In the proof of Proposition 3.27 of Görtz and Wedhorn
A non empty scheme X is integral if and only if it is reduced and irreducible.
(Actually Görtz and Wedhorn define integral schemes as those which are reduced and irreducible and proves that the ring $\Gamma(U, \mathscr{O}_X)$ will be an integral domain; but here I will take the definition of integral that the sections form an integral domain). But in any case the proof begins roughly as follows.
First assume $X$ is irreducible and reduced. Then we have that any open subschemes of are also both reduced and irreducible so it is enough to show that $\Gamma(X, \mathscr{O}_X)$ is an integral domain. To do this, we consider $f,g \in \Gamma(X, \mathscr{O}_X) \text{ s.t. } fg=0$.
I am confused by the following line:
Then $X = V(f) \cup V(g) $, so by the irreducibility we get, say, $X =V(f)$.
My question is why does it make sense to take $V(f)$ and $V(g)$ in the case where $X$ isn’t necessarily affine?
Best Answer
Here's a more global way to see the explanation in the comments.
Let $f \in \Gamma(X, \mathcal{O}_X)$ be a function. We define $V(f) = \{P \in X\;|\;f_P \in \mathfrak{m}_P \subseteq \mathcal{O}_{P, X}\}.$ If $U$ is an affine open subset, we see that $V(f) \cap U = V(f|_U)$ so that this agrees with the notion for affine schemes. Moreover, this is closed since the complement $X_f$ is the union of all $D(f|_{U_i})$ where $\{U_i\}$ is any affine open cover of $X$. See Hartshorne exercise II.2.16, which shows that $\Gamma(X_f, \mathcal{O}_{X_f}) \cong \Gamma(X, \mathcal{O}_X)_f$ when $X$ is noetherian.
Exercise: If $\mathcal{L}$ is a line bundle on $X$ and $f$ is a global section of $\mathcal{L}$, then $X_f: = \{x \in X\;|\;f_P \notin \mathfrak{m}_P\mathcal{L}_P\}$ is open.