Projection from affine line with infinitely many double points diagram

Consists of schemes $X$, $S$ and a morphism $f:X \to S$, satisfying the following properties:

• $f$ regular, not finite type, unramified, locally of finite type, formally etale, etale, not universally closed, not finite presentation, formally smooth, not proper, not affine, not quasi-finite, locally of finite presentation, smooth, degrees of fibres bounded by 2, not isomorphism, not immersion, open, formally unramified, takes closed points to closed points, finite fibres, universally bounded fibres, not closed immersion, flat, surjective, not finite, not quasi-compact, not separated
• $S$ irreducible, finite dimensional, all stalks of the structure sheaf are domains, all stalks of the structure sheaf are irreducible, connected, integral, separated, dimension 1, normal, noetherian, regular, pure dimension 1, excellent, affine, quasi-compact, reduced, cohen-macaulay, locally noetherian
• $X$ irreducible, finite dimensional, all stalks of the structure sheaf are domains, all stalks of the structure sheaf are irreducible, connected, integral, not separated, dimension 1, normal, not noetherian, regular, pure dimension 1, excellent, not affine, not quasi-compact, reduced, cohen-macaulay, locally noetherian

By the theorems, we also have {"f": {"quasi-separated": true}, "S": {"quasi-separated": true}}

References

• An example of an etale morphism to a noetherian scheme with finite fibres but which is not quasi-compact. Let K be an infinite field and S= A^1_K. Take a copy of S, then for each element of K glue on another copy of S at the complement of that point. Call the big glued thing X, with the obvious projection f: X -> S. Then f is etale, and all fibres have cardinality 1 or 2, but is not quasi-compact.
• Proof that f is not quasi-compact: For each element of $K$ there are two points of $X$ over it, one from the starting copy of S, and an extra one glued on. For p in K, let's call the first point p_0 and the second p_1. Let U_p \sub X be the union of the first copy of S with the point p_1; in other words, U_p is just `S with p doubled'. This U_p is open in X, and together the U_p cover X. But because K is infinite no finite sub-cover will cover X. \QED
• Proof that f is not closed: With the above notation, the complement of any U_p is closed, and it's image in S is exactly the rational points that are not p; this is clearly not closed. \QED

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