Let $f: X \to S$ be a morphism of algebraic spaces, where $S$ is a scheme. If $f$ is separated and étale then Knutson's criterion says that $X$ is actually a scheme.

I have a some closely related questions.

  1. Why does one require separatedness in Knutson's result? What is an example of an étale cover of a scheme which is not a scheme?

I cannot visualise at all what a non-separated étale morphism could look like.

  1. What is an example of a non-separated étale morphism $f: X \to S$ with $X$ an algebraic space? Can there even exist such a morphism with $X$ a scheme?

To answer question 2 (with $X$ a scheme), just take for $S$ the affine line over a field $k$, for $X$ the usual ``$S$ with the origin doubled'' (two copies of $S$ glued along $S\smallsetminus\{0\}$) and for $f$ the obvious projection (which is the identity on each copy).

For question 1, the best example I know is a ``Galois twist'' of the above, where $f$ is still an isomorphism above $S\smallsetminus\{0\}$ but $f^{-1}(0)$ is the spectrum of a $\mathbb{Z}/2$-extension $k'$ of $k$, instead of two $k$-rational points.

To achieve this, start with the constant $S$-group scheme $G:=(\mathbb{Z}/2\mathbb{Z})_S$. Remove the nontrivial point of $G$ above $0\in S$: you get an open subgroup $H$ of $G$. Fixing a $\mathbb{Z}/2$-extension $k'$ of $k$, the $S$-scheme $S':=S\times_kk'$ has a natural action of $G$, induced by the Galois action on $k'$, and the quotient $S'/G$ is of course $S$.
Now the required example is $X:=S'/H$, with the natural projection on $S$. Clearly the $H$-action defines an étale equivalence relation, so $X$ is an algebraic space. Next, if you extend the scalars to $k'$, you get (the $k'$-version of) the preceding example. To see that $X$ is not a scheme, look at the only point $x$ of $X$ above $0$ (with residue field $k'$). If $U$ is any open subspace of $X$ containing $x$, then $U\times_kk'$ must contain both ``funny'' points of $X\times_kk'$, so $U$ cannot be affine (not even separated).

EDIT to answer Daniel's comments:

  1. Call $X_0$ the first example (standard affine line with double origin). Then the second example $X$ can actually be obtained as a Galois twist of $X_0$: the group $\Gamma:=\mathbb{Z}/2\mathbb{Z}\cong\mathrm{Gal}(k'/k)$ acts on $X_0$ by swapping the two copies of $S$, and then $X$ is just the corresponding Galois twist, i.e. the contracted product $X_0\times_k^\Gamma\mathrm{Spec}(k')$.
  2. Even a twist of a separated scheme need not be a scheme. Hironaka has constructed proper, non-projective $k$-varieties, and you can easily tune the construction to find such a variety $Z$ having an involution $\sigma$ such that there is an orbit $Y$ which is not contained in any affine open set. Then the twist $Z':=Z\times_k^\Gamma\mathrm{Spec}(k')$ is an algebraic space but not a scheme, since the twist of $Y$ is a point of $Z'$ which cannot have an affine neighbourhood. (By the same argument, the quotient $Z/<\sigma>$ is not a scheme, and neither is the symmetric square of $Z$ over $k$).
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    $\begingroup$ This is amazing, thank you. I could not have hoped for a clearer answer. $\endgroup$ – Daniel Loughran Apr 5 '18 at 22:13
  • $\begingroup$ Is it also possible to construct your example using twisting and Galois cohomology? The map in question has an automorphism given by swapping the two copies of the origin. This gives the action of some group scheme $A$. Can we then twist by an element of $H^1(k,A)$ to obtain your example? $\endgroup$ – Daniel Loughran Apr 6 '18 at 7:41
  • $\begingroup$ I do find your example very interesting, as I always thought the the twist of a scheme was again a scheme. Is this only true for separated schemes? $\endgroup$ – Daniel Loughran Apr 6 '18 at 7:42

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