Corollary 1

Suppose that the equivalent properties of the statement of Theorem 1 hold. Let m be an integer > 0 such that, for every family $(x_i)_{1 \leq i \leq m}$ of elements of P, there exists a family $(a_i)_{1 \leq i \leq m}$ of elements of A, which are not all divisors of zero and for which $\sum_{i=1}^m a_i x_i = 0$. Then, for all $p \in \operatorname{Spec}(A)$, $r_p \leq m$.

Let p be a prime ideal of A; set $r = r_p$ and let $(y_j)_{1 \leq j \leq r}$ be a basis of the free $A_p$-module $P_p$. There exist elements $x_j$ ($1 \leq j \leq r$) of P and $s \in A - p$ such that $y_j = x_j / s$ for all $j$. Then for every family $(a_j)_{1 \leq j \leq r}$ of elements of A such that $\sum_{j=1}^r a_j x_j = 0$, $\sum_{j=1}^r (a_j/1) y_j = 0$ in $P_p$, whence $a_j/1 = 0$ for $1 \leq j \leq r$. As $A - p$ does not contain 0, this shows that the $a_j$ are all divisors of zero in A (§ 2, no. 1, Remark 3), hence of necessity $r \leq m$.