Yes. The category of algebraic spaces is the smallest subcategory of the category of sheaves of sets on Aff, the opposite of the category of rings, under the etale topology which (1) contains Aff, (2) is closed under formation of quotients by etale equivalence relations, and (3) is closed under disjoint unions (indexed by arbitrary sets). An abstract context for such things is written down in "Algebraization of complex analytic varieties and derived categories" by Toen and Vaquie, which is available on the archive. Toen also has notes from a "master course" on stacks on his web page with more information. It might be worth pointing out that their construction of this category also goes by a two-step procedure, although in their case it's a single construction performed iteratively (and which stabilizes after two steps). This is unlike the approach using scheme theory in the literal sense, as locally ringed topological spaces, where the two steps are completely different. After the first step in T-V, you get algebraic spaces with affine diagonal. Also worth pointing out is that their approach is completely sheaf theoretic. The only input you need is a category of local models, a Grothendieck topology, and a class of equivalence relations. You then get algebraic spaces from the triple (Aff, etale, etale). But the general machine (which incidentally I believe is not in its final form) has nothing to do with commutative rings. I think it would be interesting to plug opposites of other algebraic categories into it.
Normality
For an integral scheme, being normal (integrally closed in ones own fraction field) satisfies this property. Indeed, suppose that $a, b \in A = \Gamma(X, O_X)$ are such that $a/b$ satisfy some polynomial $p(x) \in A[x]$. Then $a|_U, b|_U$ satisfy the same polynomial after restriction to each (affine) set $U \subseteq X$.
Of course this shows up in many applications of things like Stein factorization.
Semi-normality
A reduced ring $R$ is seminormal if for any finite extension $R \subseteq S$ satisfying the following two properties is an isomorphism.
- The induced map on $\text{Spec}$'s is an isomorphism
- The induced residue field extensions $k(r) \subseteq k(s)$ are isomorphisms for all $s \in \text{Spec } S$ mapping to $r \in \text{Spec } R$.
The typical example of a seminormal ring is a node, the cusp $k[x^2,x^3]$ is not seminormal
Equivalently, $R$ is seminormal if and only if for any $a/b$ in the total ring of fractions of $R$, one has that if $(a/b)^2, (a/b)^3 \in R$ then $(a/b) \in R$ (see a paper by Swan, he might be assuming finitely many minimal primes, I forget the details). It follows similarly that seminormality satisfies this property.
Weak normality
Weak normality is similar to semi-normality. A reduced ring is called weakly normal if for any finite birational extension $R \subseteq S$ satisfying the following properties is an isomorphism:
- The induced map on $\text{Spec}$'s is an isomorphism
- The induced residue field extensions $k(r) \subseteq k(s)$ are purely inseparable for all $s \in \text{Spec } S$ mapping to $r \in \text{Spec } R$.
This can also be phrased as requiring that every birational universal homeomorphism is an isomorphism.
I do NOT know if weakly normal rings satisfy the sort of property asked for. I do not think it is in the literature (but perhaps I am wrong). I remember I convinced myself that they did not several years ago, but never wrote down an example carefully.
Best Answer
EGA 1 (Springer edition), proposition (1.6.3).