Why does $X = \bigcup_{i=1}^k X_{f_i^N}$? On page 70 of Mumford's The Red Book of Varieties and Schemes second, expanded edition (GTM 358), in the proof of Lemma 2, the author said

On the other hand, we have $$X = \bigcup_{i=1}^k X_{f_i}  = \bigcup_{i=1}^k X_{f_i^N}.$$

I don't know why this is true. Does this come from the previous fact "$b_i' f_j^N - b_j'f_i^N =0$"?
Thanks very much.
 A: Here is a more general result which applies directly to your problem.


Let $X = \operatorname{Spec} A$. Then for $f,g\in A$, $D(f) = D(g)$ iff $\sqrt{(f)} = \sqrt{(g)}$. Proof:


If $D(f) = D(g)$ then a prime $P$ doesn't contain $f$ iff it doesn't contain $g$. Negating, a prime $P$ contains $f$ iff it contains $g$.  The set of all primes containing an element is its radical. Hence taking the intersection over all these primes gives that $\sqrt{(f)} = \sqrt{(g)}$. 
Conversely if $\sqrt{(f)} = \sqrt{(g)}$ let $P$ be a prime such that $f \notin P$. Then $f \in \sqrt{(f)} = \sqrt{(g)}$ implies there is some $n$, $h\in A$ for which $f^n = gh$. Then $g \notin P$ too otherwise $f^n \in P \implies f \in P$. Hence $P \in D(g)$. Similarly we conclude $D(g) \subseteq D(f)$, hence $D(f) = D(g)$.
A: You should mention what everything is, but assuming it's some scheme (or even a locally ringed space) and $f \in \mathcal{O}_X(X)$ you should think of $X_f$ as being the set of points of $X$ where $f$ does not vanish. Vanishing means you map to zero under the homomorphism
$$
\mathcal{O}_X(X) \to \mathcal{O}_{X, p} \to \mathcal{O}_{X, p}/\mathfrak{m}_{X, p} = \kappa(p)
$$
which "evaluates" functions at $p$. The target is a field, so if $f^n$ vanishes then $f$ certainly does.
