Let $X$ be an algebraic variety over a field $K$. Let's assume that $X$ is integral. Now let $L$ be any field extension of $K$ and let's construct the variety:
$$X_L:= X\times_{\operatorname{Spec K}} \operatorname{Spec L}$$ Which one of the following statements is true?
- $X_L$ projective $\Rightarrow$ $X$ projective
- $X_L$ affine $\Rightarrow$ $X$ affine
- $X_L$ quasi-projective $\Rightarrow$ $X$ quasi-projective
I wasn't able to find an explicit answer to those question, but it seems that 1. and 2. are true, whereas 3. is false.
Assertion 1. is true by EGA II, Corollaire 6.6.5, page 132.
The proof assumes that $[L:K]\lt \infty$ but Grothendieck-Dieudonné claim that the result remains true for an infinite extension field, and that this will be proved in Chapter V.
That chapter never was published, but fortunately Ulrich has given a proof on mathoverflow here.
Assertion 2. is true and proved in Görtz-Wedhorn's Algebraic Geometry I by : Proposition 14.51 (6), page 442.
Assertion 3. Our ever attentive friend Alex Youcis remarks that I had overlooked that Grothendieck-Dieudonné also prove that the result for Assertion 1. is true with "projective" replaced by "quasi-projective", so that your implication 3. is true.
The proof they give is valid only for finite extension fields $K\subset L$ and they refer to the future Chapter V for the general case.
Since that tome never appeared you'll have to check whether Ulrich's proof on mathoverflow can be adapted from the projective to the quasi-projective case.