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 $k[{X}_{1},\dots ,{X}_{n}]\text{),}$ $G$ 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 $\to $ graded" is generally easy; the passage "graded $\to $ local" less easy, it requires at least the "associated graded".
Situation: We are given a local regular ring $S\text{,}$ and a finite group of automorphisms $G\text{,}$ acting on $S\text{.}$ We denote by $R={S}^{G}$ the subring of $S$ formed by the invariant elements under $G\text{.}$ The problem is to know under which hypotheses the ring $R$ is regular.
- | In the case where $S=k[{X}_{1},\dots ,{X}_{n}],$ one must determine in which cases $R=k{[{X}_{1},\dots ,{X}_{n}]}^{G}$ is an algebra of polynomials. |
- | In the language of scheme, the problem is formulated as: Given an affine scheme $\text{Spec}\hspace{0.17em}S,$ regular over $\text{Spec}\hspace{0.17em}k$ $\text{(}S$ being a local ring with residue field $k\text{),}$ and $G$ a finite group of automorphisms acting on $\text{Spec}\hspace{0.17em}S,$ in which cases is $\text{Spec}\left(S\right)/G$ 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 $k\text{.}$ The following three assertions are equivalent:
1. | $\text{Spec}\hspace{0.17em}R$ is regular (as a scheme over $\text{Spec}\hspace{0.17em}k\text{);}$ |
2. | ${R}_{\U0001d52a}$ is regular $\text{(}\U0001d52a$ denotes the maximal ideal formed by the homogeneous elements of positive degree); |
3. | $R$ is a graded algebra of polynomials. |
"Graded case" signifies: $S$ is a graded algebra of polynomials over a field $k\text{;}$ $S=k[{X}_{1},\dots ,{X}_{n}]\text{.}$
Definition. Let $G$ be a finite group $G\subseteq GL(n,k)\text{.}$ $G$ acts on ${\left(k\right)}^{n}\text{.}$ Assume $(\text{card}\hspace{0.17em}G,p)=1,$ where $p$ denotes the characteristic exponent of $k\text{.}$ One says that an element $g$ of $G$ is a pseudo-reflection if $\text{im}(1-g)$ is of rank less than or equal to 1, (i.e. if $g\left(x\right)=x+\lambda \left(x\right)e,$ for all $x\in {\left(k\right)}^{n},$ where $e$ is a fixed vector and $\lambda $ is a linear form on ${k}^{n}\text{).}$
Example of a pseudo-reflection: $$\begin{array}{cc}{\displaystyle g=\left(\begin{array}{c}1\\ & 1& & 0\\ & & \ddots \\ & 0& & 1\\ & & & & \mu \end{array}\right)}& \text{(1)}\end{array}$$
If $(\text{card}\hspace{0.17em}G,p)=1,$ the every pseudo-reflection $g\in G$ can be put in the form (1) where $\mu $ is a root of unity.
Theorem 1. Let $S=k[{X}_{1},\dots ,{X}_{n}]$ and $G$ a finite group of automorphisms of $S$ such that $(\text{card}\hspace{0.17em}G,p)=1,$ where $p$ is the characteristic exponent of $k\text{.}$ The following assertions are equivalent:
(a) | $R={S}^{G}$ 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 $k=\u2102\text{.}$
In the direction (a) $\Rightarrow $ (b), their proof is applicable to a field of arbitrary characteristic $p\ne 0\text{.}$
In the direction (b) $\Rightarrow $ (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 $S=k[{X}_{1},\dots ,{X}_{n}]$ and $G$ a finite group of automorphisms of $S\text{;}$ then: $\text{(}R={S}^{G}$ is a polynomial algebra) implies $\text{(}G$ is generated by pseudo-reflections).
The converse is false starting at the case $n=4\text{.}$ For $k={\mathbb{F}}_{q},$ and $G$ the orthogonal group for the quadratic form $2{X}_{1}{X}_{2}+2{X}_{3}X-4\text{.}$
Outline of the proof of SHEPHARD-TODD. The theorems of Shephard-Todd over $\u2102$ pass to characteristic 0, even to characteristic $p$ such that $(\text{card}\hspace{0.17em}G,p)=1\text{.}$
Thus, let $G\subseteq GL(n,\u2102)\text{.}$
1st case: There exists a real structure on ${\u2102}^{n}$ invariant under $G$ (i.e. $G$ imbeds in $GL(n,\mathbb{R}),$ 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 $(\text{card}\hspace{0.17em}G,p)=1$ (even here $p=0,$ where it is always the case), $g$ was put in the form $$g=\left(\begin{array}{c}1\\ & 1& & 0\\ & & \ddots \\ & 0& & 1\\ & & & & \mu \end{array}\right)$$ where $\mu $ was a root of unity. Since $G\subseteq GL(n,\mathbb{R})$ we thus have $\mu =\pm 1$ 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 $G\text{.}$
For this, they send $GL(n,\u2102)$ into $PGL(n,\u2102)\text{.}$ In $PGL(n,\u2102),$ 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 $GL(n,\u2102)$ generated by reflections it suffices to lift these groups into $GL(n,\u2102)\text{.}$
Among the irreducible groups $G\text{,}$ one finds two kinds:
${1}^{o}$ | 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 $m\text{.}$ These groups are variants of the symmetric group ${S}_{n}\text{.}$ In effect, they acts as follows: $$g\left({x}_{i}\right)={x}_{i}^{\prime}={\theta}^{{\nu}_{i}}{x}_{\sigma \left(i\right)},\phantom{\rule{2em}{0ex}}\text{where}\hspace{0.17em}\sigma \in {S}_{n},$$ $\theta $ is a primitive $m\text{th}$ root of unity, and $\sum {\nu}_{i}\equiv 0\hspace{0.17em}\hspace{0.17em}\left(p\right)\text{.}$ |
${2}^{o}$ | 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: | |
$\phantom{\rule{2em}{0ex}}$ ${G}_{168}\times \{\pm 1\},$ where ${G}_{168}={SL}_{3}\left({\mathbb{F}}_{2}\right)\text{;}$ | |
$\phantom{\rule{2em}{0ex}}$ Two groups of order 648 and 1296 which correspond to the same projective group; | |
$\phantom{\rule{2em}{0ex}}$ One group of order 2160 (which is a 6-fold cover of its projective group which, itself, is isomorphic to ${GL}_{6}\text{).}$ | |
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 $72\times 6!$ | |
In dimension $n=6$, one finds one group of order $108\times 9!$ |
In the following, we make the following assumptions:
$S$ is a regular local ring with maximal ideal ${\U0001d52a}_{S}\text{;}$ $G$ a finite subgroup of $\text{Aut}\left(S\right)\text{.}$ $R$ denotes the subring ${S}^{G}$ 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 $R\text{;}$ and |
(iii) | $R$ and $S$ have the same residue field $k$ of characteristic $p$ (totally ramified case) |
Let $V={\U0001d52a}_{S}/{\U0001d52a}_{S}^{2}$ be the Zariski tangent space relative to $S\text{.}$ Then $G$ acts linearly on $V\text{,}$ and we denote by $\epsilon $ the map $\epsilon :G\to \text{Aut}\left(V\right)\text{.}$
Theorem $1\prime \text{.}$ If $\text{gcd}(\text{card}\left(G\right),\text{char}\left(k\right))=1,$ then $R$ is regular if and only if $\epsilon \left(G\right)$ is generated by pseudoreflections.
Theorem $2\prime \text{.}$ If $R$ is regular then $\epsilon \left(G\right)$ is generated by pseudoreflections.
Proof of Theorem $2\prime \text{.}$
Let $H$ be the subgroups of $H$ generated by the elements $g\in G$ such that
$\epsilon \left(g\right)$ is a pseudoreflection. Let
$R\prime ={S}^{H}\text{;}$ thus we have the following inclusions:
$$S\supseteq R\prime \supseteq R\text{.}$$
We will show that $R\prime =R,$ which will lead to, by Galois theory, that
$G=H,$ and Theorem $2\prime \text{.}$
Lemma. $R\prime $ 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 ${I}_{G/H}$ of all prime ideals $\U0001d52e\prime $ of height 1 over $R\prime $ is zero $\text{(}G/H$ is the Galois group of the extension $R\prime $ over $R\text{).}$ For this, we show that if $\U0001d52d$ is a prime ideal of height 1 in $S$ such that $\U0001d52e\prime =\U0001d52d\cap R\prime $ $\text{(}\U0001d52e\prime $ is a prime ideal of height 1 in $R\prime ,$ since $S$ is and integral domain, integral over $R\prime ,$ and $R\prime $ is integrally closed), we have ${I}_{G}={I}_{H},$ which will give the desired result.
Now ${I}_{G}\text{,}$ the inertia group of $\U0001d52d$ in $G\text{,}$ is formed by the elements $g\in G$ such that $g\left(\U0001d52d\right)=\U0001d52d$ and that $g$ acts trivially on $S/\U0001d52d$ (same definition for ${I}_{H}\text{).}$ Clearly we have the inclusion ${I}_{H}\subseteq {I}_{G}\text{.}$
It remains to prove the reverse inclusion. So, let $g\in {I}_{G},$ so that $g\left(x\right)\equiv x\left(\text{mod}\hspace{0.17em}\U0001d52d\right),$ for all $x\in S$ and $g$ acts trivially on $S/\U0001d52d\text{.}$
Let $x\in {\U0001d52a}_{S},$ $$\begin{array}{cc}{\displaystyle g\left(x\right)=x+y,\phantom{\rule{2em}{0ex}}\text{where}\hspace{0.17em}y\in {\U0001d52a}_{S}\cap \U0001d52d=\U0001d52d}& \text{(2)}\end{array}$$ Now the image of $\U0001d52d$ in ${\U0001d52a}_{S}/{\U0001d52a}_{S}^{2}$ (i.e. $(\U0001d52d+{\U0001d52a}_{S})/{\U0001d52a}_{S}^{2}\text{)}$ is a $k\text{-vector}$ space of dimension $0$ or $1$. In the case that ${\text{dim}}_{k}((\U0001d52d+{\U0001d52a}_{S})/{\U0001d52a}_{S}^{2})\ne 0\text{;}$ there exists $\stackrel{\u203e}{x}\ne 0,$ $\stackrel{\u203e}{x}\in (\U0001d52d+{\U0001d52a}_{S})/{\U0001d52a}_{S}^{2}\text{;}$ equivalently, there exists $x\in \U0001d52d,$ $x\notin {\U0001d52a}_{S}^{2}$ such that the ideal ${S}_{X}$ is prime.
(In effect, $S$ being of finite type over a noetherian $R\text{,}$ it itself is noetherian; $S$ being local, noetherian, regular, and $x\ne 0,$ $x\in {\U0001d52a}_{S}$ and $x\notin {\U0001d52a}_{S}^{2},$ then $S/xS$ is regular and so $S$ is an integral domain, and $xS$ is a prime ideal.)
Now $Sx\subset \U0001d52d\text{.}$ Since $S$ is the prime ideal of height 1, it follows that $\U0001d52d=Sx\text{.}$ Thus $k\stackrel{\u203e}{x}=(\U0001d52d+{\U0001d52a}_{S})/{\U0001d52a}_{S}^{2}$ (taking images in ${\U0001d52a}_{S}/{\U0001d52a}_{S}^{2}\text{),}$ and thus ${\text{dim}}_{k}((\U0001d52d+{\U0001d52a}_{S})/{\U0001d52a}_{S}^{2})=1\text{.}$ Reducing relation (2) modulo ${\U0001d52a}_{S}^{2},$ we obtain $$g\left(\stackrel{\u203e}{x}\right)-\stackrel{\u203e}{x}\in (\U0001d52d+{\U0001d52a}_{S})/{\U0001d52a}_{S}^{2},\phantom{\rule{2em}{0ex}}\text{for all}\hspace{0.17em}\stackrel{\u203e}{x}\in {\U0001d52a}_{S}/{\U0001d52a}_{S}^{2}\text{.}$$ Thus $\text{Im}(\epsilon \left(g\right)-1)$ is of dimension $\le 1,$ and $\epsilon \left(g\right)$ is a pseudoreflection. It follows that $g\in {I}_{H}\text{.}$
C.Q.F.D.
This lemma being proved, using the purity theorem of "Nagata-Auslander" [1], since $R\prime $ is normal (as it is a ring of invariants), and $R$ is regular by hypothesis, the nonramification of $R\prime $ over $R$ in codimension 1 implies the nonramification of $R\prime $ over $R\text{.}$ Since $R$ and $R\prime $ 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 $R\prime \text{.}$ Thus $G=H$ and $\epsilon \left(G\right)$ is generated by pseudoreflections.
Proof of Theorem $1\prime \text{.}$
Remark. The problem occurs equally when one considers the quotient of a bounded domain $D$ by a discrete subgroup $\mathrm{\Gamma}\text{.}$ If $x$ denotes a point of $D\text{,}$ and ${\mathrm{\Gamma}}_{x}$ its stabilizer (this is a finite group), then:
$\text{(}D/\mathrm{\Gamma},$ an analytic variety without singularities) $\iff $ (The ${\mathrm{\Gamma}}_{x}$ are generated by pseudo-reflections).
Example of the singular case: Fix the point $(x,y)\text{.}$ $(x,y)\text{.}$ If $\mathrm{\Gamma}$ is generated by the symmetries $\left\{\begin{array}{ccc}x& \mapsto & -x\\ y& \mapsto & -y\end{array}\right\}$ then $\mathrm{\Gamma}$ 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 $(\text{card}\hspace{0.17em}G,p)=1$ and if $\epsilon \left(G\right)$ is generated by pseudoreflections, then $R$ is regular (the other implication being given by theorem $2\prime \text{).}$
Lemma. Fix $R$ and $S$ two local rings such that $R$ is contained in $S\text{,}$ $S$ is of finite type over $R$ and integral over $R\text{,}$ $R$ is noetheriam and $S$ regular. Then the following assertions are equivalent:
(a) | $R$ is regular; |
(b) | $S$ is a free $R\text{-module.}$ |
Proof. | |
(a) $\Rightarrow $ (b) As $R$ is a local regular ring, and $S$ an $R\text{-module}$ of finite type, one has: $${\text{prof}}_{R}S+{\text{dim proj}}_{R}S=\text{dim}\hspace{0.17em}R=\text{dim}\hspace{0.17em}S$$ (since $S$ is integral over $R\text{).}$ Now, we are in the following situation: $R\subseteq S,$ $R$ and $S$ local noetherian, $S$ integral over $R\text{,}$ the injection homomorphism $R\to S$ is a local homomorphism; thus, as $S$ is an $R\text{-module}$ of finite type, we have $${\text{prof}}_{R}S={\text{prof}}_{S}S\text{.}$$ Further, $S$ being local regular, is a Macaulay ring, whence $${\text{prof}}_{S}S=\text{dim}\hspace{0.17em}S\text{.}$$ By comparing the relations (3) and (4), we find: $${\text{dim proj}}_{R}S=0\text{.}$$ Therefore, $S$ is a projective $R\text{-module;}$ since, by hypothesis, $S$ is an $R\text{-module}$ of finite type and $R$ is a local noetherian ring, this implies that $S$ is a free $R\text{-module.}$ (b) $\Rightarrow $ (a). Assume $S$ is a free $R\text{-module.}$ We will show that $R$ is regular, and for that that $\text{dim coh}\hspace{0.17em}R$ is finite. Now, $S$ being free over $R\text{,}$ is flat over $R\text{,}$ whence $$\text{dim coh}\left(R\right)\le \text{dim coh}\left(S\right)\text{.}$$ $S$ being regular, has finite cohomological dimension, and thus also $R\text{.}$ $\square $ |
It remains to show that $S$ is a free $R\text{-module,}$ and theorem $1\prime $ will be proved.
For this, it suffices to show that $T={\text{Tor}}_{1}(S,k)$ is zero (since $S$ is of finite type over $R\text{).}$ Now $T$ is an $S\text{-module.}$ Thus $G$ acts on it. Further, $T$ is an $S\text{-module}$ of finite type (since it is a finite dimensional vector space over $k\text{).}$ Thus, if one shows that $T\prime =T/\left({\U0001d52a}_{S}T\right)=0,$ by Nakayama's lemma this will imply that $T=0\text{.}$ But if:
(a) | Every element of $T\prime $ invariant under $G$ is zero, and |
(b) | $G$ acts trivally on $T\prime $ (i.e. every pseudo-reflection of $G$ acts trivially on $T\prime ,$ since $G$ is generated by pseudo-reflections), then we will indeed have $T\prime =0\text{.}$ |
(a) Because $(\text{card}\hspace{0.17em}G,p)=1,$ $\text{card}\hspace{0.17em}G$ is invertible in $R\text{.}$ This thus makes the functor "invariants" exact, that is to say that the exact sequence of $G\text{-homomorphisms:}$ $$\begin{array}{cc}{\displaystyle 0\u27f6{\U0001d52a}_{S}T\u27f6T\stackrel{u}{\u27f6}T\prime \u27f60}& \text{(5)}\\ {\displaystyle 0\u27f6{\left({\U0001d52a}_{S}T\right)}^{G}\u27f6{T}^{G}\stackrel{u}{\u27f6}{\left(T\prime \right)}^{G}\u27f60\text{.}}& \text{(6)}\end{array}$$
(In fact, the only point to show is the surjectivity of $u\text{.}$ Thus, let $y\in {\left(T\prime \right)}^{G}\text{.}$ Because $y\in T\prime ,$ there exists $x\in T$ such that $u\left(x\right)=y$ (exact sequence (5)). If we then consider $${x}_{1}=\frac{1}{\text{card}\hspace{0.17em}G}\sum _{s\in G}s\left(x\right),$$ ${x}_{1}\in {T}^{G}$ and $u\left({x}_{1}\right)=y,$ since $s\left(y\right)=y$ for all $s\in G\text{.)}$
Note $${T}^{G}={\left({\text{Tor}}_{1}^{R}(S,k)\right)}^{G}={\text{Tor}}_{1}^{R}({S}^{G},k)={\text{Tor}}_{1}^{R}(R,k)=0$$ (since $R$ is free over itself). Thus ${T}^{G}=0$ and due to the exact sequence (6), ${\left(T\prime \right)}^{G}=0\text{.}$
(b) One may reduce to the case where $R$ and $S$ are complete. (In effect, for a local noetherian ring $S\text{,}$ we have the equivalence: $S$ noetherian $\iff $ $\stackrel{\u02c6}{S}$ noetherian.)
Then, if an element $g$ of $G$ is a pseudo-reflection, there exist elements ${\xi}_{1}^{1},\dots ,{\xi}_{n}^{1}$ of ${\U0001d52a}_{S}/{\U0001d52a}_{S}^{2}$ such that $$\begin{array}{rcl}{\displaystyle g\left({\xi}_{1}^{1}\right)}& =& {\displaystyle {\xi}_{1}^{1}}\\ {\displaystyle}& \vdots & {\displaystyle}\\ {\displaystyle g\left({\xi}_{n-1}^{1}\right)}& =& {\displaystyle {\xi}_{n-1}^{1}}\\ {\displaystyle g\left({\xi}_{n}^{1}\right)}& =& {\displaystyle w{\xi}_{n}^{1},}\end{array}$$ where $w$ is a root of unity in $k\text{.}$ And these $n$ elements form a basis of the $k\text{-vector}$ space ${\U0001d52a}_{S}/{\U0001d52a}_{S}^{2}$ (see the definition of a pseudo-reflection).
We may lift the $\left({\xi}_{i}^{1}\right)$ into elements $\left({\xi}_{i}^{2}\right)$ of ${\U0001d52a}_{S}/{\U0001d52a}_{S}^{3},$ and then further and further, to elements $\left({\xi}_{i}^{\ell -1}\right)$ of ${\U0001d52a}_{S}/{\U0001d52a}_{S}^{\ell}$ satisfying: $$\begin{array}{rcl}{\displaystyle g\left({\xi}_{i}^{\ell -1}\right)}& =& {\displaystyle {\xi}_{i}^{\ell -1},\phantom{\rule{2em}{0ex}}\text{for all}\hspace{0.17em}1\le i\le n-1,}\\ {\displaystyle g\left({\xi}_{n}^{\ell -1}\right)}& =& {\displaystyle w\left({\xi}_{n}^{\ell -1}\right),}\end{array}$$ and this, for any fixed positive integer $\ell \ge 3\text{.}$
In effect, for example, we will sow how to execute the lift of ${\U0001d52a}_{S}/{\U0001d52a}_{S}^{2}$ to ${\U0001d52a}_{S}/{\U0001d52a}_{S}^{3}\text{.}$
We consider the following diagram where $N$ is the kernel of the following map $g-1\text{:}$ $${\U0001d52a}_{S}/{\U0001d52a}_{S}^{3}\stackrel{g-1}{\u27f6}\text{im}({\U0001d52a}_{S}/{\U0001d52a}_{S}^{3})\u27f60\text{.}$$
$$\begin{array}{ccccccc}& & 0& & 0& & 0\\ & & \downarrow & & \downarrow & & \downarrow \\ 0& \u27f6& N\prime =({\U0001d52a}_{S}^{2}/{\U0001d52a}_{S}^{3})\cap N& \u27f6& {\U0001d52a}_{S}^{2}/{\U0001d52a}_{S}^{3}& \stackrel{g-1}{\u27f6}& \text{Im}({\U0001d52a}_{S}^{2}/{\U0001d52a}_{S}^{3})& \u27f6& 0\\ & & \downarrow & & \downarrow & & \downarrow \\ 0& \u27f6& N& \u27f6& {\U0001d52a}_{S}/{\U0001d52a}_{S}^{3}& \stackrel{g-1}{\u27f6}& \text{Im}({\U0001d52a}_{S}/{\U0001d52a}_{S}^{3})& \u27f6& 0\\ & & \downarrow & & \downarrow & & \downarrow \\ 0& \u27f6& N\u2033=N/(({\U0001d52a}_{S}^{2}/{\U0001d52a}_{S}^{3})\cap N)& \u27f6& {\U0001d52a}_{S}/{\U0001d52a}_{S}^{2}& \stackrel{g-1}{\u27f6}& \text{Im}({\U0001d52a}_{S}/{\U0001d52a}_{S}^{2})& \u27f6& 0\\ & & \downarrow & & \downarrow & & \downarrow \\ & & 0& & 0& & 0\end{array}$$ This diagram is formed of exact rows and columns, since $$(g-1)({\U0001d52a}_{S}/{\U0001d52a}_{S}^{2})\cong ({\U0001d52a}_{S}/{\U0001d52a}_{S}^{3})/[N+({\U0001d52a}_{S}^{2}/{\U0001d52a}_{S}^{3})],$$ and this is indeed the same diagram as in Cartan-Eilenberg ([3], exercise 1, chap. I), where $A={\U0001d52a}_{S}/{\U0001d52a}_{S}^{3},$ ${A}_{1}={\U0001d52a}_{S}^{2}/{\U0001d52a}_{S}^{3}$ and ${A}_{2}=N\text{.}$
Next, the elements ${\xi}_{j}^{1},$ $1\le j\le n-1,$ which belong in fact to $N\u2033,$ can be lifted to elements $\left({\xi}_{j}^{2}\right)$ $(1\le j\le n-1)$ of $N\text{,}$ thus of ${\U0001d52a}_{S}/{\U0001d52a}_{S}^{3},$ satisfying, $$g\left({\xi}_{j}^{2}\right)={\xi}_{j}^{2}\text{.}$$ On may execute the same operation for the element ${\xi}_{n}^{1},$ by replacing $g-1$ by the map $g-w\xb7\text{Id},$ and writing the analogous diagram.
Finally, as $S$ is complete, this proves to us that there exist elements ${x}_{1},\dots ,{x}_{n-1}\in {\U0001d52a}_{S}$ such that $$\begin{array}{rcl}{\displaystyle g\left({\xi}_{1}^{1}\right)}& =& {\displaystyle {\xi}_{1}^{1}}\\ {\displaystyle}& \vdots & {\displaystyle}\\ {\displaystyle g\left({\xi}_{n-1}^{1}\right)}& =& {\displaystyle {\xi}_{n-1}^{1}\text{.}}\end{array}$$ As, in addition, $S/{\U0001d52a}_{S}=R/{\U0001d52a}_{R}=k,$ one may lift $w\text{,}$ a root of unity in $k\text{,}$ to an element $\omega $ of $R\text{.}$ Thus there exists ${x}_{n}\in {\U0001d52a}_{S}$ and $\omega \in R$ such that $$g\left({x}_{n}\right)=\omega {x}_{n}\text{.}$$
We see then that $S=R[{x}_{1},\dots ,{x}_{n}]\text{.}$ In effect, let $x\in S\text{.}$ If $x\notin {\U0001d52a}_{S},$ let $\stackrel{\u203e}{x}$ be the class of $x$ in $S/{\U0001d52a}_{S}=k=R/({\U0001d52a}_{S}\cap R)\text{.}$ $\stackrel{\u203e}{x}\ne 0,$ thus there exists $y\in R\subseteq S$ such that $\stackrel{\u203e}{y}=\stackrel{\u203e}{x}\text{.}$ Then $x-y\in {\U0001d52a}_{S},$ and $x=y+z$ where $z\in {\U0001d52a}_{S}\text{.}$ One is thus brought to study the case where $x\in {\U0001d52a}_{S}\text{.}$
The case $x\in {\U0001d52a}_{S},$ $x\notin {\U0001d52a}_{S}^{2},$ can be treated as follows: Since $S$ is complete: $x=(\stackrel{\u203e}{{x}_{1}},\stackrel{\u203e}{{\xi}_{2}},\dots ,\stackrel{\u203e}{{\xi}_{n}},\dots ),$ where $\stackrel{\u203e}{{\xi}_{n}}=\text{cl}\hspace{0.17em}x$ (mod ${\U0001d52a}_{S}^{n+1}\text{).}$ We put $$y=(\stackrel{\u203e}{{x}_{1}},\stackrel{\u203e}{{x}_{2}},\dots ),$$ where $\stackrel{\u203e}{{x}_{1}}$ is the class of $x$ in ${\U0001d52a}_{S}/{\U0001d52a}_{S}^{2}$ (then $\stackrel{\u203e}{{x}_{1}}=\sum _{j=1}^{n}{\xi}_{j}^{1}{\alpha}_{j},$ where ${\alpha}_{j}\in k\text{)}$ and $\stackrel{\u203e}{{x}_{2}}$ the lift of $\stackrel{\u203e}{{x}_{1}}$ obtained by lifting the $\left({\xi}_{j}^{1}\right),$ etc.
This shows us, after the completion of $S$ and from the fact that $S$ and $R$ have the same residue field, that $$y=\sum _{j=1}^{n}{r}_{j}{x}_{j},\phantom{\rule{2em}{0ex}}\text{where}\hspace{0.17em}{r}_{j}\in R\text{.}$$ Now $y-x\in {\U0001d52a}_{S}^{2}\text{;}$ this brings us to study the case $x\in {\U0001d52a}_{S}^{2}\text{.}$ Now, in all cases, if $x\in {\U0001d52a}_{S},$ we have $$x=\sum _{\text{finite}}{y}_{1}^{j}\cdots {y}_{q\left(j\right)}^{j},\hspace{0.17em}\text{where the}\phantom{\rule{1em}{0ex}}{y}_{q\left(j\right)}^{j}\{\begin{array}{l}\in {\U0001d52a}_{S},\\ \notin {\U0001d52a}_{S}^{2}\text{.}\end{array}$$ Next, for all $j\text{,}$ $x$ can be put in the form $$x={y}_{j}+{t}_{j},\phantom{\rule{2em}{0ex}}\text{where}\phantom{\rule{1em}{0ex}}{y}_{j}\in R[{x}_{1},\dots ,{x}_{n}]\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{t}_{j}\in {\U0001d52a}_{S}^{j+1}\text{.}$$ By passing to the limit, since $S$ is complete, we see thus that $x$ is written as a formal power series in ${x}_{1},\dots ,{x}_{n}$ with coefficients in $R\text{.}$ But $R$ being complete and the ${x}_{i}$ integral over $R\text{,}$ $x$ thus belongs in fact to $R[{x}_{1},\dots ,{x}_{n}]\text{.}$ Next, we indeed have $S=R[{x}_{1},\dots ,{x}_{n}]\text{.}$
We now consider $S\prime =S/\left({x}_{n}S\right)\text{.}$ $g$ acts trivially on $S\prime $ (by construction of the ${x}_{i}\text{).}$ In other words: $\text{Im}(g-1)\subseteq {x}_{n}S,$ and the map $g-1:S\to S$ factorizes: $$\begin{array}{ccccc}S& & \stackrel{g-1}{\u27f6}& & S\\ \multicolumn{2}{c}{{}_{\phi}\searrow}& & \multicolumn{2}{c}{{\nearrow}_{\left({x}_{n}\right)}}\\ & & S\end{array}$$ where $\left({x}_{n}\right)$ denotes multiplication by ${x}_{n}$ in $S\text{.}$ Now the maps, $g-1,$ $\phi ,$ $\left({x}_{n}\right)$ are $R\text{-linear}$ maps which operate on $T={\text{Tor}}_{1}^{R}(S,k),$ and we also have the following diagram $$\begin{array}{ccccc}T& & \stackrel{g-1}{\u27f6}& & T\\ \multicolumn{2}{c}{{}_{\phi}\searrow}& & \multicolumn{2}{c}{{\nearrow}_{\left({x}_{n}\right)}}\\ & & T\end{array}$$ Whence the inclusions $(g-1)\left(T\right)\subseteq {x}_{n}T\subseteq {\U0001d52a}_{S}T\text{.}$ As $T\prime =T/\left({\U0001d52a}_{S}T\right),$ the image of $g-1$ acting on $T\prime $ is zero, that is to say, $g$ acts trivially on $T\prime $ (b) is thus proved, which permits us to conclude that $T\prime $ is zero, and thus $T\text{,}$ 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.
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.