Exercise 2.8

Proof of $i)$. Suppose $f:L^{\prime }\to L$ is an injective $A$-module homomorphism. Then, since $M,N$ are flat, $f\otimes 1_M:L^{\prime }\otimes M\to L\otimes M$ is injective, and so is $(f\otimes 1_M)\otimes 1_N:(L^{\prime }\otimes M)\otimes N\to (L\otimes M)\otimes N$. By the associativity of the tensor product (Prop.  $2.4$), $f\otimes (1_M\otimes 1_N):L^{\prime }\otimes (M\otimes N)\to L\otimes (M\otimes N)$ is injective, and thus $M\otimes N$ is flat by Prop.  $2.19$. □

Proof of $\mathit{ii})$. Suppose $f:L^{\prime }\to L$ is an injective $A$-module homomorphism. Since $B$ is a flat $A$-algebra, $f{\text{⊗}}_A{1}_B:L^{\prime }{\text{⊗}}_{A}B\to L{\text{⊗}}_{A}B$ is an injective $A$-module homomorphism. This is also an injective $B$-module homomorphism by using the $B$-module structure $b^{\prime }(\ell \otimes b)=\ell \otimes (b^{\prime }b)$ where $\ell \in L^{\prime }$ or $L$. Since $N$ is a flat $B$-module, $(f{\text{⊗}}_A{1}_B){\text{⊗}}_B{1}_N:(L^{\prime }{\text{⊗}}_{A}B){\text{⊗}}_{B}N\to (L{\text{⊗}}_{A}B){\text{⊗}}_{B}N$ is injective. By Exercise $2.15$ in the text, we can associate these tensor products differently to get that $f{\text{⊗}}_A(1_B{\text{⊗}}_B{1}_N):L^{\prime }{\text{⊗}}_A(B{\text{⊗}}_{B}N)\to L{\text{⊗}}_A(B{\text{⊗}}_{B}N)$ is injective. But by using the isomorphism $B{\text{⊗}}_{B}N\simeq N$ in Prop.  $2.14$, we then have $f{\text{⊗}}_A{1}_N:L^{\prime }{\text{⊗}}_{A}N\to L{\text{⊗}}_{A}N$ is injective, and so $N$ is flat by Prop.  $2.19$. □