A quasi-compact image stable under specialization is closed

If $g(Z)$ is closed and $x\in g(Z)$, then $\overline{\{x\}}\subseteq g(Z)$. Thus every specialization of a point of $g(Z)$ is again in $g(Z)$, so the image is stable under specialization.

Assume that $g(Z)$ is stable under specialization, and let $y\in \overline{g(Z)}$. Choose an affine open $U=\mathrm{Spec}A\subseteq Y$ containing $y$ by [F3]. Since $U$ is open, every open neighbourhood of $y$ in $U$ is also open in $Y$; therefore $y$ lies in the closure in $U$ of $g(Z)\cap U$.

The affine open $U$ is quasi-compact by [F2], so [F1] makes $Z_U=g^{-1}(U)$ quasi-compact. It is an open subscheme of $Z$ by [F4]. Its affine open neighbourhoods cover it by [F3], and [F5] gives a finite affine open cover $Z_U={\bigcup }_{i=1}^n{W}_i$; empty members may be discarded.

The image $g(Z)\cap U$ is the finite union of the images $g(W_i)$. Some $g(W_i)$ has $y$ in its closure in $U$: otherwise, for each of the finitely many $i$ there would be an open neighbourhood of $y$ disjoint from $g(W_i)$, and the finite intersection of those neighbourhoods would miss their union, contradicting step 1.2. Fix such an $i$ and write $W_i=\mathrm{Spec}B$. By [F9] the restricted morphism is induced by a ring map $\phi :A\to B$, and its image in $U$ has $y$ in its closure.

Let $\mathfrak{p}\subset A$ be the prime corresponding to $y$, and put $S=\phi (A\setminus \mathfrak{p})$. Suppose $S^{-1}B=0$. By [F13], $0\in S$, so $\phi (f)=0$ for some $f\mathrm{\notin }\mathfrak{p}$. Every prime $\mathfrak{q}\subset B$ then contains $\phi (f)$, and [F10] shows its image lies outside $D(f)$. But $D(f)$ is a basic open neighbourhood of $y$ by [F6, F7], contradicting that $y$ lies in the closure of the image of $\mathrm{Spec}B$. Thus $S^{-1}B$ is nonzero.

The zero ideal of the nonzero commutative ring $S^{-1}B$ is proper. By [F14, F15], AC supplies a maximal ideal $\mathfrak{m}$ of $S^{-1}B$, which is prime by [F16]. Its contraction $\mathfrak{q}\subset B$ along $B\to S^{-1}B$ is prime by [F10]. It avoids $S$, since each $s\mathrm{/}1$ with $s\in S$ is a unit by [F12] and a proper prime ideal contains no unit. Hence ${\mathfrak{p}}^{\mathrm{\prime }}:={\phi }^{-1}(\mathfrak{q})$ is a prime contained in $\mathfrak{p}$: every $a\mathrm{\notin }\mathfrak{p}$ has $\phi (a)\in S$ and so is not in $\mathfrak{q}$.

The point ${\mathfrak{p}}^{\mathrm{\prime }}$ lies in the image of $\mathrm{Spec}B$. Since ${\mathfrak{p}}^{\mathrm{\prime }}\subseteq \mathfrak{p}$, every basic open $D(f)$ containing $\mathfrak{p}$ also contains ${\mathfrak{p}}^{\mathrm{\prime }}$, by [F6, F7]. Thus $\mathfrak{p}\in \stackrel{‾}{\{{\mathfrak{p}}^{\mathrm{\prime }}\}}$ in $U$, so $y$ is a specialization of the image point in $Y$ as well: every open neighbourhood of $y$ in $Y$ restricts to one in the open set $U$. Stability under specialization then gives $y\in g(Z)$.

Every point of $\overline{g(Z)}$ belongs to $g(Z)$ by steps 1.2--6.1, so $g(Z)$ is closed. Together with step 1.1 this proves both directions. The empty source has empty image and satisfies both conditions; if an affine chart is the zero ring it is empty and can be discarded. A zero localization in step 4.1 is excluded by the closure argument. If $A$ is a field, then $\mathfrak{p}=(0)$ is its only prime, so the inclusion ${\mathfrak{p}}^{\mathrm{\prime }}\subseteq \mathfrak{p}$ in step 5.1 is equality and the same proof handles a one-point affine target. The only AC use is prime existence in step 5.1; only a finite affine cover and a single chart from its finite list are used.