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I'm reading Categories and Modules with K-Theory in View by A. J. Berrick, M. E. Keating, on Flat Modules and there's one part that I'm not quite sure that I fully grasp it.

It's on page 155.

3.2.9. The Flat Test

A left R-module N is flat if only if $-\otimes_R N$ respects injectivity of all inclusions $\mu:\mathfrak{a} \rightarrow R$ of finitely generated right ideals in R. The corresponding statement holds with left and right interchanged.


Proof (I'll shorten the proof to where I don't get it)

Necessity of the condition is immediate.

Sufficiency:

  1. First, he claims that $-\otimes_R N$ respects injectivity of all inclusions $\mu:\mathfrak{a} \rightarrow R$ (where $\mathfrak{a}$ needs not be finitely generated).

  2. Then, he proves $-\otimes_R N$ respects injectivity of all inclusions $\mu:M \rightarrow R^n$.

  3. Then, he shows that $-\otimes_R N$ respects injectivity of all inclusions $\mu:M \rightarrow F$, where $F$ is a free module. This is where I don't get it.

Again, it's enough to check the case where M is finitely generated. But then, when $(\mu \otimes \mbox{id})x = 0$ in $F \otimes_R M$, [[[there's a finitely generated free sub-module G of the codomain F, such that $(\mu \otimes \mbox{id})x = 0$ in $G \otimes_R M$ already, since any member of the relation group $B(F;M)$ (3.1.2) can involve only finitely many generators of F. By the previous step, $x = 0$]]].

What I don't get is the [[[...]]] part. I think it should be pretty staright forward, so if you can point me to a theorem, or a lemma, that should be enough.

Thanks everyone a lot,

And have a great day, :*

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This argument is essentially identical to the argument for why it suffices to check finitely generated $M$:

Remember that $F \otimes_R M$ can be given as the quotient of a free $\mathbb Z$-module, with basis the set $F \times M$, by the ideal $B(F;M)$ generated by all the bilinear relations. $$F \otimes_R M = \mathbb Z[F \times M]/B(F;M)$$ If an element $a$ is zero in $F \otimes_R M$ that means it's contained in $B(F;M)$. That means there's some linear combination of finitely many bilinear relations that comes out equal to $a$. Each of the bilinear relations involves finitely many basis elements from $F \times M$. Each of those finitely many basis elements is of the form $(f, m) \in F \times M$ where $f$ is a sum of finitely many basis elements in $F$. Taking all of those basis elements of $F$ needed to generate all of the relations needed to generate $a$, we still have taken only finitely many.

If you let $G$ be the free submodule generated by these finitely many basis elements then in $$G \otimes_R M = \mathbb Z[G \times M]/B(G;M)$$ we have $a = 0$ because we have ensured that the linear combination that gives $a$ as an element in $B(F;M)$ is valid as a linear combination that gives an element in $B(G;M)$.

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  • $\begingroup$ Thank you for your kind, and thorough explanation, I think I get it now. Thank you sssssoooo mmmuuucccchhhh. B-) :* $\endgroup$ – user49685 Apr 2 '13 at 11:04
  • $\begingroup$ One last question, but I think it's either that the book misprint or I'm missing something big here, should the first line on the last paragraph in my first post read "$(\mu \otimes \mbox{id}) x = 0$ in $F \otimes_RN $", instead of "... in $F \otimes_RM $"? If the book does misprint, why should we only consider $M$ to be finitely generated, since I cannot see where that piece of information is used. :( $\endgroup$ – user49685 Apr 2 '13 at 11:50
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    $\begingroup$ Yes, it should be $F \otimes_R N$. You use that $M$ is finitely generated to show that $M \to F$ restricts to $M \to G$ where $G$ has finite rank. You do this by adding to the basis elements of $G$ any generators needed to construct the images of the generators of $M$ (of which there are finitely many). $\endgroup$ – Jim Apr 2 '13 at 17:30

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