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Let $Z_n = \operatorname{ker} \partial_n$ and $B_n = \operatorname{im}\partial_{n+1}$. We know that the short exact sequence

$$0\to Z_n \to C_n \to B_n \to 0$$ is exact. The surjectivity of $$g\colon C_n\to B_n$$ implies surjectivity of $g\otimes \operatorname{id} \colon C_n \otimes G \to B_n\otimes G$. However, it seems that we can't conclude analogously that the injectivity of $$i\colon Z_n\to C_n$$ implies injectivity of $$i\otimes \operatorname{id}\colon Z_n\otimes G \to C_n \otimes G.$$

Is that correct? If so, can someone elaborate why the latter function does not need to be injective even though $i:Z_n\to C_n$ is injective?

The context of this is that i am supposed to prove the exactness of $$0\to Z_n\otimes G \to C_n\otimes G \to B_n\otimes G\to 0.$$ But the injectivity of $Z_n\otimes G\to C_n\otimes G$ does apparently not follow from the observation that the inclusion $Z_n\to C_n$ is injective.

Any help is appreciated!

Zest
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1 Answers1

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$\Bbb{Z}\otimes_\Bbb{Z} \Bbb{F}_p= \Bbb{F}_p$ while $\Bbb{Q}\otimes_\Bbb{Z} \Bbb{F}_p= 0$

reuns
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  • Thanks but i dont see what $\mathbb{Q}$ is supposed to tell me in your statement. – Zest Jun 28 '20 at 02:21
  • Isn't that your question, some example of injection of modules where the tensored ones don't inject ? – reuns Jun 28 '20 at 02:27
  • Ah, i see. My question was more on a general note why (or if) the tensored map is injective. The response of Carlo was basically what i was looking for. Nonetheless, thank you very much! – Zest Jun 28 '20 at 02:41