Lemma 1

Assume that A is factorial (Commutative Algebra, Chap. VII, § 3, no. 1, Def. 1). (i) The ring $A[P]$ is factorial. (ii) If $u, v$ are non-proportional elements of P, the elements $1 - e^u, 1 - e^v$ of $A[P]$ are relatively prime.

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 $A[P]$.