# About projective modules, tensor products and injective homomorphisms.

Let $R$ be a ring with identity. Let $P$ be a projective left $R$-module and $M,N$ two right $R$-modules. Prove that for every injective homomorphism $u:M\rightarrow N$, the homomorphism $$u\otimes1_P:M\otimes P\rightarrow N\otimes P$$ is injective, where $1_P$ is the identity map.

• What is your question? – Matthias Klupsch Sep 7 '16 at 11:40
• Which definition of projective are you using? And have you seen that projective implies flat? – Tobias Kildetoft Sep 7 '16 at 11:43
• @MatthiasKlupsch it was poorly written. Now I think it's good. – Rafael Holanda Sep 7 '16 at 11:45
• So now the question is: Can you show the claim when $P$ is a free module? – Tobias Kildetoft Sep 7 '16 at 11:47
• So you need to show that this also holds when you take a direct summand of a module for which it holds. – Tobias Kildetoft Sep 7 '16 at 11:58

Fact: A morphism of abelian groups $A\to B$ is a monomorphism if and only if the dual morphism $$\text{Hom}_{\mathbb Z}(B,{\mathbb Q}/{\mathbb Z})\longrightarrow \text{Hom}_{\mathbb Z}(A,{\mathbb Q}/{\mathbb Z})$$ is an epimorphism.
Applying this to your example, it is therefore sufficient to show that $$\text{Hom}_{\mathbb Z}(N\otimes_R P,{\mathbb Q}/{\mathbb Z})\to \text{Hom}_{\mathbb Z}(M\otimes_R P,{\mathbb Q}/{\mathbb Z})$$ is surjective. By the adjunction of $\otimes$ and $\text{Hom}$, this is equivalent to $$\text{Hom}_R(P, \text{Hom}_{\mathbb Z}(N, {\mathbb Q}/{\mathbb Z}))\to \text{Hom}_R(P, \text{Hom}_{\mathbb Z}(M, {\mathbb Q}/{\mathbb Z}))$$ being surjective. Now, $$\text{Hom}_{\mathbb Z}(N,{\mathbb Q}/{\mathbb Z})\longrightarrow \text{Hom}_{\mathbb Z}(M,{\mathbb Q}/{\mathbb Z})$$ is surjective by the assumption that $M\to N$ is injective, and $\text{Hom}_R(P, -)$ preserves epimorphisms by definition.