Finite and finite type etale schemes over an algebraically closed field

Local structure at an 'etale point. Let $X$ be 'etale and of finite type over $k$ (this covers the finite case of claim 1, since a finite morphism is of finite type), let $f\text{\hspace{0.17em}⁣}:X\to \mathrm{Spec}k$ be the structure morphism and let $x\in X$. By [F1] $f$ is locally of finite presentation and flat at $x$ and is smooth of relative dimension $0$ at $x$; since this holds at every point, $f$ is locally of finite presentation [F11]. By [F2] in the forward direction $f$ is unramified at $x$, so ${\mathrm{\Omega }}_{X\mathrm{/}k,x}=0$. The base $\mathrm{Spec}k$ has the single point $s$ with residue field $\kappa (s)=k$ and maximal ideal ${\mathfrak{m}}_s=0$, so [F3] (AC) gives that $\kappa (x)\mathrm{/}k$ is finite separable and that ${\mathfrak{m}}_x={\mathfrak{m}}_s{\mathcal{O}}_{X,x}=0$. Thus ${\mathcal{O}}_{X,x}$ is a field and ${\mathcal{O}}_{X,x}=\kappa (x)$.

Converse: finite disjoint unions of copies of $\mathrm{Spec}k$ are finite 'etale. Let $X={\coprod }_{i=1}^r\mathrm{Spec}k=\mathrm{Spec}(k^r)$ over $k$; this is a scheme over $k$ by [F9], and it is finite since $k^r$ is a module-finite $k$-algebra [F9]. The ring $k^r$ is a free $k$-module of rank $r$, hence flat [F10], and of finite type over $k$, hence finitely presented since $k$ is Noetherian [F10], so the structure morphism is locally of finite presentation [F10]. For the differentials, let $D\text{\hspace{0.17em}⁣}:k^r\to M$ be a $k$-derivation into a $k^r$-module $M$ and let $e_1,\dots ,e_r$ be the standard idempotents; for $i\mathrm{\ne }j$ one has $0=D(e_i{e}_j)=e_{i}D(e_j)+e_{j}D(e_i)$, and multiplying by $e_i$ gives $e_{i}D(e_j)=0$; also $D(e_i)=D(e_i^2)=2e_{i}D(e_i)$, and multiplying by $e_i$ resp. by $e_j$ with $j\mathrm{\ne }i$ gives $e_{i}D(e_i)=0$ and $e_{j}D(e_i)=0$; hence $D(e_i)={\sum }_j{e}_{j}D(e_i)=0$ for all $i$, so $D=0$. By [F8] this says ${\mathrm{Hom}}_{k^r}({\mathrm{\Omega }}_{k^r\mathrm{/}k},M)=0$ for every $M$, and taking $M={\mathrm{\Omega }}_{k^r\mathrm{/}k}$ and the identity gives ${\mathrm{\Omega }}_{k^r\mathrm{/}k}=0$. For étaleness, use the definition directly on each open component: its structural map is the identity of $\mathrm{Spec}k$, a flat finitely presented map; after every extension $K\mathrm{/}k$ its fibre is $\mathrm{Spec}K$, whose sole local ring is a field, regular of dimension zero. Thus [F1] makes each component, and hence the union, étale without the AC-qualified criterion. For $r=0$ this is the empty scheme, which is finite and 'etale vacuously, and the same computation with no idempotents gives ${\mathrm{\Omega }}_{0\mathrm{/}k}=0$.

The residue field is $k$. Let $\alpha \in \kappa (x)$. By step 1.1 $\kappa (x)\mathrm{/}k$ is finite, so $\kappa (x)$ is a finite-dimensional $k$-vector space and the subfield $k(\alpha )$ is a $k$-subspace of it, hence finite-dimensional; by [F4] $\alpha$ is algebraic over $k$, so there is a nonzero $P\in k[T]$ with $P(\alpha )=0$, and we choose such $P$ of least positive degree. Were $\mathrm{deg}P\ge 2$, then $P$ is nonconstant, so by [F4] it has a root $\lambda \in k$ and the factor theorem [F4] writes $P=(T-\lambda )Q$ with $Q\mathrm{\ne }0$ of degree $\mathrm{deg}P-1\ge 1$; minimality of $\mathrm{deg}P$ gives $Q(\alpha )\mathrm{\ne }0$, hence $0=P(\alpha )=(\alpha -\lambda )Q(\alpha )$ forces $\alpha =\lambda \in k$, and then $T-\lambda$ is a nonzero polynomial of degree $1<\mathrm{deg}P$ vanishing at $\alpha$, contradicting minimality. Therefore $\mathrm{deg}P=1$ and $\alpha \in k$, so $\kappa (x)=k$; with step 1.1, ${\mathcal{O}}_{X,x}=k$.

Affine charts split as a finite product of copies of $k$. Let $U=\mathrm{Spec}A\subseteq X$ be an affine chart with $A$ a finite type $k$-algebra; such charts exist since $f$ is of finite type [F11], and in the finite case $X=\mathrm{Spec}B$ with $B$ module-finite over $k$ is itself such a chart [F9]. If $A=0$, the chart is empty and is the product with zero factors; hence assume $A\mathrm{\ne }0$. For a prime $\mathfrak{p}\subseteq A$ the local ring is $A_{\mathfrak{p}}={\mathcal{O}}_{X,\mathfrak{p}}=k$ by step 2.1, a field whose only prime is $0=\mathfrak{p}{A}_{\mathfrak{p}}$; by the localization correspondence [F5] the primes of $A$ contained in $\mathfrak{p}$ correspond to the primes of $A_{\mathfrak{p}}$, so the only prime of $A$ contained in $\mathfrak{p}$ is $\mathfrak{p}$ itself: every prime of $A$ is minimal, and consequently there is no strict chain of primes, so $\dim A=0$ [F5]. The ring $A$ is Noetherian [F5]; since by [F6] every prime is contained in a maximal ideal and a strict inclusion $\mathfrak{p}⊊\mathfrak{m}$ would be a strict chain, every prime of $A$ is maximal, so $A$ is Artinian by [F6]. The structure theorem [F7] then gives $A\cong {\prod }_{\mathfrak{p}}{A}_{\mathfrak{p}}={\prod }_{\mathfrak{p}}k=k^{r_U}$, the product over the finitely many primes of $A$, and hence $U=\mathrm{Spec}(k^{r_U})$ is a finite disjoint union of copies of $\mathrm{Spec}k$.

Global conclusion for finite type $X$. Since $f$ is of finite type it is quasi-compact [F11], so $X$ admits a finite cover by affine charts $U_1,\dots ,U_n$ as in step 3.1 (in the finite case one chart suffices by [F9]). Each $U_i$ is a finite disjoint union of copies of $\mathrm{Spec}k$, so each of its points is open in $U_i$ and hence in $X$; thus the underlying space of $X$ is discrete, and quasi-compactness makes it finite, say $X=\{x_1,\dots ,x_r\}$. For each $i$ the open subscheme $\{x_i\}$ has local ring ${\mathcal{O}}_{X,x_i}=k$ by step 2.1 and lies in some chart $U_j$ that is a disjoint union of copies of $\mathrm{Spec}k$, so $\{x_i\}\cong \mathrm{Spec}k$; therefore $X\cong {\coprod }_{i=1}^r\mathrm{Spec}k$ by [F9]. This proves claims 1 and 2, and since $k^r$ is a finite $k$-module, $X$ is finite over $k$ [F9].

The finite type hypothesis is essential. By [F9] the disjoint union $X={\coprod }_{n\ge 1}\mathrm{Spec}k$ is a scheme whose components are open and isomorphic to $\mathrm{Spec}k$, with structure morphism $f\text{\hspace{0.17em}⁣}:X\to \mathrm{Spec}k$ induced by the identity on each component; at a point $x$ lying in one component, the germ of $f$ is the identity $\mathrm{Spec}k\to \mathrm{Spec}k$, which is flat, finitely presented (the algebra $k\cong k[T]\mathrm{/}(T)$ is finitely presented [F10]) and after every field extension its fibre is the spectrum of that field, regular of dimension zero; thus it is étale directly by [F1], as in the identity-component argument of step 1.2; by the germ property [F1], $f$ is 'etale at $x$. As $x$ was arbitrary, $f$ is 'etale over $k$. The underlying set of $X$ is infinite, one point per component [F9], while a finite disjoint union ${\coprod }_{i=1}^r\mathrm{Spec}k$ has exactly $r$ points, so $X$ is not isomorphic to any such finite union: the finite type hypothesis of claims 1 and 2 cannot be dropped. The components form an open cover of $X$ with no finite subcover, since each component is nonempty and the components are pairwise disjoint, so $X$ is not quasi-compact; were $X$ finite over $k$, its inverse image $X=f^{-1}(\mathrm{Spec}k)$ would be affine [F9], hence quasi-compact [F9], a contradiction; and were $X$ of finite type over $k$ it would be quasi-compact [F11]. Hence $X$ is 'etale over $k$ but neither finite nor of finite type over $k$. The Axiom of Choice [F12] is assumed in the Statement and used exactly through [F3] in step 1.1, [F6] and [F7] in step 3.1 and [F2] in step 1.1. These identity-component arguments do not invoke the AC-qualified criterion and require no choice.

Conclusion. Claims 1 and 2 are steps 1.1, 2.1, 3.1 and 4.1; claim 3 is the converse step 1.2; the uniqueness statement of claim 4 is immediate from the global step since an isomorphism of $k$-schemes induces a bijection of underlying sets and $r$ is the number of points; and claim 5 is the final step.