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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?

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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 $\operatorname{Tor}_1^R(A, F/K) = 0$.

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  • $\begingroup$ Good luck :) Before resorting to Tor I thought briefly about if there was an elementary way of doing this and didn't come up with anything. It must be possible though. $\endgroup$ – Jim May 22 '13 at 17:20
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    $\begingroup$ There is a superficial problem that not all free $R$-modules have finite rank, but noticing this is only superficial is part of the exercise. A method avoiding Tor but using direct limits is in Lam's Lectures on Modules and Rings, 4.12 page 125. It shows how to view this as the dual to Baer's criterion for injective modules. $\endgroup$ – Jack Schmidt May 22 '13 at 17:24
  • $\begingroup$ The first part of (b) is just the rank 1 case, which I've already done, so I have a feeling induction is the way to go for the elementary method. $\endgroup$ – Avi Steiner May 22 '13 at 17:37
  • $\begingroup$ @JackSchmidt I posted an answer below on the first part of (b) that uses direct limits, I'm not sure if that's what you meant. $\endgroup$ – user38268 May 23 '13 at 3:09
  • $\begingroup$ @JackSchmidt Nice answer +1. I tried to use an argument using Tor and direct limits yesterday but somehow it was not apparent to me we had to assume that $A \otimes I \to A \otimes R$ is injective for all ideals $I \subseteq R$. $\endgroup$ – user38268 May 23 '13 at 3:10
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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.

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  • $\begingroup$ +1 The only thing I would be careful about is that in general $A \otimes N \to A \otimes M$ being injective does not imply that $\operatorname{Tor}_1^R(A, M/N) = 0$. In this case it follows because $M = R$ is free but it's a little sneaky to say that this is "by assumption" ;) $\endgroup$ – Jim May 23 '13 at 6:05
  • $\begingroup$ @Jim Thanks for your kind words. I edited my answer and added that $R$ free implies that $\operatorname{Tor}_1^R(A,R) = 0$. $\endgroup$ – user38268 May 23 '13 at 8:09

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