Proposition 7

Let $A$ be a reduced local ring, $m$ its maximal ideal, $(p_i)_{i \in I}$ the family of minimal prime ideals of $A$, $K_i$ the field of fractions of $A/p_i$ and $M$ a finitely generated $A$-module. For $M$ to be free it is necessary and sufficient that

If M is free, clearly the two sides of (1) are equal to the rank of M for all $i \in I$. Suppose now that the condition is satisfied and denote by $n$ the common value of the two sides of (1); by Corollary 2 to Proposition 4 M has a system of $n$ generators $x_j$ ($1 \leq j \leq n$). Suppose first that $A$ is an integral domain, in which case $p_i = 0$ for all $i \in I$. The elements $1 \otimes x_j$ ($1 \leq j \leq n$) generate the vector space $K \otimes M$ over the field of fractions K of $A$; but as by hypothesis this space is of dimension $n$ over K, the elements $1 \otimes x_j$ are linearly independent over K. It follows (Algebra, Chapter II, § 1, no. 13, Remark 1) that the $x_j$ are linearly independent over $A$ and hence form a basis of M.

Passing to the general case, there exists a surjective homomorphism $v$ from $L = A''$ onto M. Consider the commutative diagram

where $u$ (resp. $u'$) is the mapping $x \mapsto (\phi_i(x))$ (resp. $y \mapsto (\psi_i(y))$),

(resp. $\psi_i : M \to (A/p_i) \otimes M$) being the canonical mapping, and $v'$ is the product of the $1_{A/p_i} @ v_e$. Then $(A/p)/(m/p_i) \otimes_{A/p_i} ((A/p_i) \otimes_A M) = (A/m) \otimes_A M$ and, as $A/p_i$ is a local integral domain, it follows from the first part of the argument that each of the $1_{A/p_i} \otimes v$ is an isomorphism; then so is $v'$. On the other hand, as $A$ is reduced, $\bigcap_i p_i = (0)$ ($§ 2$, no. 6, Proposition 13), whence $\bigcap_i (p_i L) = 0$ since $L$ is free (Algebra, Chapter II, $§ 3$, no. 7, Remark); as $p_i L$ is the kernel of $\phi_i$, this shows that $u$ is injective. It follows that $v' \circ u = u' \circ v$ is injective, hence $v$ is injective and, as $v$ is surjective by definition, this shows that $M$ is free.