M60 1QD, England E-mail address: psymonds@umist.ac.uk

This paper is available via http://nyjm.albany.edu:8000/PacJ/2000/195-1-10.html.