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:

$f$ not quasi-compact, smooth, open, locally of finite type, formally etale, not universally closed, not affine, not separated, flat, locally of finite presentation, regular, formally unramified, etale, degrees of fibres bounded by 2, takes closed points to closed points, not finite type, not proper, formally smooth, not finite presentation, universally bounded fibres, not isomorphism, not quasi-finite, not finite, not immersion, unramified, finite fibres, surjective, not closed immersion

$S$ quasi-compact, dimension 1, cohen-macaulay, normal, finite dimensional, connected, integral, all stalks of the structure sheaf are irreducible, locally noetherian, pure dimension 1, noetherian, affine, regular, irreducible, reduced, excellent, separated, all stalks of the structure sheaf are domains

$X$ not quasi-compact, dimension 1, cohen-macaulay, normal, finite dimensional, connected, integral, all stalks of the structure sheaf are irreducible, locally noetherian, pure dimension 1, not noetherian, not affine, regular, irreducible, reduced, excellent, not separated, all stalks of the structure sheaf are domains

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