Indeed, S can be identified with the tensor product $K[x_1] \otimes \cdots \otimes K[x_m]$, provided with the total grading. The Poincaré series of $K[x_i]$ is

and it suffices to apply (2).

Under the assumptions of Prop. 1, we shall say that S is a graded polynomial algebra over K.

Indeed, the inverse of $P_S(T)$ is the polynomial $N(T) = \prod_{i=1}^m (1 - T^{d_i})$, which is thus uniquely determined. If q is an integer $\geq 1$ and if $\zeta \in \mathbf{C}$ is a primitive qth root of unity, the multiplicity of the root $\zeta$ of $N(T)$ is equal to the number of $d_i$ that are multiples of $q$. This number is zero for $q$ sufficiently large. The number of $d_i$ equal to $q$ is thus determined uniquely by descending induction.

The integers $d_i$ are called the characteristic degrees of S. The number of them is equal to the transcendence degree of S over K when K is a field; we shall also call it the transcendence degree of S over K in the general case. It is the multiplicity of the root 1 of the polynomial $N(T)$.

Let $S = \bigoplus_{n \geq 0} S_n$ be a commutative graded K-algebra, and $R = \bigoplus_{n \geq 0} R_n$ a graded subalgebra of S. Assume that each $R_n$ is free and of finite type, and that the R-module S admits a finite basis consisting of homogeneous elements $z_1, z_2, \ldots, z_N$ of degrees $f_1, f_2, \ldots, f_N$. Then, if M denotes the graded K-module $\sum_{j=1}^N Kz_j$, the graded K-module S is isomorphic to $R \otimes_K M$, so each $S_n$ is free and of finite type and

(i) R and S have the same transcendence degree r over K. (ii) Let $p_1, \ldots, p_r$ (resp. $q_1, \ldots, q_r$) be the characteristic degrees of S (resp. R). Then

(iii) $N p_1 p_2 \ldots p_r = q_1 q_2 \ldots q_r$.

Formula (4) shows first of all that the multiplicity of the root 1 is the same in the polynomials $P_S(T)^{-1}$ and $P_R(T)^{-1}$ and, taking (3) into account, proves both (i) and (ii).

It follows from (ii) that

Putting $T = 1$ in this relation gives (iii).

a) Since every submodule of $S/(R_+S)$ is free over $R_0 = K$, it is enough to show (in view of Algebra, Chap. II, §11, no. 4, Prop. 7) that the canonical homomorphism from $R_+ \otimes_R S$ to S is injective. For any R-module E, denote by T(E) the R-module $\mathrm{Ker}(R_+ \otimes_R E \to E)$ (\*in other words, $T(E) = \mathrm{Tor}_1^R(R/R_+, E)$\*). If E, $E'$ are two R-modules and $u$ is a homomorphism from E to $E'$, the homomorphism $1 \otimes u$ from $R_+ \otimes E$ to $R_+ \otimes E'$ defines by restriction to T(E) a homomorphism from T(E) to T($E'$) that we denote by $T(u)$. If $u'$ is a homomorphism from $E'$ to an R-module $E''$, we have $T(u' \circ u) = T(u') \circ T(u)$. Thus, if G acts R-linearly on E, then G acts on T(E).

b) The group G acts R-linearly on S, and hence also on T(S). Moreover, T(S) has a natural structure of graded S-module. We show first that, if $g \in G$, then $g$ transforms every element $x$ of T(S) into an element congruent to $x$ modulo $S_1 T(S)$. It is enough to do this when $g$ is a pseudo-reflection. Then there exists a non-zero vector $v$ in V such that $g(x) - x \in K v$ for all $x \in V$. Since V generates S, it follows that $g_S$ acts trivially on $S/Sv$. Thus, for any $y \in S$, there exists an element $h(y)$ in S such that

Since S is integral and $v$ is non-zero, this element is determined uniquely by $y$; it is immediate that $h$ is an endomorphism of degree – 1 of the R-module S. Thus, $g_S - 1_S = m_v \circ h$, where $m_v$ denotes the homothety with ratio $v$ in S. Hence,

the image of which is contained in $vT(S)$, proving our assertion.

c) We show that any element of T(S) invariant under G is zero. Indeed, let Q be the endomorphism of the R-module S defined by

for all $y \in S$. Then $Q(S) = R$. We can thus write $Q = i \circ Q'$, where $Q'$ is a homomorphism from the R-module S to the R-module R and $i$ denotes the canonical injection of R into S. Thus $T(Q) = T(i) \circ T(Q')$ and $T(Q') = 0$ since $T(R) = \mathrm{Ker}(R_+ \otimes R \to R) = 0$. Thus

But $q^{-1} \sum_{g \in G} T(g_S)$ leaves fixed the elements of T(S) invariant under G. These elements are therefore zero.

d) Assume that $T(S) \neq 0$. There exists in T(S) a homogeneous element $u \neq 0$ of minimum degree. By b), $u$ is invariant under G. By c), $u = 0$. This is a contradiction, so $T(S) = 0$. Q.E.D.

(i) There exists a graded vector subspace of S forming a complement to $R_+ S$ in S and stable under G.

(ii) Let U be such a complement. The canonical homomorphism from $U \otimes_K R$ to S is an isomorphism of G-modules, and the representation of G in U (resp. S) is isomorphic to the regular representation of G on K (resp. R).

Indeed, for any integer $n \geq 0$, the K-vector spaces $S_n$ and $(R_+ S) \cap S_n$ are stable under G, and it follows from Maschke’s theorem (Appendix, Prop. 2) that there exists a G-stable complement $U_n$ of $(R_+ S) \cap S_n$ in $S_n$. Then $\sum_{n \geq 0} U_n$ is a G-stable complement of $R_+ S$ in S, hence (i).

Let U be a graded vector subspace of S forming a complement of $R_+ S$ in S and stable under G. By Remark 1, every basis of the K-vector space U is also a basis of the R-module S, and consequently is a basis of the field of fractions N of S over the field of fractions L of R. Thus, the L-vector space N can be identified with $U \otimes_K L$. Since U is stable under G, this identification is compatible with the action of G. The group algebra $L[G]$ of G over L can be identified with the algebra $K[G] \otimes_K L$. The Galois extension N of L admits a normal basis (Algebra, Chap. V, § 10, no. 9, Th. 6), which can be interpreted as saying that N, considered as an $L[G]$-module, is isomorphic to the module of the regular representation of G over L. Since U is finite dimensional over K, it follows from the Appendix, Prop. 1, that the $K[G]$-module U is isomorphic to the regular representation of G over K. Our assertions follow immediately from this.

The assumption made about the $k_i$ is superfluous, but is irrelevant for the applications to finite Coxeter groups, since then $K = \mathbf{R}$. Cf. no. 5, where we shall give another proof of Th. 3.

Th. 3 follows from Prop. 2 (i), Th. 1 and the following lemma:

By Algebra, Chap. II, § 11, no. 4, Prop. 7, the assumption made about the $\alpha_i$ is equivalent to saying that they are homogeneous and that their images in the K-vector space $R_+/ (R_+)^2$ form a basis of this space. This condition is invariant under extension of the base field; we can thus reduce to the case where the latter is perfect.

The family $(\alpha_1, \ldots, \alpha_s)$ generates the algebra R by Commutative Algebra, Chap. III, § 1, no. 2, Prop. 1. We argue by contradiction and assume that this family is not algebraically independent over K.

1) We show first of all that there exist families

of homogeneous elements of S with the following properties:

Let $X_1, \ldots, X_s$ be indeterminates and give $K[X_1, \ldots, X_s]$ with the graded algebra structure for which $X_i$ has degree $k_i$. There exist non-zero homogeneous elements $H(X_1, \ldots, X_s)$ of $K[X_1, \ldots, X_s]$ such that $H(\alpha_1, \ldots, \alpha_s) = 0$; choose H to be of minimum degree. If $\partial H / \partial X_i \neq 0$, the polynomial $\frac{\partial H}{\partial X_i}(\alpha_1, \ldots, \alpha_s)$ is a non-zero homogeneous element of R; if $p \neq 1$, H is not of the form $H_1^p$ with $H_1 \in K[X_1, \ldots, X_s]$. Take

Since K is perfect, the polynomials $\partial H / \partial X_i \in K[X_1, \ldots, X_s]$ are not all zero (Algebra, Chap. V, § 1, no. 3, Prop. 4); in view of the assumption made about the $k_i$, neither are the $\beta_i$.

On the other hand, S can be identified with a graded algebra of polynomials $K[x_1, \ldots, x_r]$ for appropriate indeterminates $x_1, \ldots, x_r$ with suitable degrees $m_i > 0$. Let $D_k$ be the partial derivative with respect to $x_k$ on S. Take d_{ik} = k_i^{-1} D_k(\alpha_i). Then the equality (10) holds because its left-hand side is $D_k(H(\alpha_1, \ldots, \alpha_s))$. On the other hand, if we put $y_1 = m_1 x_1, \ldots, y_r = m_r x_r$, the equality (9) follows from the equality (5) of no. 1.

2) Let $b$ be the ideal of $R$ generated by the $\beta_i$; there exists a subset $J$ of

such that $(\beta_i)_{i \in J}$ is a minimal system of generators of $b$. Then $J \neq \varnothing$ since $b \neq 0$. We shall deduce from (9) and (10) that, if $i \in J$, $\alpha_i$ is an $R$-linear combination of the $\alpha_j$ for $j \neq i$, which will contradict the minimality of $(\alpha_1, \ldots, \alpha_s)$ and will complete the proof.

There exist homogeneous elements $\gamma_{ji}$ of $R$ ($i \in J, j \in I - J$) such that

Taking (11) into account, formula (10) gives Put Thus Write $u_{ik} = \sum_{\lambda \in \Lambda} \delta_{ik\lambda} z_\lambda$, where the $\delta_{ik\lambda}$ belong to $R$. Relation (14) implies that If one of the $\delta_{ik\lambda}$ had a non-zero homogeneous component of degree 0, the preceding equality would imply that one of the $\beta_i$ ($i \in J$) is an $R$-linear combination of the others, contradicting the minimality of $(\beta_i)_{i \in J}$. Thus $\delta_{ik\lambda} \in R_+$ and consequently $u_{ik} \in R_+ S$ for all $i$ and $k$. Thus, there exist $u_{ikh} \in S$ such that $u_{ik} = \sum_{h=1}^s u_{ikh} \alpha_h$, in other words, by (13), Multiply both sides of (15)$_{ik}$ by $y_k$ and add for $i$ in $J$ fixed and $k = 1, 2, \ldots, r$; in view of (9), we find that Take the homogeneous components of degree $k_i$ of both sides. Since $\deg y_k > 0$, $\alpha_i$ is an $S$-linear combination of the $\alpha_j$ for $j \neq i$. Since $S$ is free over $R$ and $\alpha_1, \ldots, \alpha_s \in R$, it follows that $\alpha_i$ is an $R$-linear combination of the $\alpha_j$ with $j \neq i$ (Commutative Algebra, Chap. I, § 3, no. 5, Prop. 9 d)).

Indeed, $\mathrm{rk}_R(S) = \mathrm{Card}(G)$ (formula (6), no. 2). The characteristic degrees of S are equal to 1. Then the corollary follows from no. 1, Prop. 2 (iii).

Extending the base field if necessary, we can assume that K is algebraically closed. Let $(e_1, \ldots, e_r)$ be a basis of V with respect to which the matrix of s is lower triangular, and let $\lambda_1, \ldots, \lambda_r$ be the diagonal elements of this matrix. With respect to the basis $(e_1^{i(1)} \ldots e_r^{i(r)})_{i(1) + \cdots + i(r) = n}$ of $S_n$, ordered lexicographically, the matrix of $s^{(n)}$ is lower triangular and its diagonal elements are the products $\lambda_1^{i(1)} \ldots \lambda_r^{i(r)}$. Thus

and consequently

Indeed, the endomorphism $f = q^{-1} \sum_{g \in G} g^{(n)}$ is a projection of $S_n$ onto $R_n$, so $\mathrm{Tr}(f) = \dim_K S_n^G$. Thus, the Poincaré series of R is

and it suffices to apply Lemma 2.

By Prop. 3 of the Appendix, we can assume that K is algebraically closed. For any $g \in G$, let $\lambda_1(g), \ldots, \lambda_l(g)$ be its eigenvalues. Since every $g \in G$ is diagonalizable (Appendix, Prop. 2), $g = 1$ if and only if all of the $\lambda_i(g)$ are equal to 1, and $g \in H$ if and only if the number of $\lambda_i(g)$ equal to 1 is $l - 1$ (we then denote by $\lambda(g)$ the eigenvalue distinct from 1). Prop. 1 of no. 1 and Lemma 3 prove that

in $K[[T]]$, and hence in $K(T)$. Consequently, we have in $K(T)$

which can be written

It follows that $q - \prod_{i=1}^{l} (1 + T + \cdots + T^{k_i-1})$ vanishes for $T = 1$, so $q = k_1 k_2 \ldots k_l$, which we knew already by the Cor. of Th. 3. This granted, let Q(T) be the polynomial $(1 - T)^{-1}(q - \prod_{i=1}^{l} (1 + T + \cdots + T^{k_i-1}))$. Differentiating the equality $(1 - T)Q(T) = q - \prod_{i=1}^{l} (1 + T + \cdots + T^{k_i-1})$ and putting $T = 1$, we see that $-Q(1)$ is the value for $T = 1$ of

so

Returning to (17), we have on the other hand

Thus,

Now the elements of G that leave fixed the points of a given hyperplane leave stable a complementary line of the hyperplane (Appendix, Prop. 2), and so form a cyclic subgroup G' of G by Algebra, Chap. V, § 11, no. 1, Th. 1. Let t be the order of G'; the values of $\lambda(g)$ for $g \in G'$ are $\theta, \theta^2, \ldots, \theta^{t-1}$ with $\theta$ a primitive tth root of unity. We have $\frac{1}{1-\theta^i} + \frac{1}{1-\theta^{t-i}} = 1$, so $\sum_{g \in G', g \neq 1} \frac{1}{1-\lambda(g)} = \frac{1}{2}(t-1) = \frac{1}{2}\mathrm{Card}(H \cap G')$. The equality (18) thus proves the proposition.

Let $f$ be the automorphism of the algebra S that extends the automorphism $-1$ of V. Then $f(z) = (-1)^{\deg z} z$ for all homogeneous $z$ in S. Thus, if $-1 \in G$, every homogeneous element of odd degree of R is zero, and the $k_i$ are all even. Conversely, if the $k_i$ are all even, every element of R is invariant under $f$, and Galois theory shows that $-1 \in G$.

(ii) Identify S with the polynomial algebra $K[X_1, \ldots, X_l]$ by choosing a basis $(X_1, \ldots, X_l)$ of V, and let $(P_1, \ldots, P_l)$ be algebraically independent homogeneous elements of S generating the algebra R (Th. 3). Then the jacobian $J = \det \left( \frac{\partial P_i}{\partial X_j} \right)$ is of the form $\lambda D$, where $\lambda \in K^*$.

a) With the notation in (ii), we have

so, for all $g \in G$,

hence J is anti-invariant under G. On the other hand, the field of fractions N of S is a Galois extension of the field of fractions E of R (no. 2); if $\Delta$ is a derivation of E with values in an extension field $\Omega$ of N, $\Delta$ extends to a derivation of N with values in $\Omega$ (Algebra, Chap. V, § 16, no. 4, Th. 3); since the $P_i$ are algebraically independent, it follows that $dP_1 \wedge \ldots \wedge dP_l \neq 0$, hence $J \neq 0$.

b) Let z be an element of S anti-invariant under G. We show that z is divisible by D in S. Let a be a non-zero vector in V. The elements of G that leave Ka stable leave stable a complementary hyperplane L (Appendix, Prop. 2); an element of G leaving Ka stable is 1 or a pseudo-reflection with vector a if and only if it induces 1 on L; the pseudo-reflections with vector a belonging to G thus constitute, together with 1, a cyclic subgroup $G'$ of G; let t be its order. There exists a basis $(X_1, \ldots, X_l)$ of V such that $a = X_1$, $X_2 \in L, \ldots, X_l \in L$, and z can be identified with a polynomial $P(X_1, \ldots, X_l)$ with coefficients in K. From $g(z) = (\det g)^{-1} z$ for $g \in G'$, we see that $X_1$ only appears in $P(X_1, \ldots, X_l)$ with exponents congruent to -1 modulo t. Thus, $P(X_1, \ldots, X_l)$ is divisible by $X_1^{t-1} = a^{t-1}$. Now D is, up to a scalar factor, the product of the $a^{t-1}$ for those $a \in V$ such that $t > 1$, and these elements of S are mutually coprime. Since S is factorial, z is divisible by D.

c) By a) and b), J is divisible by D in S. Now

(Prop. 3), so $\deg J = \deg D$, hence $J = \lambda D$ with $\lambda \in K$. Since $J \neq 0, \lambda \in K^*$. This proves (ii).

d) Parts a) and c) of the proof show that D is anti-invariant under G. Next, if $y \in R$, it is clear that $yD$ is anti-invariant under G. Finally, if $z \in S$ is anti-invariant under G, we have seen in b) that there exists $y \in S$ such that $z = yD$. Since S is integral, $y \in R$. This proves (i).

To say that B is ramified over R means that its inertia group $G^T(B)$ does not reduce to the identity, in other words that there exists $g \neq 1$ in G such that $g(z) \equiv z \pmod{B}$ for all $z \in S$. Since S is a factorial ring, B is a principal ideal Sa, and a must divide all the elements $g(z) - z$ ($z \in S$); now, for $z \in V$, these elements are homogeneous of degree 1 and are all non-zero (since $g \neq 1$); thus, a must be homogeneous of degree 1, in other words $a \in V$. Then there exists a linear form f on V such that $g = s_{a,f}$. Conversely, if g is a pseudo-reflection $s_{a,f}$ different from 1, then $g(z) \equiv z \pmod{Sa}$ for all $z \in S$, so g belongs to the inertia group of the prime ideal $B = Sa$. This proves the first assertion of the lemma and the characterizations of $G^Z(B)$ and $G^T(B)$. Since q is coprime to the characteristic p of K (which is also that of S(B)), the extension S(B) of $R(p)$ is separable (Commutative Algebra, Chap. V, § 2, no. 2, Cor. of Prop. 5). The equality $e(B/p) = \operatorname{Card}(H_a)$ follows from this (Commutative Algebra). Since $e(B/p)$ is coprime to p, the coefficient of B in $\operatorname{div}(D_{S/R})$ is $e(B/p) - 1$ (Commutative Algebra). This proves the lemma.

4 In this number, we use results from chapters in preparation in the book on Commutative Algebra. We indicate them by the symbol "Commutative Algebra". 5 This means that, for all $i \in \{1, 2, \ldots, s\}$, the canonical image of $\alpha_i$ in the ring

is not a zero divisor in this ring.

When S is an R-module of finite type, b) is a consequence of a).

See Commutative Algebra for the proof.

(i) G is generated by pseudo-reflections; (ii) S is a free graded R-module; (iii) R is a graded polynomial algebra over K.

The equivalence of (ii) and (iii) follows from no. 2 and Lemma 5. The implication (i) $\Longrightarrow$ (ii) follows from Th. 1.

We show that (iii) $\Longrightarrow$ (i). Let G' be the subgroup of G generated by the pseudo-reflections belonging to G, and let R' be the subalgebra of S consisting of the elements invariant under G'. We have R $\subset$ R' $\subset$ S. By Lemma 4, div(D_{S/R}) = div(D_{S/R'}), so div(D_{R'/R}) = 0. Assume then that R is a graded polynomial algebra. Since this is also the case for R' (since G' is generated by pseudo-reflections), Lemma 5 shows that the R-module R' admits a homogeneous basis (Q_1, ..., Q_m); let q_i = deg(Q_i). Put

The fact that div(D_{R'/R}) is zero shows that div(d) = 0 (Commutative Algebra), which means that d belongs to K*. On the other hand Tr_{R'/R}(Q_i Q_j) is a homogeneous element of degree q_i + q_j, and d is homogeneous of degree $2 \sum_i q_i$. Thus, $\sum_i q_i = 0$, i.e. $q_i = 0$ for all i, which means that R' = R, and so G' = G by Galois theory. This proves that G is generated by pseudo-reflections. Q.E.D.

(i) The different of S over R is the principal ideal SD.

(ii) Identify S with the algebra $K[X_1, \ldots, X_l]$ by choosing a basis $(X_1, \ldots, X_l)$ of V, and let $P_1, \ldots, P_l$ be algebraically independent homogeneous elements generating the algebra R. Then the jacobian $J = \det \left( \frac{\partial P_i}{\partial X_j} \right)$ is of the form $\lambda D$ where $\lambda \in K^*$.

(iii) $\sum_{i=1}^l (\deg(P_i) - 1) = \mathrm{Card}(H)$.

(iv) The set of anti-invariant elements of S is RD.

Assertion (i) follows from Lemma 4. Assertion (ii) follows from the fact that SJ is the different of S over R. (Commutative Algebra). Assertion (iii) is obtained by writing down the fact that the homogeneous polynomials D and J are of the same degree. The proof of (iv) is then the same as that given in no. 4 (proof of Prop. 5, parts b) and d)).