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We assume all rings are commutative. Let $A$ be a ring. Let $I$ be a small filtered category. Let $F\colon I \rightarrow A$-alg be a functor, where $A$-alg is the category of $A$-algebras. We write $X = Spec(A)$, $B_i = F(i)$ for each $i \in I$, $B =$ colim $B_i$, $Y_i = Spec(B_i)$ for each $i\in I$, $Y = Spec(B)$. Let $f\colon Y \rightarrow X$, $f_i\colon Y_i \rightarrow X$ be the canonical morphisms.

Then $f(Y) = \bigcap f_i(Y_i)$?

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It suffices to prove that $\bigcap f_i(Y_i) \subset f(Y)$. Let $x \in \bigcap f_i(Y_i)$. Suppose $x \in X - f(Y)$. Then $f^{-1}(x) = \emptyset$. Since $f^{-1}(x) =$ Spec $(B \otimes_A \kappa(x)), B \otimes_A \kappa(x) = 0$. Since $B \otimes_A \kappa(x) =$ colim $(B_i \otimes_A \kappa(x))$, there exists $i \in I$ such that $B_i \otimes_A \kappa(x) = 0$. Hence $f_i^{-1}(x) = \emptyset$. This is a contradiction. QED

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