$\def\sF{\mathcal{F}} \def\sO{\mathcal{O}}$Today I wondered is every finite scheme affine? and I found a proof using Serre's criterion on affiness; however, I don't know if I may be using a sledgehammer to crack a nut. My questions are:
Is there an easier way of proving this?
Does this result show up somewhere in the literature? (I could not find anything on google.)
If one deals with finite discrete schemes, then it's easy (one uses that singleton schemes are affine + a finite union of disjoint open affine subschemes is affine), but in the general case I don't know if there's an simpler argument.
[EDIT 1/9/23: thanks to Alex Kruckman for spotting the mistake on my proof (see his comment). The argument below (I edited it slightly) now shows "if $X$ is a finite scheme with $n$ points, then $H^i(X,\mathcal{F})=0$ for all $i\geq n$ and all quasi-coherent sheaves $\mathcal{F}$ over $X$." (This is a weaker form of 0A3G(1); note that $\dim X< n$.)]
Here's how the proof goes: Suppose $X$ is a finite scheme. We proceed by induction on the cardinal $|X|=n$. For $n=1$ it's trivial, so suppose $n>1$. If there is $x\subset X$ which has a unique open neighborhood (namely, $X$), then we are done (by the vanishing of quasi-coherent sheaves on affines). So suppose there is no such $x$. Let $\sF$ be a quasi-coherent sheaf of $\sO_X$-modules. Using the induction hypothesis and Mayer-Vietoris, to show $H^i(X,\sF)=0$ for all $i\geq n$, it suffices to construct open cover $X=U\cup V$, where $U,V\neq X$. For each $x\in X$, denote $U_x$ to the intersection of all open neighborhoods of $x$ (which is open). We assumed $U_x\neq X$ for all $x\in X$. One has $$ U_x=\{ y\in X\mid y\rightsquigarrow x \} $$ (this formula holds in all topological spaces). Now, let $x_1,\dots,x_r$ be the closed points of $X$ (i.e., the maximal points in the specialization order) and call $U_i=U_{x_i}$. Then $x_r\notin U_1\cup\cdots\cup U_{r-1}=:V$. Calling $U=U_r$, we have that $X=U\cup V$ is the desired open cover. $\square$
There's a three-point scheme that is not affine: glue two copies of $\operatorname{Spec}(R)$ for $R$ a DVR along the generic point.