a flatness criterion I'm having trouble with part (b) of Exercise 10.5.25 from Dummit & Foote (the goal of the problem is to prove that $A$ is a flat $R$-module iff $A\otimes_R I\to A\otimes_R R$ is one-to-one for all finitely-generated ideals $I$ of $R$). 
Part (b) has 3 parts. The first and third I've done, but I'm stuck on the second. Here're the first two parts of (b):

If $A\otimes_R I\to A\otimes_R R$ is injective for every finitely-generated ideal $I$, prove that $A\otimes_R I\to A\otimes_R R$ is injective for every ideal $I$. Show that if $K$ is any submodule of a finitely generated free module $F$, then $A\otimes_R K\to A\otimes_R F$ is injective.

Here's what I've tried:
Let $e_1,\ldots,e_n$ be a basis for $F$. Let $x\in \ker(1\otimes\iota)$, where $\iota\colon K\to F$ is inclusion. Then
$$x=\sum_{i=1}^m a_i\otimes k_i$$
for some $a_i\in A$, $k_i\in K$. Express each $k_i$ in terms of the basis: $k_i=\sum_j r_{ij}e_j$. Then 
$$0 = 1\otimes \iota(x)=\sum_i\sum_j a_i\otimes r_{ij}e_j=\sum_j\left(\sum_i a_i r_{ij}\right)\otimes e_j.$$
So, using the fact that $A\otimes_R F\cong A^n$ via $\sum_j b_j\otimes e_j \mapsto (b_1,\ldots,b_j)$, we get 
$$\sum_i a_i r_{ij}=0$$
for all $j$. My issue is that I don't know how to lift this back to $A\otimes_R K$. Also, I think I should be using the first part of (b) somewhere.
Any hints? Am I even on the right track?
 A: Since Jack Schmidt has mentioned direct limits here is how we can prove the first part of (b) using direct limits. Let $I$ be any ideal of $R$; we can write $I = \operatorname{colim} I_\alpha$ where the colimit is over all finitely generated ideals contained in $I$. Now consider the ses
$$0 \to I\to R  \to R/I \to 0$$ 
from which we get a long exact sequence in Tor 
$$\ldots \to 0 \to \text{Tor}_1^R(A,R/I) \to A \otimes_R I\to A \otimes_R R \to A \otimes_R R/I \to 0$$
where the zero on the left appears because $R$ a free module over itself implies $\text{Tor}_1^R(A,R) = 0$. Now I claim that in fact $\text{Tor}_1^R(A,R/I) = 0$. Indeed we have
$$\begin{eqnarray*} \text{Tor}_1^R(A,R/I)&=& \text{Tor}_1^R(A,\left(R/\operatorname{colim}I_\alpha\right)) \\
&=& \text{Tor}_1^R\left(A,\operatorname{colim} (R/I_\alpha)\right) \hspace{1cm} (\text{Colimits are right exact}) \\
&=&  \operatorname{colim} \text{Tor}_1^R\left(A, (R/I_\alpha)\right) \hspace{1cm} (\text{Tor commutes with colimits})\\
&=& 0
\end{eqnarray*}$$
where we get $0$ at last because for all $\alpha$, $\text{Tor}_1^R\left(A, (R/I_\alpha)\right)  = 0$ by assumption. Thus the first part of (b) is proven.
A: First note that that the condition $A \otimes_R K \to A \otimes_R F$ is injective for all free $R$-modules $F$ and submodules $K$ is equivalent to $A$ being flat.  So you will definitely need to assume that $A \otimes_R I \to A \otimes_R R$ is injective for all ideals $I \subseteq R$.
We will prove that $\operatorname{Tor}_1^R(A, F/K) = 0$ for any such $F$ and $K$ and this will give you injectivity of the map using the long exact sequence coming from Tor.  You're going to do this by induction on the rank of $F$.  When $F$ has rank $1$ it's simply $R$ and $K$ is some ideal $I$.  Look at the exact sequence $I \to R \to R/I$.  Write down the corresponding long exact sequence for $\operatorname{Tor}_1^R(A, -)$.  Use injectivity of $A \otimes_R I \to A \otimes_R R$ and that $R$ is free to get $\operatorname{Tor}_1^R(A, R/I) = 0$.
Next the inductive step.  Let $e_1, \ldots, e_n$ be a basis for $F$.  Let $N = \langle e_1, \ldots, e_{n-1}\rangle$ and consider the exact sequence
$$\frac{N}{N \cap K} \to F/K \to \frac{F/N}{(K + N)/N}$$
Both $N$ and $F/N$ are free of smaller rank.  So use the inductive hypothesis along with the long exact sequence of $\operatorname{Tor}_1^R(A, -)$ again to get that $\operatorname{Tor}_1^R(A, F/K) = 0$.
