Say $V$ be an irreducible affine variety contained in $\mathbb{A}^n_k$, where k is algebraically closed
Since, the coordinate ring $\Gamma(V)=\frac{k[X_1,X_2,...,X_n]}{I(V)}$ of $V$ is an integral domain , we can form its field of fractions , say $k(V)$. Now, an $f\in k(V)$ is said to be defined at $P\in V$ if $f$ can be expressed in the form $\frac{a}{b}$ , with $a, b \in \Gamma(V)$ and $b(P)\neq0$ . I need to prove that any such $f$ is completely determined by the values $f(P)$ it takes at the points where it is defined.
My thoughts are :
Say $g\in k(V)$ be anything other than $f$ so that $g$ is defined at all points where $f$ is defined and $f(P)=g(P)$ at these points. Now at some $P$ where $f$ is defined we must have different local expressions $\frac{a}{b}$ and $\frac{c}{d}$ for $f$ and $g$ respectively but $\frac{a(P)}{b(P)}= \frac{c(P)}{d(P)}$. I don't know how to proceed after this.
Suppose we have $a(p)/b(p)=c(p)/d(p)$ for $p\in (D(b)\cap D(d))\cap V(I)$ Then we have $(ad-bc)(p)=0$ for all $p\in (D(b)\cap D(d))\cap V(I)$. Now since $V(ad-bc)\cap V(I)$ is closed and dense in $V(I)$ , then we have $V(ad-bc)=V(I)$. Therefore, we have $ac-bd\in I$ as $I$ is prime. Therefore, $a/b=c/d$ in $k(V)$