The base field will be the field of complex numbers. I have a slightly technical problem concerning the resolution of singularities of a certain variety. Basically, I want to to know if it is projective.


Let $Y$ be a normal projective variety with only quotient singularities. Let $C$ be a smooth projective curve.


Let $p:Y\longrightarrow C$ be a flat projective morphism.


Let $f:X\longrightarrow Y$ be a resolution of singularities. Is $X$ necessarily projective over the base field? I know this is true in dimension 2. Are there any counterexamples in dimension 3?



Is the desingularization of a normal variety with only quotient singularities projective
The following paper might be useful. 


*Nagata*, "Existence theorems for nonprojective complete algebraic varieties", Illinois J. Math. 2 1958 490--498.


At the end of the paper Nagata says the following:


"Hironaka recently proved the following: If $V$ is a nonsingular projective variety of dimension not less than 3, then there exists a complete nonsingular nonprojective variety $V'$ such that (1) $V$ is birationally equivalent to $V$, and (2) $V'$ dominates $V$."


So for 3-folds, then this seems to give a negative answer to your question (assuming one can track down the reference, I didn't seem to find it). 


In your setting, simply take $Z \to Y$ to be any resolution of singularities. Then take $V'$ as in Hironaka's result. It should also be a resolution of singularities (assuming I am reading it right). 


