Exercise 22, § 2

Let E be a left A-module. For every right ideal a of A and every element $a \in A$, denote by $a : a$ the set of $x \in A$ such that $ax \in a$ and by $aE : a$ the set of $y \in E$ such that $ay \in aE$. Then clearly $(a : a)E \subset aE : a$. Show that, for E to be flat, it is necessary and sufficient that, for every right ideal a of A and every element $a \in A$, $(a : a)E = aE : a$. (To see that the condition is necessary, consider the exact sequence of right A-modules $0 \to (a : a)/a \xrightarrow{\psi} A_a/a \xrightarrow{\phi} A_a/a$, where $\psi$ is the canonical injection and $\phi$ the mapping obtained by taking quotients under left multiplication by a. To see that the condition is sufficient apply the criterion of Corollary 1 to Proposition 13 of no. 11: starting with a relation $\sum_{i=1}^n a_i x_i = 0$, where $a_i \in A, x_i \in E$, apply the hypothesis to the ideal $a_2 = \sum_{t=a}^n a_t A$ and the element $a$, and argue by induction on $n$.)