Note: I am doing an introductory course in Commutative Algebra and not Algebraic Geometry; This is just a quick application I am trying to understand. Also, rings are commutative with $1$.
When discussing localizations $S^{-1}A$ for a ring $A$ and $S\subset A$, we made the following example.
Def. Let $k$ be a field, and $P$ be a prime ideal in $A=k[X_1,\dots, X_n]$. Let $V=\{(a_1,\dots,a_n)\in k^n:f(a_1,\dots,a_n)=0, \forall f\in P\}$ be the variety of $P$, and let $S=A\setminus P$.
Then $S^{-1}A$ is denoted $A_P$, and we have $A_P=\{\dfrac{f}{g}:f\in A,g\not\in P\}$. In class, this was called the ring of regular functions on $V$ (EDIT: I learned this is apparently not how the ring of regular functions on V is defined, see the comments). These are apparently well defined rational functions $k^n\to k$ since $g$ does not vanish on $V$, i.e. it seems we have
$g\not\in P\Rightarrow g(a_1,\dots a_n)\not =0$ for all $(a_1,\dots,a_n)\in V$.
But, I think it's also evident that we have
$g\in P\Rightarrow g(a_1,\dots,a_n)=0$ for all $(a_1,\dots,a_n)\in V$
which is logically equivalent to $(g\not\in P) \vee (g(a_1,\dots,a_n)=0$ for all $(a_1,\dots,a_n)\in V)$.
It just feels like I'm missing something obvious, since I can't seem to deduce a contradiction from $g\not\in P$, and $g(a_1,\dots,a_n)=0$ for some point of $V$.
Question: Why couldn't we have $g\not\in P$, but still have $g(a_1,\dots,a_n)=0$, for some $(a_1,\dots,a_n)\in V$?
We absolutely can have $g\notin P$ and $g(a_1,\dots,a_n)=0$ for some $(a_1,\dots,a_n)\in V$.
Take $n=2$ and the prime ideal $P=(x)\subset k[x,y]$, so that $V=\{(a,b)\in k^2\mid a=0\}$. The function $g=y$ will vanish at the point $(0,0)\in V$, but it will not vanish at any other point $(0,b)$ with $b\neq0$.