Lemma 4

Let $E$ be a right $A$-module and $M$ a left $A$-module such that $E$ is $M$-flat. If $N$ is either a submodule or a quotient module of $M$, then $E$ is $N$-flat.

The case in which $N$ is a submodule is easy, as, if $N'$ is a submodule of $N$, the composite homomorphism

is injective, hence so is $E \otimes_A N' \to E \otimes_A N$. Suppose then that $N$ is a quotient module of $M$, that is there exists an exact sequence $0 \to R \xrightarrow{i} M \xrightarrow{p} N \to 0$.

Let $N'$ be a submodule of $N$ and $M' = \overline{p}(N')$. Let $i'$ denote the mapping of $R$ to $M'$ with the same graph as $i$, $p'$ the surjection $M \to N'$ with the same graph as the restriction of $p$ to $M'$, $r$ the identity mapping of $\mathbf{R}$ to $\mathbf{R}$, $m$ the canonical injection $M' \to M$ and $n$ the canonical injection $N' \to N$. The diagram

is commutative and its rows are exact.

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 left $A$-module homomorphism $v$. The diagram

is commutative and its rows are exact by Lemma 1 of no. 1. Moreover, since $E$ is $M$-flat, the homomorphism $T(m)$ is injective. As $T(r)$ and $T(p')$ are surjective, it follows from § 1, no. 4, Corollary 2 to Proposition 2 that $T(n)$ is injective, which proves the lemma.