Let $I$ be a small filtered category. Let $F\colon I \rightarrow \textbf{CRng}$ be a functor, where $\textbf{CRng}$ is the category of commutative rings. We write $A_i = F(i)$ for $i \in I$, $A =$ colim $A_i$. Suppose $I$ has an initial object $0$. Let $M_0, N_0$ be $A_0$-modules. Suppose $M_0$ is of finite presentation. Then colim $\textrm{Hom}_{A_i}(M_0\otimes_{A_0} A_i, N_0\otimes_{A_0} A_i)$ is canonically isomorphic to $\textrm{Hom}_{A}(M_0\otimes_{A_0} A, N_0\otimes_{A_0} A)$?
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Yes. The universal property of scalar extension says $$\textrm{Hom}_{A_i}(M_0 \otimes_{A_0} A_i, N_0 \otimes_{A_0} A_i) \cong \textrm{Hom}_{A_0} (M_0, N_0 \otimes_{A_0} A_i)$$ naturally, so $$\varinjlim \textrm{Hom}_{A_i}(M_0 \otimes_{A_0} A_i, N_0 \otimes_{A_0} A_i) \cong \varinjlim \textrm{Hom}_{A_0} (M_0, N_0 \otimes_{A_0} A_i) \cong \textrm{Hom}_{A_0} (M_0, \varinjlim N_0 \otimes_{A_0} A_i)$$ where in the last step we have used the fact that $M_0$ is of finite presentation; and since $N_0 \otimes_{A_0} (-)$ is a left adjoint, $$\varinjlim N_0 \otimes_{A_0} A_i \cong N_0 \otimes_{A_0} \varinjlim A_i \cong N_0 \otimes_{A_0} A$$ and therefore $$\varinjlim \textrm{Hom}_{A_i}(M_0 \otimes_{A_0} A_i, N_0 \otimes_{A_0} A_i) \cong \textrm{Hom}_{A_0}(M_0, N_0 \otimes_{A_0} A) \cong \textrm{Hom}_A(M_0 \otimes_{A_0} A, N_0 \otimes_{A_0} A)$$ as claimed.
Zhen Lin
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Could you explain how the last step of the second equation is proved by using the assumption that $M_0$ is of finite presentation? – Makoto Kato Dec 28 '12 at 15:47
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$\textrm{Hom}_{A_0}(M_0, -)$ preserves filtered colimits if (and only if) $M_0$ is a finitely-presented $A_0$-module. – Zhen Lin Dec 28 '12 at 16:21
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A proof, please? – Makoto Kato Dec 28 '12 at 16:26
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The proof I posted here covers this. – Zhen Lin Dec 28 '12 at 16:31