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Let $R=\mathbb{Z}[x]$. Define $F: \text{Mod}_R \to \text{Ab}$ by $F(M)= \{x \cdot m : m \in M \} \subseteq M$, it can be seen that $F$ defines a covariant functor.

Next, I must check that the functor is exact, but in order to talk about exact functors we must have $F: \text{Mod}_R \to \text{Mod}_S$, i.e $F$ maps modules to modules. So, my question is, what module is $F(M)$? Is it considered as a $\mathbb{Z}$-module here? I am confused because we can also regard it as a $\mathbb{Z}[x]$-module.

Hujiik
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  • That's not true; you can talk about exact functors between abelian categories (at least). In this case you can take $S = \mathbb{Z}$. – Qiaochu Yuan Oct 03 '15 at 01:40
  • I see, I have not studied abelian categories yet. What do you mean can? This is a question I am working with and I am wondering what the author means, does he mean $S= \mathbb{Z}$? – Hujiik Oct 03 '15 at 02:00
  • Yes, that's what the author means. – Qiaochu Yuan Oct 03 '15 at 07:28

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