I <a href="http://mathoverflow.net/questions/163185/is-the-big-cell-a-principal-open-set" target="_blank">asked a question</a> on mathoverflow, and while researching I stumbled upon a theorem which I did not know before. I consider it quite powerful and the proof is short and sweet, I copied it from <a href="http://math.arizona.edu/~cais/Papers/Expos/AbVar.pdf" target="_blank">this expository paper</a> by <a href="http://math.arizona.edu/~cais/" target="_blank">Bryden Cais</a>: <b>Theorem.</b> Let $X$ be a normal, separated, Noetherian scheme and $U\subseteq X$ a nonempty affine open subset. Then, $X\setminus U$ is pure of codimension one. <b>Proof.</b> We need to show that for each generic point $y\in Y=X\setminus U$, the local ring $\mathcal{O}_{X,y}$ has dimension one. Let $y$ be the generic point of a component of $Y$. Denote by $\mathfrak{m}$ the maximal ideal of the local ring $A:=\mathcal{O}_{X,y}$. Let $S:=\mathrm{Spec}(A)$. The inclusion morphism $f:S\to X$ is affine because $X$ is separated (<a href="#l1">Lemma 1</a>). Thus, $V:=f^{-1}(U)=S\setminus\{ \mathfrak{m} \}$ is an affine, open subscheme of $S$. The morphism on global sections $A=\mathcal{O}_S(S)\to\mathcal{O}_{V}(V)$, corresponding to the inclusion $V\to S$, is therefore injective. It can not be surjective because $V\ne S$. Pick some $a\in\mathcal{O}_{V}(V)\setminus A$. If we now had $\dim(A)>1$, then the prime ideals of height one $\mathfrak{p}\subseteq A$ would satisfy $\mathfrak{p}\ne\mathfrak{m}$, i.e. they would correspond to points contained in $V$. Consequently, we would have $a\in \mathcal{O}_{V,\mathfrak{p}} = A_{\mathfrak{p}}$. Because $A$ is a normal Noetherian domain, it is the intersection of all localizations at prime ideals of height one - this is Corollary 11.4 in Eisenbud's book. This yields the contradition that $a\in A$. <a name="l1"></a><b>Lemma 1.</b> If $f:X\to Y$ is a morphism of schemes with $Y$ separated and $X$ affine, then $f$ is an affine morphism. <b>Proof.</b> This is exercise 3.3.6 in Qing Liu's book. Let $V\subseteq Y$ be an open affine subset. Proposition 3.9(f) in the same book tells you that there is a closed immersion $f^{-1}(V)\cong X\times_Y V\to X\times V$, so $f^{-1}(V)$ is isomorphic to a closed subscheme of an affine scheme, hence affine.