From Weibel's "An Introduction to Homological Algebra":

Exercise 3.2.1: An $R$-module $B$ is flat if Tor$_i^R(A,B) = 0$ for every $R$-module A.

It seems to me that the obvious way to do this would be to use the definition:

Tor$_i^R(A,B) = $H$_i(P.\otimes B)$ where $P.$ is a projective resolution of A.

We need to show that for an exact sequence:

$...\rightarrow A_{n+1} \rightarrow A_{n} \rightarrow A_{n-1} \rightarrow ...$ ,

$...\rightarrow A_{n+1} \otimes B \rightarrow A_{n} \otimes B \rightarrow A_{n-1} \otimes B \rightarrow ...$ is exact.

Since we can only access exactness via the definition of Tor, it seems to me that we might have to construct a projective resolution around this point in the sequence, however I am currently struggling.

Any help is appreciated.

  • $\begingroup$ I think it's enough to show that for any short exact sequence $$0\to A_1\to A_2\to A_3\to 0$$ we have $$0\to A_1\otimes B\to A_2\otimes B\to A_3\otimes B\to 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with).. $\endgroup$
    – Arthur
    Jan 7, 2019 at 14:53
  • $\begingroup$ Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks $\endgroup$
    – Daven
    Jan 7, 2019 at 14:59
  • 1
    $\begingroup$ Check the horseshoe lemma, then the snake lemma. $\endgroup$
    – Arthur
    Jan 7, 2019 at 15:01
  • $\begingroup$ That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions. $\endgroup$
    – Daven
    Jan 7, 2019 at 15:05
  • $\begingroup$ It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them? $\endgroup$
    – Arthur
    Jan 7, 2019 at 15:08

1 Answer 1


We want to show that $\_ \otimes B$ is exact. Now Lets take for this a short exact sequence $$0 \to X \to Y \to Z \to 0$$ Now lets apply the functors $\mathrm{Tor}^i(\_,B)$ to that sequence. Now this gives by the definition of $\mathrm{Tor}^i(\_,B)$ as a homological functor a long exact sequence $$ ... \to \mathrm{Tor}^1(X,B) \to \mathrm{Tor}^1(Y,B) \to \mathrm{Tor}^1(Z,B) \to\mathrm{Tor}^0(X,B) \to \mathrm{Tor}^0(Y,B) \to \mathrm{Tor}^0(Z,B) \to 0 $$ Now since we have a natural isomorphism $\mathrm{Tor}^0(Z,B) \cong Z\otimes B$ we may rewrite the top sequence as: $$ \mathrm{Tor}^1(X,B) \to \mathrm{Tor}^1(Y,B) \to \mathrm{Tor}^1(Z,B) \to X\otimes B \to Y\otimes B \to Z \otimes B \to 0 $$ But since $ \mathrm{Tor}^1(Z,B)=0$ this becomes: $$0\to X\otimes B \to Y\otimes B \to Z \otimes B \to 0 $$ as desired. (for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $\mathrm{Tor}$-terms vanish)

  • $\begingroup$ This is the kind of nice answer I was looking for, thanks so much! $\endgroup$
    – Daven
    Jan 7, 2019 at 17:04
  • $\begingroup$ you are very welcome! one just has to love long exact sequences! $\endgroup$
    – Felix
    Jan 8, 2019 at 8:06
  • 2
    $\begingroup$ @Daven In some sense, the $\operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $\operatorname{Tor}$ gives you $0$. Similarily, the $\operatorname{Ext}$ functor measures the non-exactness of $\operatorname{hom}(B, -)$, in the other direction (and even for $\operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way). $\endgroup$
    – Arthur
    Jan 8, 2019 at 11:44

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