See  [ GW1 ] Proposition/Definition 10.5.

For the interesting implication (iii) \(\Rightarrow \) (i), use that an open subset of a scheme is quasi-compact if and only it can be covered by finitely many affine open subschemes. Indeed, take a cover \(S=\bigcup _j V_j\) as in (iii) and let \(V \subseteq S\) be any quasi-compact open. Cover it by finitely many principal opens of the \(V_j\). It is then enough to show that these principal opens have quasi-compact inverse image. But by assumption \(f^{-1}(V_j)\) can be covered by finitely many affine opens in \(X\); thus the inverse image of a principal open therein can be covered by one principal open in each of these affine schemes.

Omitted (See  [ GW1 ] Sections (10.1), (10.2).)

This is a version of Hilbert’s Nullstellensatz: Namely, this theorem says that for a \(k\)-algebra \(A\) of finite type and maximal ideal \({\mathfrak m}\subset A\), the field \(A/{\mathfrak m}\) is a finite extension of \(k\). This shows (i) \(\Rightarrow \) (ii). Furthermore, the implication (ii) \(\Rightarrow \) (iii) is obvious.

Now assume (iii), and let \(U \subseteq X\) be an affine open neighborhood of \(x\), say \(U = \operatorname{Spec}(A)\), and \(x\) corresponds to the prime ideal \({\mathfrak p}\subset A\). Then \(A/{\mathfrak p}\subseteq \kappa (x)\) and the assumption implies that the ring homomorphism

is an injective integral homomorphism between domains. Since the source is a field, so is the target. Hence \(x\) is closed in \(U\). Since this holds for all affine open neighborhoods of \(x\) in \(X\), it follows that \(x\) is closed in \(X\).

Let \(x\) be a closed point of \(X\). We obtain an inclusion \(\kappa (f(x)) \subseteq \kappa (x)\) between the residue class fields (which is a \(k\)-homomorphism). Since \(\kappa (x)\) is algebraic over \(k\) by Theorem 8.6, so is \(\kappa (f(x))\), and invoking the theorem again, we obtain the desired conclusion.

By the theorem, and since algebraically closed fields have no non-trivial algebraic extensions, the set of closed points in \(X\) equals the set of points with residue class field \(k\). By Proposition 4.22, this set can be identified with the set \(X(k)\) of \(k\)-valued points of \(X\).

The following implications are easy:

Let us show that (iv) implies (v). Let \(\emptyset \ne L \subseteq X\) be locally closed,say \(L = F\setminus F'\) for \(F' \subsetneq F \subseteq X\) closed. Then \(F'\cap Z\ne F\cap Z\), and this implies \(L\cap Z\ne \emptyset \). Next we show that (v) implies (ii). Take \(F, F' \subseteq X\) closed with \(F\cap Z = F'\cap Z\), equivalently \(((F\cup F') \setminus (F'\cap F))\cap Z = \emptyset \). But \((F\cup F') \setminus (F'\cap F)\) is locally closed, so (v) implies that this set is empty, whence \(F=F'\). At this point we have seen that (i), (ii), (iv) and (v) are all equivalent.

It remains to show that these properties also imply (iii). For this, first note that for any locally closed subset \(L \subseteq X\),

(Clearly, we have \(\subseteq \). Furthermore, the set

is open in \(\overline{L}\), hence locally closed in \(X\), but has empty intersection with \(Z\), and thus is empty in view of (v). But \(L\) is dense in \(\overline{L}\), so it intersects every non-empty open subset, so \(\overline{L}\setminus \overline{L\cap Z} = \emptyset \).) This shows, that for locally closed subsets \(L, L'\) in \(X\) with \(L\cap Z = Ö'\cap Z\) we have \(\overline{L} = \overline{L'}\); call this closed set \(F\).

It follows easily from the characterization (ii) that \(Z\cap F\) is very dense in \(F\). Since \(L\) and \(L'\) are open in \(F\), and have same intersection with \(Z\cap F\), this implies \(L=L'\) in view of the characterization (i).

We check that every non-empty locally closed subset \(Z\) of \(X\) contains a closed point. Shrinking \(Z\), if necessary, we may assume that it is closed in an affine open subscheme \(U = \operatorname{Spec}(A)\) of \(X\). We may view \(Z\) as a closed subscheme of the form \(\operatorname{Spec}(A/{\mathfrak a})\) for some ideal \({\mathfrak a}\). But then \(A/{\mathfrak a}\) is non-zero, because the subset is non-empty, hence contains a maximal ideal, and this corresponds to a closed point of \(\operatorname{Spec}(A/{\mathfrak a})\), hence also of \(U\).

Since \(U\) and \(X\) are locally of finite type over \(k\), Theorem 8.6 implies that this closed point of \(U\) is also closed in \(X\).

We need to show that the equalizer \({\rm Eq}(f,g)\) (cf. the proof of Proposition 7.20) equals \(X\). But \({\rm Eq}(f,g)\) is a subscheme of \(X\) and by assumption \({\rm Eq}(f,g)(k) = X(k)\). By the proposition this implies that this subscheme has the same underlying topological space as \(X\). In particular, it is a closed subscheme. Because \(X\) is reduced, it has to be equal to \(X\).