Why this set is the pole set of $z$? Suppose $V\subset \mathbb P$ a projective variety, $z$ a rational function on $V$ and let's define $J_z=\{F\in k[X_1,\ldots, K_{n+1}\mid \overline Fz\in \Gamma_h(V)\}$ and $S_z$ the polo set of $z$, i.e., the set of those points where $z$ is not defined. I would like to prove $V(J_z)=S_z$.
I'm having problems to prove $S_z\subset V(J_z)$.
My attempt 
Let $P\in S_z$, i.e., $P$ is a polo of $z$. We can write $z=\frac{\overline a}{\overline b}$, where $\overline a,\overline b$ are the images in $\Gamma_h(V)$ of the polynomials $a$ and $b$, resp. In this way, we have $\overline a(P)\neq \overline 0$ e $\overline b(P)=\overline 0$. In order to show $P\in V(I_z)$, we will show for every $F\in I_z$, we have $F(P)=0$. 
If $F\in I_z$, then $\overline Fz\in \Gamma_h(V)$, i.e., there exists $G\in k[X_1,\ldots,X_{n+1}]$ such that $\overline a\overline F=\overline b \overline G$. Therefore, we have $aF-bg\in I(V)$. In this way, evaluating at $P$ in the classes of  $\Gamma_h(V)$, we have $\overline a(P)\overline F(P)-\overline b(P)\overline g(P)=\overline 0$. We know that $\overline b(P)=\overline 0$. Thus,  $\overline a(P)\overline F(P)=\overline 0$. Since $\overline a(P)\neq \overline 0$, we have $\overline F(P)=\overline 0$.
Troubles
The problem with this solution is I proved $\overline F(P)=\overline 0$, but I have to prove $F(P)=0$. 
I don't know how to prove $F(P)=0$. I need help to finish this proof.
Thanks
 A: I think your confusion boils down to this: what is the definition of $\overline{F}(P)$?  Now $\overline{F} \in \Gamma_h(V) = k[X_1, \ldots, X_{n+1}]/I(V)$ is not a polynomial; it is a coset, i.e., $\overline{F} = F + I(V)$.  So what does it mean to evaluate a coset at a point?  Well, the first idea that comes to mind is to choose a representative $f \in F + I(V)$, and define $\overline{F}(P) = f(P)$.  But what if we chose a different representative?  Well, if $f_1, f_2 \in F + I(V)$, then $f_1 - f_2 \in I(V)$.  Then $f_1 - f_2 = g$ for some $g \in I(V)$, so $g$ vanishes on $V$.  Since $P \in V$, then
$$
f_1(P) = (f_2 + g)(P) = f_2(P) + g(P) = f_2(P)
$$
so we can give a well-defined meaning to $\overline{F}(P)$.
Now, back to your original question.  We showed above that the definition of $\overline{F}(P)$ was independent of the choice of representative.  Well, we certainly have $F \in F + I(V)$, so $0 = \overline{F}(P) = F(P)$ by definition.  Or if you don't like that approach: Choose a representative $f \in F + I(V)$.  Then $F = f + g$ for some $g \in I(V)$.  Since $P \in V$, then
$$
0 = \overline{F}(P) = f(P) = F(P) - g(P) = F(P) \, .
$$
Here's how I might phrase a (very detailed) conclusion to your proof.  The evaluation map
\begin{align*}
\varphi = \text{eval}_P : k[X_1, \ldots, X_{n+1}] &\to k\\
G &\mapsto G(P)
\end{align*}
is a homomorphism.  Since $P \in V$, then $\varphi(I(V)) = \{0\}$, so $\varphi$ descends to a (well-defined) homomorphism $\overline{\varphi} : \Gamma_h(V) = k[X_1, \ldots, X_{n+1}]/I(V) \to k$.  Letting $\pi : k[X_1, \ldots, X_{n+1}] \to \Gamma_h(V)$ denote the quotient map, then $\varphi = \overline{\varphi} \circ \pi$.  Then
$$
0 = \overline{F}(P) = \overline{\varphi}\left(\overline{F}\right) = \overline{\varphi}(\pi(F)) = \varphi(F) = F(P) \, .
$$
Just as a remark: although the coordinate ring is defined as a quotient, it's probably better to just think of its elements as polynomial functions restricted to $V$, where $\overline{F}$ corresponds to $F|_V$.  Then all the above argument really amounts to is $F(P) = F|_V(P) = \overline{F}(P) = 0$.
