Isomorphism of function field Let $X$ be a prevariety. The function field $k(X)$ of $X$ is defined to be the inverse limit of $O_{X}(U)$, where  $U$ is non-empty open set of $X$. Question:how to give the explicit isomorphism between $k(X)$ and $k(V)$, where $V$ is any open affine set in $X$? What I know is that the intersection of any two open sets in $X$ is non-empty. But I don't know how to use this to prove my question. Could any one help me? Thanks a lot!
 A: What do you mean exactly with prevariety? Anyway, if you know what is a scheme, the right condition is $X$ integral (to ensure that intersection of non-empty open sets is non-empty and that you are dealing with domains). And the limit is direct, not inverse!
You have $k(X)=\varinjlim_{U\subseteq X} O_X(U)$ and $k(V)=\varinjlim_{U'\subseteq V} O_V(U')=\varinjlim_{U'\subseteq V} O_X(U')$.
Now, the first limit is made along the direct system of open sets in $V$, wich you can see as a subsystem of the system of open sets in $X$. This subsystem is cofinal, meaning that for all open sets $U\subseteq X$ there exists an open set $U'\subseteq V$ such that $U'\subseteq U$. To find $U'$, you can take the intersection $U\cap V$, that is non-empty. Using the universal property of the limit, it's an easy exercise to show that if you have a cofinal sub-system, this defines the same limit.
In fact, let $J\subseteq I$ be two direct systems of indices, $J$ cofinal in $I$, and $\{R_i\}_{i\in I}$ an associated direct system of rings, with morphisms $\phi_{hi}:R_i\to R_h$ for all $i\leq h$, with $\phi_{hi}\circ\phi_{ik}=\phi_{hk}$. If $R$ is the limit along $I$ with morphisms $\phi_j:R_j\to R$, you can show that it satisfies the universal property also on $J$.
Take a ring $S$ and morphisms $\{\psi_{j}:R_j\to S\}_{j\in J}$: using the fact that $J$ is cofinal in $I$, you can extend $\psi_i$ on all $i\in I$. Now, you use the universal property of $R$ on $I$, and it gives you the desired morphism $\psi:R\to S$ such that $\psi_j=\psi\circ\phi_j$
