Value of rational function well-defined? Suppose $f\in O_p(V)$ is a rational function on $V$ that has a value at $p$. Then write $f=a/b=a'/b'$ where $a,b,a',b'\in \Gamma(V)$, the coordinate ring of $V$. Want to show the value of $f$ at $p$ is well-defined, i.e. $a(p)/b(p)=a'(p)/b'(p)$.
So since $a/b=a'/b'$ are the same equivalence class, there is some non-zero poly $x\in \Gamma(V)$ such that $x(ab'-a'b)=0$. Then $x(p)(a(p)b'(p)-a'(p)b(p))=0$. Then what? How do we know $x(p)\neq0$?

So $x\neq \bar{0} \in \Gamma(V)$. Since $\Gamma(V)$ is an integral domain, $a'b-ab'=\bar{0}$. Actually one can remove $x$ from the definition of fraction field altogether.
