Finite groups of automorphisms of local regular rings

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last updated: 21 January 2015


It is advisable to view the action of certain finite groups of automorphisms on local regular rings or on graded algebras of polynomials over a field (i.e. on k[X1,,Xn]), G acts linearly.

One often has, in effect, a correspondence between the theorems over local regular rings and those overs graded rings of finite type over a field. The passage "local graded" is generally easy; the passage "graded local" less easy, it requires at least the "associated graded".

Situation: We are given a local regular ring S, and a finite group of automorphisms G, acting on S. We denote by R=SG the subring of S formed by the invariant elements under G. The problem is to know under which hypotheses the ring R is regular.

- In the case where S=k[X1,,Xn], one must determine in which cases R=k[X1,,Xn]G is an algebra of polynomials.
- In the language of scheme, the problem is formulated as: Given an affine scheme SpecS, regular over Speck (S being a local ring with residue field k), and G a finite group of automorphisms acting on SpecS, in which cases is Spec(S)/G a regular scheme?

The passage of translating from one to the other is given by the following lemma:

Lemma. Let R be a graded algebra of finite type over a field k. The following three assertions are equivalent:

1. SpecR is regular (as a scheme over Speck);
2. R𝔪 is regular (𝔪 denotes the maximal ideal formed by the homogeneous elements of positive degree);
3. R is a graded algebra of polynomials.

Statement and outline of proof of the theorem in the graded case (historically first resolved)

"Graded case" signifies: S is a graded algebra of polynomials over a field k; S=k[X1,,Xn].

Definition. Let G be a finite group GGL(n,k). G acts on (k)n. Assume (cardG,p)=1, where p denotes the characteristic exponent of k. One says that an element g of G is a pseudo-reflection if im(1-g) is of rank less than or equal to 1, (i.e. if g(x)=x+λ(x)e, for all x(k)n, where e is a fixed vector and λ is a linear form on kn).

Example of a pseudo-reflection: g= ( 1 1 0 0 1 μ ) (1)

If (cardG,p)=1, the every pseudo-reflection gG can be put in the form (1) where μ is a root of unity.

Theorem 1. Let S=k[X1,,Xn] and G a finite group of automorphisms of S such that (cardG,p)=1, where p is the characteristic exponent of k. The following assertions are equivalent:

(a) R=SG is an algebra of polynomials;
(b) G is generated by pseudo-reflections.

This theorem has been proved by SHEPHARD-TODD, in [6], in the case k=.

In the direction (a) (b), their proof is applicable to a field of arbitrary characteristic p0.

In the direction (b) (a), their proof is the following: they make a list of the all the groups having this property, and they check that R is indeed a polynomial algebra.

In 1955, CHEVALLEY has given a case free proof of the theorem.

Theorem 2. Let S=k[X1,,Xn] and G a finite group of automorphisms of S; then: (R=SG is a polynomial algebra) implies (G is generated by pseudo-reflections).

The converse is false starting at the case n=4. For k=𝔽q, and G the orthogonal group for the quadratic form 2X1X2+2X3X-4.

Outline of the proof of SHEPHARD-TODD. The theorems of Shephard-Todd over pass to characteristic 0, even to characteristic p such that (cardG,p)=1.

Thus, let GGL(n,).

1st case: There exists a real structure on n invariant under G (i.e. G imbeds in GL(n,), and the elements of G may be written as matrices with real entries).

We have seen that, if g was a pseudo-reflection, in the case (cardG,p)=1 (even here p=0, where it is always the case), g was put in the form g= ( 1 1 0 0 1 μ ) where μ was a root of unity. Since GGL(n,) we thus have μ=±1 forced here.

Thus the elements of H are in this case reflections (a reflection is a pseudo-reflection of order 2).

These groups are the groups studied in the works of Elie CARTAN, and are, with a few exceptions, the Weyl groups of semisimple Lie groups.

When G is irreducible, it is represented by a Coxeter diagram; and the list of possible diagrams is well known.

2nd case: G is truly a "complex" group.

SHEPHARD and TODD then draw up the list of all possible G.

For this, they send GL(n,) into PGL(n,). In PGL(n,), these transformations correspond to "homologies".

Then MITCHELL, in 1914, has made the list of finite groups generated by homologies. To obtain the finite subgroups of GL(n,) generated by reflections it suffices to lift these groups into GL(n,).

Among the irreducible groups G, one finds two kinds:

1o The imprimitives: These are the groups G(m,p,n), where n= the dimension of the vector space, and m is an integer such that p divides m. These groups are variants of the symmetric group Sn. In effect, they acts as follows: g(xi)=xi= θνixσ(i), whereσSn, θ is a primitive mth root of unity, and νi0(p).
2o The primitives:
(a) In dimension 2, there are a large number, around 12.
(b) In dimension n>2, one finds only a finite number.
In dimension 3, one finds:
G168×{±1}, where G168=SL3(𝔽2);
Two groups of order 648 and 1296 which correspond to the same projective group;
One group of order 2160 (which is a 6-fold cover of its projective group which, itself, is isomorphic to GL6).
In dimension n=4, one finds two groups of orders 46080 and 7680, and one group of order 25920.
In dimension n=5, one finds one group of order 72×6!
In dimension n=6, one finds one group of order 108×9!

Statement and proof of the theorems in the local regular case

In the following, we make the following assumptions:

S is a regular local ring with maximal ideal 𝔪S; G a finite subgroup of Aut(S). R denotes the subring SG of invariant of S by G (we know that R is local), and assume in addition:

(i) R is noetherian;
(ii) S is of finite type over R; and
(iii) R and S have the same residue field k of characteristic p (totally ramified case)

Let V=𝔪S/𝔪S2 be the Zariski tangent space relative to S. Then G acts linearly on V, and we denote by ε the map ε:GAut(V).

Theorem 1. If gcd(card(G),char(k))=1, then R is regular if and only if ε(G) is generated by pseudoreflections.

Theorem 2. If R is regular then ε(G) is generated by pseudoreflections.

Proof of Theorem 2.
Let H be the subgroups of H generated by the elements gG such that ε(g) is a pseudoreflection. Let R=SH; thus we have the following inclusions: SRR. We will show that R=R, which will lead to, by Galois theory, that G=H, and Theorem 2.

Lemma. R is unramified over R in codimension 1 (i.e. divisorially unramified).

Said another way, after localization by the prime ideals of height 1, the extension is unramified.

Said another way again, the ramification locus of the cover is of codimension greater that or equal to 2.

We show that the inertia group IG/H of all prime ideals 𝔮 of height 1 over R is zero (G/H is the Galois group of the extension R over R). For this, we show that if 𝔭 is a prime ideal of height 1 in S such that 𝔮=𝔭R (𝔮 is a prime ideal of height 1 in R, since S is and integral domain, integral over R, and R is integrally closed), we have IG=IH, which will give the desired result.

Now IG, the inertia group of 𝔭 in G, is formed by the elements gG such that g(𝔭)=𝔭 and that g acts trivially on S/𝔭 (same definition for IH). Clearly we have the inclusion IHIG.

It remains to prove the reverse inclusion. So, let gIG, so that g(x)x(mod𝔭), for all xS and g acts trivially on S/𝔭.

Let x𝔪S, g(x)=x+y, wherey𝔪S𝔭=𝔭 (2) Now the image of 𝔭 in 𝔪S/𝔪S2 (i.e. (𝔭+𝔪S)/𝔪S2) is a k-vector space of dimension 0 or 1. In the case that dimk((𝔭+𝔪S)/𝔪S2)0; there exists x0, x(𝔭+𝔪S)/𝔪S2; equivalently, there exists x𝔭, x𝔪S2 such that the ideal SX is prime.

(In effect, S being of finite type over a noetherian R, it itself is noetherian; S being local, noetherian, regular, and x0, x𝔪S and x𝔪S2, then S/xS is regular and so S is an integral domain, and xS is a prime ideal.)

Now Sx𝔭. Since S is the prime ideal of height 1, it follows that 𝔭=Sx. Thus kx=(𝔭+𝔪S)/𝔪S2 (taking images in 𝔪S/𝔪S2), and thus dimk((𝔭+𝔪S)/𝔪S2)=1. Reducing relation (2) modulo 𝔪S2, we obtain g(x)-x (𝔭+𝔪S)/𝔪S2, for allx𝔪S /𝔪S2. Thus Im(ε(g)-1) is of dimension 1, and ε(g) is a pseudoreflection. It follows that gIH.


This lemma being proved, using the purity theorem of "Nagata-Auslander" [1], since R is normal (as it is a ring of invariants), and R is regular by hypothesis, the nonramification of R over R in codimension 1 implies the nonramification of R over R. Since R and R are both local rings having the same residue field (since, by hypothesis, S and R have the same residue field) it follows that R is R. Thus G=H and ε(G) is generated by pseudoreflections.

Proof of Theorem 1.

Remark. The problem occurs equally when one considers the quotient of a bounded domain D by a discrete subgroup Γ. If x denotes a point of D, and Γx its stabilizer (this is a finite group), then:

(D/Γ, an analytic variety without singularities) (The Γx are generated by pseudo-reflections).

Example of the singular case: Fix the point (x,y). (x,y). If Γ is generated by the symmetries {x-xy-y} then Γ is not generated by pseudo-reflections, and the quotient space is a quadratic cone, which thus has a singularity at the origin.

It is thus necessary to prove that, if (cardG,p)=1 and if ε(G) is generated by pseudoreflections, then R is regular (the other implication being given by theorem 2).

Lemma. Fix R and S two local rings such that R is contained in S, S is of finite type over R and integral over R, R is noetheriam and S regular. Then the following assertions are equivalent:

(a) R is regular;
(b) S is a free R-module.


(a) (b) As R is a local regular ring, and S an R-module of finite type, one has: profRS+dim projRS= dimR=dimS (since S is integral over R). Now, we are in the following situation: RS, R and S local noetherian, S integral over R, the injection homomorphism RS is a local homomorphism; thus, as S is an R-module of finite type, we have profRS=profSS. Further, S being local regular, is a Macaulay ring, whence profSS=dimS. By comparing the relations (3) and (4), we find: dim projRS=0. Therefore, S is a projective R-module; since, by hypothesis, S is an R-module of finite type and R is a local noetherian ring, this implies that S is a free R-module.

(b) (a). Assume S is a free R-module. We will show that R is regular, and for that that dim cohR is finite. Now, S being free over R, is flat over R, whence dim coh(R)dim coh(S). S being regular, has finite cohomological dimension, and thus also R.

It remains to show that S is a free R-module, and theorem 1 will be proved.

For this, it suffices to show that T=Tor1(S,k) is zero (since S is of finite type over R). Now T is an S-module. Thus G acts on it. Further, T is an S-module of finite type (since it is a finite dimensional vector space over k). Thus, if one shows that T=T/(𝔪ST)=0, by Nakayama's lemma this will imply that T=0. But if:

(a) Every element of T invariant under G is zero, and
(b) G acts trivally on T (i.e. every pseudo-reflection of G acts trivially on T, since G is generated by pseudo-reflections), then we will indeed have T=0.

(a) Because (cardG,p)=1, cardG is invertible in R. This thus makes the functor "invariants" exact, that is to say that the exact sequence of G-homomorphisms: 0𝔪ST TuT0 (5) 0(𝔪ST)G TGu (T)G0. (6)

(In fact, the only point to show is the surjectivity of u. Thus, let y(T)G. Because yT, there exists xT such that u(x)=y (exact sequence (5)). If we then consider x1=1cardG sGs(x), x1TG and u(x1)=y, since s(y)=y for all sG.)

Note TG=(Tor1R(S,k))G =Tor1R(SG,k)= Tor1R(R,k)=0 (since R is free over itself). Thus TG=0 and due to the exact sequence (6), (T)G=0.

(b) One may reduce to the case where R and S are complete. (In effect, for a local noetherian ring S, we have the equivalence: S noetherian Sˆ noetherian.)

Then, if an element g of G is a pseudo-reflection, there exist elements ξ11,,ξn1 of 𝔪S/𝔪S2 such that g(ξ11) = ξ11 g(ξn-11) = ξn-11 g(ξn1) = wξn1, where w is a root of unity in k. And these n elements form a basis of the k-vector space 𝔪S/𝔪S2 (see the definition of a pseudo-reflection).

We may lift the (ξi1) into elements (ξi2) of 𝔪S/𝔪S3, and then further and further, to elements (ξi-1) of 𝔪S/𝔪S satisfying: g(ξi-1) = ξi-1, for all1in-1, g(ξn-1) = w(ξn-1), and this, for any fixed positive integer 3.

In effect, for example, we will sow how to execute the lift of 𝔪S/𝔪S2 to 𝔪S/𝔪S3.

We consider the following diagram where N is the kernel of the following map g-1: 𝔪S/𝔪S3g-1 im(𝔪S/𝔪S3)0.

0 0 0 0 N=(𝔪S2/𝔪S3)N 𝔪S2/𝔪S3 g-1 Im(𝔪S2/𝔪S3) 0 0 N 𝔪S/𝔪S3 g-1 Im(𝔪S/𝔪S3) 0 0 N=N/((𝔪S2/𝔪S3)N) 𝔪S/𝔪S2 g-1 Im(𝔪S/𝔪S2) 0 0 0 0 This diagram is formed of exact rows and columns, since (g-1) (𝔪S/𝔪S2) (𝔪S/𝔪S3)/ [N+(𝔪S2/𝔪S3)], and this is indeed the same diagram as in Cartan-Eilenberg ([3], exercise 1, chap. I), where A=𝔪S/𝔪S3, A1=𝔪S2/𝔪S3 and A2=N.

Next, the elements ξj1, 1jn-1, which belong in fact to N, can be lifted to elements (ξj2) (1jn-1) of N, thus of 𝔪S/𝔪S3, satisfying, g(ξj2)=ξj2. On may execute the same operation for the element ξn1, by replacing g-1 by the map g-w·Id, and writing the analogous diagram.

Finally, as S is complete, this proves to us that there exist elements x1,,xn-1𝔪S such that g(ξ11) = ξ11 g(ξn-11) = ξn-11. As, in addition, S/𝔪S=R/𝔪R=k, one may lift w, a root of unity in k, to an element ω of R. Thus there exists xn𝔪S and ωR such that g(xn)=ωxn.

We see then that S=R[x1,,xn]. In effect, let xS. If x𝔪S, let x be the class of x in S/𝔪S=k=R/(𝔪SR). x0, thus there exists yRS such that y=x. Then x-y𝔪S, and x=y+z where z𝔪S. One is thus brought to study the case where x𝔪S.

The case x𝔪S, x𝔪S2, can be treated as follows: Since S is complete: x=(x1,ξ2,,ξn,), where ξn=clx (mod 𝔪Sn+1). We put y=(x1,x2,), where x1 is the class of x in 𝔪S/𝔪S2 (then x1=j=1nξj1αj, where αjk) and x2 the lift of x1 obtained by lifting the (ξj1), etc.

This shows us, after the completion of S and from the fact that S and R have the same residue field, that y=j=1nrj xj,whererj R. Now y-x𝔪S2; this brings us to study the case x𝔪S2. Now, in all cases, if x𝔪S, we have x=finitey1j yq(j)j,where the yq(j)j { 𝔪S, 𝔪S2. Next, for all j, x can be put in the form x=yj+tj, whereyjR [x1,,xn] andtj 𝔪Sj+1. By passing to the limit, since S is complete, we see thus that x is written as a formal power series in x1,,xn with coefficients in R. But R being complete and the xi integral over R, x thus belongs in fact to R[x1,,xn]. Next, we indeed have S=R[x1,,xn].

We now consider S=S/(xnS). g acts trivially on S (by construction of the xi). In other words: Im(g-1)xnS, and the map g-1:SS factorizes: S g-1 S φ (xn) S where (xn) denotes multiplication by xn in S. Now the maps, g-1, φ, (xn) are R-linear maps which operate on T=Tor1R(S,k), and we also have the following diagram T g-1 T φ (xn) T Whence the inclusions (g-1)(T)xnT𝔪ST. As T=T/(𝔪ST), the image of g-1 acting on T is zero, that is to say, g acts trivially on T (b) is thus proved, which permits us to conclude that T is zero, and thus T, and complete the proof.


[1] AUSLANDER (Maurice). – On the purity of the branch locus, Amer. J. of Math., v. 84, 1962, p. 116-125.

[2] BOURBAKI (Nicolas). – Algèbre commutative. Chapitre 2: Localisation. – Paris, Hermann, 1961 (Act. scient. et ind., 1290; Bourbaki, 27)

[3] CARTAN (Henri) and EILENBERG (Samuel). – Homological algebra. – Princeton, Princeton University Press, 1956 (Princeton mathematical Series, 19).

[4] GROTHENDIECK (Alexander) and DIEUDONN\'E (Jean). – Eléments de géométrie algébrique. Chapitre 4: Etude locale des schémas et des morphismes de schémas. – Paris, Presses Universitaires de France, 1964 (Institut des hautes Etudes scientifiques. Publications mathématiques, 20).

[5] JACOBSON (Nathan). – Lie algebras. – New York, Interscience Publishers, 1962 (Interscience Tracts in pure and applied Mathematics, 10).

[6] SHEPHARD (G.C.) and TODD (J.A.). – Finite unitary reflection groups, Canad. J. of Math., v. 6, 1954, p. 274-304.

Notes and References

This is a translation of the paper Ecole Normale Supérieure de Jeunes Filles
Colloque d'Algèbre [1967. Paris], no. 8, 11 p. Lecture Notes taken by Marie-José BERTIN
by J. P. Serre. Translation to English by Arun Ram, January 2015.

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