Let $f: R \to A$ be a ring morphism ($R, A$ a commutative rings) and
$$ 0 \to M_1 \to M_2 \to M_3 \to 0 $$
a ex. sequence of $R$ modules. Assume, that $M_3 = R^n$ is free (or
say projective). I want to show that of we tensor this sequence
with $A$ the sequence we obtain is still exact. That's always true that
the part
$$ M_1 \otimes A \to M_2 \otimes A \to M_3 \otimes A \to 0 $$
is exact. What we have to show is that the left induced map
$M_1 \otimes A \to M_2 \otimes A$ is injective. Since
$M_3= R^n$, after tesnor $ M_3 \otimes A= A^n$ is still free and
we have a splitting $id_{A^n} = A^n \to M_2 \otimes A \to A^n $. But
I not know why this shows that the left map $M_1 \otimes A \to M_2 \otimes A$
is injective.
Best Answer
Denote by $\alpha$ the injection $M_1\to M_2$ and by $\beta$ the surjection $M_2\to M_3$. Since $\beta$ is surjective, projectivity of $M_3$ gives us a map $\gamma:M_3\to M_2$ making the diagram below commute.
$$\require{AMScd} \require{cancel} \def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1} \lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}} \begin{CD} && M_2\\ & \diaguparrow{\exists \gamma} @VV\beta V \\ M_3 @>>\text{id}_{M_3}> M_3 \end{CD}$$ In particular, the map $\beta$ is split, and so the splitting lemma gives us $\delta:M_2\to M_1$ a retraction of $\alpha$: ie, a map such that $\delta\circ\alpha=\text{id}_{M_1}$. Then the map $\delta\otimes\text{id}_A:M_2\otimes A\to M_1\otimes A$ is a left inverse to $\alpha\otimes\text{id}_A$, whence $\alpha\otimes\text{id}_A$ is injective, as desired.