Let $A$ be a ring, $P$ a projective left $A$-module and $E,F$ two
right $A$-modules. If $u:E\rightarrow F$ is an injective homomorphism,
the homomorphism $$u\otimes 1_P:E\otimes_A P\rightarrow F\otimes_A P$$
is injective.
Attempt:
Since $P$ is projective, there exists a free $A$-module $L$ with submodules $R,Q$ such that $\phi:R\oplus Q\rightarrow L,\,(r,q)\mapsto r+q,$ is an $A$-module isomorphism and there exists an $A$-module isomorphism $f:P\rightarrow R$. Thus the mapping
$$h:P\oplus Q\rightarrow L,\,(p,q)\mapsto f(p)+q,$$
is an $A$-module isomorphism. Furthermore, the mapping
$$g:E\otimes_A(P\oplus Q)\rightarrow(E\otimes_AP)\oplus(E\otimes_A Q)$$
such that $g(x\otimes(p,q))=(x\otimes p,x\otimes q)$, for $x\in E$ and $(p,q)\in P\oplus Q$, is a $\mathbf{Z}$-module isomorphism. Similarly
$$g':F\otimes_A(P\oplus Q)\rightarrow(F\otimes_A P)\oplus(F\otimes_A Q)$$
such that $g'(y\otimes(p,q))=(y\otimes p,y\otimes q)$, for $y\in F$ and $(p,q)\in P\oplus Q$, is a $\mathbf{Z}$-module isomorphism.
On the other hand $u\otimes 1_P:E\otimes_A P\rightarrow F\otimes_A P$ and $u\otimes 1_Q:E\otimes_A Q\rightarrow F\otimes_A Q$ are $\mathbf{Z}$-module homomorphisms. Hence
$$(u\otimes 1_P)\oplus(u\otimes 1_Q):(E\otimes_A P)\oplus(E\otimes_A Q)\rightarrow(F\otimes_A P)\oplus(F\otimes_A Q)$$
is a $\mathbf{Z}$-linear mapping. It follows that $\left[(u\otimes 1_P)\oplus(u\otimes 1_Q)\right]\circ g=g'\circ(u\otimes1_{P\oplus Q})$.
I don't know how much of this is useful. What should be the strategy here?
Best Answer
Recall that every projective $A$-module is flat, hence by definition of flatness, we have that $E \otimes_A P \to F \otimes_A P$ is an injective group homomorphism. We will prove it in two steps.
First, we prove that for any collection $\{M_i\}_{i \in I}$ of (left) $A$-modules, we have that $M = \oplus_{i \in I} M_i$ is flat if and only if $M_i$ is flat for all indices $i.$
Proof. Given any (right) $A$-module $N,$ we have a natural isomorphism $N \otimes_A (\oplus_{i \in I} M_i) \cong \oplus_{i \in I} (N \otimes_A M_i).$ Consequently, if $\varphi : E \to F$ is an injective map of (right) $A$-modules, then the following diagram commutes. $$\require{AMScd} \begin{CD} E \otimes_A (\oplus_{i \in I} M_i) @>\varphi \otimes_A 1_M>> F \otimes_A (\oplus_{i \in I} M_i) \\ @VVV @VVV \\ \oplus_{i \in I} (E \otimes_A M_i) @>\oplus_{i \in I}(\varphi \otimes_A 1_{M_i})>> \oplus_{i \in I} (F \otimes_A M_i) \end{CD}$$ Considering that the vertical maps are isomorphisms, it follows that $M$ is flat if and only if $\varphi \otimes_A 1_M$ is injective if and only if $\oplus_{i \in I}(\varphi \otimes_A 1_{M_i})$ is injective if and only if $\varphi \otimes_A 1_{M_i}$ is injective for all indices $i$ if and only if $M_i$ is flat for all indices $i.$
We may now prove that every projective $A$-module is flat.
Proof. Certainly, $A$ is a flat $A$-module. By the previous claim, we have that $\oplus_{i \in I} A$ is flat. Consequently, every free $A$-module is flat. Considering that projective $A$-modules are direct summands of free $A$-modules, it follows that projective modules are flat.