Proposition 3

Let $A$ be a ring and $E$ a right $A$-module.

(i) Suppose that $E$ is flat. For every element $a \in A$ which is not a right divisor of $0(*)$, the relations $x \in E,\ xa = 0$ imply $x = 0$.

(*) Recall that a right (resp. left) divisor of 0 in a ring $A$ is an element $b \in A$ such that the mapping $x \mapsto xb$ (resp. $x \mapsto bx$) is not injective.

(ii) Suppose that $A$ is an integral domain in which every finitely generated ideal is principal (for example a principal ideal domain (Algebra, Chapter VII, , no. 1)). Then for $E$ to be flat it is necessary and sufficient that $E$ be torsion-free

We prove (i). Let $v : A, \to A$, be the left $A$-module homomorphism $t \mapsto ta$; the hypothesis implies that $v$ is injective. As $E$ is flat, the homomorphism $1_E \otimes v : E \otimes_A A, \to E \otimes_A A$, is also injective. When $E \otimes_A A$, is canonically identified with $E$, $1, \otimes v$ becomes the endomorphism $x \mapsto xa$ of $E$. Thus the relation $xa = 0$ implies $x = 0$.

We prove (ii). By (i), if $E$ is flat, $E$ is torsion-free. Conversely, let $E$ be a torsion-free $A$-module; we verify that, for every finitely generated ideal $a$ of $A$, the canonical homomorphism $E \otimes_A a \to E$ is injective (no. 3, Remark 1). This assertion is obvious if $a = (0)$; otherwise, by hypothesis $a = Aa$ for some $a \in A, a \neq 0$, and $t \mapsto ta$ is then an isomorphism $v$ of $A$ onto $a$; using $i$ to denote the canonical injection $a \to A$, $i \circ v$ is the homothety with ratio $a$ on $E$ and is injective since $E$ is assumed to be torsion-free. Then $1, \otimes (i \circ v) = (1, \otimes i) \circ (1_E \otimes v)$; as $1_E \otimes v$ is an isomorphism, $1, \otimes i$ is injective, which completes the proof.