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Suppose that I have an $A$ - module $N$ with $A$ commutative and I take a projective resolution of $N$:

$$\ldots \rightarrow P_2 \rightarrow P_1 \rightarrow P_0 \rightarrow N \rightarrow 0.$$

Suppose $M$ is some other $A$ - module. Now why is it the case that

$$\ldots \rightarrow P_2 \otimes_A M \rightarrow P_1 \otimes_A M \rightarrow P_0 \otimes_A M \rightarrow N \otimes_A M \rightarrow 0 $$

is not exact? I know that the tensor product is not in general left exact. However if the projective resolution is an infinite one then there is no "left" so why should the sequence above not be exact?

There has to be some problem with my understanding for then we always have $\textrm{Tor}_i^A(M,N) = 0$ for all $i$.

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  • $\begingroup$ Thanks for the questions! I guess this is one of the "fundamental questions" for learning Tor and Ext. $\endgroup$
    – Rookiecookie
    Jul 15, 2021 at 19:38

2 Answers 2

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The exactness of a long exact sequence is equivalent to the exactness of lots of short exact sequences. Precisely, if:

$$\cdots\to X\stackrel{f}{\to}Y\stackrel{g}{\to}Z\to\cdots$$

is part of a long exact sequence, then the short sequence:

$$0\to\operatorname{im}{f}\stackrel{\iota}{\to}Y\stackrel{g}{\to}{\operatorname{im}{g}}\to0$$

is exact, where $\iota$ is the inclusion. Conversely, if each such short sequence (one for each object of the long sequence) is exact, then the long sequence is exact. As the tensor product may fail to take one of the short exact sequences to a short exact sequence, due to the failure of left exactness, it also may not preserve the exactness of the long sequence.

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  • $\begingroup$ Right. However if I understand correctly, the tensored projective resolution is still a chain complex, which is why it makes sense to take its homology. $\endgroup$
    – user38268
    Oct 25, 2012 at 13:11
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    $\begingroup$ Yes, it's easy to check that $\partial^2=0$ still holds in a tensored chain complex, basically because you define the new boundary map componentwise and $0\otimes$ anything is $0$. $\endgroup$
    – Kevin Carlson
    Oct 25, 2012 at 13:13
  • $\begingroup$ Yes, it is still a chain complex. I think the general statement should be that functors preserve composition and zero morphisms, so the image of any chain complex under a functor is a chain complex. I don't immediately see the zero-morphism statement, but maybe I'm being slow - in any case, you're working with a $k$-linear category of modules, so functors must induce $k$-linear maps on Hom-spaces, which gives you $F(0)=0$ in this case. $\endgroup$
    – mdp
    Oct 25, 2012 at 13:25
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Example: $A = k[x]/x^2$, $k$ some field, $M = N = k$, the trivial module.

A resolution of $N$ is given by $$ \cdots A \to A \to A \to A \to k $$ where the maps $A\to A$ are multiplication by $x$, and the map $A \to k$ sends $1$ to $1$ and $x$ to $0$.

I claim all the maps $A\otimes _A k \to A \otimes _A k$ are zero. This is true because $1 \otimes _A 1 \mapsto x \otimes _A 1 = 1 \otimes _A x\cdot 1 = 1\otimes _A 0 = 0$. The new sequence fails to be exact since it consists of non-zero modules, but all of the maps (except the last) are zero.

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