If X is a quasi-separated algebraic space and Spec k -> X is an etale presentation, then X is isomorphic to Spec k' for a field k'. (This is also true if X is Zariski locally quasi-separated.) The examples of algebraic spaces I know where this fails have diagonals that are not immersions.

Is this also true for weakly locally separated algebraic spaces?

An algebraic space is locally weakly separated if the diagonal is an immersion (not necessarily quasi-compact). In the literature, the term locally separated seems to have been reserved for quasi-compact immersions.

(This question is asked by Johan in the stacks project (currently Remark 32.4.8))

  • $\begingroup$ Can you easily explain one of your non-locally-weakly-separated examples? As soon as you have that your algebraic space has a dense open subspace which is a scheme, you should be done, but I think you need to have quasi-compact diagonal for that result. $\endgroup$ Oct 2, 2009 at 4:37

1 Answer 1


Coincidently, I thought about this a few weeks ago (without any conclusion). I think that I can prove that a "weakly locally separated" algebraic space X with an étale cover Spec(k)->X is of the form Spec(k') if X lives over a field k_0 such that k/*k*_0 is algebraic. If X is not locally separated, this condition does not always hold (take A^1/Z where Z acts by translation and restrict this action to the generic point).

Let K be the algebraic closure of k. Let R=Spec(K) x_ X Spec(K). By assumption

j : R -> Spec(K) x_{k_0} Spec(K)

is an immersion and it is enough to show that this is a closed immersion (since fpqc morphisms descend closed immersions).

We can replace k_0 with its perfect closure. This follows from the observation that R is reduced.

Now, the right-hand scheme is a group scheme over Spec(K). Indeed, it is the fundamental group scheme \pi_0(k_0). It is totally disconnected and all its residue fields are K and the group of K-points is the pro-finite group Gal(K/*k*_0).

R is also totally disconnected and all its points have residue field K. The map

j(K) : R(*K*) -> Gal(K/*k*_0)

is injective and locally closed. Since R(*K*) => K(*K*) is an equivalence relation, it follows that j(K) identifies R(*K*) with a subgroup of Gal(K/*k*_0).

Lemma: A locally closed subgroup H of a topological group G is closed.

pf: The closure of H is a subgroup so we can assume that H is open. It is then easily seen that the complement of H is open.

Thus, R(*K*) is a closed subgroup of Gal(K/*k*_0). In particular, j is a closed immersion.

Remark: If K/*k*_0 is not algebraic, then (if K is algebraically closed) we still have a group structure on the K-points of the fiber product of K over k_0. R(*K*) will be a closed subgroup of this group but it is not clear whether this implies that j is closed.


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