**Background:** Properness is a much more robust notion than projectiveness. For example, properness descends along arbitrary fpqc covers (see, for example, Vistoli's [Notes on Grothendieck topologies, fibered categories and descent theory](http://arxiv.org/abs/math/0412512), Proposition 2.36). This is far from true for projectiveness. In fact, projectiveness doesn't even descend along Zariski covers! The standard example of a proper non-projective morphism is locally projective over the base. (pictured in an appendix of [Hartshorne](http://books.google.com/books?id=3rtX9t-nnvwC&printsec=frontcover&dq=hartshorne+algebraic+geometry&ei=xDHeSvL_E4GEkgSSzqCVAQ#v=onepage&q=&f=false); page 443, unfortunately not on Google books)




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So how delicate is projectivity? Suppose X is a scheme over a field k, and suppose K is a field extension of k such that XK is projective over K. Does it follow that X is projective over k?


The obvious thing to do (I think) is to pick a very ample line bundle L on XK and try to descend it to X. If K is a finite extension of k, then the *norm* of L will be an ample line bundle on X (if I haven't misunderstood [EGA II](http://www.numdam.org/numdam-bin/item?id=PMIHES_1961__8__5_0), section 6.5 and Corollary 6.6.2). But could it be that there is an *infinite* extension K of k such that XK is projective over K, but X is not projective over k?



Does projectiveness descend along field extensions?
For finite extensions, this is stated explicitly as Corollary 6.6.5 in EGA II and it is also stated there that the result is true for arbitrary extensions. One may reduce the general case to the finite case as follows:


First, we may asume that K is finitely generated since any projective scheme is defined by finitely many equations. This allows us to find a finitely generated integral domain A over k with quotient K and a projective scheme Y over Spec(A) with generic fibre equal isomorphic to X\_K. Let Y' = X x Spec(A) and p the projection from Y' to Spec(A). By construction the generic fibre of the two schemes above over Spec(A) are isomorphic. By properness it easily follows that there is an open set U of Spec(A) such that the two schemes become isomorphic over U. U has rational points over a finite extensions of k so we reduce to the finite extension case.


