Let $R$ be a commutative Ring with $1$ and $M$ be a finitely generated free $R$-Module. Show that $M$ is flat.
By definition, $M$ is flat $\iff$ For every injective $R$-Module Homomorphism $\varphi: N \to L$, the homomorphism $\varphi \otimes \operatorname{id}_M: N \otimes_R M \to L \otimes_R M$ is injective.
Ok, so let $\varphi$ be an injective $R$-Module homomorphism, $(x_1, … , x_n)$ be a basis of $M$ and $a \otimes b \in \ker(\varphi \otimes \operatorname{id}_M)$. Without loss of generality let $b \neq 0$. Then there exist $r_i$ such that $b = \sum_{i=1}^n r_i x_i$. Then it follows
$$\varphi\otimes \operatorname{id}_M(a \otimes b) = \varphi(a) \otimes b = \sum_{i=1}^nr_i(\varphi(a)\otimes x_i) = 0.$$
How do I advance from here? Since $b \neq 0$ not all $r_i$ are zero, so $\varphi(a) \otimes x_i$ must be zero right? But how do I get $\varphi(a) = 0$ now? Advice greatly appreciated.
Best Answer
It may be easiest to simply use the fact that $M\cong R^n$ for some $n,$ and the tensor product distributes over direct sums. Explicitly, for any $R$-module $A$ we have $M\otimes A\cong A^n.$ Writing it like this, any map $\varphi:N\to L$ becomes $$\widetilde\varphi:N^n\to L^n\\ \widetilde\varphi(a_1,\ldots,a_n)=(\varphi(a_1),\ldots, \varphi(a_n))$$ under the tensor product with $M.$ Looking at it like this, injectivity is trivial to show.