In Gathmann's notes, there is a lemma that if $X$ is a projective variety, then $O_X(X) = k$.
There is one step in the proof saying that if $\varphi\in O_X(X)$, then we can suppose $\varphi = \frac{f}{g}$ where $f,g \in S(X)^d$ with $g\neq 0$. If $\deg g >0$, then $Z_p(g)\cap X\neq \emptyset$.
I don't understand why $Z_p(g)\cap X \neq \emptyset$. Here is my attempt:
Suppose $X = Z_p(I)\subset \mathbb{P}^n$. If $Z_p(g)\cap X = Z_p(g, I) = \emptyset$, then we have $Z_a(g, I) = \emptyset$ or $\{0\}$. The first case is impossible since $Z_a(g,I) = \emptyset$ implies $\sqrt{(g, I)} =(1)$. Then $1\in (g,I)$ and hence there is $c$ such that $cg+I = 1+I$. Therefore $g+I$ is invertible in $k[x_0,\ldots, x_n]/I$ which implies $\deg g = 0$. The second case implies $\sqrt{(g+I)} = (x_0,\ldots, x_n)$. I don't know how to get a contradiction from here.
Suppose that $Z_p(g)\cap X \neq \emptyset$ is false. Let $p, q$ be any two distinct points in $X$ and pick a polynomial $g'$ such that $g'(p) = 0$ and $g'(q) = 1$ (why can we find such a function?). Define the morphism $\varphi \colon X \rightarrow \mathbf{P}^1$ by $x \mapsto [g(x), g'(x)]$ (why is this a morphism?). The image of $\varphi$ is closed because projective spaces are complete and therefore a point (why?). But $\varphi(p) \neq \varphi(q)$ and we have a contradiction.