Let $(p_1, p_2, \ldots, p_l)$ be a basis of P, and $X_1, X_2, \ldots, X_l$ be indeterminates. The A-linear map from $A[X_1, \ldots, X_l, X_1^{-1}, \ldots, X_l^{-1}]$ to $A[P]$ that takes $X_1^{n_1} X_2^{n_2} \ldots X_l^{n_l}$ (where $n_1, n_2, \ldots, n_l \in \mathbf{Z}$) to $e^{n_1 p_1 + \cdots + n_l p_l}$ is an isomorphism of rings. Now $A[X_1, \ldots, X_l]$ is a factorial ring (Commutative Algebra, Chap. VII, § 3, no. 5), and $A[X_1, \ldots, X_l, X_1^{-1}, \ldots, X_l^{-1}]$ is the ring of fractions of $A[X_1, \ldots, X_l]$, hence is also factorial.

Let $P'$ (resp. $P''$) be the set of elements of P of which some multiple belongs to $\mathbf{Z}u + \mathbf{Z}v$ (resp. $\mathbf{Z}u$). Then the groups $P/P'$ and $P'/P''$ are torsion-free, so there exists a complement of $P''$ in $P'$ and a complement of $P'$ in P. Consequently, there exists a basis $(z_1, z_2, \ldots, z_l)$ of the $\mathbf{Z}$-module P and rational integers $j, m, n$ such that $u = jz_1, v = mz_1 + nz_2, j > 0, n > 0$. Putting $X_i = e^{z_i}$ for $1 \leq i \leq l$, we then have $1 - e^u = 1 - X_1^j, 1 - e^v = 1 - X_1^m X_2^n$. Let K be an algebraic closure of the field of fractions of A, so that $A[P]$ can be identified with a subring of the ring $B = K[X_1, \ldots, X_l, X_1^{-1}, \ldots, X_l^{-1}]$. For any $j$th root of unity $z$, $1 - zX_1$ is extremal in

moreover, the ideal generated by $1 - zX_1$ contains no monomial in the $X_i$. We conclude that the ideal $(1 - zX_1)B$ of B is a prime ideal of height 1

(Commutative Algebra, Chap. VII, § 1, no. 6), hence that $1 - zX_1$ is extremal in B. The extremal factors of $1 - X_1^j$ in B are thus of the form $1 - zX_1$. Now none of these factors divide $1 - X_1^m X_2^n$ in B (for the homomorphism $f$ from B to B such that $f(X_1) = z^{-1}, f(X_i) = X_i$ for $i \geq 2$, satisfies

Thus, $1 - X_1^j$ and $1 - X_1^m X_2^n$ are relatively prime in B. Consequently, any common divisor of $1 - X_1^j$ and $1 - X_1^m X_2^n$ in $A[P]$ is invertible in B and so, up to multiplication by an element of the form $X_1^{k_1} X_2^{k_2} \ldots X_l^{k_l}$, is equal to an element $a$ of A; in other words, $a$ divides 1 in A, hence is invertible in A. Thus, finally, $1 - X_1^j$ and $1 - X_1^m X_2^n$ are relatively prime in $A[P]$.

Put $x = \sum_p x_p e^p, y = \sum_r y_r e^r$ and $xy = \sum_t z_t e^t$. Then $r \leq q$ for all $r \in P$ such that $y_r \neq 0$ and $z_t = \sum_{p+r=t} x_p y_r$.

If $t = p + q = p' + r$ with $p \in X$ and $x_{p'} y_r \neq 0$, then $r \leq q$, hence $p' \geq p$ and consequently $p' = p$. Thus $z_{p+q} = x_p y_q = x_p \neq 0$. This shows that $X + q$ is contained in the support of the product $xy$.

On the other hand, if $t = p' + r$ with $x_{p'} y_r \neq 0$, there exists $p \in X$ such that $p' \leq p$ and we have $t \leq p + q$. The maximal support of $xy$ is therefore contained in $X + q$. Since no two elements of $X + q$ are comparable, it follows that $X + q$ is exactly the maximal support of $xy$ and we have seen above that $z_{p+q} = x_p$ for $p \in X$, which completes the proof of the lemma.

The weights $w(p)$ for $w \in W$ and $p \in P \cap C$ are pairwise distinct. It follows that the $J(e^p)$ for $p \in P \cap C$ are linearly independent.

On the other hand, let $x = \sum_p x_p e^p$ be an anti-invariant element of $A[P]$. If $p_0$ belongs to a wall, it is invariant under a reflection $s \in W$ and

It follows that $2x_{p_0} = 0$, so $x_{p_0} = 0$. Since every element that does not belong to any wall can be written uniquely in the form $w(p)$ with $w \in W$ and $p \in P \cap C$, we thus have Since $w(x) = \sum_p x_p e^{w(p)} = \varepsilon(w) \sum_p x_p e^p$, $x_{w(p)} = \varepsilon(w)x_p$ and we deduce from (2) that which completes the proof.

Consider now the element $d$ of the algebra $A[\frac{1}{2}P]$ defined by

Since $\rho \in P$, $d \in A[P]$.

We know that, for $1 \leq i \leq l$, the reflection $s_i = s_{\alpha_i}$ leaves stable the set of positive roots other than $\alpha_i$ and that $s_i(\alpha_i) = -\alpha_i$ (§ 1, no. 6, Cor. 1 of Prop. 17). Hence,

Since the $s_i$ generate $W$, this proves the first assertion in (i). The second assertion in (i) follows immediately from (3) and Lemma 2, noting that 1 is the unique maximal term of $1 - e^{-\alpha}$ for $\alpha \in R, \alpha > 0$.

Assume now that $A = \mathbf{Z}$. By Prop. 1,

On the other hand, it is clear that

If $p \in P \cap C$ with $p \neq \rho$, then $p > \rho$ and the coefficient of $e^p$ in $d$ is zero by (5). Thus, $c_p = 0$. Moreover, comparison of the coefficients of $e^\rho$ in (4) and (5) shows that $c_\rho = 1$ and hence that $d = J(e^\rho)$.

We continue to assume that $A = \mathbf{Z}$. Let $p \in P, \alpha \in \mathbf{R}$ and $M$ be a system of representatives of the right cosets of $W$ with respect to the subgroup $\{1, s_\alpha\}$. Then,

Now $s_\alpha w(p) = w(p) - \langle \alpha^*, w(p) \rangle \alpha = w(p) + n_w \alpha$, with $n_w \in \mathbf{Z}$. Thus,

If $n_w \geq 0$, it is clear that $1 - e^{n_w \alpha}$ is divisible by $1 - e^\alpha$ and this is also true when $n_w < 0$ since $1 - e^{n_w \alpha} = -e^{n_w \alpha}(1 - e^{-n_w \alpha})$. Hence, $J(e^p)$ is divisible by $1 - e^\alpha$ in $\mathbf{Z}[P]$.

By Lemma 1, $\mathbf{Z}[P]$ is factorial and the elements $1 - e^\alpha$ for $\alpha \in \mathbf{R}$ and $\alpha > 0$ are mutually prime. It follows that $J(e^p)$ is divisible in $\mathbf{Z}[P]$ by the product $\prod_{\alpha > 0} (1 - e^\alpha)$, and hence also by $d = e^{-\rho} \prod_{\alpha > 0} (1 - e^\alpha)$.

Returning to the general case, by extension of scalars from $\mathbf{Z}$ to $A$, we deduce from the above that $d = J(e^\rho)$ and that every element $J(e^p)$ is divisible by $d$. Since $e^\rho$ is the unique maximal term of $d$, the Remark of no. 2 shows that there exists a unique element $y \in A[P]$ such that $J(e^p) = dy$ and it follows immediately that $y$ is invariant under $W$, and hence that $d$ and $J(e^p)$ are anti-invariant. This proves (i) and (ii).

Finally, if 2 is not a zero divisor in $A$, the Remark of no. 2 and Prop. 1 imply (iii).

from which (b) follows. Hence $J' \in \mathfrak{S}$, a contradiction. Thus, $J = I$ and the lemma is proved.

We now prove Prop. 3. We apply Lemma 4 with $I = P \cap \overline{C}$. Let $q \in I$, and let $I_q$ be the set of $p \in I$ such that $p \leq q$. If $p \in I_q$, the relations

imply that and hence that The set $I_q$ is thus bounded. Since $I$ is discrete, it follows that $I_q$ is finite, and it is clear that $I$ satisfies the condition (MIN). On the other hand, for all $p \in I$, so by (6). The proposition now follows from Lemmas 3 and 4.

Lemma 2 implies that the image under $\varphi$ of the monomial $X_1^{n_1} \ldots X_l^{n_l}$ is an element with unique maximal term $e^{n_1 \overline{\omega}_1 + \cdots + n_l \overline{\omega}_l}$. Since every element of $P \cap \overline{C}$ can be written uniquely in the form $n_1 \overline{\omega}_1 + \cdots + n_l \overline{\omega}_l$, Prop. 3 shows that the images under $\varphi$ of the monomials $X_1^{n_1} \ldots X_l^{n_l}$ are a basis of $A[P]^W$, hence the theorem.