Proving that free modules are flat (without appealing projective modules)

Suppose $R\neq 0$ is a commutative ring with $1$. Let $M$ be a free $R$-module. I would like to prove that $M$ is a flat $R$-module. Everywhere I have looked (mostly online) this is proved by first proving that every free module is projective, and then proving that every projective module is flat. Unfortunately, Atiyah & Macdonald's "Introduction to Commutative Algebra" (Chapter 2) does not discuss projective modules. But the result that every free module is flat comes very handy in the exercises.

So my question is,

Is it possible to prove that every free module is flat just by definitions and without appealing to projective modules?

Thanks!

Solution 1:

You can just show that the functor $M\otimes_R(-)$ is left-exact. Write $M=\bigoplus_{i\in I} R$ and let $$0\to N'\xrightarrow{\quad\iota\quad} N\to N''\to 0$$ be an exact sequence of $R$-modules. Note that $M\otimes N=\bigoplus_{i\in I} R\otimes_R N = \bigoplus_{i\in I} N$, so the sequence

$$0 \to M\otimes N' \xrightarrow{\quad\mathrm{id}\otimes\iota\quad} M\otimes N \to M\otimes N'' \to 0$$

is the same as

$$0 \to \bigoplus_{i\in I} N' \xrightarrow{\textstyle\quad\bigoplus_{i\in I} \iota\quad} \bigoplus_{i\in I}N \to \bigoplus_{i\in I}N'' \to 0$$

and the morphism $\bigoplus_{i\in I} \iota$ is clearly injective if $\iota$ is injective.

Solution 2:

1. The identity functor is exact.

2. If $\{F_i\}$ is a family of exact functors, then $\oplus_i F_i$ is also exact.

Now apply this to $M \otimes - $ for a free module $M$. 1. deals with the case $M=R$, and 2. generalizes this to $M = \oplus_i R$.