Invariant Theory of Finite Groups

RWTH Aachen, SS 2004

Friedrich-Schiller-Universitat Jena, WS 2004

Jurgen Muller

Abstract

This lecture is concerned with polynomial invariants of finite groups which comefrom a linear group action. We introduce the basic notions of commutativealgebra needed and discuss structural properties of invariant rings.

Keywords are: polynomial rings, graded commutative algebras, finite generationproperty, integrality, Hilbert series, prime ideals, dimension theory, Noethernormalization, Cohen-Macaulay property, primary and secondary invariants,symmetric groups, pseudoreflection groups, ...

Contents

1 Commutative algebras and invariant rings . . . . . . . . . . . . . . . . 12 Noether’s degree bound . . . . . . . . . . . . . . . . . . . . . . . . . . 63 Molien’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 Polynomial invariant rings . . . . . . . . . . . . . . . . . . . . . . . . . 125 Permutation groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 Prime ideals in commutative rings . . . . . . . . . . . . . . . . . . . . 227 Noether normalisation . . . . . . . . . . . . . . . . . . . . . . . . . . . 288 Cohen-Macaulay algebras . . . . . . . . . . . . . . . . . . . . . . . . . 339 Invariant theory live: the icosahedral group . . . . . . . . . . . . . . . 3810 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4311 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

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1 Commutative algebras and invariant rings

We introduce our basic objects of study, commutative algebras having a grading,and invariant rings as a special case.

(1.1) Definition. a) Let F be a field, let X := X1, . . . , Xn, for n ∈ N0, bea finite set of algebraically independent indeterminates over F , and let F [X ] :=F [X1, . . . , Xn] be the corresponding polynomial ring. Hence F [X ] is the freecommutative F -algebra with free generators X , i. e. each map X → R, whereR is a commutative F -algebra, can be extended uniquely to a homomorphismF [X ]→ R of F -algebras.

b) Let V1, V2 be F -vector spaces. An F -bilinear map ⊗ : V1 × V2 → V , whereV is an F -vector space, is called a tensor product of V1 and V2, if for eachF -bilinear map α : V1 × V2 → W , where W is an F -vector space, there is aunique F -linear map β : V → W such that α = ⊗ · β. By Exercise (10.1), ifa tensor product exists, then V =: V1 ⊗F V2 is unique up to isomorphism ofF -vector spaces. Moreover, if dimF (Vi) ∈ N0 then V1 ⊗F V2 exists and we havedimF (V1 ⊗F V2) = dimF (V1) · dimF (V2).

c) Let V be an F -vector space such that n = dimF (V ) ∈ N0. For d ∈ N the d-thsymmetric power S[V ]d := V ⊗d/V ′d of V is defined as the quotient F -space ofthe d-th tensor power space V ⊗d := V ⊗F V ⊗F · · ·⊗F V , with d tensor factors,with respect to the F -subspace

V ′d := 〈v1 ⊗ · · · ⊗ vd − v1π−1 ⊗ · · · ⊗ vdπ−1 ; vi ∈ V, π ∈ Sd〉F ≤ V ⊗d.

The symmetric algebra S[V ] over V is defined as S[V ] :=⊕

d≥0 S[V ]d, wherewe let S[V ]0 := 1 · F ∼= F . It becomes a finitely generated commutative F -algebra, where multiplication is inherited from concatenation of tensor prod-ucts and where each F -basis b1, . . . , bn ⊆ V = S[V ]1 of V is an F -algebragenerating set of S[V ].

(1.2) Definition. a) A commutative F -algebra R is called graded, if we haveR =

⊕d≥0Rd as F -vector spaces, such that R0 = 1 ·F ∼= F and dimF (Rd) ∈ N0

as well as RdRd′ ⊆ Rd+d′ , for d, d′ ≥ 0. Note that R is a direct sum, i. e. for0 6= r = [rd; d ≥ 0] ∈ R there is k ∈ N0 minimal such that rd = 0 for alld > k; the elements rd ∈ Rd are called the homogeneous components anddeg(r) = k ∈ N0 is called the degree of r. For convenience we let Rd := 0for d < 0. The F -subspace Rd ≤ R, for d ∈ Z, is called the d-th homogeneouscomponent of R. Let R+ :=

⊕d>0Rd C R be the irrelevant ideal, i. e. the

unique maximal homogeneous ideal of R.

b) Let R be a graded F -algebra. An R-module M is called graded, if we haveM =

⊕d≥dM Md as F -vector spaces, for some dM ∈ Z, such that dimF (Md) ∈

N0 as well as MdRd′ ⊆ Md+d′ , for d ≥ dM and d′ ≥ 0. For 0 6= m = [md; d ≥0] ∈ M there is k ≥ dM minimal such that md = 0 for all d > k; the elements

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md ∈ Rd are called the homogeneous components and deg(m) = k ∈ Z iscalled the degree of m.

For convenience we let Md := 0 for d < dM . The F -subspace Md ≤ M , ford ∈ Z, is called the d-th homogeneous component of M . An R-submoduleM ′ ≤M is called homogeneous, if M ′ =

⊕d≥dM M ′d, where M ′d := M ′ ∩Md.

Note that in this case M ′ is a graded R-module as well, the grading beinginherited from M . Moreover M ′ is homogeneous if and only if M ′ is as anR-module generated by homogeneous elements.

Let M and M ′ be graded R-modules. Then we conclude HomR(M,M ′) =∏d∈Z

∏d′∈ZHomR(Md,M

′d′), and HomR(M,M ′)c :=

∏d∈ZHomR(Md,M

′d+c)

are called its homogeneous components, for c ∈ Z. Hence HomR(M,M ′)0

is the set of homomorphisms of graded R-modules from M to M ′.

c) The Hilbert series (Poincare series) HR ∈ Z[[T ]] ⊆ Q((T )) of a gradedF -algebra R is the formal power series defined by HR(T ) :=

∑d≥0 dimF (Rd)T d.

The Hilbert series (Poincare series) HM ∈ Q((T )) of a graded R-moduleM is the formal Laurent series defined by HM (T ) :=

∑d≥dM dimF (Md)T d.

(1.3) Example. a) Let X := X1, . . . , Xn be algebraically independent in-determinates, and let δ := [d1, . . . , dn] ⊆ N be a degree vector. Then F [X ]becomes a graded F -algebra by letting degδ(Xi) := di. The standard gradingdeg = degX of F [X ] is given by the degree vector [1, . . . , 1].

By Exercise (10.2) the Hilbert series of F [X ] with respect to the grading degδis given as Hδ

F [X ] =∏ni=1

11−Tdi ∈ Q(T ) ⊆ Q((T )).

b) The symmetric algebra S[V ] :=⊕

d≥0 S[V ]d is graded, the d-th homogeneouscomponent being S[V ]d. In particular, for 0 6= v ∈ V we have deg(v) = 1.

(1.4) Proposition. Let dimF (V ) = n ∈ N0. Then we have S[V ] ∼= F [X ] =F [X1, . . . , Xn] as graded F -algebras, for the standard grading of F [X ].

Proof. Let b1, . . . , bn ⊆ V be an F -basis of V . As F [X ] is the free com-mutative F -algebra on X , there is an F -algebra homomorphism α : F [X ] →S[V ] : Xi 7→ bi, for i ∈ 1, . . . , n. Conversely, for d ∈ N there is an F -linearmap βd : V ⊗d → F [X ] : bi1 ⊗ bi2 ⊗ · · · ⊗ bid 7→

∏dk=1Xik , for ik ∈ 1, . . . , n.

Since F [X ] is commutative, we have V ′d ≤ ker(βd), and hence there is an F -linearmap β :=

∑d≥0 βd : S[V ] → F [X ]. Moreover, β is a ring homomorphism, and

we have αβ = idF [X ] and βα = idS[V ]. Finally we have deg(bi) = 1 = deg(Xi)for i ∈ 1, . . . , n, hence deg(fα) = deg(f) for f ∈ F [X ]. ]

(1.5) Definition. Let G be a group and let DV : G→ GL(V ) ∼= GLn(F ) be anF -representation of G. Hence the F -vector space V becomes an FG-module,where FG denotes the group algebra of G over the field F . The representationDV is called faithful if ker(DV ) = 1.

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By diagonal G-action, V ⊗d becomes an FG-module, for d ∈ N. As the Sd-actionand the G-action on V ⊗d commute, we conclude that V ′d ≤ V ⊗d is an FG-submodule, and hence the quotient F -space S[V ]d = V ⊗d/V ′d is an FG-moduleas well. Letting G act trivially on S[V ]0 ∼= F , we have S[V ] =

⊕d≥0 S[V ]d as

FG-modules.

As the group G acts by F -algebra automorphisms on S[V ], the set

S[V ]G := f ∈ S[V ]; fπ = f for all π ∈ G ⊆ S[V ]

of G-invariants forms an F -subalgebra of S[V ], called the corresponding in-variant ring. Moreover, we have S[V ]G =

⊕d≥0 S[V ]Gd , where S[V ]Gd =

S[V ]G ∩S[V ]d, and in particular S[V ]G0 = S[V ]0 ∼= F . Hence S[V ]G is a gradedF -algebra, the grading being inherited from S[V ].

Moreover, the group G acts by field automorphisms on the field of fractionsS(V ) := Quot(S[V ]) of S[V ]. Hence the corresponding invariant field is givenas

S(V )G = Quot(S[V ])G := f ∈ S(V ); fπ = f for all π ∈ G ⊆ S(V ).

In particular, we have S[V ]G = S(V )G ∩ S[V ].

(1.6) Definition. Let R be a commutative ring. An R-module M is calledNoetherian if each chain M0 ≤ M1 ≤ · · ·Mi ≤ · · · ≤ M of R-submodulesstabilises, i. e. there is k ∈ N0 such that Mi = Mk for all i ≥ k. The ring R iscalled Noetherian, if the regular R-module RR is Noetherian.

Some basic properties of Noetherian modules are collected in Exercise (10.3):Let N ≤ M be R-modules. If M is Noetherian, then so are N and M/N andif conversely both N and M/N are Noetherian, then so is M . Moreover, M isNoetherian if and only if each submodule of M is finitely generated. Hence ifR is Noetherian, then M is Noetherian if and only if M is finitely generated.

(1.7) Theorem: Hilbert’s Basis Theorem, 1890.Let R be a Noetherian commutative ring. Then the polynomial ring R[X] inthe indeterminate X is Noetherian as well.

Proof. Let I ER[X], and for d ∈ N0 let Jd := lc(f); 0 6= f ∈ I,deg(f) = d.∪

0, where by lc(f) ∈ R we denote the leading coefficient of 0 6= f ∈ R[X].Hence we have JdER and Jd ⊆ Jd+1. As R is Noetherian, let k ∈ N0 such thatJk = Jd for d ≥ k, and let Jd = 〈rd,1, . . . , rd,nd〉R E R for d ∈ 0, . . . , k andsome nd ∈ N. Let fd,i ∈ I such that deg(fd,i) = d and lc(fd,i) = rd,i ∈ R.

We show that I = 〈fd,i; d ∈ 0, . . . , k, i ∈ 1, . . . , nd〉R[X] E R[X]. Let 0 6=f ∈ I such that deg(f) = d, and use induction on d ∈ N0. If d > k thenJd = 〈rk,i; i ∈ 1, . . . , nk〉R E R, hence for some ci ∈ R we have f ′ := f −∑nki=1 ciX

d−kfk,i ∈ I such that deg(f ′) < d. If d ≤ k then for some ci ∈ R wehave f ′ := f −

∑ndi=1 cifd,i ∈ I such that deg(f ′) < d. By induction we have

f ′ ∈ 〈fd,i; d ∈ 0, . . . , k, i ∈ 1, . . . , nd〉R[X], hence the assertion follows. ]

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(1.8) Corollary. A finitely generated commutative F -algebra is Noetherian.

Proof. A finitely generated commutative F -algebra is the epimorphic image ofa polynomial ring F [X1, . . . , Xn], for some n ∈ N0. ]

(1.9) Definition. Let R ⊆ S be an extension of commutative rings, i. e. thering S is an R-algebra, and in particular we have 1R = 1S .

An element s ∈ S is called integral over R, if there is 0 6= f ∈ R[X] monic, suchthat f(s) = 0. By Exercise (10.4) an element s ∈ S is integral over R, if andonly if there is an R-subalgebra of S containing s, which is finitely generated asan R-module. The ring extension R ⊆ S is called integral, if each element ofS is integral over R.

The ring extension R ⊆ S is called finite, if S is a finitely generated R-algebraand integral over R. By Exercise (10.4) the ring extension R ⊆ S is finite, ifand only if S is a finitely generated R-module.

By Exercise (10.4) the subset RS

:= s ∈ S; s integral over R ⊆ S is a subringof S, called the integral closure (normalisation) of R in S. If R

S= R holds,

then R is called integrally closed (normal) in S. If R is a domain and R isintegrally closed in its field of fractions Quot(R), then R is called integrallyclosed (normal).

(1.10) Proposition. Let G be a finite group. Then for the invariant fieldS(V )G = (Quot(S[V ]))G we have S(V )G = Quot(S[V ]G), and the field exten-sion S(V )G ⊆ S(V ) is finite Galois with Galois group G/ ker(DV ).

Proof. We clearly have Quot(S[V ]G) ⊆ S(V )G. Conversely let f = gh ∈

S(V )G, for g, h ∈ S[V ]. By extending with∏

1 6=π∈G hπ ∈ S[V ] we may as-sume that h ∈ S[V ]G, and hence g ∈ S[V ]G as well, thus f ∈ Quot(S[V ]G).By Artin’s Theorem, see [10, Thm.6.1.8], the field extension S(V )G ⊆ S(V ) isGalois. ]

(1.11) Theorem: Hilbert, 1890; Noether, 1916, 1926.Let G be a finite group. Then the invariant ring S[V ]G is a finitely generated,integrally closed F -algebra, and the ring extension S[V ]G ⊆ S[V ] is finite.

Proof. Let S := S[V ]. If f ∈ Quot(SG) = Quot(S)G ⊆ Quot(S) is integralover SG, it is also integral over S. As S is a unique factorisation domain, it isintegrally closed in Quot(S), see Exercise (10.4). Hence we have f ∈ S, thusf ∈ Quot(S)G ∩ S = SG.

For s ∈ S let fs :=∏π∈G(X − sπ) ∈ SG[X]. Hence we have fs(s) = 0 and

fs is monic. Let s1, . . . , sk ⊆ S be an F -algebra generating set of S, and letR ⊆ SG ⊆ S be the F -algebra generated by the coefficients of the polynomials

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fs1 , . . . , fsk ⊆ SG[X]. As all si are integral over R, the F -algebra S is integralover R. As S is a finitely generated F -algebra, it is also a finitely generatedR-algebra, and hence the ring extension R ⊆ S is finite. In particular, the ringextension SG ⊆ S is finite as well.

Moreover, as R is a finitely generated F -algebra, it is Noetherian. As the R-module S is finitely generated, it hence is a Noetherian R-module. Thus theR-submodule SG ≤ S also is Noetherian, hence SG is a finitely generated R-module. As R is a finitely generated F -algebra, SG is a finitely generatedF -algebra as well. ]

(1.12) Remark. Note that there are F -subalgebras of S[V ] ∼= F [X ] which arenot finitely generated. For example the F -subalgebra of F [X,Y ] generated byXY i; i ∈ N is not finitely generated, see Exercise (10.5).

The statement on finite generation in (1.11) does not hold for arbitrary groups.There is a famous counterexample by Nagata (1959), see [4, Ex.2.1.4]. Butit holds for so-called linearly reductive groups, see (2.6). Actually, Hilbertworked on linearly reductive groups, although this notion has only been coinedlater, while Noether developed the machinery for finite groups.

The proof of (1.11) is purely non-constructive. At Hilbert’s times this led to thefamous exclamation of Gordan, then the leading expert on invariant theory anda dogmatic defender of the view that mathematics must be constructive: Dasist Theologie und nicht Mathematik! (This is theology and not mathematics!)Actually, Hilbert’s ground breaking work now is recognised as the beginningand the foundation of modern abstract commutative algebra.

Finally, the statement on finite generation in (1.11) is related to Hilbert’s 14thproblem: If K ⊆ S(V ) is a subfield, is K ∩ S[V ] a finitely generated algebra?As we have S(V )G ∩ S[V ] = S[V ]G for any group G, Nagata’s counterexamplegives a negative answer to this problem as well.

(1.13) Example. Let k ∈ N such that char(F ) 6 | k, and let ζk ∈ F be aprimitive k-th root of unity. Let G := 〈δ〉 ∼= Ck be the cyclic group of order k,and let DV : G→ GL1(F ) : δ 7→ ζk. We identify S[V ] ∼= F [X]. As δ : X 7→ ζkXwe have F [X]G = F [Xk]. Hence S[V ]G ∼= F [X]G is polynomial having degreevector [k].

We identify S[V ⊕V ] ∼= F [X,Y ]; the ring S[V ⊕V ]G ∼= F [X,Y ]G is called a ringof vector invariants. As δ : X 7→ ζkX,Y 7→ ζkY , for d ∈ N such that k 6 | dwe have F [X,Y ]Gd = 0, while if k | d we have F [X,Y ]Gd = F [X,Y ]d. Hencewe have F [X,Y ]G = F 〈Xk, Xk−1Y, . . . ,XY k−1, Y k〉, where F 〈· · · 〉 denotes F -algebra generation.

As dimF (F [X,Y ]d) = d + 1, the Hilbert series of F [X,Y ]G is HF [X,Y ]G =∑i≥0(ik + 1)T ik = ∂

∂T (T ·∑i≥0 T

ik) = ∂∂T ( T

1−Tk ) = 1+(k−1)Tk

(1−Tk)2 ∈ Q((T )).Indeed, S := F [Xk, Y k] ⊆ F [X,Y ]G is a polynomial F -subalgebra having degree

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vector [k, k], and F [X,Y ]G = (1 · S) +∑k−1i=1 (Xk−iY i · S) as S-modules.

Let R := (1 · S) ⊕⊕k−1

i=1 (Xk−iY i · S) be the free S-module generated by1, Xk−1Y, . . . ,XY k−1. As we have H1·S = 1

(1−Tk)2 and HXk−iY i·S = Tk

(1−Tk)2 ,we conclude HR = HF [X,Y ]G . As F [X,Y ]G is an epimorphic image of R as agraded F -algebra, we conclude that F [X,Y ]G = (1 · S)

⊕(⊕k−1

i=1 (Xk−iY i · S)).

2 Noether’s degree bound

We address the question whether it is possible to give a bound on the degreesof the elements of a homogeneous algebra generating set of an invariant ring,the main result being the Noether-Fleischmann Theorem.

(2.1) Definition. Let H ≤ G such that [G : H] < ∞, and let T ⊆ G be aright transversal of H in G, i. e. a set of representatives of the right cosetsH|G of H in G. The relative transfer map TrGH is defined as as the F -linearmap

TrGH : S[V ]H → S[V ]G : f 7→∑π∈T

fπ.

If |G| < ∞, then the F -linear map TrG := TrG1 : S[V ] → S[V ]G is called thetransfer map.

It is clear that TrGH is well-defined and independent of the choice of the transver-sal T . Moreover, we have TrGH |S[V ]Hd

: S[V ]Hd → S[V ]Gd , for d ∈ N0, andfor f ∈ S[V ]G ⊆ S[V ]H and g ∈ S[V ]H we have TrGH(gf) = TrGH(g) · f .Thus TrGH is a degree-preserving homomorphism of S[V ]G-modules. As wehave TrGH(1) = [G : H] · 1, we conclude TrGH |S[V ]G : f 7→ [G : H] · f . We haveim (TrGH)ES[V ]G, where possibly im (TrGH) 6= S[V ]G, see Exercise (10.20). Notethat TrGH extends to TrGH : S(V )H → S(V )G.

(2.2) Proposition. Let G be a finite group acting faithfully on V , and letH ≤ G. Then we have im (TrGH) 6= 0.

Proof. Since TrGH TrH = TrG, it is sufficient to prove that im (TrG) 6= 0.The elements of G = π1, π2, . . . induce pairwise distinct field automorphismsof S(V ), which by Dedekind’s Independence Theorem, see [10, Thm.8.11.19],are algebraically independent over S(V ), i. e. since S(V ) is an infinite field,whenever f ∈ S(V )[Y1, . . . , Y|G|] such that f(gπ1, gπ2, . . .) = 0 for all g ∈ S(V ),then f = 0. In particular, these automorphisms are S(V )-linearly independent,hence we have

∑π∈G π 6= 0 ∈ End(S[V ]) = EndF (V ), where End(S[V ]) denotes

the degree-preserving F -algebra endomorphisms of S[V ]. ]

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(2.3) Definition. Let char(F ) 6 | [G : H]. The relative Reynolds operatorRGH is the projection of S[V ]G-modules defined by

RGH :=1

[G : H]· TrGH : S[V ]H → S[V ]G.

If char(F ) 6 | |G| < ∞, using the Reynolds operator RG := RG1 : S[V ] →S[V ]G we in particular obtain S[V ] = S[V ]G

⊕ker(RG) as S[V ]G-modules.

For G finite, the case char(F ) 6 | |G| is called the non-modular case, otherwiseit is called the modular case.

(2.4) Definition. The ideal of S[V ] generated by the homogeneous invariantsof positive degree IG[V ] := S[V ]G+ · S[V ] = (S[V ]+ ∩ S[V ]G) · S[V ] C S[V ] iscalled the Hilbert ideal of S[V ] with respect to G.

Hence IG[V ]C S[V ] is a homogeneous ideal, and as S[V ] is Noetherian, IG[V ]is finitely generated by homogeneous invariants.

(2.5) Theorem. Let G be a finite group such that char(F ) 6 | |G|. Let IG[V ] =∑ri=1 fiS[V ] C S[V ], where the fi ∈ S[V ]G are homogeneous. Then we have

S[V ]G = F 〈f1, . . . , fr〉.

Proof. Let h ∈ S[V ]G be homogeneous such that deg(h) = d. We proceed byinduction on d. The case d = 0 is clear, hence let d > 0. Let h =

∑ri=1 figi, for

gi ∈ S[V ]d−deg(fi). We have h = RG(h) =∑ri=1 fi ·RG(gi). As RG(gi) ∈ S[V ]G

and deg(RG(gi)) = d− deg(fi) < d, we are done by induction. ]

(2.6) Remark. Note that in (2.5) only the property of RG : S[V ] → S[V ]G

being a degree preserving S[V ]G-module projection is used. More generally, thegroups possessing a generalised Reynolds operator sharing these propertiesare called linearly reductive. Hence for these groups the assertion of (2.5)remains valid.

(2.7) Proposition: Benson’s Lemma, 2000.Let G be a finite group such that char(F ) 6 | |G|, and let I E S[V ] be an FG-invariant ideal. Then we have I |G| ⊆ (I ∩ S[V ]G) · S[V ]E S[V ].

Proof. Let S := S[V ] and fπ;π ∈ G ⊆ I. Then we have∏π∈G(fππσ−fπ) =

0, for σ ∈ G. Expanding the product and summing over σ ∈ G yields

∑M⊆G

(−1)|G\M | ·

(∑σ∈G

∏π∈M

fππσ

)·

∏π∈G\M

fπ

= 0,

where the sum runs over all subsets M ⊆ G. If M 6= ∅, then we have(∑σ∈G

∏π∈M fππσ) ∈ I ∩ SG, and thus the corresponding summand of the

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above sum is an element of (I ∩ SG) · S. Hence for M = ∅ we from this obtain|G| ·

∏π∈G fπ ∈ (I ∩ SG) · S, hence

∏π∈G fπ ∈ (I ∩ SG) · S. ]

(2.8) Theorem: Noether, 1916; Fleischmann, 2000.Let G be a finite group such that char(F ) 6 | |G|. Then there is a homogeneousgenerating set of IG[V ]C S[V ] consisting of invariants of degree ≤ |G|.

Proof. Let S := S[V ] ∼= F [X ]. Let Xα ∈ S, for α = [α1, . . . , αn] ⊆ N0,be a monomial such that deg(Xα) =

∑ni=1 αi ≥ |G|. By (2.7), applied to

the irrelevant ideal S+ C S, we have Xα ∈ (S+ ∩ SG) · S = IG[V ] C S. Ifdeg(Xα) > |G|, let Xα = Xα′ · Xα′′ such that deg(Xα′) = |G|. Hence wealready have Xα′ ∈ IG[V ].

Let IG[V ] =∑ri=1 fiSCS, where f1, . . . , fr ⊆ SG is a minimal homogeneous

generating set and where deg(fr) > |G|, say. By the above we conclude thatfr ∈

∑j,deg(fj)≤|G| fjS C S, a contradiction. ]

(2.9) Corollary. Let G be a finite group such that char(F ) 6 | |G|. Then thereis a homogeneous F -algebra generating set of S[V ]G consisting of elements ofdegree ≤ |G|.

(2.10) Remark. In the non-modular case, an F -algebra generating set ofS[V ]G is contained in

⊕|G|d=1 S[V ]Gd = RG(

⊕|G|d=1 S[V ]d). Hence evaluating the

Reynolds operator at monomials Xα, where deg(Xα) ≤ |G|, yields an F -algebragenerating set of S[V ]G. For an example see Exercise (10.8).

Actually, Noether proved the degree bound only for the case char(F ) = 0, butthe proof is valid for the case |G|! 6= 0 ∈ F as well. Noether’s degree boundis best possible in the sense that no improvement is possible in terms of thegroup order alone, see (1.13). For more involved improvements of Noether’sdegree bound see [4, Ch.3.8]. Actually, in most practical non-modular casesboth Noether’s degree bound and its improvements are far from being sharp. Inthe modular case, Noether’s degree bound in general does not hold, and neitherdoes Benson’s Lemma, see Exercise (10.9). For some known but unrealisticdegree bounds, see [4, Ch.3.9].

3 Molien’s Formula

The show how to use character theory of finite groups to determine the Hilbertseries of invariant rings. Hence some familiarity with the ordinary and modularrepresentation theory of finite groups is needed.

(3.1) Theorem: Molien’s formula, 1897.Let F ⊆ C, let G be a finite group and let V be an FG-module.

9

a) Then for π ∈ G we have∑d≥0

TrS[V ]d(π) · T d =1

detV (1− Tπ)∈ F ((T )),

where TrS[V ]d(π) ∈ F denotes the usual matrix trace, and where 1−T ·DV (π) ∈F [T ]n×n and detV (1− Tπ) := detF [T ]n×n(1− T ·DV (π)) ∈ F [T ].b) We have

HS[V ]G =1|G|·∑π∈G

1detV (1− Tπ)

∈ F ((T )).

In particular, HS[V ]G ∈ Q(T ) = F (T ) ∩Q((T )) is a rational function.

Proof. a) We may assume Q[ζ|G|] ⊆ F , where ζ|G| := exp 2π√−1

|G| ∈ C is aprimitive |G|-th root of unity. Hence DS[V ]d(π) is diagonalisable, for d ≥ 0.In particular, let λ1, . . . , λn ∈ F be the eigenvalues of DV (π). Hence we havedetV (1− Tπ) =

∏ni=1(1− λiT ) ∈ F [T ].

The eigenvalues of DS[V ]d(π) are given as the products∏ni=1 λ

αii ∈ F , where

α = [α1, . . . , αn] ∈ Nn0 such that∑ni=1 αi = d. Thus

∑d≥0 TrS[V ]d(π) · T d =∑

α∈Nn0

∏ni=1(λiT )αi =

∏ni=1

∑j≥0(λiT )j =

∏ni=1

11−λiT ∈ F ((T )).

b) The Reynolds operator RG projects S[V ]d onto S[V ]Gd , for d ≥ 0. Using thiswe obtain dimF (S[V ]Gd ) = TrS[V ]d(RG) = 1

|G|∑π∈G TrS[V ]d(π) ∈ F . ]

(3.2) Remark. Molien’s Formula is remains valid in the non-modular case:

Let G be a finite group, and let F be a finite field such that char(F ) 6 | |G|.Let Q ⊆ K ⊆ Q ⊆ C be an algebraic number field, having a discrete valuationring R ⊆ K with maximal ideal πR CR such that : R/πR ∼= F . Let V be anFG-module. Thus the projective FG-module V by [9, Ch.1.14] has a uniquelydefined lift to an R-free RG-module V , which is RG-projective anyway. LetVK := V ⊗R K, which is a KG-module.

For d ∈ N let the RG-submodule V ′d ≤ V ⊗d be defined as in (1.1), and letS[V ]d := V ⊗d/V ′d as well as S[V ] :=

⊕d≥0 S[V ]d, where S[V ]0 := R is the

trivial RG-module. Note that for the trivial FG-module F we have F ∼= R.

By the right exactness of tensor products we for d ∈ N0 have (S[V ]d)K ∼=(V ⊗d)K/(V ′d)K ∼= S[VK ]d as KG-modules, and S[V ]d ∼= V ⊗d/V ′d

∼= S[V ]d ∼=S[V ]d as FG-modules. As dimK(S[VK ]d) =

(n+d−1

d

)= dimF (S[V ]d), we con-

clude that V ′d ≤ V ⊗d is R-pure, hence by [9, Ch.1.17] the homogeneous com-ponent S[V ]d is R-free. Generalising (1.1), the ring S[V ] is a graded R-algebrawhose homogeneous components are R-free of finite rank.

(3.3) Proposition. We have HS[V ]G = HS[VK ]G ∈ Q(T ).

10

Proof. There is a Reynolds operator RG : S[V ]→ S[V ]G : f 7→ 1|G| ·

∑π∈G fπ,

which is a degree preserving RG-module projection. Let d ∈ N0, hence S[V ]Gd ≤S[V ]d is an RG-direct summand, hence an R-direct summand, and thus we have

(S[V ]Gd )K ∼= S[VK ]Gd as KG-modules, as well as S[V ]Gd ∼= (S[V ]d)G ∼= S[V ]Gd asFG-modules. Thus we have dimK(S[VK ]Gd ) = rkR(S[V ]Gd ) = dimF (S[V ]Gd ). ]

(3.4) Remark. To evaluate Molien’s Formula we use character theory:

We have detV (1 − Tπ) =∏ni=1(1 − λiT ) = Tn ·

∏ni=1(T−1 − λi) for π ∈ G,

where λ1, . . . , λn ∈ F ⊆ C are the eigenvalues of DV (π). Using the elementarysymmetric polynomials en,i(X ) ∈ F [X ] = F [X1, . . . , Xn], for i ∈ 1, . . . , n, see(5.1), we get detV (1 − Tπ) = Tn ·

(T−n +

∑ni=1(−1)ien,i(λ1, . . . , λn)T i−n

)=

1 +∑ni=1(−1)ien,i(λ1, . . . , λn)T i; note that deg(en,i) = i.

By the Newton identities, see Exercise (10.10), the elementary symmetric poly-nomials en,i(X ) ∈ F [X ], for i ∈ 1, . . . , n, can be determined from the powersums pn,k(X ) :=

∑ni=1X

ki ∈ F [X ], for k ∈ 1, . . . , n. Finally, we have

pn,j(λ1, . . . , λn) =∑ni=1 λ

ji = TrV (πj) = χV (πj) ∈ C, where χV ∈ ZIrrC(G)

denotes the ordinary character of G afforded by V .

Hence Molien’s Formula can be evaluated once χV ∈ ZIrrC(G) and the powermaps pj : Cl(G) → Cl(G), for j ∈ 1, . . . , n, on the conjugacy classes Cl(G) ofG are known.

(3.5) Example. For k ∈ N let G = 〈δ, σ〉 ∼= D2k be the dihedral group of order2k. Let F ⊆ C or let F be a finite field, such that char(F ) 6 | 2k and having ak-th primitive root of unity ζk ∈ F , and let

DV : G→ GL2(F ) : δ 7→[ζk .. ζ−1

k

], σ 7→

[. 11 .

].

We have G = δi, σδi; i ∈ 0, . . . , k−1, where δi ∈ G has eigenvalues ζik, ζ−ik ∈

F , and where σδi ∈ G has eigenvalues 1,−1 ∈ F . Thus by Molien’s formula forthe case F ⊆ C we obtain

HS[V ]G =12k·

(k

(1− T ) · (1 + T )+k−1∑i=0

1(1− ζikT ) · (1− ζ−ik T )

)∈ F ((T )).

Using Exercise (10.11), where the rightmost summand is evaluated, we obtainHS[V ]G = 1

(1−T 2)·(1−Tk)∈ Q(T ). By (3.3) we conclude that this result also holds

for the finite field case.

Hence we are led to conjecture that S[V ]G is a polynomial ring having degreevector [2, k]. Indeed, let f1 := X1X2 ∈ S[V ]G and f2 := Xk

1 + Xk2 ∈ S[V ]G.

We apply the Jacobian criterion, see (3.7): Indeed, for the determinant of thecorresponding Jacobian matrix we find

det J(f1, f2) = det([

X2 X1

k ·Xk−11 k ·Xk−1

2

])= k · (Xk

2 −Xk1 ) 6= 0,

11

thus f1, f2 ⊆ S[V ]G is algebraically independent, and hence the Hilbert seriesof F [f1, f2] ⊆ S[V ]G is given as HF [f1,f2] = 1

(1−T 2)·(1−Tk)∈ Q(T ). Thus we

conclude that indeed F [f1, f2] = S[V ]G. ]

We prove the Jacobian criterion used above, to decide whether an n elementsubset of a polynomial ring F [X ] = F [X1, . . . , Xn] is algebraically independent;this will be used again in Section 4, see also Exercises (10.12) and (10.13).

(3.6) Definition. For f1, . . . , fn ⊆ F [X ] = F [X1, . . . , Xn] the Jacobianmatrix is defined as J(f1, . . . , fn)=JX (f1, . . . , fn) := [ ∂fi∂Xj

]i,j=1,...,n∈F [X ]n×n.Its determinant det J(f1, . . . , fn) ∈ F [X ] is called the Jacobian determinant.

(3.7) Proposition: Jacobian criterion.Let f1, . . . , fn ⊆ F [X ] = F [X1, . . . , Xn].a) Let F be a perfect field. If detJ(f1, . . . , fn) 6= 0, then f1, . . . , fn is alge-braically independent.b) Let char(F ) = 0. If f1, . . . , fn is algebraically independent, then we havedet J(f1, . . . , fn) 6= 0.

Note that the condition char(F ) = 0 in (b) is necessary: If char(F ) = p > 0,then Xp ⊆ F [X] is algebraically independent, while det J(Xp) = 0 ∈ F [X].

Proof. a) Let 0 6= h ∈ F [X ] of minimal degree such that h(f1, . . . , fn) = 0.Differentiation ∂

∂Xjusing the chain rule yields the following system of linear

equations over F (X ) = Quot(F [X ]):[∂h

∂Xi(f1, . . . , fn)

]i=1,...,n

· J(f1, . . . , fn) = 0.

Assume we have ∂h∂Xi

= 0 for all i ∈ 1, . . . , n. As deg(h) > 0 this im-plies char(F ) = p > 0, and as F is perfect we have h = (h′)p for someh′ ∈ F [X ]. Hence deg(h′) < deg(h) and h′(f1, . . . , fn) = 0, a contradiction.Hence there is i ∈ 1, . . . , n such that ∂h

∂Xi6= 0. As deg( ∂h

∂Xi) < deg(h), we

have ∂h∂Xi

(f1, . . . , fn) 6= 0. Hence the above system of linear equations has anon-trivial solution, thus det J(f1, . . . , fn) = 0, a contradiction.

b) Let f1, . . . , fn be algebraically independent. As trdeg(F (X )) = n, see [10,Ch.8.1], for k ∈ 1, . . . , n the sets Xk, f1, . . . , fn are algebraically depen-dent. Let 0 6= hk ∈ F [X0, X1, . . . , Xn] = F [X+] of minimal degree such thathk(Xk, f1, . . . , fn) = 0. Differentiation ∂

∂Xjyields[

∂hk∂Xi

(Xk, f1, . . . , fn)]k,i=1,...,n

·J(f1, . . . , fn) = diag[− ∂hk∂X0

(Xk, f1, . . . , fn)]k=1,...,n

.

As f1, . . . , fn is algebraically independent, the indeterminate X0 occurs inhk, hence we have degX0

(hk) > 0. As char(F ) = 0 we get ∂hk∂X0

6= 0, and as

12

degX+( ∂hk∂X0) < degX+(hk) we have ∂hk

∂X0(Xk, f1, . . . , fn) 6= 0. Hence we conclude

det diag[− ∂hk∂X0

(Xk, f1, . . . , fn)] 6= 0, and thus det J(f1, . . . , fn) 6= 0. ]

4 Polynomial invariant rings

We address the question whether we can characterise the representations whoseinvariant ring is a polynomial ring, the main result being the Shephard-Todd-Chevalley Theorem.

(4.1) Proposition. Let G be a finite group and let S[V ]G = F [f1, . . . , fr] bepolynomial, where the fi are homogeneous such that deg(fi) = di.a) Then we have r = n.b) Let additionally S[V ]G = F [f ′1, . . . , f

′n], where the f ′i are homogeneous such

that deg(f ′i) = d′i. Then there is π ∈ Sn such that d′iπ = di, for i ∈ 1, . . . , n.

Proof. a) As we have S[V ]G ∼= F [X ] := F [X1, . . . , Xr] with degree vectorδ := [d1, . . . , dr], we conclude S(V )G ∼= F (X ), and by (1.10) we have r =trdeg(S(V )G) = trdeg(S(V )) = n.

b) We also have S[V ]G ∼= F [X ′] := F [X ′1, . . . , X′n] with degree vector δ′ :=

[d′1, . . . , d′n]. Differentiation yields idn = JX (X ) = JX ′(X ) · JX (X ′). Hence we

have det JX ′(X ) 6= 0, and thus there is π ∈ Sn such that∏ni=1

∂Xi∂X′iπ

6= 0. Hencedi ≥ d′iπ, for i ∈ 1, . . . , n, and thus

∑ni=1 di ≥

∑ni=1 d

′i. As detJX (X ′) 6= 0,

we also have∑ni=1 d

′i ≥

∑ni=1 di. Hence

∑ni=1 di =

∑ni=1 d

′i, thus d′iπ = di. ]

(4.2) Definition. Let G be a finite group and let S[V ]G = F [f1, . . . , fn] bepolynomial, where the fi are homogeneous. Then f1, . . . , fn is called a set ofbasic invariants (fundamental invariants).

While basic invariants are in general not uniquely defined, even not up reorderingand multiplication by scalars, their degrees di = deg(fi) are uniquely definedup to reordering. They are called the polynomial degrees of G with respectto V , where we may assume d1 ≤ · · · ≤ dn.

(4.3) Definition. Let νc : C(T )∗ → Z, for c ∈ C, denote the discrete val-

uation of C(T ) at the place T = c, i. e. for 0 6= H = (T−c)a′·H′

(T−c)a′′ ·H′′ ∈ C(T ),where a′, a′′ ∈ N0, and H ′,H ′′ ∈ C[T ] such that H ′(c),H ′′(c) 6= 0, we haveνc(H) = a′ − a′′ ∈ Z; for H = 0 we let νc(H) :=∞. Hence νc(H) is the orderof the zero T = c of H. By evaluating at T = c we obtain the degree of0 6= H ∈ C(T ) as degc(H) :=

((c− T )−νc(H) ·H

)(c) ∈ C∗.

(4.4) Example. Let F [X ] = F [X1, . . . , Xn] with degree vector δ = [d1, . . . , dn].As we have Hδ

F [X ] =∏ni=1

11−Tdi ∈ Q(T ), we conclude ν1(Hδ

F [X ]) = −n and

13

thus we obtain deg1(HδF [X ]) =

((1− T )n ·Hδ

F [X ]

)(1) =

∏ni=1

(1−T

1−Tdi

)(1) =∏n

i=1

(1∑di−1

j=0 T j

)(1) =

∏ni=1

1di

.

(4.5) Definition. An element π ∈ G is called a pseudoreflection with re-spect to the representation DV , if for its fixed point space FixV (π) ≤ V wehave dimF FixV (π) = n − 1, where n = dimF (V ); if additionally DV (π2) = 1then π is called a reflection. If π is a pseudoreflection, then FixV (π) ≤ V iscalled its reflecting hyperplane. Let NG,V ∈ N0

.∪ ∞ be the number of

pseudoreflections with respect to V in G.

(4.6) Theorem. Let G be a finite group, let F ⊆ C or let F be a finite fieldsuch that char(F ) 6 | |G|, and let V be faithful, where n = dimF (V ).a) Then there is H ∈ Q(T ) such that ν1(H) ≥ −(n− 2) and

HS[V ]G =1|G|· (1− T )−n +

NG,V2 · |G|

· (1− T )−(n−1) + H ∈ Q(T ).

b) Let S[V ]G = F [f1, . . . , fn] be polynomial, where the fi are homogeneoussuch that deg(fi) = di. Then

∏ni=1 di = |G| and

∑ni=1(di − 1) = NG,V .

Proof. a) We first assume that Q[ζ|G|] ⊆ F ⊆ C, and let λ1, . . . , λn ∈ F be theeigenvalues of DV (π), for π ∈ G. As detV (1−Tπ) =

∏ni=1(1−λiT ) ∈ F [T ], we

have ν1(detV (1 − Tπ)) ≤ n, while ν1(detV (1 − Tπ)) = n if and only if π = 1.Thus ν1(HS[V ]G) = −n and deg1(HS[V ]G) =

((1− T )n ·HS[V ]G

)(1) = 1

|G| .

Let H ′ := HS[V ]G − 1|G| · (1− T )−n ∈ Q(T ). Hence we have ν1(H ′) ≥ −(n− 1).

We have ν1(detV (1 − Tπ)) = n − 1 if and only if π 6= 1 has the eigenvalue 1with multiplicity n − 1, hence if and only if π is a pseudoreflection. In thiscase, let 1 6= λ ∈ F be the non-trivial eigenvalue of π, and hence we have(

(1−T )n−1

detV (1−Tπ)

)(1) = 1

1−λ . As 11−λ + 1

1−λ−1 = 1, pairing each pseudoreflection

with its inverse, where for a reflection we have λ = −1 and hence 11−λ = 1

2 ,and summing over all the pseudoreflections in G, yields

((1− T )n−1 ·H ′

)(1) =

NG,V2·|G| . Thus we obtain H := H ′ − NG,V

2·|G| · (1− T )−(n−1) ∈ Q(T ) as asserted.

Let secondly F be a finite field, where we assume that there is a |G|-th primitiveroot of unity ζ|G| ∈ F . Let R ⊆ K and VK be as in (3.2). We have HS[V ]G =HS[VK ]G , and as V is faithful this also holds for VK . Moreover, considering thenatural linear F -representation of the cyclic group of order |G|, we conclude

that there is a |G|-th primitive root of unity ζ|G| ∈ K such that ζ|G| = ζ|G|.Hence DV (π), for π ∈ G, we conclude that π is a pseudoreflection with respectto V if and only if π is a pseudoreflection with respect to VK .

b) We have HS[V ]G =∏ni=1

11−Tdi , and by (4.4) we obtain deg1(HS[V ]G) =∏n

i=11di

= 1|G| . Moreover, we have (1−T )n ·H ′ = − 1

|G|+∏ni=1

1∑di−1j=0 T j

= NG,V2·|G| ·

14

(1−T )+(1−T )n ·H, where ν1(H) ≥ −(n−2). Differentiation ∂∂T and evaluation

at T = 1, on the left hand side yields −(∏n

i=11∑di−1

j=0 T j

)·(∑n

i=1

∑di−1j=1 jT j−1∑di−1j=0 T j

)evaluated at T = 1, thus −

(∏ni=1

1di

)·∑ni=1

di(di−1)2di

. On the right hand side

we obtain −NG,V2·|G| . Hence using∏ni=1 di = |G| we get NG,V =

∑ni=1(di − 1). ]

For a statement related to (4.6)(b) see Exercise (10.32). We next first prove theharder direction of the Shephard-Todd-Chevalley Theorem (4.9): If G is a finitegroup generated by pseudoreflections, then S[V ]G is polynomial.

(4.7) Lemma. LetG be a finite group such that char(F ) 6 | |G|, being generatedby pseudoreflections with respect to V , and let R := S[V ]G ⊆ S[V ] =: S. Leth1, . . . , hr ∈ R and h ∈ R\(

∑ri=1 hiR), as well as p, p1, . . . , pr ∈ S homogeneous

such that hp =∑ri=1 hipi. Then we have p ∈ I := IG[V ]C S.

Proof. We proceed by induction on deg(p). Let deg(p) = 0 and assume p 6= 0.Using the Reynolds operator we have hp = RG(hp) =

∑ri=1 hi · RG(pi) ∈∑r

i=1 hiR. Hence h ∈∑ri=1 hiR, a contradiction. Hence p = 0 ∈ I.

Let deg(p) > 0. Let π ∈ G be a pseudoreflection, and we may assume thatDV (π) = diag[λ, 1, . . . , 1] ∈ Fn×n, where 1 6= λ ∈ F . Hence we have Xiπ = Xi,for i ≥ 2, and thus Xα(π − 1) = 0 for all monomials Xα ∈ F [X ] such thatα1 = 0. As X1(π − 1) = (λ − 1)X1 6= 0 we conclude that there are p′, p′i ∈ Shomogeneous such that p(π − 1) = X1p

′ ∈ S and pi(π − 1) = X1p′i ∈ S, for

i ∈ 1, . . . , r. In particular we have deg(p′) < deg(p). Applying π − 1 tothe equation hp =

∑ri=1 hipi we obtain X1 · hp′ = X1 ·

∑ri=1 hip

′i. Hence by

induction we have p′ ∈ I, and thus p(π − 1) = X1p′ ∈ I as well.

Let : S → S/I denote the natural epimorphism of F -algebras. As I C Sis an FG-submodule, G acts on the quotient F -algebra S/I. By the abovewe have pπ = p for all pseudoreflections π ∈ G, and as G is generated bypseudoreflections, we have pπ = p for all π ∈ G. Since RG(p) ∈ R+ ⊆ I weconclude p+ I = RG(p) + I = 0 + I ∈ S/I, hence p ∈ I. ]

(4.8) Theorem. Let F be a perfect field, and let G be a finite group suchthat char(F ) 6 | |G|, being generated by pseudoreflections with respect to V . LetS[V ]G = F 〈f1, . . . , fr〉, where f1, . . . , fr is minimal and the fi are homoge-neous such that char(F ) 6 | deg(fi). Then f1, . . . , fr is algebraically indepen-dent, thus S[V ]G is polynomial.

Proof. Let R := S[V ]G ⊆ S[V ] =: S and I := IG[V ] =∑ri=1 fiSCS. By (2.5)

the set f1, . . . , fr is a minimal generating set of I. Assume to the contrarythat there is 0 6= h ∈ F [Y] := F [Y1, . . . , Yr] such that h(f1, . . . , fr) = 0. Lettingdi := deg(fi), we may assume h is homogeneous with respect to the grading

15

of F [Y] given by the degree vector δ := [d1, . . . , dr], and that d := degδ(h) ischosen minimal.

Let hi := ∂h∂Yi

(f1, . . . , fr) ∈ R, for i ∈ 1, . . . , r. Hence we have hi = 0 ordeg(hi) = d − di. Assume, we have hi = 0 for all i ∈ 1, . . . , r, then byminimality we have ∂h

∂Yi= 0 ∈ F [Y], for all i ∈ 1, . . . , r. As d > 0 this

implies char(F ) = p > 0, and as F is perfect we have h = (h′)p for someh′ ∈ F [Y], hence h′(f1, . . . , fn) = 0, contradicting the minimality of h. Thusthere is i ∈ 1, . . . , r such that hi 6= 0. We may assume by reordering that∑r′

i=1 hiR =∑ri=1 hiRER, where 1 ≤ r′ ≤ r is minimal. For j > r′ let gij ∈ R,

for i ∈ 1, . . . , r′, be homogeneous such that hj =∑r′

i=1 higij ∈ R, hencegij = 0 or deg(gij) = deg(hj)−deg(hi) = di−dj . Differentiation ∂

∂Xkusing the

chain rule yields

0 =r∑i=1

hi ·∂fi∂Xk

=r′∑i=1

hi ·

∂fi∂Xk

+r∑

j=r′+1

gij ·∂fj∂Xk

∈ S.Let pi,k := ∂fi

∂Xk+∑rj=r′+1 gij ·

∂fj∂Xk∈ S, for i ∈ 1, . . . , r′ and k ∈ 1, . . . , n.

Hence the pi,k are homogeneous such that pi,k = 0 or deg(pi,k) = di−1. By theminimality of r′, from (4.7) we conclude that p1,k ∈ I, thus p1,k =

∑rj=1 fjqj,k,

for qj,k ∈ S homogeneous such that qj,k = 0 or deg(qj,k) = d1 − dj − 1 ≥ 0;hence in particular we have q1,k = 0.

From this we obtain∑nk=1Xk ·

(∂f1∂Xk

+∑rj=r′+1 g1j · ∂fj∂Xk

)=∑nk=1Xkp1,k =∑n

k=1(Xk ·∑rj=2 fjqj,k) =

∑rj=2(fj ·

∑nk=1Xkqj,k) =

∑rj=2 fjq

′j ∈ S, where

q′j =∑nk=1Xkqj,k ∈ S is homogeneous such that q′j = 0 or deg(q′j) = d1 − dj ,

for j ∈ 2, . . . , r. By the Euler identity∑nk=1Xk · ∂f

∂Xk= deg(f) · f , for

f ∈ S, the left hand side equals d1 · f1 +∑rj=r′+1 dj · g1jfj . Hence comparing

both sides we conclude d1 · f1 ∈∑rj=2 fjS. As char(F ) 6 | d1, we have have

f1 ∈∑rj=2 fjS C S, a contradiction. ]

Note that the condition on the degrees deg(fi) is fulfilled anyway if char(F ) = 0or, by Noether’s degree bound, if char(F ) > |G|. Moreover, under the assump-tions of (4.8), the polynomiality of S[V ]G by (4.6) already implies deg(fi) | |G|,and thus indeed char(F ) 6 | deg(fi). This seems to indicate that the abovecondition on the degrees deg(fi) might not necessary, see (4.10).

(4.9) Theorem: Shephard-Todd, 1954; Chevalley, 1955.Let F ⊆ C and let G be a finite group acting faithfully on V . Then S[V ]G ispolynomial if and only if G is generated by pseudoreflections with respect to V .

Proof. By (4.8) it remains to show that if S[V ]G is polynomial, then G isgenerated by pseudoreflections. Let S[V ]G = F [f1, . . . , fn], where the fi arehomogeneous such that di := deg(fi). Let H ≤ G be the subgroup generated

16

by the pseudoreflections in G. Hence by (4.8) we have S[V ]G ⊆ S[V ]H =F [g1, . . . , gn] ⊆ S[V ], where the gi are homogeneous such that ei := deg(gi).Hence there are hi ∈ F [X ] such that fi = hi(g1, . . . , gn), for i ∈ 1, . . . , n.We may assume that the hi are homogeneous such that degδ(hi) = di, withrespect to the grading of F [X ] given by the degree vector δ := [e1, . . . , en].Differentiation ∂

∂Xkand the chain rule yields

J(f1, . . . , fn) =[∂hi∂Xj

(g1, . . . , gn)]i,j=1,...,n

· J(g1, . . . , gn) ∈ F [X ]n×n.

The Jacobian criterion yields det J(f1, . . . , fn) 6= 0, hence det[ ∂hi∂Xj(g1, . . . , gn)] 6=

0 as well. Hence by renumbering f1, . . . , fn if necessary, we may assume that∏ni=1

∂hi∂Xi

(g1, . . . , gn) 6= 0. Thus we have di ≥ ei, for i ∈ 1, . . . , n. Since by(4.6) we have

∑ni=1(di − 1) = NG,V = NH,V =

∑ni=1(ei − 1), we have ei = di

for i ∈ 1, . . . , n. Finally, we have |G| =∏ni=1 di =

∏ni=1 ei = |H|. ]

(4.10) Remark. Shephard-Todd (1954) proved (4.9) by first classifying thefinite groups generated by pseudoreflections with respect to some irreduciblefaithful complex representation, see (4.11), and then by a case-by-case analysisverified (4.8). Later, Chevalley (1955) gave a conceptual proof.

As was proved by Serre (1967), (4.9) remains valid in the non-modular case;even in the modular case, if an invariant ring is polynomial, then the group isgenerated by pseudoreflections. But the converse does not hold, see Exercise(10.14). Using the classification of the finite groups generated by pseudoreflec-tions with respect to some irreducible faithful module in prime characteristic(Kantor, 1979; Wagner, 1978, 1980; Zalesskii-Serezkin, 1976, 1981), the classi-fication of polynomial invariant rings in the irreducible modular case has beengiven by Kemper-Malle (1997).

(4.11) Remark: The Shephard-Todd classification.Let F ⊆ C and let G = 〈π1, . . . , πt〉 be a finite group, where the πi are pseudore-flections with respect to the faithful FG-module V . By Maschke’s Theorem letV =

⊕kj=1 Vj as FG-modules, where the Vj are irreducible. By considering the

eigenvalues of the pseudoreflections πi it follows that DVj (πi) 6= idVj for a uniquej ∈ 1, . . . , k. Hence letting Gj := 〈πi; i ∈ 1, . . . , t, DVj (πi) 6= idVj 〉 ≤ G,we have G ∼=

∏kj=1Gj , where Gj acts trivially on Vj′ , for j′′ 6= j, while Gj

is generated by pseudoreflections with respect to the faithful and irreducibleFGj-module Vj .

Hence we are reduced to the case where V is irreducible, and by enlarging thefield F if necessary we may assume that V is absolutely irreducible. Moreover,let π ∈ G be a pseudoreflection with respect to V , and let 1 6= λ ∈ F be its non-trivial eigenvalue. By abuse of notation let λ : H := 〈π〉 → GL1(F ) : π 7→ λ alsodenote the corresponding linear representation. Hence by Frobenius reciprocitywe have dimF HomFG(λG, V ) = dimF HomFH(λ, VH) = 1. As the Schur index

17

of V divides dimF HomFG(λG, V ), we conclude that the former also equals 1,thus V is realizable over its character field, which hence is its unique minimalrealization field.

The classification of the finite groups G being generated by pseudoreflectionswith respect to some faithful, absolutely irreducible FG-module V , for someF ⊆ C, is given in Table 1. The fields F given in the last but one columnare the character fields, where ζk := exp 2π

√−1k ∈ C is a k-th primitive root of

unity, for k ∈ N. The last column indicates the real groups, i. e. the groupswhose character field is a subfield of R, and their Dynkin type; in particularthese groups are generated by reflections. The classes 1, 2a, 2b and 3 consistof infinite series, while the 34 groups G4, . . . , G37 are called the exceptionalgroups.

The groups in class 1, being real of Dynkin type An, are the symmetric groupsSn+1 = 〈(1, 2), . . . , (n, n+ 1)〉 acting via the deleted permutation represen-tation: The group Sn+1 is generated by reflections with respect to its naturalpermutation representation DW : Sn+1 → GLn+1(Q). As Sn+1 acts multiplytransitively, we have dimQ EndQSn+1(W ) = 2. Thus we have W ∼= F ⊕ V ,where F is the trivial representation, and V is an absolutely irreducible faithfulQSn+1-module, with respect to which Sn+1 is generated by reflections. Hencefor W ′ :=

⋂ni=1 FixW ((i, i + 1)) we have W ′ ∼= F and hence V ∼= W/W ′.

As we have χW = 1Sn+1Sn ∈ ZIrrC(Sn+1), using the parametrisation of the

irreducible ordinary characters of Sn+1 by the partitions of n + 1, we getχV = χ[n,1] ∈ ZIrrC(Sn+1). For a set of basic invariants being derived fromthe elementary symmetric polynomials in Q[W ]Sn+1, see Exercise (10.18).

The groups in class 2a are denoted by G(m, k, n), for m ≥ 2 and k | m. LettingT := diag[ζaim ; i ∈ 1, . . . , n] ∈ GLn(C); k |

∑ni=1 ai, we have G(m, k, n) ∼=

Sn n T , where Sn → GLn(C) is the natural permutation representation. Thegroup G(m, k, n) is real if and only if m = 2; in this case k = 1 is the Dynkintype Bn, and k = 2 is the Dynkin type Dn.

The groups in class 2b are the dihedral groups D2m for m ≥ 3, see (3.5); notethat the Dynkin type G2 is equal to I2(6). The groups in class 3 are the cyclicgroups Cm, see (1.13).

5 Permutation groups

The symmetric group Sn = 〈(1, 2), . . . , (n− 1, n)〉 is generated by pseudoreflec-tions with respect to its natural permutation representation. Hence for F = C

we conclude that F [X ]Sn is polynomial. Actually, this turns out to be truefor arbitrary fields F , a set of basic invariants being the elementary symmetricpolynomials.

For the relation of the elementary symmetric polynomials to a set of basicinvariants of the deleted permutation representation, see (4.11), on which Snhence also acts faithfully as a group generated by pseudoreflections see Exercise

18

Table 1: Shephard-Todd classification.

i dimF |G| polynomial degrees F

1 n (n+ 1)! 2, . . . , n+ 1 Q An2a n mn

k · n! m, . . . , (n− 1)m, nmk Q(ζm)2b 2 2m 2,m Q(ζm + ζ−1

m ) I2(m)3 1 m m Q(ζm)4 2 24 4, 6 Q(ζ3)5 2 72 6, 12 Q(ζ3)6 2 48 4, 12 Q(ζ12)7 2 144 12, 12 Q(ζ12)8 2 96 8, 12 Q(ζ4)9 2 192 8, 24 Q(ζ8)

10 2 288 12, 24 Q(ζ12)11 2 576 24, 24 Q(ζ24)12 2 48 6, 8 Q(

√−2)

13 2 96 8, 12 Q(ζ8)14 2 144 6, 24 Q(ζ3,

√−2)

15 2 288 12, 24 Q(ζ24)16 2 600 20, 30 Q(ζ5)17 2 1200 20, 60 Q(ζ20)18 2 1800 30, 60 Q(ζ15)19 2 3600 60, 60 Q(ζ60)20 2 360 12, 30 Q(ζ3,

√5)

21 2 720 12, 60 Q(ζ12,√

5)22 2 240 12, 20 Q(ζ4,

√5)

23 3 120 2, 6, 10 Q(√

5) H3

24 3 336 4, 6, 14 Q(√−7)

25 3 648 6, 9, 12 Q(ζ3)26 3 1296 6, 12, 18 Q(ζ3)27 3 2160 6, 12, 30 Q(ζ3,

√5)

28 4 1152 2, 6, 8, 12 Q F4

29 4 7680 4, 8, 12, 20 Q(ζ4)30 4 14400 2, 12, 20, 30 Q(

√5) H4

31 4 46080 8, 12, 20, 24 Q(ζi)32 4 155520 12, 18, 24, 30 Q(ζ3)33 5 51840 4, 6, 10, 12, 18 Q(ζ3)34 6 39191040 6, 12, 18, 24, 30, 42 Q(ζ3)35 6 51840 2, 5, 6, 8, 9, 12 Q E6

36 7 2903040 2, 6, 8, 10, 12, 14, 18 Q E7

37 8 696729600 2, 8, 12, 14, 18, 20, 24, 30 Q E8

19

(10.18). For the invariant ring F [X ]An of the alternating group An C Sn seeExercise (10.17).

(5.1) Definition. For n ∈ N let Sn denote the symmetric group on n letters,and let V ∼= Fn be its natural permutation module. Hence Sn acts on S[V ] ∼=F [X1, . . . , Xn] = F [X ] by permuting the indeterminates X1, . . . , Xn, andthus induces permutation representations on the F [X ]d, for d ≥ 0, where thepermutation F -bases consist of the monomials of degree d, respectively. Theelements of the invariant ring F [X ]Sn are called symmetric polynomials.

Let Y be an indeterminate over F [X ]. Then we have∏ni=1(Y − Xi) = Y n +∑n

i=1(−1)ien,i(X )Y n−i ∈ F [X ]Sn [Y ] ⊆ F [X , Y ], using the elementary sym-metric polynomials en,i(X ) :=

∑1≤j1<j2<···<ji≤n(

∏ik=1Xjk) ∈ F [X ]Sn , for

i ∈ 1, . . . , n. Note that the en,i ∈ F [X ]Sn are homogeneous such thatdeg(en,i) = i, and that in particular we have en,1(X ) =

∑ni=1Xi and en,n(X ) =∏n

i=1Xi.

Next to the standard grading of F [X ], given by the degree vector [1, . . . , 1], weconsider the grading of F [X ] given by the degree vector ε := [1, . . . , n]. Notethat degε(Xi) = i = deg(en,i).

(5.2) Theorem. Let f ∈ F [X ]Sn such that deg(f) = d. Then there is g ∈ F [X ]such that degε(g) ≤ d and f = g(en,1, . . . , en,n).

Proof. We proceed by induction on n. The case n = 1 is clear, hence let n > 1.We in turn proceed by induction on d. The case d = 0 is clear, hence let d > 0.We consider the F -algebra homomorphism F [X , Y ] → F [X ′, Y ], where X ′ :=X1, . . . , Xn−1, defined by Y 7→ Y and Xi 7→ Xi, for i ∈ 1, . . . , n− 1, whileXn 7→ 0. Using this we obtain Y n +

∑ni=1(−1)ien,i(X1, . . . , Xn−1, 0)Y n−i =

Y ·∏n−1i=1 (Y − Xi) = Y n +

∑n−1i=1 (−1)ien−1,i(X ′)Y n−i ∈ F [X ′][Y ]. Hence we

have en,i(X1, . . . , Xn−1, 0) = en−1,i(X ′) for i ∈ 1, . . . , n − 1, while of courseen,n(X1, . . . , Xn−1, 0) = 0.

We have f(X1, . . . , Xn−1, 0) ∈ F [X ′]Sn−1 , and hence by induction there is g′ ∈F [X ′] such that degε(g′) ≤ d and f(X1, . . . , Xn−1, 0) = g′(en−1,1, . . . , en−1,n−1).We consider g′(en,1, . . . , en,n−1) ∈ F [X ], and as the en,i ∈ F [X ] are homoge-neous and deg(en,i) = i, we conclude that deg(g′(en,1, . . . , en,n−1)) ≤ d. Let-ting f ′ := f − g′(en,1, . . . , en,n−1) ∈ F [X ], we have deg(f ′) ≤ d as well. Asf ′(X1, . . . , Xn−1, 0) = f(X1, . . . , Xn−1, 0) − g′(en−1,1, . . . , en−1,n−1) = 0, weconclude that f ′ ∈ F [X ′][Xn] is divisible by Xn. As f ′ ∈ F [X ]Sn and theXi ∈ F [X ] are pairwise non-associate primes, we have f ′ = en,n · f ′′ ∈ F [X ].Hence f ′′ ∈ F [X ]Sn as well, and we have deg(f ′′) ≤ d−n < d. Thus by inductionthere is g′′ ∈ F [X ] such that degε(g′′) ≤ d−n and f ′′ = g′′(en,1, . . . , en,n). Hencef = g′(en,1, . . . , en,n−1) + f ′ = g′(en,1, . . . , en,n−1) + en,n · g′′(en,1, . . . , en,n),where degε(Xng

′′) ≤ n+ (d− n) = d and hence degε(g′ +Xng′′) ≤ d. ]

Note that the proof of (5.2) is constructive: Given f ∈ F [X ]Sn , a polynomial

20

g ∈ F [X ] such that f = g(en,1, . . . , en,n) can be explicitly computed, see alsoExercise (10.16). Moreover, (5.3) shows that g is uniquely determined:

(5.3) Theorem. The invariant ring F [X ]Sn ∼= F [en,1, . . . , en,n] is polynomial.

Proof. It remains to show that en,1, . . . , en,n ⊆ F [X ] is algebraically inde-pendent. We proceed by induction on n. The case n = 1 is clear, hence letn > 1. Assume to the contrary, that there is 0 6= f ∈ F [X ] of minimal degree,such that f(en,1, . . . , en,n) = 0. Let f =

∑di=0 fi(X ′)Xi

n ∈ F [X ′][Xn], whereX ′ := X1, . . . , Xn−1. Assume we have f0 = 0. Then we have f = Xn · f ′, andhence f ′(en,1, . . . , en,n) = 0, where f ′ 6= 0 and deg(f ′) < deg(f), a contradic-tion. Thus we have 0 6= f0 ∈ F [X ′].We consider the F -algebra homomorphism F [X ]→ F [X ′], defined by Xi 7→ Xi,for i ∈ 1, . . . , n − 1, and Xn 7→ 0. Hence using the notation of the proofof (5.2), we from 0 = f(en,1, . . . , en,n) =

∑di=0 fi(en,1, . . . , en,n−1)ein,n obtain

0 =∑di=0 fi(en−1,1, . . . , en−1,n−1) · 0i = f0(en−1,1, . . . , en−1,n−1). By induction

this a contradiction to the algebraic independence of en−1,1, . . . , en−1,n−1 ⊆F [X ′]Sn−1 . ]

We turn to the description of F -algebra generating sets of invariant rings F [X ]G

of permutation groups G ≤ Sn; for a slightly improved treatment for intransitivegroups see [4, Ch.3.10]; see also Exercise (10.19). As G ≤ Sn induces permuta-tion representations on the F [X ]Gd , for d ≥ 0, where the permutation F -basesconsist of the monomials of degree d, respectively, F [X ]G has an F -basis consist-ing of orbit sums of monomials. The main result is Gobel’s Theorem exhibitingan F -algebra generating set consisting of orbit sums; actually, the statementholds true for any commutative base ring.

(5.4) Lemma. Let G be a finite group, and let Ω be a finite G-set. For ω ∈ Ωlet ω+ :=

∑ω∈ω·G ω =

∑π∈StabG(ω)|G ω · π = TrGStabG(ω)(ω) ∈ FΩ be the

corresponding orbit sum. Then ω+ ∈ FΩ;ω ∈ Ω is an F -basis of the F -space (FΩ)G of G-fixed points in FΩ.

Proof. It is clear that ω+ ∈ (FΩ)G, and that the set ω+ ∈ FΩ;ω ∈ Ω isF -linearly independent. Moreover, if v =

∑ω∈Ω vω · ω ∈ FΩ, where vω ∈ F , we

for ω ∈ ω ·G have vω = vω, hence v ∈ 〈ω+ ∈ FΩ;ω ∈ Ω〉F . ]

(5.5) Definition. a) A partition λ = [λ1, . . . , λl] ∈ Pd, where l = lλ is thelength of λ, is called special (column 2-regular), if λi − λi+1 ≤ 1 for i ∈1, . . . , l, where we let λl+1 := 0. A special partition λ is called k-special,if lλ < k, for k ∈ N. Note that if λ ∈ Pd is k-special, then d ≤

∑k−1i=1 i =(

k2

). If λ ∈ Pd is not special, then we have i;λi − λi+1 > 1 6= ∅, and thus

s = sλ := mini;λi − λi+1 > 1 ∈ N is well-defined. Using this, the partitionλ := [λ1− 1, . . . , λs− 1, λs+1, . . . , λl] ∈ Pd−s is called the reduction of λ ∈ Pd.

21

Let λ1 ∈ Pd1 and λ2 ∈ Pd2 . Then λ1Eλ2 in the dominance partial order, if∑ij=1 λ1,j ≤

∑ij=1 λ2,j , for i ≥ 1. Note that λ1Eλ2 and λ2Eλ1 imply λ1 = λ2.

b) Let α = [α1, . . . , αn] ∈ Nn0 and let σ = σα ∈ Sn such that for ασ :=[α1σ−1 , . . . , αnσ−1 ] ∈ Nn0 we have α1σ−1 ≥ · · · ≥ αnσ−1 ≥ 0. Hence we mayconsider ασ ∈ Pd, where d = dα :=

∑α. The vector α is called (k-)special, if

ασ ∈ Pd is (k-)special. Note that ασ, and hence being (k-)special, is independentfrom the original ordering of the parts of α. If ασ ∈ Pd is not special, and hasreduction ασ ∈ Pd−s, where s = sα := sασ , then α := (ασ)σ

−1 ∈ Nn0 , is calledthe reduction of α.

Let α1, α2 ∈ Nn0 . Then α1Eα2 in the dominance relation, if for the associatedpartitions we have ασ1

1 Eασ22 . Hence we have α1 ≡ α2, i. e. α1Eα2 and α2Eα1,

if and only if α2 is obtained from α1 by reordering parts.

(5.6) Lemma. We keep the notation of (5.5), let G ≤ Sn and let orbits sumsbe formed with respect to G. Let α ∈ Nn0 be not special.a) Let β ∈ Nn0 such that X β occurs in (Xα)+ · en,sα(X ) ∈ F [X ]. Then β E α.b) If X β occurs in (Xα)+, then X β also occurs in (Xα)+ · en,sα(X ). Moreover,the X β occurring in (Xα)+ are precisely the X β occurring in (Xα)+ · en,sα(X )such that β ≡ α.

Proof. a) We may assume that parts of α are sorted weakly decreasing, hencewe have i ∈ 1, . . . , n;αi 6= αi = 1, . . . , s. As X β occurs in (Xα)+ ·en,s(X ), where s = sα, the vector β is obtained from α by first reducingthe entries 1, . . . , s by 1, then permuting the entries by π ∈ G, and fi-nally increasing the entries j1, . . . , js by 1. As the summands of en,s(X ) =∑

1≤j1<···<js≤n(∏sk=1Xjk) consist of a single Sn-orbit, it is sufficient to consider

βπ−1

, hence we may assume that X β = Xα ·∏sk=1Xjk . Let J := j1, . . . , js,

then for i ∈ 1, . . . , n we have 4 cases: If i ≤ s and i 6∈ J , then βi = αi − 1;if i ≤ s and i ∈ J , then βi = αi; if i > s and i 6∈ J , then βi = αi; ifi > s and i ∈ J , then βi = αi + 1. Note that

∑α = (

∑α) − s, and hence∑

β = (∑α) + s =

∑α.

Let σ ∈ Sn such that the parts of βσ are sorted weakly decreasing. As theparts of α are sorted weakly decreasing, where αs − 1 ≥ αs+1 + 1, and 0 ≤αi − βi ≤ 1, for i ≤ s, and 0 ≤ βi − αi ≤ 1, for i > s, respectively, we haveβi ≥ βj , for i ≤ s and j > s, and hence we may choose σ such that 1, . . . , sand s + 1, . . . , n are left invariant, and that for each σ-orbit i1, . . . , ik ⊆1, . . . , n we have αi1 = · · · = αik . Hence for i ≤ s we have

∑ij=1(βσ)j ≤∑i

j=1 αj , and∑sj=1(βσ)j =

∑sj=1 βj = (

∑sj=1 αj) − |j ≤ s; j 6∈ J |, where

moreover |j ≤ s; j 6∈ J | = s − |j ≤ s; j ∈ J |. Analogously, for i > s wehave

∑ij=s+1(βσ)j ≤ (

∑ij=s+1 αj) + |j > s; j ∈ J | and thus

∑ij=1(βσ)j ≤

(∑sj=1 αj) − |j ≤ s; j 6∈ J | + (

∑ij=s+1 αj) + |j > s; j ∈ J | = (

∑ij=1 αj) −

s+ |j ≤ s; j ∈ J |+ |j > s; j ∈ J | =∑ij=1 αj , and thus β E α.

22

b) If X β occurs in (Xα)+, then clearly X β occurs in (Xα)+ · en,s(X ), andmoreover β is obtained from α by reordering parts, thus β ≡ α. Conversely,if X β occurs in (Xα)+ · en,s(X ), then the vector β is obtained from α by firstreducing the entries i1, . . . , is by 1, then permuting the entries by π ∈ G,and finally increasing the entries j1, . . . , js by 1. If moreover β ≡ α, then theresult is a reordering of α, and we conclude that j1, . . . , js = i1, . . . , is ·π−1,and hence β = απ, thus X β occurs in (Xα)+. ]

(5.7) Theorem: Gobel, 1995.Let G ≤ Sn. Then we have F [X ]G = F 〈en,n(X ), (Xα)+;α ∈ Nn0 n-special〉.

Proof. Let R := F 〈en,n(X ), (Xα)+;α ∈ Nn0 n-special〉; we show F [X ]G ⊆ R.Let α ∈ Nn0 be not n-special; by induction on d = dα =

∑α and on the

dominance relation we show that (Xα)+ ∈ R. Let σ = σα ∈ Sn such thatασ ∈ Pd. If (ασ)n ≥ 1, then en,n(X ) =

∏ni=1Xi | Xα, and hence we may

assume that (ασ)n = 0. As hence α is not special, let 1 ≤ s = sα < n be as in(5.5). Let f := (Xα)+ − (Xα)+ · en,s(X ) ∈ F [X ]G.

As s < n, the monomials∏sk=1Xjk , for 1 ≤ j1 < j2 < · · · < js ≤ n, are n-

special. As the summands of en,s(X ) =∑

1≤j1<j2<···<js≤n(∏sk=1Xjk) consist of

a single Sn-orbit, and thus are a union of G-orbits, we hence have en,s(X ) ∈ R.Moreover, as

∑α = (

∑α) − s we by induction have (Xα)+ ∈ R. Finally, for

all monomials X β , for β ∈ Nn0 , occurring in f we by (5.6) have β C α, and thusby induction we have f ∈ R. ]

(5.8) Corollary: Gobel’s degree bound.Let G ≤ Sn. Then there is a homogeneous F -algebra generating set of F [X ]G

consisting of elements of degree ≤ maxn,(n2

).

6 Prime ideals in commutative rings

We turn our attention to structural properties of general commutative rings, andintroduce some of the basic notions and theorems of commutative algebra, whichare also important in other fields, such as algebraic geometry. Our objective isto study the properties of prime ideals in Noetherian commutative rings. Webegin by introducing a basic and at the same time powerful tool, localisation.

(6.1) Definition. Let R be a commutative ring.a) Let I E R. Then

√I := f ∈ R; fn ∈ I for some n ∈ N E R is called the

radical of I; note that I ⊆√I. In particular, N (R) :=

√0ER is the set of

nilpotent elements of R. If N (R) = 0 then R is called reduced.

b) Let M be an R-module. A subset U ⊆ R is called multiplicatively closed,if 1 ∈ U and for f, g ∈ U we have fg ∈ U as well. Let ∼ denote the equivalencerelation on M × U defined by [m,u] ∼ [m′, u′], for m,m′ ∈ M and u, u′ ∈ U ,

23

if there is v ∈ U such that (mu′ −m′u)v = 0 ∈ M . Let the localisation MU

be defined as MU := (M ×U)/ ∼, the ∼-equivalence class of [m,u] ∈M ×U isdenoted by m

u ∈MU .

c) Some properties of localisations are collected in Exercise (10.23): The mapν : R→ RU : r 7→ r

1 is a homomorphism of commutative rings. Since for JERUwe have (ν−1(J))U = J , the contraction map ι : J E RU → I E R : J 7→ν−1(J) is an inclusion-preserving and intersection-preserving injection, mappingprime ideals to prime ideals. In particular, if R is Noetherian, then RU isNoetherian as well.

Moreover, for an ideal IER we have I ⊆ ν−1(IU ) = f ∈ R; fu ∈ I for some u ∈U E R. In particular, if U ∩ I = ∅, then for the extended ideal we haveIU 6= RU ; and a prime ideal P CR is in the image of ι if and only if U ∩P = ∅.In particular, if P CR is a prime ideal, then the set R\P ⊆ R is multiplicativelyclosed, and RR\P is a local ring, i. e. has a unique maximal ideal, namely PR\P .

(6.2) Proposition. Let R be a commutative ring and let I C R. Then wehave

√I =

⋂I ⊆ P C R;P prime. In particular, we have N (R) =

√0 =⋂

P CR;P prime.

Proof. Let f ∈√I and P ∈ P := I ⊆ P C R;P prime. Then we have

fn ∈ I ⊆ P for some n ∈ N, and thus f ∈ P , hence f ∈⋂P. Conversely, let

f 6∈√I, and consider the multiplicatively closed set U := fn;n ∈ N0 ⊆ R, as

well as J := I ⊆ J C R;U ∩ J = ∅. As U ∩ I = ∅ we have I ∈ J 6= ∅, andhence by Zorn’s Lemma we choose a maximal element J ∈ J . Note that theuse of Zorn’s Lemma can be avoided if R is Noetherian.

As U∩J = ∅ we have JUCRU . Since for any ideal JCRU we have ν−1(J)∩U = ∅,and as the contraction map is injective, we conclude that JU CRU is maximal,thus a prime ideal. Hence ν−1(JU ) C R is a prime ideal as well, and sinceJ ⊆ ν−1(JU ) we have I ⊆ J = ν−1(JU )C R, and thus J ∈ P such that f 6∈ J ,hence f 6∈

⋂P. ]

(6.3) Lemma: Prime avoidance.Let R be a commutative ring, let P1, . . . , PnCR be prime ideals, for n ∈ N, andlet I C R be an ideal such that I ⊆

⋃ni=1 Pi. Then there is i ∈ 1, . . . , n such

that I ⊆ Pi.

Proof. We proceed by induction on n ∈ N. As the case n = 1 is clear, welet n > 1. By induction we may assume that for j ∈ 1, . . . , n there is fj ∈I \

⋃i 6=j Pi. Hence we have fj ∈ Pj . Thus

∏nj=2 fj ∈ (

⋂ni=2 Pi) \ P1 and

f1 ∈ P1 \⋃ni=2 Pi, and thus f1 +

∏nj=2 fj 6∈

⋃ni=1 Pi, a contradiction. ]

We set out to study the relationship between prime ideals in a Noetheriancommutative ring and actions on modules, leading to the notion of associated

24

prime ideals, and to a finiteness property. Actually this is only the beginning ofa long story, related to the notion of primary decomposition, which has first beenexamined by Lasker (1905), but whose modern description is original work byNoether (1921), in particular underlining the importance of the ascending chaincondition, for that reason nowadays carrying Noether’s name. For the relationbetween non-zerodivisors and annihilators, leading to an interesting dichotomy,see Exercise (10.24).

(6.4) Definition. Let R be a commutative ring.a) Let M be an R-module. Then the annihilator of a subset M ⊆ M isdefined as annR(M) := f ∈ R;Mf = 0 E R. An element 0 6= f ∈ R iscalled a zerodivisor on M , if there is 0 6= m ∈ M such that f ∈ annR(m).A prime ideal P C R is called associated to M , if there is m ∈ M such thatannR(m) = P ; note that we hence have mR ∼= R/P as R-modules. Let assR(M)denote the set of prime ideals associated to M .b) Let I C R be an ideal. Then the prime ideals associated to I are definedas ass(I) := assR(R/I); note that annR(R/I) = I. Note moreover that ifP C R is a prime ideal, then we have annR(f + P ) = P , for f ∈ R \ P , henceass(P ) = assR(R/P ) = P.

(6.5) Proposition. Let R be Noetherian and let M 6= 0 be a finitely gener-ated R-module.a) Then assR(M) is a finite non-empty set, whose minimal elements are preciselythe prime ideals of R minimal over annR(M), and (

⋃P∈assR(M) P ) \ 0 are

precisely the zerodivisors on M .b) Let additionally R be a graded F -algebra, let M be a graded R-module.Then the elements of assR(M) are homogeneous ideals.

Proof. a) Let 0 6= m ∈ M such that annR(m) C R is maximal amongstannR(x) C R; 0 6= x ∈ M, and let f, g ∈ R such that fg ∈ annR(m) andg 6∈ annR(m). Since annR(m) ⊆ annR(mg) C R we conclude f ∈ annR(mg) =annR(m), thus annR(m) C R is a prime ideal, hence assR(M) 6= ∅. From thisthe assertion on zerodivisors also follows.

Let M ′ ≤ M be an R-submodule, let P ∈ assR(M) and let R/P ∼= M ′′ ≤ M .If M ′′ ∩M ′ 6= 0, then we have P ∈ assR(M ′); while if M ′′ ∩M ′ = 0, thenwe have R/P ∼= (M ′′ + M ′)/M ′ ≤ M/M ′, and thus P ∈ assR(M/M ′). Hencewe conclude assR(M) ⊆ assR(M ′) ∪ assR(M/M ′). Thus we pick P ′ ∈ assR(M)and let M ′ ∼= R/P ′, hence we have assR(M ′) = P ′ and if M/M ′ = 0 we aredone, otherwise we pick P ′′ ∈ assR(M/M ′) and proceed by iteration. Since M isNoetherian, the ascending chain of R-submodules of M thus obtained stabilises.

We have P ∈ assR(M) if and only if PR\P ∈ assRR\P (MR\P ); this is seen asfollows: Let firstly 0 6= m ∈ M such that P = annR(m). Hence we haveannRR\P (m1 ) = fu ∈ RR\P ;mfv = 0 for some v ∈ R \ P = annR(m)R\P =PR\P . Conversely, let 0 6= m

u ∈ MR\P such that PR\P = annRR\P (mu ) =

25

annRR\P (m1 ) = annR(m)R\P , hence we have annR(m) ⊆ ν−1(annR(m)R\P ) =ν−1(PR\P ) = P . Moreover, we may assume that 0 6= m ∈ M is chosen suchthat annR(m) is maximal amongst annR(mv); v ∈ R \P. Let f ∈ P , then wehave mf

1 = 0, hence fv ∈ annR(m) for some v ∈ R \ P , thus f ∈ annR(mv) =annR(m). Hence we have P = annR(m).

Using this, let P C R be a prime ideal minimal over annR(M). Hence PR\P CRR\P is a prime ideal minimal over annR(M)R\P CRR\P , and hence the uniqueprime ideal over annR(M)R\P . Since M is a finitely generated R-module wehave annRR\P (MR\P ) = annR(M)R\P , hence assRR\P (MR\P ) 6= ∅, and thusassRR\P (MR\P ) = PR\P , and hence P ∈ assR(M). Finally, for any P ∈assR(M) we have annR(M) ⊆ P anyway.

b) Let 0 6= m =∑ri=1mi ∈ M , where the mi ∈ Mdi are the homogeneous

components such that d1 < · · · < dr, and we show by induction on r ≥ 1 thatif annR(m) C R is a prime ideal, then annR(m) is homogeneous. Let r = 1,and let 0 6= f =

∑sj=1 fj ∈ annR(m), where the fj ∈ Rej are the homogeneous

components such that e1 < · · · < es. Hence from mf = m1f = 0 we obtainm1fj = 0, hence fj ∈ annR(m), for j ∈ 1, . . . , s. Let r ≥ 2, we show thatf1 ∈ annR(m), and then proceed by induction on s ≥ 1. We have m1f1 = 0, andthus annR(m) ⊆ annR(mf1) = annR(

∑ri=2mif1). If annR(m) = annR(mf1),

then annR(mf1) is a prime ideal, hence by induction annR(mf1) = annR(m) ishomogeneous anyway, and thus f1 ∈ annR(m). Otherwise, let g ∈ annR(mf1) \annR(m), hence f1g ∈ annR(m), thus f1 ∈ annR(m). ]

(6.6) Corollary. Let R be Noetherian and let I CR be an ideal.a) Then there are only finitely many prime ideals of R minimal over I.b) Let additionally R be a graded F -algebra and let I C R be homogeneous.Then the prime ideals of R minimal over I are homogeneous as well.

We introduce one of the most important notions of commutative algebra, havingutmost importance for algebraic geometry. The leading theme is the considera-tion of chains of prime ideals. This immediately leads to one of the bread-and-butter theorems of commutative algebra, Krull’s Principal Ideal Theorem.

(6.7) Definition. Let R 6= 0 be a commutative ring.a) For a prime ideal P C R let the height ht(P ) ∈ N0

.∪ ∞ of P be defined

as the supremum of the lengths r ∈ N0 of strictly ascending chains P0 ⊂ P1 ⊂· · · ⊂ Pr = P ⊂ R of prime ideals PiCR ending in P . The (Krull) dimensiondim(R) ∈ N0

.∪ ∞ of R is defined as dim(R) := supht(P );P C R prime.

Note that we hence have ht(P ) = dim(RR\P ).

For an ideal I C R let the dimension dim(I) := dim(R/I) and the heightht(I) := minht(P ); I ⊆ P CR prime. For an R-module M let the dimensiondim(M) := dim(R/annR(M)).

b) If R is not Noetherian, we easily find examples having infinite dimension:Let R := F [X1, X2, . . .] be the polynomial ring in countably infinitely many

26

indeterminates. Then ∑ij=1XjR C R; i ∈ N0, constitutes an infinite strictly

ascending chain of prime ideals. Note that even a Noetherian ring may haveinfinite dimension; an example given by Nagata (1962) is given in Exercise(10.28). Despite this, by Krull’s Principal Ideal Theorem(6.8), for R Noetherianand ICR we have ht(I) <∞. For a converse of Krull’s Principal Ideal Theoremsee Exercise (10.25).

Let R := F [X ] = F [X1, . . . , Xn], then by the same argument as above wefor i ∈ 0, . . . , n have ht(

∑ij=1XjR) ≥ i. Moreover, by Krull’s Principal Ideal

Theorem we have ht(∑ij=1XjR) ≤ i as well, hence equality holds. In particular,

we have ht(∑nj=1XjR) = n, and thus dim(R) ≥ n. Astonishingly enough, it is

rather difficult to prove the converse inequality, see (7.2).

(6.8) Theorem: Krull’s Principal Ideal Theorem, 1928.Let R be Noetherian, let r ∈ N and f1, . . . , fr ∈ R, and let P C R be a primeideal minimal over

∑ri=1 fiR. Then we have ht(P ) ≤ r.

Proof. We proceed by induction on r, and let r = 1 and f = f1. We show thatfor any prime ideal QCR such that Q ⊂ P we have ht(Q) = 0; this then impliesht(P ) ≤ 1. Note that since P is a prime ideal minimal over fR, we have f 6∈ Q.

By considering the localisation RR\P we may assume that R is a local ringhaving maximal ideal P . As P C R is the unique prime ideal containing fR,we have N (R/fR) = P/fR, and as P is finitely generated there is n ∈ N suchthat Pn ⊆ fR. Hence we have a filtration R/fR ⊇ P/fR ⊇ (P 2 + fR)/fR ⊇· · · ⊇ (Pn−1 + fR)/fR ⊇ (Pn+ fR)/fR = 0, whose subquotients are finitelygenerated R/P -vector spaces. By refining we obtain a finite filtration of R/fRwhose subquotients are simpleR/P -modules, hence are simpleR-modules. Thusthis is a finite R-module composition series of R/fR.

Let ν : R → RR\Q and let Q(i) := ν−1(QiR\Q) = g ∈ R; gu ∈ Qi for some u ∈R \ Q C R, for i ∈ N, be the i-th symbolic power of Q. As by the Jordan-Holder Theorem each finite chain of R-submodules of R/fR can be refined toa finite composition series, we conclude that the descending chain Q(1) + fR ⊇Q(2) + fR ⊇ · · · of R-submodules of R/fR stabilises; let m ∈ N such thatQ(m)+fR = Q(m+1)+fR. Hence for g ∈ Q(m) there are g′ ∈ R and g′′ ∈ Q(m+1)

such that g = g′′ + fg′. Thus fg′ ∈ Q(m), and as f ∈ R \Q we have g′ ∈ Q(m).Hence we conclude that Q(m) = Q(m+1)+fQ(m), and since f ∈ P the NakayamaLemma, see Exercise (10.21), implies Q(m) = Q(m+1). Thus localisation at R\Qagain yields QmR\Q = (Q(m))R\Q = (Q(m+1))R\Q = Qm+1

R\Q . As RR\Q is a localring with maximal ideal QR\Q, the Nakayama Lemma implies QmR\Q = 0.Hence we have QR\Q ⊆ N (RR\Q), thus QR\QCRR\Q is the unique prime ideal,and hence ht(Q) = dim(RR\Q) = 0.

Let r > 1. Again by considering the localisation RR\P we may assume that Ris a local ring having maximal ideal P , and as above there is n ∈ N such thatPn ⊆

∑ri=1 fiR. Let Q C R be maximal amongst the prime ideals of R being

27

properly contained in P . By construction we may assume that fr 6∈ Q. HenceP is a prime ideal minimal over Q+frR, thus the unique prime ideal containingQ + frR, and hence there is m ∈ N such that Pm ⊆ Q + frR. In particularthere are gi ∈ Q and g′i ∈ R such that fmi = gi + frg

′i, for i ∈ 1, . . . , r − 1.

We show that Q C R is a prime ideal minimal over I :=∑r−1i=1 giR C R. As I

is generated by r − 1 elements, we then by induction have ht(Q) ≤ r − 1, andthus ht(P ) ≤ r.Indeed, by the above construction there is k ∈ N such that P k ⊆ I+frR. HenceP/(frR+ I) ⊆ N (R/(frR+ I)), thus P/(frR+ I)CR/(frR+ I) is the uniqueprime ideal. Hence P/ICR/I is the unique prime ideal containing (frR+ I)/I,thus ht(P/I) ≤ 1, and hence ht(Q/I) = 0. ]

Finally we study the behaviour of prime ideals under integral ring extensions.This already indicates the way to the Noether Normalisation Theorem. Theconverse notion of going down is not discussed here. For further statementsconcerning the relation between prime ideals and integral ring extensions seeExercise (10.26).

(6.9) Theorem: Cohen-Seidenberg, 1946.Let R ⊆ S be an integral extension of commutative rings.a) Let P CR be a prime ideal. Then there is a prime ideal QCS (lying over),such that Q∩R = P . Moreover, if J C S is an ideal such that J ∩R ⊆ P , thenthe prime ideal QC S can be chosen (going up), such that J ⊆ Q.b) Let Q 6= Q′ C S be prime ideals such that Q ∩ R = Q′ ∩ R. Then we have(incomparability) Q 6⊂ Q′ 6⊂ Q.

Proof. a) By considering R/(J∩R) ⊆ S/J we may assume J = 0, and hencehave to show the existence of a prime ideal Q C S such that Q ∩ R = P . Byconsidering the localisations RR\P ⊆ SR\P , and noting that the ideal QCS weare looking for fulfils Q ∩ (R \ P ) ⊆ R ∩ (R \ P ) = ∅, we may assume that R isa local ring having maximal ideal P .

We consider PS E S, and assume PS = S. Let 1 =∑ri=1 pisi, for pi ∈ P and

si ∈ S, and let S′ ⊆ S be the R-subalgebra generated by s1, . . . , sr. Hence S′

is a finitely generated R-algebra and integral over over R, hence it is a finitelygenerated R-module. By construction we have PS′ = S′, and since P = rad(R)we by the Nakayama Lemma, see Exercise (10.21), conclude that S′ = 0,a contradiction. Thus we have PS C S, and hence there is a maximal idealPS ⊆ QC S. Since P ⊆ Q ∩R and P CR are maximal, we have P = Q ∩R.

b) Assume we have Q ⊂ Q′, and consider R/(Q∩R) ⊆ S/Q. As the elements ofS are zeroes of monic polynomials over R, the ring extension R/(Q∩R) ⊆ S/Qis integral as well. As Q C S is a prime ideal, we may assume that R ⊆ S isan integral extension of domains and Q′ ⊃ Q = 0 = Q ∩ R = Q′ ∩ R. Let0 6= s ∈ Q′, and let f =

∑di=1 fiX

i ∈ R[X] be monic such that f(s) = 0, where,

28

as S is a domain, we may assume that f0 6= 0. Hence we have f0 ∈ sS ∩ R ⊆Q′ ∩R = 0, a contradiction. ]

(6.10) Corollary. Let R ⊆ S be an integral extension, let J C S be an idealand I := J ∩ R C R. Then we have dim(J) = dim(I). In particular, we havedim(R) = dim(S).

Proof. Let I ⊆ P0 ⊂ · · · ⊂ Pr ⊂ R be a strictly ascending chain of prime idealsPi CR. By going up and lying over, there is a chain J ⊆ Q0 ⊆ · · · ⊆ Qr ⊂ S ofprime ideals Qi C S, such that Qi ∩R = Pi for i ∈ 0, . . . , r. Hence the latterchain is strictly ascending, and we have dim(I) ≤ dim(J).

Conversely, let J ⊆ Q0 ⊂ · · · ⊂ Qr ⊂ S be a strictly ascending chain of primeideals QiCS. Then by incomparability the chain I = J ∩R ⊆ (Q0∩R) ⊆ · · · ⊆(Qr∩R) ⊂ R of prime ideals Qi∩RCR, for i ∈ 0, . . . , r, is strictly ascending,and hence we have dim(I) ≥ dim(J). ]

7 Noether normalisation

We return to finitely generated commutative F -algebras, and prove a majorstructure theorem, a rather general version of the Noether Normalisation Theo-rem, which for infinite fields has been proved by Noether (1926), while the caseof finite fields has been dealt with by Zariski (1943), and where finally Nagata(1962) has given the refined version, actually involving only a single ideal.

(7.1) Proposition. Let 0 6= f ∈ R := F [X ] = F [X1, . . . , Xn] such thatdeg(f) > 0. Then there is Y := Y1, . . . , Yn−1 ⊆ R such that f,Y ⊆ R is al-gebraically independent and R is finite over the polynomial ring S := F [Y, f ] ⊆R. Moreover, for i ∈ 1, . . . , n− 1 we have:a) We may choose e ∈ N suitably such that Yi = Xi − (Xn)e

i

.b) If F is infinite, we may choose ai ∈ F suitably such that Yi = Xi − aiXn.c) If f is homogeneous, we may choose the Yi homogeneous.

Proof. As F (Y, f) ⊆ F (X ) is an algebraic field extension, we conclude n =trdeg(F (X )) = trdeg(F (Y, f)), and hence Y, f is algebraically independent.a) Let e ∈ N be greater than any αi ∈ N0, for i ∈ 1, . . . , n, occurring inany monomial Xα, for α ∈ Nn0 , occurring in f . Letting Yi = Xi − Xei

n , fori ∈ 1, . . . , n − 1, and expanding a monomial Xα ∈ F [Y][Xn] occurring in f

with respect to the indeterminate Xn, we obtain Xα = Xαnn ·∏n−1i=1 (Yi+Xei

n )αi =

Xαnn ·(

∏n−1i=1 Y

αii )+· · ·+Xαn+

∑n−1i=1 αie

i

n , where Xαn+∑n−1i=1 αie

i

n is the unique termoccurring with highest Xn-degree. By the choice of e, the numbers αi, for i ∈1, . . . , n, are the coefficients of the e-adic representation of αn +

∑n−1i=1 αie

i ∈N0, where αn is the 0-th coefficient. Hence for different monomials Xα thehighest Xn-degrees occurring are different. Thus f ∈ F [Y][Xn] is monic with

29

respect to the indeterminate Xn. Thus for g := f(X1, . . . , Xn−1, T ) − f(X ) ∈S[T ] we have g(Xn) = 0, thus Xn is integral over S. Hence R is a finitelygenerated S-algebra and integral over S.

b) Let f =∑dj=0 fj , where the fj ∈ F [X ] are homogeneous such that deg(fj) =

j, and let fd 6= 0, hence d = deg(f). Letting Yi = Xi − aiXn, for ai ∈F and i ∈ 1, . . . , n − 1, and expanding fd ∈ F [Y][Xn] with respect to theindeterminate Xn, the coefficient of Xd

n turns out to be fd(a1, . . . , an−1, 1) ∈F . As 0 6= fd ∈ F [X ] and F is infinite, there are a1, . . . , an−1 ∈ F suchthat fd(a1, . . . , an−1, 1) 6= 0. Thus f ∈ F [Y][Xn] is monic with respect to theindeterminate Xn, with Xn-degree d. Hence R is a finitely generated S-algebraand integral over S.

c) For i ∈ 1, . . . , n − 1 we choose Yi ∈ R+ homogeneous successively suchthat for Ii := fR +

∑i−1j=1 YjR CR we have ht(Ii) = i. Since R is a domain for

i = 1 by Krull’s Principal Ideal Theorem we have ht(I1) = 1. We proceed asfollows: Let P1, . . . , Ps C R be the prime ideals of R minimal over Ii. As thePk are homogeneous, we have Pk ⊆ R+. Assume that

⋃sk=1 Pk = R+. Then by

prime avoidance we have R+ = Pk for some k ∈ 1, . . . , s, hence R+ is a primeideal minimal over Ii, and thus by Krull’s Principal Ideal Theorem we haveht(R+) ≤ i. Since ht(R+) = n, this is a contradiction. Thus we may chooseYi ∈ R+ \

⋃sk=1 Pk homogeneous. By construction and Krull’s Principal Ideal

Theorem we have i ≤ ht(Ii+1) ≤ i+ 1. Assume ht(Ii+1) = i, and let QCR be aprime ideal minimal over Ii+1 such that ht(Q) = i. Since Ii ⊆ Q and ht(Ii) = i,we conclude that Q is a prime ideal minimal over Ii, hence Yi 6∈ Q, and sinceYi ∈ Ii+1 ⊆ Q this a contradiction.

Hence we have ht(In) = n, and as ht(R+) = n we conclude that R+ C R is aprime ideal minimal over In, and thus the unique prime ideal over In. As R+CRis finitely generated, R+/In C R/In is nilpotent, and hence R/In has a finitefiltration consisting of finitely generated R/R+-modules, and since R/R+

∼= Fwe conclude that dimF (R/In) < ∞. As S = F 〈Y, f〉 is a graded F -algebra aswell, we have In = S+R. As dimF (R/S+R) < ∞, by the graded NakayamaLemma, see Exercise (10.22), we conclude that R is finitely generated as anS-module, hence R is finite over S. ]

(7.2) Theorem. Let n ∈ N0. Then we have dim(F [X1, . . . , Xn]) = n.

Proof. We proceed by induction on n. As the case n = 0 is clear, we let n ≥ 1,let R := F [X1, . . . , Xn], and let 0 = P0 ⊂ · · · ⊂ Pr, for r ∈ N0, be a strictlyascending chain of prime ideals Pk C R. As we have dim(R) ≥ n, we have toshow that r ≤ n holds. Let 0 6= f ∈ P1 and let Y and S := F 〈Y, f〉 be as in(7.1). By incomparability we conclude that 0 = S∩P0 ⊂ S∩P1 ⊂ · · · ⊂ S∩Pris a strictly ascending chain of prime ideals of S, yielding the strictly ascendingchain of prime ideals (S ∩ P1) + fS ⊂ · · · ⊂ (S ∩ Pr) + fS of S/fS ∼= F [Y]. Asby induction we have dim(F [Y]) = n− 1, we conclude r − 1 ≤ n− 1. ]

30

(7.3) Corollary. Let R := F 〈f1, . . . , fn〉, for n ∈ N0, be a finitely gener-ated commutative F -algebra. Then we have dim(R) ≤ n. Moreover, we havedim(R) = n if and only if f1, . . . , fn ⊆ R is algebraically independent.

(7.4) Theorem: Noether Normalisation Theorem.Let R := F 〈f1, . . . , fr〉, for r ∈ N0, be a finitely generated commutative F -algebra, let n := dim(R), and let I1 ⊂ · · · ⊂ Is, for s ∈ N0, be a strictlyascending chain of ideals Ik C R such that n > n1 > · · · > ns ≥ 0, wherenk := dim(Ik). Then there is Y := Y1, . . . , Yn ⊆ R algebraically independentsuch that R is finite over the polynomial ring S := F [Y] ⊆ R and S ∩ Ik =∑ni=nk+1 YiS C S, for k ∈ 1, . . . , s. Moreover for i ∈ 1, . . . , n we have:

a) If F is infinite, we may choose the Yi as F -linear combinations of f1, . . . , fr.b) If the ideals Ik are homogeneous, we may choose the Yi homogeneous.

Proof. Let R ∼= F [X1, . . . , Xr]/I, where we assume I ⊂ I1 ⊂ · · · ⊂ Is CF [X1, . . . , Xr], hence dim(I) = dim(R) = n > n1. Thus we may assume thatR = F [X ] = F [X1, . . . , Xn] is polynomial and that s ≥ 1.

Moreover, it is sufficient to find Y := Y1, . . . , Yn ⊆ R such that R is finiteover S := F [Y] and

∑ni=nk+1 YiS ⊆ S ∩ Ik, for k ∈ 1, . . . , s; this is seen

as follows: As F (Y) ⊆ F (X ) is an algebraic field extension, we have n =trdeg(F (X )) = trdeg(F (Y)), and hence Y is algebraically independent. Wehave dim(S∩Ik) = dim(Ik) = nk = dim(

∑ni=nk+1 YiS), where

∑ni=nk+1 YiSCS

is a prime ideal, hence we have∑ni=nk+1 YiS = S ∩ Ik.

We construct the Yi ∈ R successively, using auxiliary elements Y ′i ∈ R, whereinitially we let Y ′i := Xi for i ∈ 1, . . . , n. Let for some i ∈ ns + 1, . . . , n,where initially i = n, the polynomial ring Si := F [Y ′1 , . . . , Y

′i , Yi+1, . . . , Yn] ⊆ R

be known, such that R is finite over Si and∑nj=maxnk+1,i+1 YjSi ⊆ Si ∩ Ik,

for k ∈ 1, . . . , s. We then introduce the element Yi and change the Y ′j , forj ∈ 1, . . . , i− 1, suitably, retaining the above conditions, and then go over toi− 1. If finally i = ns, then we let S = Si where Yj := Y ′j for j ∈ 1, . . . , ns.Let k be minimal such that nk < i. Assume we have F [Y ′1 , . . . , Y

′i ] ∩ Ik =

0. Since∑nj=i+1 YjSi ⊆ Si ∩ Ik, we conclude

∑nj=i+1 YjSi = Si ∩ Ik. As

dim(Si ∩ Ik) = dim(Ik) = nk < i = dim(∑nj=i+1 YjSi), this is a contradiction.

Hence let 0 6= Yi ∈ F [Y ′1 , . . . , Y′i ] ∩ Ik, which if Ik and the Y ′i are homogeneous

may be chosen homogeneous as well. Moreover, by (7.1) let Y ′′1 , . . . , Y ′′i−1 ⊆F [Y ′1 , . . . , Y

′i ] algebraically independent such that F [Y ′1 , . . . , Y

′i ] is finite over

F [Y ′′1 , . . . , Y′′i−1, Yi], where if Yi is homogeneous the Y ′′j may be chosen homo-

geneous as well, while if F is infinite the Y ′′j may be chosen as F -linear com-binations of Y ′1 , . . . , Y ′i . Thus R is finite over the polynomial ring Si−1 :=F [Y ′′1 , . . . , Y

′′i−1, Yi, Yi+1, . . . , Yn] ⊆ Si, and since Yi ∈ Ik we moreover have∑n

j=maxnk+1,i YjSi−1 ⊆ Si−1 ∩ Ik, for k ∈ 1, . . . , s. ]

(7.5) Corollary. Let R be a finitely generated commutative F -algebra, whichis a domain. Then we have dim(R) = trdeg(Quot(R)).

31

Proof. Let F [Y] = F [Y1, . . . , Yn] ∼= S ⊆ R be a Noether normalisation. SinceR is finite over S, we have dim(R) = n = trdeg(F (Y)) = trdeg(Quot(S)) =trdeg(Quot(R)). ]

(7.6) Definition. a) Let R be a finitely generated graded F -algebra, andlet f1, . . . , fn ⊆ R+ homogeneous and algebraically independent, such thatF [f1, . . . , fn] ⊆ R is finite. Then f1, . . . , fn ⊆ R is called a homogeneoussystem of parameters (hsop) of R. Note that hence n = dim(R), and thatby Noether normalisation homogeneous system of parameters always exist.

The set X1, X2, . . . , Xn ⊆ F [X ] = F [X1, . . . , Xn] is a homogeneous system ofparameters. The set X2

1 , X2, . . . , Xn ⊆ F [X ] is algebraically independent aswell, and we have F [X ] = (1 · S)

⊕(X1 · S), where S := F [X2

1 , X2, . . . , Xn] ⊆F [X ]. Hence the set X2

1 , X2, . . . , Xn ⊆ F [X ] also is a homogeneous systemof parameters. Hence the degrees of the elements of a homogeneous system ofparameters are in general not uniquely defined.

b) If R = S[V ]G is an invariant ring of a finite group G, then a homogeneous sys-tem of parameters f1, . . . , fn ⊆ S[V ]G is called a set of primary invariants.If moreover S[V ]G =

∑mj=1 gjS, for m ∈ N, where S := F [f1, . . . , fn] ⊆ S[V ]G

and the gj ∈ S[V ]G are homogeneous, then g1, . . . , gm ⊆ S[V ]G generatingthe S-module S[V ]G is called a set of secondary invariants.

In particular, if S[V ]G is a polynomial ring, then a set of basic invariants is a setof primary invariants, a set of secondary invariants being given by 1 ⊆ S[V ]G.

For further comments on Noether normalisation and homogeneous systems ofparameters see Exercises (10.29) and (10.30). As we are mainly interested infinitely generated graded F -algebras and their finitely generated graded mod-ules, we consider the question how the Krull dimension is related to the Hilbertseries associated to these algebras and modules.

(7.7) Theorem: Hilbert, Serre.Let R := F 〈f1, . . . , fr〉 be a finitely generated graded F -algebra, where thefi ∈ R are homogeneous such that di := deg(fi) > 0. Then the Hilbert series ofa finitely generated graded R-module M is of the form HM (T ) = f(T )∏r

i=1(1−Tdi ) ∈Q(T ), where f ∈ Z[T±1]. In particular, HM ∈ C((T )) converges in the pointedopen unit disc z ∈ C; 0 < |z| < 1 ⊆ C.

Proof. We proceed by induction on r, and let r = 0. Hence R = F , andthus M is a finitely generated F -vector space. Hence HM (T ) ∈ Z[T±1]. Letr > 0, and for d ∈ Z consider the exact sequence of F -vector spaces induced bymultiplication with fr ∈ R:

0 →M ′d := kerMd(·fr)→Md

·fr−→Md+dr → cokMd+dr(·fr) =: M ′′d+dr → 0.

32

As R is Noetherian, M ′ :=⊕

d∈ZM′d and M ′′ :=

⊕d∈ZM

′′d are finitely gener-

ated graded R-modules. As M ′ ·fr = 0 = M ′′ ·fr, these are finitely generatedF 〈f1, . . . , fr−1〉-modules, hence by induction HM ′ and HM ′′ have the assertedform. Moreover, the exact sequence yields T drHM ′−T drHM +HM −HM ′′ = 0,and thus HM = HM′′−T

drHM′1−Tdr also has the asserted form. ]

(7.8) Definition. Let R be a finitely generated graded F -algebra and let Mbe a finitely generated graded R-module. Then the order of M is defined asord(M) := −ν1(HM ) ∈ Z, and the degree is defined as deg(M) := deg1(HM ) =(

(1− T )ord(M) ·HM (T ))

(1) = limz→1−

((1− z)ord(M) ·HM (z)

)∈ Q.

Hence by (4.4), for F [X ] = F [X1, . . . , Xn] with degree vector δ = [d1, . . . , dn]we have ord(F [X ]) = n = dim(F [X ]) and deg(F [X ]) =

∏ni=1

1di

.

For another approach, using the growth of the coefficients of Hilbert series, seeExercise (10.31).

(7.9) Theorem. Let R be a finitely generated graded F -algebra and let M bea finitely generated graded R-module.a) If M ′ ≤ M as graded R-modules, or M ′ is an epimorphic image of M asgraded R-modules, then we have ord(M ′) ≤ ord(M).b) We have ord(R) = dim(R).c) We have ord(M) = ord(R/annR(M)) = dim(R/annR(M)) =: dim(M). Notethat annR(M)ER is homogeneous, and hence R/annR(M) is a finitely generatedgraded F -algebra.

Proof. a) For d ∈ Z we have dimF (M ′d) ≤ dimF (Md). Hence for z ∈ Rsuch that 0 < z < 1 we have 0 ≤ HM ′(z) ≤ HM (z) ∈ R. Thus we have0 ≤ limz→1−((1− z)ord(M) ·HM ′(z)) ≤ limz→1−((1− z)ord(M) ·HM (z)), wherethe latter limit exists in R>0. Hence (1 − T )ord(M) · HM ′(T ) does not have apole at T = 1.

b) Let S ⊆ R be a Noether normalisation. Hence we have ord(S) ≤ ord(R).Conversely, as R is a finitely generated graded S-module, and hence an epimor-phic image of a finitely generated free graded S-module N :=

⊕ri=1 fiS, where

di := deg(fi) ∈ Z. Thus we have HN = (∑ri=1 T

di) ·HS ∈ Q(T ). Hence we haveord(R) ≤ ord(N) = ord(S), and thus ord(R) = ord(S) = dim(S) = dim(R).

c) As M is a finitely generated R/annR(M)-module, and hence an epimor-phic image of a finitely generated free R/annR(M)-module, we have ord(M) ≤ord(R/annR(M)) = dim(R/annR(M)). Conversely, let P C R be a primeideal minimal over annR(M). Hence we have P ∈ assR(M) and there isR/P ∼= M ′ ≤ M . Hence we have dim(R/P ) = ord(R/P ) ≤ ord(M), andthus dim(R/annR(M)) ≤ ord(M). ]

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(7.10) Theorem. Let S be a finitely generated graded F -algebra, and letS ⊆ R be a finite extension of graded F -algebras, such that R is a domain.Then we have deg(R) = [Quot(R) : Quot(S)] · deg(S).

Proof. As R is a finitely generated graded S-module, there is a Quot(S)-basisf1, . . . , fr ⊆ R of Quot(R) consisting of homogeneous elements, where r :=[Quot(R) : Quot(S)] < ∞. Let M :=

⊕ri=1 fiS ⊆ R, and di := deg(fi). Hence

we have HM = (∑ri=1 T

di) ·HS ∈ Q(T ), where(∑r

i=1 Tdi)

(1) = r. Moreover,there is f ∈ R homogeneous such that R ⊆ f−1M =

⊕ri=1 fif

−1S ⊆ Quot(R),hence we have Hf−1M = T− deg(f) · HM . Thus ord(R) = dim(R) = dim(S) =ord(S) = ord(M) = ord(f−1M) as well as r · deg(S) = deg(M) = deg(f−1M).Moreover, since dimF (Md) ≤ dimF (Rd) ≤ dimF ((f−1M)d) for d ∈ Z, we havedeg(M) ≤ deg(R) ≤ deg(f−1M), hence deg(R) = r · deg(S). ]

(7.11) Corollary. Let G be a finite group acting faithfully on the F -vectorspace V . Then dim(S[V ]G) = dim(S[V ]) = dimF (V ) and deg(S[V ]G) = 1

|G| .

Note that for F ⊆ C, or for finite F such that char(F ) 6 | |G|, these assertionsalso follow from Molien’s Formula.

8 Cohen-Macaulay algebras

Having Noether normalisation at hand, we are led to ask whether the finitelygenerated module for the polynomial subring has particular properties, and inparticular whether it could be a free module, leading to the notion of Cohen-Macaulay algebras. For invariant rings of finite groups, in the non-modularcase this indeed turns out to be true, the main result being the Hochster-EagonTheorem.

We only treat a special class of Cohen-Macaulay rings here, as we only allowfor homogeneous regular sequences; their behaviour is similar to the case oflocal Cohen-Macaulay rings. We begin by looking at regular sequences, whichif of appropriate length turn out to be particularly nice homogeneous systemsof parameters.

(8.1) Definition. a) Let R be a finitely generated graded F -algebra, and letM 6= 0 be a finitely generated graded R-module. A homogeneous elementf ∈ R+ is called regular (non-zerodivisor) for M , if kerM (·f) = 0, where·f : M →M : m 7→ mf . A sequence [f1, . . . , fr] ⊆ R+ of homogeneous elementsis called regular for M , if fi is regular for M/(

∑i−1j=1Mfj), for i ∈ 1, . . . , r.

For the case M = R elements and sequences are called regular.

The depth depth(M) ∈ N0

.∪ ∞ of M is the maximal length of a regular

sequence in M , and M is called Cohen-Macaulay, if depth(M) = dim(M).

b) Let F [X ] = F [X1, . . . , Xn]. Since F [X ]/(∑i−1j=1XjF [X ]) ∼= F [Xi, . . . , Xn] is

a domain, for i ∈ 1, . . . , n, we conclude that the sequence [X1, . . . , Xn] ⊆ F [X ]

34

is regular. As dim(F [X ]) = n and as the depth is bounded above by thedimension anyway, as is shown below, we conclude that the polynomial ringF [X ] is Cohen-Macaulay.

An example of a finitely generated graded F -algebra not being Cohen-Macaulayis given in Exercise (10.33). The unmixedness property of Cohen-Macaulayalgebras was found by Macaulay for polynomial rings and by Cohen for regularlocal rings, and is the reason for the terminology used today:

(8.2) Proposition: Macaulay, 1916; Cohen, 1946.Let R be a finitely generated graded F -algebra, and let M 6= 0 be a finitelygenerated graded R-module. Then depth(M) ≤ mindim(P );P ∈ assR(M) ≤dim(M).

Hence if M is Cohen-Macaulay, then we have depth(M) = dim(P ) for all P ∈assR(M) (unmixedness), and in particular assR(M) consists precisely of theminimal prime ideals over annR(M).

Proof. We proceed by induction on r := depth(M), and may assume r ≥ 1.Let [f = f1, . . . , fr] ⊆ R+ be a maximal regular sequence for M . Thus we haver−1 = depth(M/Mf) ≤ mindim(R/Q);Q ∈ assR(M/Mf). We show that foreach P ∈ assR(M) there is Q ∈ assR(M/Mf) such that P ⊂ Q; then we haver − 1 ≤ mindim(R/Q);Q ∈ assR(M/Mf) < mindim(R/P );P ∈ assR(M).As f is regular on M we have f 6∈ P . Let Mf ≤ N := m ∈M ;mP ⊆Mf ≤M . Assume we have N = Mf . As P ∈ assR(M) we have 0 6= N ′ := m ∈M ;mP = 0 ≤ N = Mf . Hence for each n ∈ N ′ there is m ∈ M suchthat n = mf , and thus, as f is regular on M , we from 0 = nP = mfPconclude mP = 0, hence m ∈ N ′. Thus N ′ = N ′f , and by the gradedNakayama Lemma, see Exercise (10.22), we have N ′ = 0, a contradiction.Hence N > Mf and thus for 0 6= N/Mf ≤M/Mf we have (N/Mf)P = 0.Hence for Q ∈ assR(N/Mf) ⊆ assR(M/Mf) we have P ⊆ Q and f ∈ Q \ P . ]

(8.3) Proposition. Let R be a finitely generated graded F -algebra, and letM 6= 0 be a finitely generated graded R-module.a) If f ∈ R is homogeneous, then we have dim(M/Mf) ≥ dim(M)−1. If f ∈ Ris regular for M , then we have dim(M/Mf) = dim(M)− 1.b) Let M be Cohen-Macaulay. Then f ∈ R+ homogeneous is regular for M ifand only if dim(M/Mf) = dim(M)− 1.c) Each regular sequence [f1, . . . , fr] ⊆ R+, for r ≤ dim(R), can be extendedto a homogeneous system of parameters. In particular, a regular sequence oflength r = dim(R) is a homogeneous system of parameters.

Proof. a) Let g ∈ annR(M/Mf), and let m1, . . . ,ms generate M as anR-module. Hence by the Cayley-Hamilton Theorem there is h(T ) = T s +∑s−1i=0 hiT

i ∈ R[T ] monic such that hi ∈ fR and h(g) ∈ annR(M). Hence we

35

have gs ∈ annR(M) + fR, thus annR(M/Mf) ⊆√

annR(M) + fR. Hence thefirst assertion follows from Krull’s Principal Ideal Theorem.

We consider the exact sequence of R-modules 0 → kerM (·f) → M·f−→

M → cokM (·f) = M/Mf → 0. As we have kerM (·f) = 0, we for thecorresponding Hilbert series obtain −T deg(f)HM + HM − HM/Mf = 0, henceHM = HM/Mf

1−Tdeg(f) , and thus dim(M) = dim(M/Mf) + 1.

b) Let 0 6= f ∈ R+ homogeneous be not regular, hence f is a zerodivisor onM , and thus there is P ∈ assR(M) such that f ∈ P . By the above we haveannR(M/Mf) ⊆

√annR(M) + fR ⊆ P , hence using unmixedness we have

dim(M) = dim(P ) ≤ dim(M/Mf) ≤ dim(M).

c) Let R := R/(∑rj=1 fjR) and let : R→ R denote the natural epimorphism of

graded F -algebras. Moreover, by Noether normalisation let g1, . . . , gs ⊆ R+

and h1, . . . , ht ⊆ R homogeneous such that g1, . . . , gs ⊆ R+ is a homoge-neous system of parameters of R, and such that h1, . . . , ht ⊆ R generates Ras an F [g1, . . . , gs]-module. Note that we have s = dim(R)− r.Let S := F 〈f1, . . . , fr, g1, . . . , gs〉 ⊆ R. Hence we have S = F [g1, . . . , gs] ⊆ R.As h1, . . . , ht generates the S-module R, by the graded Nakayama Lemmawe conclude that h1, . . . , ht generates the F -vector space R/(

∑sj=1 gjR) ∼=

R/(∑ri=1 fiR +

∑sj=1 gjR). By the graded Nakayama Lemma again we con-

clude that h1, . . . , ht generates the S-module R. Hence S ⊆ R is finite,and thus we have dim(S) = dim(R) = r + s, As S is as an F -algebra gen-erated by r + s elements, we conclude that S is a polynomial ring, hencef1, . . . , fr, g1, . . . , gs ⊆ R is a homogeneous system of parameters of R. ]

(8.4) Theorem. Let R be a finitely generated graded F -algebra such thatn = dim(R) ∈ N0. Then the following assertions are equivalent:i) The F -algebra R is Cohen-Macaulay, i. e. there is a homogeneous system ofparameters f1, . . . , fn ⊆ R which is regular.ii) Each homogeneous system of parameters f1, . . . , fn ⊆ R is regular, inde-pendently of the ordering of the fi.iii) For each homogeneous system of parameters f1, . . . , fn ⊆ R, the F -algebraR is a finitely generated free graded F [f1, . . . , fn]-module.iv) There is a homogeneous system of parameters f1, . . . , fn ⊆ R, such thatR is a finitely generated free graded F [f1, . . . , fn]-module.

Proof. i) ⇒ ii): As R is a finitely generated F [f1, . . . , fn]-module, by thegraded Nakayama Lemma R/(

∑ni=1 fiR) is a finitely generated F -vector space,

hence we have dim(R/(∑ni=1 fiR)) = 0, see also Exercise (10.29). As R is

Cohen-Macaulay, the sequence [f1, . . . , fn] is regular.

ii) ⇒ iii): As R is a finitely generated F [f1, . . . , fn]-module, by the gradedNakayama Lemma we conclude dimF (R/(

∑ni=1 fiR)) <∞. Let g1, . . . , gm ⊆

R be homogeneous such that g1, . . . , gm ⊆ R/(∑ni=1 fiR) is an F -basis of

36

R/(∑ni=1 fiR). Hence g1, . . . , gm ⊆ R is a minimal generating set of the

F [f1, . . . , fn]-module R. Assume there are hj ∈ F [X ] = F [X1, . . . , Xn], forj ∈ 1, . . . ,m, such that [h1, . . . , hm] 6= [0, . . . , 0] and

∑mj=1 gjhj(f1, . . . , fn) =

0 ∈ R. Let α1 ∈ N0 be maximal such that Xα11 divides all the hj , and let

h′j := hjXα11∈ F [X ]. As f1 ∈ R is regular, we have

∑mj=1 gjh

′j(f1, . . . , fn) =

0 ∈ R. Hence we have∑mj=1 gjh

′j(0, f2, . . . , fn) = 0 ∈ R/f1R, where by

construction [h′1(0, X2, . . . , Xn), . . . , h′m(0, X2, . . . , Xn)] 6= [0, . . . , 0]. By itera-tion this finally yields

∑mj=1 gjηj = 0 ∈ R/(

∑ni=1 fiR), where ηj ∈ F and

[η1, . . . , ηm] 6= [0, . . . , 0], a contradiction.

iii) ⇒ iv): Trivial.

iv)⇒ i): Let g1, . . . , gm ⊆ R homogeneous such that as F [f1, . . . , fn]-moduleswe have R =

⊕mj=1(gj ·F [f1, . . . , fn]). As F [f1, . . . , fn] is a domain, the element

f1 ∈ R is regular, and R/f1R =⊕m

j=1(gj · F [f2, . . . , fn]). By iteration, thesequence [f1, . . . , fn] is regular. ]

(8.5) Remark. Let R be Cohen-Macaulay, let f1, . . . , fn ⊆ R be a homoge-neous system of parameters, where n = dim(R), and let g1, . . . , gm ⊆ R be aminimal homogeneous generating set of the F [f1, . . . , fn]-module R. Hence thecorresponding Hironaka decomposition of R as graded F [f1, . . . , fn]-modulesis given as R =

⊕mj=1(gj · F [f1, . . . , fn]). Then the Hilbert series of R is given

as

HR =

∑mj=1 T

deg(gj)∏ni=1(1− T deg(fi))

∈ Q(T ).

In particular, as dim(R) = n we have deg(R) = m∏ni=1 deg(fi)

.

Hence, if a homogeneous system of parameters f1, . . . , fn has been found,the Hilbert series can be used to find the degrees of the elements of a minimalhomogeneous generating set g1, . . . , gm ⊆ R of the F [f1, . . . , fn]-module R.

(8.6) Theorem. Let G be a finite group acting faithfully on the FG-module V ,where dimF (V ) = n. Let f1, . . . , fn ⊆ S[V ]G be a set of primary invariantssuch that di := deg(fi), and let g1, . . . , gm ⊆ S[V ]G be a minimal set ofsecondary invariants.a) Then |G| |

∏ni=1 di, and we have

∏ni=1 di ≤ m · |G|.

b) The invariant ring S[V ]G is Cohen-Macaulay, if and only if∏ni=1 di = m · |G|.

c) If∏ni=1 di = |G|, then we have S[V ]G = F [f1, . . . , fn], hence the invariant

ring S[V ]G is polynomial. Note that if conversely S[V ]G = F [f1, . . . , fn], thenwe have |G| = 1

deg(S[V ]G)=∏ni=1 di.

Proof. a) As both the ring extensions S := F [f1, . . . , fn] ⊆ S[V ]G ⊆ S[V ]are finite, the set f1, . . . , fn is a homogeneous system of parameters of S[V ].Let h1, . . . , hr ⊆ S[V ] be a minimal homogeneous S-module generating set ofS[V ]. As the polynomial ring S[V ] is Cohen-Macaulay, we have deg(S[V ]) =

r∏ni=1 di

. As deg(S[V ]) = 1 we conclude r =∏ni=1 di.

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We have Quot(S) = F (f1, . . . , fn). As the S-module S[V ] is free of rank r, theset h1, . . . , hr is a Quot(S)-linearly independent generating set of S(V ), hencewe have [S(V ) : Quot(S)] = r =

∏ni=1 di. As we have [S(V ) : S(V )G] = |G|, we

conclude [S(V )G : Quot(S)] =∏ni=1 di|G| . As g1, . . . , gm generates S(V )G as a

Quot(S)-vector space, we have m ≥∏ni=1 di|G| .

b) If we have m =∏ni=1 di|G| , then g1, . . . , gm is Quot(S)-linearly independent,

hence in particular S[V ]G is a free S-module. Conversely, if S[V ]G is Cohen-Macaulay, then we have 1

|G| = deg(S[V ]G) = m∏ni=1 di

.

c) We have [S(V )G : Quot(S)] =∏ni=1 di|G| = 1. Hence S ⊆ S[V ]G ⊆ Quot(S). As

the ring extension S ⊆ S[V ]G is integral and S is a unique factorisation domain,and hence integrally closed, see Exercise (10.4), we conclude S = S[V ]G. ]

(8.7) Theorem: Broer’s degree bound, 1997.Let G be a finite group and let S[V ]G be Cohen-Macaulay. Let f1, . . . , fn ⊆S[V ]G, where n = dimF (V ), be a set of primary invariants such that di :=deg(fi), and let g1, . . . , gm ⊆ S[V ]G be a minimal set of secondary invariantssuch that ej := deg(gj). Then we have ej ≤

∑ni=1(di−1), for all j ∈ 1, . . . ,m.

Proof. Let S := F [f1, . . . , fn] ⊆ S[V ]G ⊆ S[V ], and let h1, . . . , hr ⊆ S[V ]be a minimal homogeneous S-module generating set of S[V ] such that ck :=deg(hk), where we assume c1 ≤ · · · ≤ cr. We have S[V ] =

⊕rk=1 hkS as graded

S-modules. Using Hilbert series we conclude HS[V ] = 1(1−T )n =

∑rk=1 T

ck∏ni=1(1−Tdi ) .

Thus we have∑rk=1 T

ck =∏ni=1

1−Tdi1−T , and hence cr =

∑ni=1(di−1). Moreover,

the elements ψ ∈ HomS(S[V ], S) are determined by [ψ(h1), . . . , ψ(hr)] ⊆ S.Thus we conclude that HomS(S[V ], S)d = 0 for d < −cr = −

∑ni=1(di − 1).

We have S[V ]G =⊕m

j=1 gjS as graded S-modules, where we may assume thate1 ≤ · · · ≤ em. Hence by the same argument as above there is 0 6= ϕ ∈HomS(S[V ]G, S)−em . Using the transfer map TrG ∈ HomS(S[V ], S[V ]G)0, weconsider the S-homomorphism ϕ TrG ∈ HomS(S[V ], S)−em . We may assumethat V is faithful. Hence 0 6= TrG : S(V ) → S(V )G is surjective, and let 0 6=f, g ∈ S[V ] such that TrG( fg ) = 1. As the finite field extension Quot(S) ⊆ S(V )is generated by h1, . . . , hr ⊆ S[V ], we may assume that g ∈ S. Hence wehave 0 6= TrG(f) = g ∈ S. Let h ∈ S[V ]G such that ϕ(h) 6= 0. Hence wehave ϕ(TrG(fh)) = ϕ(TrG(f) · h) = ϕ(g · h) = g · ϕ(h) 6= 0. Thus we conclude−em ≥ −

∑ni=1(di − 1). ]

(8.8) Theorem: Hochster-Eagon, 1971.Let F be a field, let G be a finite group and let H ≤ G such that char(F ) 6 |[G : H]. If the invariant ring S[V ]H is Cohen-Macaulay, then the invariant ringS[V ]G is Cohen-Macaulay as well. In particular, if char(F ) 6 | |G|, then S[V ]G

is Cohen-Macaulay.

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Proof. Let f1, . . . , fn ⊆ S[V ]G be a set of primary invariants, where n =dimF (V ). Hence we have S := F [f1, . . . , fn] ⊆ S[V ]G ⊆ S[V ]H ⊆ S[V ]. As bothring extensions S ⊆ [V ]G ⊆ S[V ] are finite, the ring S[V ] is a finitely generatedS-module as well. As S is Noetherian, the S-submodule S[V ]H ⊆ S[V ] also isa finitely generated S-module, hence f1, . . . , fn is a set of primary invariantsof S[V ]H as well. Thus the S-module S[V ]H is free.

The relative Reynolds operator RGH : S[V ]H → S[V ]G in particular is a pro-jection of graded S-modules. Hence S[V ]G is a direct summand of the finitelygenerated free S-module S[V ]H , thus is a finitely generated projective gradedS-module, and hence by Exercise (10.22) is a finitely generated free S-module.Thus S[V ]G is Cohen-Macaulay. ]

(8.9) Example. Let F be a field such that char(F ) 6= 2 and let F [X ]An be theinvariant ring of the alternating group An ≤ Sn, see Exercise (10.17). Hence wehave F [X ]An = (1 · F [X ]Sn)

⊕(∆n · F [X ]Sn), where F [X ]Sn ∼= F [en,1, . . . , en,n]

is a polynomial ring, and ∆n ∈ F [X ] is the discriminant polynomial, see Ex-ercise (10.16), note that ∆2

n ∈ F [X ]Sn . Hence the set of elementary symmet-ric polynomials en,1, . . . , en,n ⊆ F [X ]An is a set of primary invariants, and1,∆n ⊆ F [X ]An is a minimal set of secondary invariants, hence the abovedecomposition of F [X ]An is the corresponding Hironaka decomposition.

Further examples are given in Exercises (10.36), (10.37) and (10.38). Moreover,in Section 9 we present an elaborated classical example, the invariants of theicosahedral group.

(8.10) Remark. a) Under the assumptions of (8.6), it suffices to assume thatthe homogeneous set f1, . . . , fn ⊆ S[V ]G is algebraically independent; it thenfollows by [4, Thm.3.7.5] that it already is a set of primary invariants.b) Let G ≤ Sn, and let char(F ) 6 | |G|. As en,1, . . . , en,n ⊆ F [X ]Sn ⊆ F [X ]G ⊆F [X ] = F [X1, . . . , Xn] is a set of primary invariants of F [X ]G, Broer’s degreebound yields

∑ni=1(deg(en,i) − 1) =

∑n−1i=0 i =

(n2

). This coincides with the

degree bound found by Gobel’s Theorem.c) In the modular case, invariant rings in general are not Cohen-Macaulay,see Exercises (10.34) and (10.35), where the latter invariant ring actually is aunique factorisation domain (Bertin, 1973). The structure of invariant rings inthe modular case turns out to be substantially more complicated than in thenon-modular case, and currently is in the focus of intensive study.

9 Invariant theory live: the icosahedral group

We present an elaborated classical example, the invariants of the icosahedralgroup, due to Klein (1884) and Molien (1897). This shows invariant theory atwork, and in particular how geometric features are related to invariant theory.The computations carried out below can easily be reproduced using either of

39

Table 2: Character table of A5.

order 1 2 3 5 5′

power2 1 2 3 5′ 5# 1 15 20 12 12χ1 1 1 1 1 1χ2 3 −1 . ζ ζ ′

χ3 3 −1 . ζ ′ ζχ4 4 . 1 −1 −1χ5 5 1 −1 . .

the computer algebra systems GAP or MAGMA; we include some GAP code atthe end.

Let I ⊆ R1×3 be the regular icosahedron, one of the platonic solids. The facesof I consist of regular triangles, where at each vertex 5 triangles meet. Letf, e, v ∈ N be the number of faces, edges, and vertices of I, respectively. Henceby Euler’s Polyhedron Theorem we have f − e + v = 2. Since we have e = 3f

2

and v = 3f5 , we conclude f = 20 and e = 30 as well as v = 12.

Let G := π ∈ O3(R); Iπ = I ≤ O3(R) be the symmetry group of I, where weassume I ⊆ R1×3 to be centred at the origin, and where O3(R) is the isometrygroup of the Euclidean space R3. Let H = G∩SO3(R) be the group of rotationalsymmetries of I, where SO3(R) := π ∈ O3(R); det(π) = 1 ≤ O3(R).

By the regularity of I, the group H acts transitively on the vertices of I, wherethe corresponding point stabilisers have order 5. Hence we have |H| = 60.The rotational axes of the elements of H are given by the lines joining oppositevertices, midpoints of opposite edges, and midpoints of opposite faces of I. Thisyields 12

2 · 4 elements of order 5, and 302 elements of order 2, as well as 20

2 · 2elements of order 3, respectively, accounting for all the non-trivial elements ofH. Using basic group theory it straightforwardly follows that H ∼= A5

Moreover, as for the inversion σ ∈ O3(R) with respect to the origin we haveσ ∈ G \H. As σ ∈ Z(O3(R)), we have G = H × 〈σ〉 ∼= A5 × C2, in particular|G| = 120. Note that the eigenvalues of σ are −1,−1,−1, hence σ is nota reflection. As the elements of H are rotations, its elements of order 2 haveeigenvalues 1,−1,−1, hence are not reflections either. As H is generated byits elements of order 2, we conclude that G is generated by the set of reflectionsπσ ∈ G; 1 6= π ∈ H;π2 = 1. Thus G is a reflection group; actually it isthe exceptional irreducible pseudoreflection group G23 in the Shephard-Toddclassification, having Dynkin type H3.

The ordinary character table of H ∼= A5 is given in Table 9, where ζ5 :=exp 2π

√−1

5 ∈ C is a primitive 5-th root of unity, as well as ζ := ζ5 + ζ45 ∈ R ⊆ C

40

and ζ ′ := ζ25 + ζ3

5 = −1 − ζ ∈ R. Moreover, the conjugacy classes are denotedby the corresponding element orders, and the second power map as well as thecardinality of the conjugacy classes are also given.

Hence R3 is an absolutely irreducible RH-module, affording the character χ2 ∈ZIrrC(H), say, and by Molien’s Formula we findHS[R3]H = 1−T 2−T 3+T 6+T 7−T 9

(1−T 2)2·(1−T 3)·(1−T 5) .By the Hochster-Eagon Theorem the invariant ring HS[R3]H is Cohen-Macaulay,and by the Shephard-Todd-Chevalley Theorem it is not polynomial. TakingTheorem (8.6) into account we guess the following form of the Hilbert series:

HS[R3]H =1 + T 15

(1− T 2) · (1− T 6) · (1− T 10)∈ Q(T ).

Hence we conjecture that there are primary invariants f1, . . . , f3 such thatdeg(f1) = 2, deg(f2) = 6 and deg(f3) = 10, and secondary invariants g1, g2such that g1 := 1 and deg(g2) = 15, such that the corresponding Hironakadecomposition of the invariant ring is S[R3]H =

⊕2j=1(gj · R[f1, . . . , f3]).

As G is a pseudoreflection group, the invariant ring S[R3]G is polynomial. Us-ing Molien’s Formula we find HS[R3]G = 1

(1−T 2)·(1−T 6)·(1−T 10) , leading to theconjecture that the polynomial degrees of G are [2, 6, 10], and that we moreoverhave S[R3]G = R[f1, . . . , f3] ⊆ S[R3]H .

Let H := 〈α, β, γ〉, where α2 = 1, β3 = 1, γ5 = 1 and αβ = γ. Moreover, letF := Q(ζ) ⊆ R be the real number field generated by ζ ∈ R, let V := F 3 andlet DV : H → GL3(F ) ≤ GL3(R) be given as

DV : α 7→

. 1 .1 . .. . −1

, β 7→

. . 1ζ 1 −1−1 . −1

, γ 7→

ζ 1 −1. . 11 . 1

.Hence we indeed have χV = χ2 ∈ ZIrrC(H). Moreover, as H acts by isometriesof the Euclidean space R3, there is an H-invariant scalar product on V . As Hacts absolutely irreducibly on V , it is uniquely defined up to scalars, and itsmatrix turns out to be given as

Φ :=

1 ζ′

2 − 12

ζ′

2 1 12

− 12

12 1

∈ F 3×3.

Let S[V ] ∼= F [X ] = F [X1, . . . , X3], let X := [X1, . . . , X3] ∈ F [X ]3, and let·D denote the action of D ∈ GL3(F ) on F [X ]. For a column vector w ∈F 3×1 we then have (X · w)D = X · Dtrw. As the bilinear form describedby Φ is H-invariant, for D = DV (π), where π ∈ H, we have DΦDtr = Φ,hence D−trΦ−1D−1 = Φ−1, and thus DtrΦ−1D = Φ−1 as well. From thiswe obtain (XΦ−1Xtr)D = (XΦ−1)D · (XD)tr = (X · DtrΦ−1) · (XDtr)tr =XDtrΦ−1DXtr = XΦ−1Xtr. Thus f1 := XΦ−1Xtr ∈ S[V ]H is homogeneoussuch that deg(f1) = 2.

41

The group H permutes the 6 lines joining opposite vertices of I transitively. Avector 0 6= v1 ∈ V on one of these lines is found as an eigenvector of γ ∈ Hwith respect to the eigenvalue 1. Note that γ has order 5, and as γ is a rotationthe corresponding eigenspace has F -dimension 1. It turns out that the H-orbit±v1, . . . ,±v6 ⊆ V of v1 has length 12. We choose v1, . . . , v6 ⊆ V , one fromeach of 2-element sets ±vi, and let f2 :=

∏6k=1 vk ∈ S[V ] homogeneous such

that deg(f2) = 6. As H permutes ±v1, . . . ,±v6, we conclude that 〈f2〉F ∈S[V ]6 is a 1-dimensional FH-submodule. As H is a perfect group, we concludethat f2 ∈ S[V ]H .

Analogously, the group H permutes the 10 lines joining the midpoints of oppo-site faces of I transitively. We consider the eigenspace of β ∈ H with respect tothe eigenvalue 1, which again has F -dimension 1, and where β has order 3. Thisleads to an H-orbit of length 20, and as above we get f3 ∈ S[V ]H homogeneoussuch that deg(f3) = 10.

By the Jacobian Criterion we find that det J(f1, . . . , f3) 6= 0 ∈ S[V ], hencef1, . . . , f3 ⊆ S[V ]H is algebraically independent. To prove that f1, . . . , f3 isa set of primary invariants of S[V ]H , we argue as follows: As σ ∈ G\H has eigen-values −1,−1,−1 and the deg(fi) are even, we have f1, . . . , f3 ⊆ S[V ]G.Since we have HS[V ]G = 1

(1−T 2)·(1−T 6)·(1−T 10) this shows S[V ]G = F [f1, . . . , f3],and hence f1, . . . , f3 ⊆ S[V ]G is a set of basic invariants of S[V ]G. Hence weconclude that f1, . . . , f3 ⊆ S[V ]H is a set of primary invariants of S[V ]H .

As HS[V ]H = (1+T 15)·HS[V ]G , we are left to find a secondary invariant of S[V ]H

of degree 15. As H is a perfect group, we conclude that detV is the trivial H-representation, hence by Exercise (10.12) we have 0 6= g2 := det J(f1, . . . , f3) ∈S[V ]H homogeneous such that deg(g2) = 15. As the deg(fi) and deg(g1) = 0are even, while deg(g2) is odd, we conclude that g1, g2 is F [f1, . . . , f3]-linearlyindependent, and hence we have S[V ]H =

⊕2j=1(gj · F [f1, . . . , f3]).

Note that detV (σ) = −1, hence detV is not the trivial G-representation, andthus the assumption of Exercise (10.12) is not fulfilled. Indeed, we have alreadyproved that det J(f1, . . . , f3) ∈ S[V ]H \S[V ]G. Moreover, note that analogouslyto f2, f3 ∈ S[V ]H we can find a homogeneous H-invariant of degree 15 by usingsuitable vectors lying on the lines joining the midpoints of opposite edges ofI. As dimF (S[V ]H)15 = 1, the invariant thus found is a scalar multiple ofg2 ∈ S[V ]H . ]

tbl:=CharacterTable("A5");hs:=MolienSeries(tbl,Irr(tbl)[2]);# ( 1-z^2-z^3+z^6+z^7-z^9 ) / ( (1-z^5)*(1-z^3)*(1-z^2)^2 )MolienSeriesWithGivenDenominator(hs,[2,6,10]);# ( 1+z^15 ) / ( (1-z^10)*(1-z^6)*(1-z^2) )

z:=E(5)+E(5)^4;

42

rep:= # representation of A5 over Z[z], on (2,3,5)-triple[ [ [ 0, 1, 0 ], [ 1, 0, 0 ], [ 0, 0,-1 ] ],[ [ 0, 0, 1 ], [ z, 1,-1 ], [-1, 0,-1 ] ],[ [ z, 1,-1 ], [ 0, 0, 1 ], [ 1, 0, 1 ] ] ];

h:=Group(rep);

polring:=PolynomialRing(Cyclotomics,["X_1","X_2","X_3"]);indets:=IndeterminatesOfPolynomialRing(polring);x1:=indets[1];x2:=indets[2];x3:=indets[3];x:=[x1,x2,x3];

f:= # the A5-invariant scalar product[ [ 1, 1/2*(-1-z), -1/2 ],[ 1/2*(-1-z), 1, 1/2 ],[ -1/2, 1/2, 1 ] ];

f1:=x*f^(-1)*x; # invariant of degree 2

v:=NullspaceMat(rep[3]-rep[3]^0)[1];vorb:=Orbit(h,v);PermList(List(vorb,x->Position(vorb,-x)));# (1,12)(2,11)(3,10)(4,9)(5,6)(7,8)vvecs:=vorb[1,2,3,4,5,7];# [ [ -E(5)-2*E(5)^2-2*E(5)^3-E(5)^4,# -E(5)-2*E(5)^2-2*E(5)^3-E(5)^4, 1 ],# [ -E(5)-2*E(5)^2-2*E(5)^3-E(5)^4,# -E(5)-2*E(5)^2-2*E(5)^3-E(5)^4, -1 ],# [ E(5)+E(5)^4, -E(5)-2*E(5)^2-2*E(5)^3-E(5)^4, -1 ],# [ -E(5)-2*E(5)^2-2*E(5)^3-E(5)^4, E(5)+E(5)^4, 1 ],# [ -E(5)-E(5)^4, E(5)+E(5)^4, 1 ],# [ -E(5)-E(5)^4, E(5)+E(5)^4, -E(5)+E(5)^2+E(5)^3-E(5)^4 ] ]f2:=Product(List(vvecs,i->x*i)); # invariant of degree 6

w:=NullspaceMat(rep[2]-rep[2]^0)[1];worb:=Orbit(h,w);PermList(List(worb,x->Position(worb,-x)));# (1,20)(2,19)(3,17)(4,14)(5,18)(6,15)(7,16)(8,11)(9,12)(10,13)wvecs:=worb[1,2,3,4,5,6,7,8,9,10];f3:=Product(List(wvecs,i->x*i)); # invariant of degree 10

prim:=[f1,f2,f3];jac:=List(prim,p->List(x,i->Derivative(p,i)));g2:=DeterminantMat(jac); # invariant of degree 15

43

10 Exercises

(10.1) Exercise: Tensor products.Let V1 and V2 be F -vector spaces.a) If a tensor product⊗ : V1×V2 → V exists, show that V is uniquely determinedup to isomorphism of F -vector spaces.b) Let nk := dimF (Vk) ∈ N0 and let bk,1, . . . , bk,nk ⊆ Vk be F -bases of Vk.Show that a tensor product ⊗ : V1 × V2 → V =: V1 ⊗F V2 exists, and thatb1,i ⊗ b2,j ; i ∈ 1, . . . , n1, j ∈ 1, . . . , n2 is an F -basis of V1 ⊗F V2.c) Let V3 be an F -vector space such that dimF (V3) ∈ N0. Show that V1⊗F V2

∼=V2 ⊗F V1 and (V1 ⊗F V2)⊗F V3

∼= V1 ⊗F (V2 ⊗F V3) as F -vector spaces.

(10.2) Exercise: Hilbert series of polynomial rings.Let F [X ] := F [X1, . . . , Xn], for n ∈ N0.a) For the standard grading of F [X ] show that dimF (F [X ]d) =

(n+d−1

d

), for

d ∈ N0, and conclude that HF [X ] = 1(1−T )n ∈ Q(T ) ⊆ Q((T )).

b) For the grading of F [X ] given by the degree vector δ = [d1, . . . , dn] ⊆ N showthat Hδ

F [X ] =∏ni=1

11−Tdi ∈ Q(T ) ⊆ Q((T )).

(10.3) Exercise: Noetherian rings and modules.Let R be a commutative ring and let M be an R-module.a) Let N ≤M be an R-submodule. Show that if M is Noetherian, then so areN and M/N . Show that if conversely both N and M/N are Noetherian, thenso is M .b) Show that M is Noetherian if and only if each submodule of M is finitelygenerated. Conclude that if R is Noetherian, then M is Noetherian if and onlyif M is finitely generated.

Proof. See [1, Ch.1.2], [13, Ch.2.1] or [10, Ch.6]. ]

(10.4) Exercise: Integral ring extensions.Let R ⊆ S be an extension of commutative rings.a) Show that an element s ∈ S is integral over R, if and only if there is anR-subalgebra of S containing s, which is finitely generated as an R-module.b) Show that the ring extension R ⊆ S is finite, if and only if S is a finitelygenerated R-module.c) Show that the integral closure R

S:= s ∈ S; s integral over R ⊆ S of R in

S is a subring of S, and that RSS

= RS

holds.d) Let R be a unique factorisation domain. Show that R is integrally closed.

Proof. See [10, Ch.9]. ]

44

(10.5) Exercise: Finite generation.a) Give an example of an F -subalgebra of the polynomial ring F [X,Y ] whichis not finitely generated.b) Show that the polynomial ring F [X1, X2, . . .] in countably many indetermi-nates is not Noetherian.

(10.6) Exercise: Invariant rings.a) Let G be a group, let H E G be a normal subgroup, and let V be an FG-module. Show that S[V ]G = (S[V ]H)G/H holds.b) Let G and H be groups, and let V and W be an FG-module and an FH-module, respectively. Show that V ⊕W becomes an F [G×H]-module, and thatS[V ⊕W ] ∼= S[V ]⊗F S[W ] and S[V ⊕W ]G×H ∼= S[V ]G ⊗F S[W ]H holds.c) Let G be a finite group and let V be an FG-module. For d ∈ N0 show thatS[V ]Gd 6= 0 only if |DV (G) ∩ Z(GL(V ))| divides d.

Proof. See [14, Prop.1.5.1, Prop.1.5.2, Prop. 2.2.4]. ]

(10.7) Exercise: Cyclic groups.For k ∈ N such that char(F ) 6 | k let ζk ∈ F be a primitive k-th root of unity.Let k1, . . . , kn ⊆ N be pairwise relatively prime such that char(F ) 6 | ki, letk :=

∏ni=1 ki, let G := 〈δ〉 ∼= Ck be the cyclic group of order k, and let

DV : δ 7→ diag[ζki ; i ∈ 1, . . . , n] ∈ GLn(F ).

Show that S[V ]G ∼= F [Xk11 , . . . , Xkn

n ].

Proof. See [14, Ex.1.5.1]. ]

(10.8) Exercise: Generators of invariant rings.Let G = 〈δ〉 ∼= C3 be the cyclic group of order 3, let F be a field such thatchar(F ) 6= 3, and let

DV : G→ GL2(F ) : δ 7→[

. 1−1 −1

].

Compute a minimal F -algebra generating set of S[V ]G, and show that Noether’sdegree bound is attained in this case.

Proof. See [13, Ex.2.3.1]. ]

(10.9) Exercise: Noether’s degree bound.Let G = 〈δ〉 ∼= C2 be the cyclic group of order 2, let F be a field such thatchar(F ) = 2, and let

DW : G→ GL2(F ) : δ 7→[. 11 .

].

45

For the following FG-modules V determine the Hilbert ideal IG[V ] C S[V ],decide whether Benson’s Lemma holds for S[V ]+ C S[V ], compute a minimalF -algebra generating set of S[V ]G, and decide whether Noether’s degree boundholds:a) V := W as FG-modules.b) V := W ⊕W as FG-modules.c) V := W ⊕W ⊕W as FG-modules.

Proof. See [4, Ex.3.5.7], [4, Rem.3.8.7] and [13, Ex.2.3.2]. ]

(10.10) Exercise: Newton identities.Let F be a field and let F [X ] := F [X1, . . . , Xn].a) For k ∈ N let pn,k :=

∑ni=1X

ki ∈ F [X ] be the power sums, let en,1, . . . , en,n ∈

F [X ] be the elementary symmetric polynomials, where deg(en,i) = i, and leten,0 := 1 ∈ F [X ]. Show that ken,k =

∑ki=1(−1)i−1pn,ien,k−i, for k ∈ 1, . . . , n.

b) Let char(F ) > n. Determine all solutions [x1, . . . , xn] ∈ F 1×n of the systemof n− 1 equations

∑ni=1 xi = 0,

∑ni=1 x

2i = 0, . . . ,

∑ni=1 x

n−1i = 0.

Proof. a) See [13, p.81]. ]

(10.11) Exercise: Molien’s formula.For k ∈ N let G = 〈δ〉 ∼= Ck be the cyclic group of order k, let ζk ∈ C be a k-thprimitive root of unity, and let

DV : G→ GL2(C) : δ 7→[ζk .. ζ−1

k

].

a) Show that S[V ]G ∼=⊕k−1

i=0 ((X1X2)i ·F [Xk1 , X

k2 ]). Conclude that the Hilbert

series of S[V ]G is given as HS[V ]G = 1−T 2k

(1−T 2)·(1−Tk)2 ∈ C((T )).

b) Prove the identity 1k ·∑k=1i=0

1(1−ζikT )·(1−ζ−ik T )

= 1(1−Tk)2 ·

∑k−1i=0 T

2i ∈ C((T )).

c) Evaluate the sum∑k−1i=0

1|1−ζik|2

∈ C.

Proof. See [13, Ex.3.1.1, Ex.3.1.2] and [4, Ex.5.6.1]. ]

(10.12) Exercise: Jacobian and Hessian determinant.Let G be a group, let V be an FG-module and n = dimF (V ). Moreover, letdetV : G → F : π 7→ detV (π) denote the corresponding determinant repre-sentation.a) For f1, . . . , fn ⊆ S[V ]G let J(f1, . . . , fn) := [ ∂fi∂Xj

]i,j=1,...,n ∈ F [X ]n×n bethe corresponding Jacobian matrix. Show that if detV is the trivial represen-tation, then we have det J(f1, . . . , fn) ∈ S[V ]G.

46

b) For f ∈ S[V ]G let H(f) := J([ ∂f∂Xi ]i=1,...,n) ∈ F [X ]n×n denote the corre-sponding Hessian matrix. Show that if (detV )2 is the trivial representation,then we have detH(f) ∈ S[V ]G.

(10.13) Exercise: Jacobian criterion.Let F be a field such that char(F ) = 0 and let F [X ] := F [X1, . . . , Xn]. Letpn,k :=

∑ni=1X

ki ∈ F [X ], for k ∈ N, be the power sums, and let en,1, . . . , en,n ∈

F [X ] be the elementary symmetric polynomials, where deg(en,i) = i. Show thatpn,1, . . . , pn,n and en,1, . . . , en,n are algebraically independent.

Proof. See [7, Exc.3.10]. ]

(10.14) Exercise: Modular pseudoreflection groups.Let p be a prime, let

G :=

1 . a+ b b. 1 b b+ c. . 1 .. . . 1

∈ GL4(Fp); a, b, c ∈ Fp

≤ GL4(Fp),

and let V := F1×4p be the natural FG-module.

a) Show that G is generated by pseudoreflections with respect to V , and that|G| = p3.b) Show that S[V ]G is not a polynomial ring.

Proof. See [4, Ex.3.7.7]. ]

(10.15) Exercise: Elementary symmetric polynomials.Let F be a field and F [X ] := F [X1, . . . , Xn], and let Sn act on F [X ] by per-muting the indeterminates.a) Let the monomials Xα ∈ F [X ], for α ∈ N

n0 , be totally ordered lexico-

graphically by letting X1 > · · · > Xn, and let Xα be the leading monomialof f ∈ F [X ]Sn , for α = [α1, . . . , αn] ∈ Nn0 . Show that α1 ≥ · · · ≥ αn, and that∏ni=1 e

αi−αi−1n,i (X ) ∈ F [X ]Sn , where α0 := 0, has leading monomial Xα.

b) Give an algorithm to write a symmetric polynomial as a polynomial in theelementary symmetric polynomials en,1, . . . , en,n ⊆ F [X ]Sn .

Proof. See [4, Thm.3.10.1]. ]

(10.16) Exercise: Elementary symmetric polynomials.Let F be a field and F [X ] := F [X1, . . . , Xn]. Let ∆n :=

∏1≤i<j≤n(Xi −Xj) ∈

F [X ] be the discriminant polynomial, and let pn,k :=∑ni=1X

ki ∈ F [X ], for

k ∈ N, be the power sums.

Show that ∆2n and pn,k are symmetric, and write ∆2

3, as well as pn,2, pn,3 andpn,4 as polynomials in the elementary symmetric polynomials en,1, . . . , en,n.

47

(10.17) Exercise: Alternating polynomials.Let F be a field such that char(F ) 6= 2. An element f ∈ F [X ] := F [X1, . . . , Xn]is called alternating, if fπ = sgn(π) · f for all π ∈ Sn.a) Show that f ∈ F [X ] is alternating, if and only if f = ∆n ·g, where g ∈ F [X ]Snand ∆n ∈ F [X ] is the discriminant polynomial, see Exercise (10.16).b) Show that F [X ]An = (1 ·F [X ]Sn)

⊕(∆n ·F [X ]Sn) as F [X ]Sn -modules. Con-

clude that F [X ]An is not a polynomial ring.

(10.18) Exercise: Reflection representations of Sn.Let n ∈ N and let W be the natural permutation QSn-module, having permu-tation Q-basis b1, . . . , bn ⊆W .a) Show that W ′ := 〈

∑ni=1 bi〉Q ≤ W is a trivial QSn-submodule, and that

V := W/W ′ is an absolutely irreducible faithful reflection representation of Sn.b) Determine algebraically independent homogeneous f1, . . . , fn−1 ⊆ S[V ]Snsuch that deg(fi) = i+ 1 and S[V ]Sn = Q[f1, . . . , fn−1].

(10.19) Exercise: Direct products of symmetric groups.Let r ∈ N and G := Sn1 × · · · × Snr , for ni ∈ N. Let F be a field, and letG act on F [X1, . . . ,Xr] := F [X1,1, . . . , X1,n1 , X2,1, . . . , Xr,nr ], where the directfactor Sni ≤ G permutes the indeterminates Xi,1, . . . , Xi,ni and leaves theother indeterminates fixed. Show that F [X1, . . . ,Xr]G is a polynomial ring anddetermine a set of basic invariants.

Proof. See [4, Thm.3.10.1]. ]

(10.20) Exercise: Transfer map.Let F be a field such that char(F ) = 2, let n ≥ 2 and let ∆n ∈ F [X ] =F [X1, . . . , Xn] be the discriminant polynomial, see Exercise (10.16). Show thatfor the transfer map holds im (TrSn) = ∆n · F [X ]Sn C F [X ]Sn .

Proof. See [13, Ex.2.2.1]. ]

(10.21) Exercise: Nakayama Lemma.Let R be a commutative ring and let M be a finitely generated R-module. TheJacobson radical rad(R) is defined as rad(R) :=

⋂J CR; J maximal.

a) Let I CR such that I ⊆ rad(R) and MI = M . Then we have M = 0.b) Let : M → M/MI be the natural epimorphism of R-modules. Show thatX ⊆ M generates M as an R-module, if and only if X ⊆ M/MI generatesM/MI as an R/I-module.

Hint for (a). Use the Cayley-Hamilton Theorem.

Proof. See [5, Cor.4.8]. ]

48

(10.22) Exercise: Graded Nakayama Lemma.Let R be a graded F -algebra and let M =

⊕d≥0Md be an N0-graded R-module.

a) Let : M → M/MR+ be the natural epimorphism of graded R-modules.Show that a set X ⊆M of homogeneous elements generates M as an R-module,if and only if X ⊆M/MR+ generates the F -vector space M/MR+.b) Let additionally M be finitely generated as an R-module. Show that M isprojective if and only if M is free.

Proof. a) See [4, La.3.5.1] or [5, Exc.4.6.6].b) See [1, La.4.1.1] or [5, Exc.4.6.11] or [12, Thm.2.5]. ]

(10.23) Exercise: Localisation.Let R be a commutative ring, let U ⊆ R be multiplicatively closed, and let Mbe an R-module.a) Show that RU is a commutative ring and that ν : R → RU : r 7→ r

1 is ahomomorphism of commutative rings. Moreover show that MU is a RU -moduleand that M →MU : m 7→ m

1 is an R-module homomorphism.b) Show that the localisation RU has the following universal property: Ifϕ : R → S is a homomorphism of commutative rings such that Uϕ ⊆ S∗, thenthere is unique ring homomorphism ϕ : RU → S such that νϕ = ϕ.c) Show that for J E RU we have (ν−1(J))U = J , and conclude that the mapι : JERU → IER : J 7→ ν−1(J) is an inclusion-preserving and intersection-preserving injection, mapping prime ideals to prime ideals.d) Show that for an ideal I E R we have I ⊆ ν−1(IU ) = f ∈ R; fu ∈I for some u ∈ U E R, conclude that if U ∩ I = ∅ then we have IU 6= RU ,and that and a prime ideal P CR is in the image of ι if and only if U ∩ P = ∅.

Proof. See [5, Ch.2.1]. ]

(10.24) Exercise: Prime avoidance.a) Let R be a commutative ring, let P1, . . . , Pn CR be prime ideals, for n ∈ N,and let I E R be an ideal such that I ⊆

⋃ni=1 Pi. Then there is i ∈ 1, . . . , n

such that I ⊆ Pi.b) Let R be Noetherian, let I E R be an ideal and let M 6= 0 be a finitelygenerated R-module. Show that either I contains a non-zerodivisor on M , orthere is 0 6= m ∈M such that I ⊆ annR(m).

Proof. a) See (6.3). b) See [5, Cor.3.2]. ]

(10.25) Exercise: Krull’s Principal Ideal Theorem.Let R be Noetherian and let P C R be a prime ideal such that ht(P ) = r, forsome r ∈ N. Show that there are f1, . . . , fr ∈ R such that P C R is a minimalprime ideal such that

∑ri=1 fiR ⊆ P .

Proof. See [5, Cor.10.5]. ]

49

(10.26) Exercise: Integral extensions.Let R ⊆ S be an integral extension of commutative rings.a) Let S be a domain. Show that R is a field if and only if S is a field.b) Let QCS be a prime ideal. Show that Q is a maximal ideal of S if and onlyif Q ∩R is a maximal ideal of R.

Proof. See [5, La.4,16, Cor.4.17]. ]

(10.27) Exercise: Integral extensions.Let R ⊆ S be an integral extension of commutative rings, let J C S be an idealand I := J ∩RCR. Show that dim(J) = dim(I) holds.

Proof. See (6.10). ]

(10.28) Exercise: Infinite dimension.Let R = F [X1, X2, . . .] the polynomial ring in countably infinitely many vari-ables, let d0 := 0 and di ∈ N such that di < di+1 and Pi :=

∑dij=di−1+1Xj CR,

for i ∈ N, and let U := R \⋃i≥1 Pi ⊆ R.

a) Show that RU is Noetherian.b) Show that dim(RU ) = supdi − di−1; i ∈ N.

Proof. See [5, Exc.9.6]. ]

(10.29) Exercise: Dimension 0.Let R be a finitely generated commutative F -algebra. Show that dim(R) = 0 ifand only if dimF (R) <∞.

(10.30) Exercise: Homogeneous systems of parameters.Let R be a finitely generated graded F -algebra and let f1, . . . , fn ⊆ R+ behomogeneous, where n = dim(R) ∈ N0. Show that the following assertions areequivalent:i) The set f1, . . . , fn is a homogeneous system of parameters.ii) We have dim(R/(

∑ni=1 fiR)) = 0.

iii) For all j ∈ 1, . . . , n we have dim(R/(∑ji=1 fiR)) = n− j.

Proof. See [4, Prop.3.3.1]. ]

(10.31) Exercise: Coefficient growth.Let H := f(T )∏r

i=1(1−Tdi ) =∑d≥0 hdT

d ∈ Q((T )), where f ∈ Z[T±1] as well as

r ≥ 1 and di ∈ N. Let k ∈ Z be minimal such that the sequence hddk∈ C; d ≥

0 ⊆ Q is bounded. If ν1(H) ≤ −1, show that k = −ν1(H)− 1.

Proof. See [1, Prop.2.1.2] and [13, Prop.3.3.3]. ]

50

(10.32) Exercise: Polynomial invariant rings.Let F ⊆ C, let G be generated by pseudoreflections, and let f1, . . . , fn ⊆S[V ]G+ be homogeneous and algebraically independent such that

∏ni=1 deg(fi) =

|G|. Show that S[V ]G = F [f1, . . . , fn].

Proof. See [7, Prop.3.12]. ]

(10.33) Exercise: Cohen-Macaulay algebras.Let F be a field and let R := F 〈X4

1 , X31X2, X1X

32 , X

42 〉 ⊆ F [X ] = F [X1, X2]

a) Show that X41 , X

42 ⊆ R is a homogeneous system of parameters.

b) Show that R is not Cohen-Macaulay.

Proof. See [4, Ex.2.5.4]. ]

(10.34) Exercise: Cohen-Macaulay property.Let p be a prime, let G := 〈π〉 ∼= Cp be the cyclic group of order p, let F be afield such that char(F ) = p, and let V = W ⊕W ⊕W as FG-modules, where

DW : G→ GL2(F ) : π 7→[

1 1. 1

].

Let S[W ] ∼= F [X,Y ] and S[V ] ∼= F [X1, Y1, X2, Y2, X3, Y3].a) For 1 ≤ i < j ≤ 3 let hij := XiYj −XjYi ∈ S[V ]. Show that hij ∈ S[V ]G.b) Show that Y1, Y2, Y3 ⊆ S[V ]G can be extended to a homogeneous systemof parameters of S[V ]G, but [Y1, Y2, Y3] is not a regular sequence for S[V ]G.

Proof. Uses Exercise (10.30), see [4, Ex.3.4.3] and also [13, Ex.5.5.2]. ]

(10.35) Exercise: Cohen-Macaulay property.Let G := 〈(1, 2, 3, 4)〉 ≤ S4 be the cyclic group of order 4, and let F2[X ] =F2[X1, . . . , X4].a) Determine an F2-algebra generating set of F2[X ]G, and show that Gobel’sdegree bound is not sharp for F2[X ]G.b) Show that F2[X ]G is not Cohen-Macaulay, by showing that [e4,1(X ), e4,2(X )+(X1X2)+, e4,3(X ), e4,4(X )] ⊆ F2[X ]G is a homogeneous system of parameters,but not a regular sequence for F2[X ]G.

Hint for (b). Consider (X21X2)+, (X3

1X2)+ ∈ F2[X ]G.

Proof. See [13, Ex.5.8.1] and [1, Ex.2]. ]

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(10.36) Exercise: Hironaka decomposition.Let G = 〈(1, 2)(3, 4), (1, 3)(2, 4)〉 ≤ S4 be the Klein group of order 4, and letQ[X ] = Q[X1, . . . , X4].a) Show that the Hilbert series of Q[X ]G is given as HQ[X ]G = 1+T 3

(1−T )·(1−T 2)3 .b) Find primary invariants f1, . . . , f4 ⊆ Q[X ]G such that deg(f1) = 1 anddeg(f2) = deg(f3) = deg(f4) = 2, and secondary invariants g1, g2 ⊆ Q[X ]G

such that deg(g1) = 0 and deg(g2) = 3, yielding the Hironaka decompositionQ[X ]G =

⊕2i=1(gi ·Q[f1, . . . , f4]).

Proof. See [4, Ex.3.3.6(a)]. ]

(10.37) Exercise: Hironaka decomposition.Let G = 〈α, β〉 ∼= C2 × C4 be the abelian group of order 8 defined by

DV : G→ GL3(Q(ζ4)) : α 7→

1 . .. 1 .. . ζ4

, β 7→

−1 . .. −1 .. . 1

,where ζ4 ∈ C is a primitive 4-th root of unity.a) Show that the Hilbert series of S[V ]G is given as HS[V ]G = 1

(1−T 2)3 ∈ Q(T ).b) Show that there is no set of primary invariants f1, . . . , f3 ⊆ S[V ]G suchthat deg(f1) = deg(f2) = deg(f2) = 2.c) Find primary invariants f1, . . . , f3 ⊆ S[V ]G such that deg(f1) = deg(f2) =2 and deg(f3) = 4, and secondary invariants g1, . . . , gm ⊆ S[V ]G for some m ∈N, yielding the Hironaka decomposition S[V ]G =

⊕mi=1(gi ·Q(ζ4)[f1, . . . , f3]).

Proof. See [4, Ex.3.3.6(b)]. ]

(10.38) Exercise: Hironaka decomposition.Let G = 〈σ, τ〉 ∼= Q8 be the quaternion group of order 8, and let

DV : G→ GL2(Q(ζ4)) : σ 7→[ζ4 .. −ζ4

], τ 7→

[. −11 .

],

where ζ4 ∈ C is a primitive 4-th root of unity.a) Show that the Hilbert series of S[V ]G is given as HS[V ]G = 1+T 6

(1−T 4)2 ∈ Q(T ).b) Find primary invariants f1, f2 ⊆ S[V ]G such that deg(f1) = deg(f2) = 4,and secondary invariants g1, g2 ⊆ S[V ]G such that deg(g1) = 0 and deg(g2) =6, yielding the Hironaka decomposition S[V ]G =

⊕2i=1(gi ·Q(ζ4)[f1, f2]).

c) Show that S[V ]G ∼= Q(ζ4)[X,Y, Z]/(Z2−X2Y +4Y 3)Q(ζ4)[X,Y, Z] as gradedQ(ζ4)-algebras, where Q(ζ4)[X,Y, Z] has degree vector δ = [4, 4, 6].

Proof. See [1, Exc.2.5] or [13, Ex.5.5.1]. ]

52

11 References

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[2] N. Bourbaki: Elements de mathematique, Fasc. XXXIV: Groupes etalgebres de Lie, Chapitre IV: Groupes de Coxeter et systemes de Tits,Chapitre V: Groupes engendres par des reflexions, Chapitre VI: systemesde racines, Actualites Scientifiques et Industrielles 1337, Hermann, 1968.

[3] W. Bruns, J. Herzog: Cohen-Macaulay rings, Cambridge Studies inAdvanced Mathematics 39, Cambridge University Press, 1993.

[4] H. Derksen, G. Kemper: Computational invariant theory, InvariantTheory and Algebraic Transformation Groups I, Encyclopaedia of Mathe-matical Sciences 130, Springer, 2002.

[5] D. Eisenbud: Commutative algebra, with a view toward algebraic geom-etry, Graduate Texts in Mathematics 150, Springer, 1995.

[6] The GAP Group: GAP-4.3 — Groups, Algorithms and Programming,Aachen, St. Andrews, 2003, http://www-gap.dcs.st-and.ac.uk/gap/.

[7] J. Humphreys: Reflection groups and Coxeter groups, Cambridge Studiesin Advanced Mathematics 29, Cambridge University Press, 1990.

[8] E. Kunz: Introduction to commutative algebra and algebraic geometry,Birkhuser, 1985.

[9] P. Landrock: Finite group algebras and their modules, London Mathe-matical Society Lecture Note Series 84, Cambridge University Press, 1983.

[10] S. Lang: Algebra, Graduate Texts in Mathematics 211, Springer, 2002.

[11] The Computational Algebra Group: MAGMA-V2.10 — The MagmaComputational Algebra System, School of Mathematics and Statistics, Uni-versity of Sydney, 2003, http://magma.maths.usyd.edu.au/magma/.

[12] H. Matsumura: Commutative ring theory, Cambridge Studies in Ad-vanced Mathematics 8, Cambridge University Press, 1989.

[13] M. Neusel, L. Smith: Invariant theory of finite groups, MathematicalSurveys and Monographs 94, American Mathematical Society, 2002.

[14] L. Smith: Polynomial invariants of finite groups, Research Notes in Math-ematics 6, Peters, 1995.