Here's a dangerous question: We all know that a variety is an integral scheme, separated and of finite type over an algebraically closed field. Now, if I remove the separated hypothesis, I get the class of schemes which are made by gluing together (finitely many) classical affine varieties and applying the famous fully faithful functor $Var_k \longrightarrow Schemes_k$. Thus, separatedness must relate somehow to the "way of gluing" together these affine varieties, (which is kind of confirmed when you look at the uberclassical example of the double line). Is there someone here that can explain what way of gluing we ban when restricting to separated schemes ?

1$\begingroup$ interesting question. 1+ $\endgroup$ – Martin Brandenburg Jul 3 '10 at 17:57
The necessary and sufficient condition on a schemes $\{U_i\}$ and gluing isomorphisms $\varphi_{ij}:U_{ij} \simeq U_{ji}$ between opens $U_{ij} \subset U_i$ ($i, j \in I$) that the gluing $X$ be separated is that the graph map $U_{ij} \rightarrow U_i \times U_j$ defined by $u \mapsto (u, \varphi_{ij}(u))$ is a closed immersion (or equivalently, has closed image) for all $i, j$. This is seen by intersecting $\Delta(X)$ with the opens $U_i \times U_j$ that cover $X \times X$.
(Taking $i=j$, this says that all $U_i$ are separated, which is automatic when all $U_i$ are affine; in such cases the closed immersion condition forces all $U_{ij}$ to also be affine, so in the context of the question one loses nothing by requiring all $U_{ij}$ to be affine open prior to stating the closed immersion condition.)

7$\begingroup$ Just to say that in other words: for all open affines $U_i$ and $U_j$, you must have both (i) $U_{ij}$ is also affine, and (ii) the coordinate ring $k[U_{ij}]$ must be generated as a $k$algebra by the union of the images of $k[U_i]$ and $k[U_j]$ via the restriction homomorphisms. $\endgroup$ – Jason Starr Mar 11 '14 at 14:22
Basically it's the sort of gluing that would violate the (discrete) valuative criterion for separatedness. See Chapter II, Section IV of Hartshorne for example.
For example, if you are gluing $U$ to $V$ (say separated noetherian schemes) along a common open subscheme $W$, if you want the result to be separated, you presumably want to require that:
For every spec of a DVR, $Z =$ { generic Pt, closed Pt } with a map $Z \to U$ such that, the generic point of $Z$ is sent into $W$, then the closed point of $Z$ is also sent to $W$ as well whenever the map {generic Pt} $\to W$ extends to a map $Z \to V$.

$\begingroup$ That does not seem like a very useful way of telling whether a given atlas defines a separated scheme. Boyarsky's answer is the "standard" criterion in terms of an atlas. $\endgroup$ – Jason Starr Mar 11 '14 at 14:24

$\begingroup$ Hi Jason, I'm not sure I agree that this answer (given in 2010) isn't useful, at least in terms of constructing examples. But perhaps I tend to think about too many things valuatively. I would agree that Boyarsky's answer is the canonical one. $\endgroup$ – Karl Schwede Mar 11 '14 at 16:57

1$\begingroup$ Somebody else revived the post. When I learned about separatedness and properness from Brian Conrad (many years ago), his philosophy was not to use the valuative criterion where a more direct argument would do. To prove properness of moduli spaces, use the valuative criterion, i.e., "semistable reduction". But almost everything that is proved using the valuative criterion in Hartshorne's book, can, in fact, be proved directly. Anyway, here is a challenge: give one example of an explicit atlas by affines where your technique is "faster" than Boyarsky's. $\endgroup$ – Jason Starr Mar 11 '14 at 17:33