Are there non-projective, but algebraic, hyperkahler varieties?

Question

Let $k$ be an algebraically closed field of characteristic zero. I am not sure what the right definition of a hyperkahler variety over $k$ is, but I think the following might be close enough.

Definition. A smooth proper connected scheme $X$ over $k$ is a hyperkahler variety (over $k$) if $\mathrm{h}^{2,0}(X) = 1$, the etale fundamental group $\pi_1^{et}(X)$ of $X$ is trivial, $\omega_X$ is trivial, and (added later) the generator of $\mathrm{H}^{2,0}(X)$ is everywhere non-degenerate.

With this definition, it is not clear whether every hyperkahler variety over $k$ is projective. If $X$ is a two-dimensional hyperkahler variety over $k$, then $X$ is projective. But what about $\dim X > 2$?

Is every hyperkahler variety over $k$ projective?

PS. Please feel free to correct my definition of a hyperkahler variety over $k$.

Answer 1

With the correct definition of hyperkähler (which as Jason said requires $H^0(X,\Omega^2_X)$ to be generated by a holomorphic symplectic form), there are examples constructed by Yoshioka in section 4.4 here. The field $k=\mathbb{C}$.

These are non-Kähler compact complex simply connected holomorphic symplectic manifolds which are bimeromorphic to a projective hyperkähler manifold via a Mukai flop. They are therefore Moishezon, but certainly not projective. They are also discussed for example in the book by Gross-Huybrechts-Joyce, example 21.7.