Proposition

If $f: A \rightarrow B$ is a ring homomorphism and $M$ is a flat $A$-module, then $M_B = B \otimes_A M$ is a flat $B$-module.

Lemma

We will show in general that a ring $A$ is a flat $A$-module. Let $f: M \rightarrow M’$ be injective. Then $f \otimes 1 : M \otimes A \rightarrow M’ \otimes A$ is equal to $(p’)^{-1} \circ f \circ p$ where $p: M \otimes A \rightarrow M$ and $p’: M’ \otimes A \rightarrow M’$ are the canonical isomorphisms. Then $f \otimes 1$ is a composition of injective maps, so $f \otimes 1$ is injective. By Proposition 2.19 [Atiyah], $A$ is a flat $A$-module.

Solution

Let $N_1, N_2$ be $B$-modules and $j: N_1 \rightarrow N_2$ be injective.

We will consider $j \otimes 1: N_1 \otimes_B M_B \rightarrow N_2 \otimes_B M_B$. We want to show that $j \otimes 1$ is injective.

$N_1 \otimes_B M_B$ is isomorphic to $N_1 \otimes_B (M \otimes_A B)$ by the canonical isomorphism. $N_1 \otimes_B (M \otimes_A B)$ is isomorphic to $(N_1 \otimes_B M) \otimes_A B$ by the canonical isomorphism [Exercise 2.15, P.27, Atiyah]. Similarly, $N_2 \otimes_B M_B$ is isomorphic to $(N_2 \otimes_B M) \otimes_A B$. Then it suffices to show that $(n \otimes m) \otimes b \mapsto (j(n) \otimes m) b$ is injective.

By Proposition 2.19, $j \otimes 1: N_1 \otimes M \rightarrow N_2 \otimes M$ is injective because $M$ is flat. By the Lemma above, $B$ is a flat $B$-module, so $j \otimes 1: (N_1 \otimes_B M) \otimes_A B \rightarrow (N_2 \otimes_B M) \otimes_A B$ is injective. Thus $(n \otimes m) \otimes b \mapsto (j(n) \otimes m) \otimes b$ is injective, so $M_B$ is a flat $B$-module.