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.


  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, :*


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 $B(G;M)$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.