Function Field of an Integral Scheme

Let $X$ be an integral scheme. Then it is irreducible so it has a generic point $\eta$. Let $\text{Spec} A$ be a nonempty affine open of $X$ (thus must contain $\eta$). Let $K(A)$ denote the fraction field of $A$.

Claim. $(O_X)_{\eta} \cong K$ canonically.

Proof. We have $\text{Spec} A$ is itself irreducible, being a nonempty open subspace of an irreducible space. Thus $\eta$ corresponds to the generic point of $\text{Spec} A$, i.e. the nilradical of $A$ (which must be prime). However, since $A$ is integral, $\sqrt{0_A} = (0)$.
Therefore we have $(O_X)_{\eta} = A_{\sqrt{0_A}} = A_{(0)} = K(A)$.
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Let $V \supset U$ be nonempty open subsets of $X$.

Claim. The restriction map $\text{res}: O_X(U) \to O_X(V)$ is an inclusion.

Proof. Let $s \in O_X(U)$ be such that $s|_V = 0$. Notice that $U$ is irreducible, being a nonempty open subset of an irreducible space, so $V$ is dense in $X$. Thus as $s$ vanishes on $V$, it also vanishes on $U$. (Warning: this is NOT enough to conclude that $s = 0$). However, as $O_X(U)$ is reduced, its sections are uniquely determined by their values at points of $U$, so $s$ must indeed be $0$.
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Claim. The map $O_X(U) \to (O_X)_{\eta}= K(A)$ is an inclusion.

Proof. Suppose $s_{\eta} = 0$ for some $s \in O_X(U)$. Then $s|_V = 0$ for some $\emptyset \neq V \subset U$. Thus by the previous claim, $s = 0$.
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Advantage of Irreducible Varieties. 

Irreducible varieties form an important class of integral schemes. We thus see that for such varieties, all $O_X(U)$'s can be embedded into the same ring $K(X)$. As restriction maps are simply inclusion, sections glue iff they are the same element of $K(X)$.
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Dominant maps of Integral Schemes induce morphism of Function Fields.

Let $f: X\dashrightarrow Y$ be a dominant rational map. Then $f$ sends $\eta_X$ to $\eta_Y$, as dominant rational map between irreducible schemes preserve generic pt.  Thus it induces a map on stalk $(O_Y)_{\eta_Y} = K(Y) \to (O_X)_{\eta_X} = K(X).$