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Assume we want to define $\mathrm{Tor}_n (M,N)$ where $M,N$ are $R$-modules and $R$ is a commutative unital ring.

We take a projective resolution of $M$:

$$ \dots \to P_1 \to P_0 \to M \to 0$$

Now does it matter whether we apply $-\otimes N$ or $N \otimes -$ to this? It shouldn't because we have $N \otimes P \cong P \otimes N$. Right? Thanks.

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$R$ is commutative? – Ehsan M. Kermani Jul 24 '12 at 7:35
Well, $M$ is a right module and $N$ is a left module. – Rasmus Jul 24 '12 at 7:37
Yes, $R$ is commutative. – Rudy the Reindeer Jul 24 '12 at 7:39
@ehsanmo The question is tagged [commutative-algebra].... – user38268 Jul 24 '12 at 9:04
@BenjaLim That tag was added in response to the comment. – Rudy the Reindeer Jul 24 '12 at 10:10
up vote 3 down vote accepted

I guess the point here is that $N \otimes_R -$ and $- \otimes_R N$ are naturally isomorphic functors. Therefore, you get an isomorphism of chain complexes $N \otimes_R P^{\bullet} \cong P^{\bullet} \otimes_R N$, which implies that the two complexes have isomorphic homology. So, it doesn't matter if you apply $N \otimes_R -$ or $- \otimes_R N$.

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Do you mean homology? The $\mathrm{Tor}$ functor is the homology of the resulting long exact sequence. – Rudy the Reindeer Jul 24 '12 at 10:11
As you use chain complexes (as opposed to cochain complexes), it should indeed be homology instead of cohomology. I'll edit my answer accordingly. – Sebastian Jul 24 '12 at 10:18

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