Proposition 1

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

(a) For a sequence $N' \xrightarrow{v} N \xrightarrow{w} N''$ of left $A$-modules to be exact, it is necessary and sufficient that the sequence

be exact.

(b) $E$ is flat and, for every left $A$-module $N$, the relation $E \otimes_A N = 0$ implies $N = 0$.

(c) $E$ is flat and, for every homomorphism $v : N' \to N$ of left $A$-modules, the relation $1_E \otimes v = 0$ implies $v = 0$.

(d) $E$ is flat and, for every maximal left ideal $m$ of $A$, $E \neq Em$.

To simplify the writing we set $T(Q) = E \otimes_A Q$ for every left $A$-module $Q$ and $T(v) = 1_E \otimes v$ for every homomorphism $v$ of left $A$-modules.

We prove first the equivalence of (a), (b) and (c).

We prove that (a) implies (b). If (a) holds, clearly $E$ is flat (§ 2, no. 3, Proposition 1). On the other hand, let $N$ be a left $A$-module such that $T(N) = 0$ and consider the sequence $0 \to N \to 0$; the hypothesis $T(N) = 0$ means that the sequence $0 \to T(N) \to 0$ is exact. By (a) the sequence $0 \to N \to 0$ is exact, whence $N = 0$.

We show that (b) implies (c). Suppose that (b) holds and let $u : N' \to N$ be a homomorphism and $I$ its image. As the image of $T(v)$ is identified with $T(I)$ (§ 2, no. 3, Remark 2), the hypothesis $T(v) = 0$ implies $T(I) = 0$, hence $I = 0$ by (b) and consequently $v = 0$.

We show that (c) implies (a). Suppose then that (c) holds and consider a sequence

of homomorphisms of left $A$-modules and the corresponding sequence

If the sequence (1) is exact, so is (2), since $E$ is flat (§ 2, no. 3, Proposition 1). Conversely, if (2) is exact, we have first $T(w \circ v) = T(w) \circ T(v) = 0$, hence $w \circ v = 0$ by hypothesis. Set $I = v(N')$ and $K = \overline{w}(0)$; then $I$ is contained in $K$ by the above. Consider the exact sequence

$i$ and $p$ being the canonical mappings. As $E$ is flat, the sequence is exact, in other words, $T(K/I)$ is isomorphic to $T(K)/T(I)$, which is 0 by hypothesis, since $T(I)$ (resp. $T(K)$) is identified with the image of $T(v)$ (resp. the kernel of $T(w)$) (§ 2, no. 3, Remark 2). But the relation $T(p) = 0$ implies $p = 0$ by hypothesis, hence $K = I$, which proves that the sequence (1) is exact.

Finally we show the equivalence of (b) and (d). If (b) holds, then

since $A_s/m \neq 0$; whence (d). Conversely, suppose that (d) holds; every left ideal $a \neq A$ of $A$ is contained in a maximal left ideal $m$ (Algebra, Chapter I, §8, no. 7, Theorem 2), then the hypothesis $E \neq Em$ implies $E \neq Ea$, in other words, $E \otimes_A (A_s/a) \neq 0$. That is to say, for every monogenous left $A$-module $N \neq 0$, $T(N) \neq 0$. If now $N$ is any left $A$-module \#0, it contains a monogenous submodule $N' \neq 0$; since $E$ is flat, $T(N')$ is identified with a subgroup of $T(N)$; we have just seen that $T(N') \neq 0$, hence $T(N) \neq 0$.