- $Tor_0^A(-,P) = - \otimes_A P$
- For any short exact sequence of $A$-modules $0 \to L \to M \to N \to 0$, you get a long exact sequence.

$\dots \to Tor_{n+1}^A(L,P) \to Tor_{n+1}^A(M,P) \to Tor_{n+1}^A(N,P)
\to Tor_n^A(L,P) \to Tor_n^A(M,P) \to Tor_n^A(N,P)
\to \dots$

which, on the right side, stops at
$\dots \to Tor_1^A(L,P) \to Tor_1^A(M,P) \to Tor_1^A(N,P)
\to L \otimes_A P \to M \otimes_A P \to N \otimes_A P \to 0$

`23:44 ` isekaijin can you describe the existence proof of Tor? :)
23:45 A projective resolution is a chain complex of projective A-modules “... -> P_{n+1} -> P_n -> ... -> P_1 -> P_0 -> 0” that is chain-homotopic to “0 -> P -> 0”.
23:45 And you need the axiom of choice to show that it exists in general.
23:45 Now, projective A-modules behave much more nicely w.r.t. the tensor product than arbitrary A-modules.
23:46 In particular, projective modules are flat, so tensoring with a projective module *is* exact.
23:47 So to compute Tor_i(M,P), you tensor M with the projective resolution, and then take its homology.
23:47 To show that this is well-defined, you need to show that Tor_i(M,P) does not depend on the chosen projective resolution of P.
23:48 bollu: just use the axiom of choice like everyone else
23:48 why do you need to take homology?
23:48 That's just the definition of Tor.
23:49 Okay, to show that Tor does not depend on the chosen projective resolution, you use the fact that any two chain-homotopic chains have the same homology.
23:49 right
23:49 Which is a nice cute exercise in homological algebra that I am too busy to do right now.
23:49 whose proof I have seen in hatcher
23:49 :)
23:49 Oh, great.
23:49 thanks, the big picture is really useful