Proposition 4

Let $E$ be a right $A$-module. The three following properties are equivalent:

(a) $E$ is flat. (b) For every exact sequence of right $A$-modules of the form

(1)

and every left $A$-module $F$, the sequence

(2)

is exact.

(c) There exists an exact sequence (1), where $H$ is flat, such that the sequence (2) is exact for every left $A$-module $F$ of the form $A_s/a$, where $a$ is a finitely generated left ideal of $A$.

We show first that (a) implies (b). The left $A$-module $F$ is isomorphic to a quotient of a free module (Algebra, Chapter II, § 1, no. 11, Proposition 20); in other words, we have an exact sequence

where L is free. Consider the diagram

(3)

It follows immediately that this diagram is commutative, and its rows and columns are exact by Lemma 1 of no. 1; moreover, as $l_G \otimes p$ and $l_H \otimes p$ are surjective (no. 1, Lemma 1), we have $G \otimes F = \operatorname{Coker}(l_G \otimes i)$,

$w \otimes 1_1$ is surjective (no. 1, Lemma 1); finally, as L is free and hence flat, $v \otimes 1_L$ is injective. Thus the snake diagram (§ 1, no. 4, Proposition 2, (iii)) can be applied to prove the existence of an exact sequence

(4)

This being so, if E is flat, $l_E \otimes i$ is injective, in other words $\operatorname{Ker}(l_E \otimes i) = 0$, and the exact sequence (4) shows that $v \otimes 1_F$ is injective, hence the sequence (2) is exact (taking account of Lemma 1 of no. 1).

As (b) obviously implies (c), it remains to prove that (c) implies (a). Consider the diagram (3) in the case $\mathbf{R} = \mathbf{a}, \mathbf{L} = \mathbf{A}_s, \mathbf{F} = \mathbf{A}_s/\mathbf{a}$ and apply the exact sequence (4). By hypothesis $v \otimes 1_F$ is injective, hence $\operatorname{Im}(d) = 0$; moreover, as H is flat, we have $\operatorname{Ker}(1, @i) = 0$; the exactness of the sequence (4) then implies $\operatorname{Ker}(l_E \otimes i) = 0$, in other words, $1, \otimes i$ is injective and this proves that E is flat (no. 3, Remark 1).