An object $C$ in an additive category admitting all filtered direct limits $\mathcal{C}$ is called "of finite type" if the canonical map $$\underrightarrow{\lim} Hom_{\mathcal{C}}(C,F(i))\to Hom_{\mathcal{C}}(C,\underrightarrow{\lim}F)$$ is injective for every $I$ directed poset for every functor $F:I\to \mathcal{C}$

In the case $\mathcal{C}$=Mod-R prove that this definition is equivalent to the definition of "finitely generated"

The exercise has this strange hint: Use the fact that if $\mathcal{F}$ is the set of finitely generated submodules of a module C then $$C/{\sum_{A\in\mathcal{F}}A}=\underrightarrow{\lim}C/A$$

I say that the hint is strange because $\displaystyle {\sum_{A\in\mathcal{F}}A}=C$


You're right in that $\sum_A A = C$, so $C/\sum_A A={\lim\limits_{\rightarrow}}_A C/A=0$, and that's exactly the colimit you need to instantiate the given condition with. The right hand side being zero, any element of $\varphi\in\text{Hom}(C,C/A)$ must eventually vanish in some $\text{Hom}(C,C/B)$ for $A\subseteq B$. What's a natural choice for $A$ and $\varphi$ to consider here?

  • $\begingroup$ $\Rightarrow$ If we choose $I=\mathcal{F}$, then we have an embedding $\displaystyle \underrightarrow{\lim}_{A\in \mathcal{F}} Hom_{\mathcal{C}}(C,C/A)\to Hom_{\mathcal{C}}(C,\underrightarrow{\lim}_{A\in \mathcal{F}}C/A)=0$ So $\underrightarrow{\lim}_{A\in \mathcal{F}} Hom_{\mathcal{C}}(C,C/A)=0$, so $[id_C]=[0]\in \underrightarrow{\lim}_{A\in \mathcal{F}}Hom_{\mathcal{C}}(C,C/A)$, which means that there is an $A\in \mathcal{F}$ s.t. the projection to $C/A$ is the zero morphism, so $A=C$. $\endgroup$ – Anselm Sep 17 '16 at 14:10
  • $\begingroup$ @DDT: Yes, exactly! $\endgroup$ – Hanno Sep 17 '16 at 14:45

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