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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.

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  • $\begingroup$ What is your question? $\endgroup$ – Matthias Klupsch Sep 7 '16 at 11:40
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    $\begingroup$ Which definition of projective are you using? And have you seen that projective implies flat? $\endgroup$ – Tobias Kildetoft Sep 7 '16 at 11:43
  • $\begingroup$ @MatthiasKlupsch it was poorly written. Now I think it's good. $\endgroup$ – Rafael Holanda Sep 7 '16 at 11:45
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    $\begingroup$ So now the question is: Can you show the claim when $P$ is a free module? $\endgroup$ – Tobias Kildetoft Sep 7 '16 at 11:47
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    $\begingroup$ So you need to show that this also holds when you take a direct summand of a module for which it holds. $\endgroup$ – Tobias Kildetoft Sep 7 '16 at 11:58
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For the fun of it, there's a way which does not make use of the non-categorical notion of free modules:

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.

Remark: While this might seem overkill, it's actually the way to go in more abstract settings. For example, when showing that K-projective complexes are K-flat, one has to use the Hom-duality (at least to my knowledge).

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  • $\begingroup$ May I please ask for a reference to the fact. $\endgroup$ – Atbey Mar 15 '18 at 21:31

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