In section 10.5 of Dummit and Foote, we are given a proof that free modules are flat that goes like this:
1). Finitely generated free modules are flat (easy)
2). Suppose now that $F$ is an arbitrary free $R$-module, and $L$ and $M$ are $R$-modules with $\psi: L \to M$ injective. We want to show that $1 \otimes \psi: F \otimes L \to F \otimes M$ is injective.
3). This part contains the step I am confused about:
Suppose that some $\sum (f_i \otimes l_i)$ is sent to $0$ by $1 \otimes \psi$. Then $\sum (f_i \otimes \psi(l_i))$ is $0$ in $F \otimes M$, so $\sum(f_i, \psi(l_i))$ can be written as a finite sum of generators for the relations that give $F \otimes M$ as a quotient of the free ($\mathbb{Z}$) module on $F \times M$.
After this, I am lost as the proof goes "because the sum is finite, all of the first coordinates of the resulting equation lie in some finitely generated free submodule $F'$ of $F$ ". How do we get this? Can someone help me write it down more explicitly?
4). Because of 1). we can conclude that $\sum (f_i \otimes l_i)$ must be $0$ in $F' \otimes L$ hence also $0$ in $F \otimes L$
Tihs can be done a lot easier, free $R-$modules are of the form $R[S]$, (freely generated module on elements of a set $S$, not polynomial algebra) we also know that the tensor product commutes with direct sums $$(\bigoplus_i M_i) \otimes_R A = \bigoplus_i (M_i \otimes_R A)$$
So tensoring with a free module $F = R[S]$ is the same as applying the functor $\bigoplus_{i \in S}$ since $R[S] = \bigoplus_{i \in S} R$ and $R \otimes_R M = M$.
Then if
$$0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$$
is exact we show that
$$0 \rightarrow \bigoplus_{i \in S} A \rightarrow \bigoplus_{i \in S} B \rightarrow \bigoplus_{i \in S} C \rightarrow 0$$
is also exact pretty easily hence proving that free modules are also flat.