Suppose $X$ is a projective variety, $f\colon X\to Y$ is a finite surjective morphism onto variety $Y$, must $Y$ be a projective variety?



Is "quotient" of projective variety projective?
No, that is not true. Let $X$ be $\mathbb{P}^3\_k$. Let $g:L\hookrightarrow X$ be a line in $X$. Let $h:C\hookrightarrow X$ be a plane conic in $X$ that is disjoint from $L$ and that contains a $k$-point. Let $i:L\to C$ be an isomorphism of $k$-schemes. Let $f:X\to Y$ be the coproduct of the two morphisms $g$ and $h\circ i$. Then $Y$ is a proper $k$-variety, and $f$ is finite and surjective.


If $\mathcal{L}$ were an ample invertible sheaf on $Y$, then the pullback $f^\*\mathcal{L}$ would be an ample invertible sheaf on $X$ whose degree on $L$ equals the degree on $C$. Every invertible sheaf on $\mathbb{P}^3$ is of the form $\mathcal{O}(d)$ for some $d\in \mathbb{Z}$. Only for $d=0$ is the degree on $L$ equals to the degree on $C$. For $d=0$, this invertible sheaf is not ample. Thus $Y$ is not projective.


