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Considering

$$0\to \Bbb Z \to \Bbb Z\to \Bbb Z/(2)\to 0$$ We want to show that the tensor product is not left exact by tensoring with $\Bbb Z/(2)$, which apparently gives us what we want.

But I did this, and I get

$$0\to \Bbb Z/(2)\overset{\sim}{\to}\Bbb Z/(2)\overset{0}{\to} \Bbb Z/(2)$$

Where I can't see why this isn't exact? Is there a reason why I can't have the zero map on the right?

1 Answers 1

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The question title is "tensoring is not left exact," so you should probably be looking for failures in exactness towards the left of the second sequence.

The map $\Bbb{Z}\rightarrow\Bbb{Z}$ in the original sequence is multiplication by $2$. You need to figure out what the induced map is after tensoring by $\Bbb{Z}/2\Bbb{Z}$. (The important thing is that it is NOT an isomorphism, as you initially thought.)

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    How are the maps changed with tensoring? Are these left unchanged, i.e. if I know how each of the maps behave, and then left tensor each term, the maps are the same? (I get that here you are saying that the maps are left unchanged, so the doubling map is now the zero map. I just meant in general) – 2017-02-15
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    I'm not sure what you mean by "unchanged." The maps have certainly changed, since the domains and ranges have changed post-tensoring! In general, the map induced from $f:M\rightarrow N$ by tensoring with $L$ is $f\otimes \mbox{id}: M\otimes L\rightarrow N\otimes L$, where $(f\otimes \mbox{id})(a\otimes b)=f(a)\otimes b$. – 2017-02-15
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    Great, thanks very much! – 2017-02-16