Let $k$ be an arbitrary field, we work with schemes $X$ of finite type over $k$. Does every irreducible projective scheme have a finite surjective morphism to a projective space $\mathbb{P}^n\_k$?. What if I just assume that $X$ is equidimensional. Does the same argument work? 


We know that a proper $k$-scheme with this property must be also be projective by formal properties of ample line bundles.


I would do this by projecting from sufficiently general points, and this probably works (maybe not in complete generality) but I can't help but think there is a cleaner argument (that also doesn't require possibly assuming that the field is infinite, algebraically closed or of characteristic $0$).


Feel free to assume our schemes have basic niceness properties.



Finite extension of projective space
Let $X$ be any projective scheme of dimension $n$ over an arbitrary field $k$. Then there exists a finite surjective morphism from $X$ to $\mathbb P^n\_k$. 


Embed $X$ in some $P=\mathbb P^N\_k$. By the homogeneous prime avoidance lemma, there exists a hypersurface $H\_0$ of $P$ which doesn't contain any generic point of $X$. Then $\dim (X\cap H\_0)=n-1$. Similarly there exists a hypersurface $H\_1$ of $P$ which doesn't contain any generic point $X\cap H\_0$. We have $\dim (X\cap H\_0\cap H\_1)=n-2$. Repeating the argument we find $n+1$ hypersurfaces $H\_0, \dots, H\_n$ such that 
$$X\cap H\_0\cap \dots \cap H\_n=\emptyset.$$
Each $H\_i$ is defined by a homogeneous polynomial $F\_i$. Remplacing $F\_i$ with some positive power (this doesn't change the property of the intersection being empty), we can suppose they all have the same degree $d$. These $n+1$ sections of $O\_P(d)$ don't have commun zeros in $X$, so they define a morphism $f: X\to \mathbb P^n\_k$. 


It remains to show $f$ is finite. I just repeat the argument from Lemma 3 in Kedlaya: *More étale covers of affine spaces in positive characteristics*, J. Alg. Geometry (2005): let $z\in \mathbb P^n\_k$. There exists $i\le n$ such that $z\in P\setminus H\_i$. So the fiber $X\_z$ is projective over $k(z)$ and also affine because it is closed in the affine scheme $P\setminus H\_i$. This implies that $X\_z$ is finite. So $f$ is quasi-finite and projective, so it is finite. It is surjective because $\dim X=n$. 


