# A module $B$ is flat if Tor $= 0$

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.

• 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).. Jan 7, 2019 at 14:53
• Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks Jan 7, 2019 at 14:59
• Check the horseshoe lemma, then the snake lemma. Jan 7, 2019 at 15:01
• 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. Jan 7, 2019 at 15:05
• It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them? Jan 7, 2019 at 15:08

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)
• @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). Jan 8, 2019 at 11:44