This is a question from an online note. Let $A$ be a two-dimensional $\mathbb C$-torus. And there is an involution on $A$: $A\to A, x\mapsto -x$. The action has 16 fixed points. Let $Y:=A/\{\pm1\}$, then $Y$ is a complex surface with 16 ordinary double points. Let $X$ be the blow up of $Y$ at all 16 singular points. After some calculations, we can see $X$ is a $K3$ surface. Then it's claimed that if $A$ is not projective then $X$ is not projective and we get an example of a non-projective $K3$ surface. But I feel confused why "if $A$ is not projective then $X$ is not projective".


I know since $A\to X$ is finite, thus if $Y$ is projective, then we can pullback an ample line bundle to an ample line to $A$. But I can't see why $X$ is projective implies $Y$ is projective? Do we have contracting a rational curve on a complex surface preserves the projectivity?



Kummer surfaces which are not projective
In fact, $X$ is projective if and only if $A$ is projective.


If $A$ is projective, then $Y$ is so, being the quotient of a projective variety by a finite group (this is a toy model of GIT, see [this](https://mathoverflow.net/questions/209695/is-quotient-of-projective-variety-projective) MO question). Then $X$ is projective, too, being the blow-up of the projective variety $Y$ at a finite number of points.


Conversely, assume $X$ projective. Then there is a double cover $\tilde{A} \to X$, where $\tilde{A}$ is the blow-up of $A$ at its $16$ points of order $2$. This shows that $\tilde{A}$ is projective, so the blow-down $A$ is projective as well (an alternative argument is noticing that $X$ projective implies $Y$ projective and so $A$ projective, since contracting a $(-2)$ curve on a projective surface preserves projectivity, as explained in abx's comment).


