I'm reading Fulton's algebraic curves book.
I'm trying to understand this solution which I found online of the question 4.17 on page 97.
What I didn't understand is why $V(J_z)$ are exactly those points where $z$ is not defined (see the last line the solution).
Maybe I'm missing something, but I couldn't understand why this is true.
Thanks
Write $z = a/b$, with $a, b \in \Gamma_{h}(V)$ forms of the same degree $d$. Note that $\overline{b}z = \overline{a} \in \Gamma_{h}(V)$, and therefore $b \in J_{z}$.
Let $P \in V(J_{z})$. Then $F(P) = 0$, for every polynomial $F \in k[X_{1}, \ldots, X_{n+1}]$ such that $\overline{F}z \in \Gamma_{h}(V)$. Since $b \in J_{z}$, we have $b(P) = 0$, and thus $z$ is not defined at $P$. This shows that $V(J_{z})$ is contained in the pole set of $z$.
Conversely, let $P$ be a pole of $z$. Then $a(P) \neq 0$ and $b(P) = 0$. Let $F \in J_{z}$. Then, $\overline{F}z = \overline{G}$, for some $\overline{G} \in \Gamma_{h}(V) = k[X_{1}, \ldots, X_{n+1}]/I(V)$. That is, $Fz - G \in I(V)$. Since $I(V)$ is an ideal, $b(Fz - G) = Fa - bG \in I(V)$. Evaluating at $P$, we obtain $0 = F(P)a(P) - b(P)G(P) = F(P)a(P)$, because $b(P) = 0$. Since $a(P) \neq 0$, we conclude that $F(P) = 0$. This shows that the pole set of $z$ is contained in $V(J_{z})$.
Therefore, the pole set of $z$ is exactly $V(J_{z})$; in particular, it is an algebraic subset of $V$, as the solution posted shows.