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This is a reinterpretation of a HW question, so I'm not totally sure if this belongs here, but I was wondering about the following question:

Let $k$ be an algebraically closed field. We define a variety to be a scheme over $k$ that is integral, separated, and finite type, and we say that a variety $X$ is projective if there exists a closed immersion $X\hookrightarrow\mathbb{P}_k^n$ for some $k$. Is it true that for any projective variety $X$, there exists some affine scheme $Y$ and a surjective morphism of $k$-schemes $Y\to X$?

In the original question, we do not require $Y$ to be irreducible (which turns it into a trivial problem), however, I did not see this, and was nonetheless able to prove the theorem in the case where $X=\mathbb{P}_k^n$, without being able to carry my arguments over to a closed subscheme of $X$. In general, I suspect that we would require $\dim Y>\dim X$, as can be seen in the case where $X$ is a nonsingular curve that is not $\mathbb{P}^1$.

My main intuition here is that this problem most likely requires a geometric rather than algebraic solution. More precisely, if we let $X\subset\mathbb{P}^n$ be of codimension $d$, then my thinking was that if we could cover $\mathbb{P}^n\setminus\{*\}$ by an affine, where $*\notin X$, then we could then do some sort of projection, but this is all informal speculation.

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  • $\begingroup$ What was your solution for $\mathbb P^n_k$? $\endgroup$
    – Asvin
    May 10, 2017 at 21:24
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    $\begingroup$ For a much stronger statement than what you're asking for, Google for the "Jouanolou trick". $\endgroup$ May 10, 2017 at 21:26
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    $\begingroup$ The answer is "yes": For $\mathbb{P}^n$, note that this is isomorphic to $PGL_{n+1}/P$, where $P$ is the stabilizer of a point; $PGL_{n+1}$ is affine, giving a surjective (quotient) map $PGL_{n+1}\to \mathbb{P}^n$. Now given $X\hookrightarrow \mathbb{P}^n$, one may base change along this map; the result is a closed subvariety of $PGL_{n+1}$ (hence affine) surjecting onto $X$ as desired. $\endgroup$ May 11, 2017 at 2:06
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    $\begingroup$ If you want, you can also do it with $\dim Y= \dim X$. For example for curves, the idea is to take some higher degree cover and remove one point in a smooth fibre (or really any fibre containing more than one point). The covering curve minus the point is affine, and the map is still surjective. You can make this work in higher dimension as well. In fact there exist surjections from $\mathbb A^n$ to $\mathbb P^n$. (But I don't know how to use this method to get $Y \to X$ generically finite with $Y$ irreducible in general. For that, Jouanolou's trick is great if $\dim Y$ is allowed to be big.) $\endgroup$ May 11, 2017 at 4:25
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    $\begingroup$ @R.vanDobbendeBruyn: To get $Y$ irreducible, one must assume $X$ irreducible; then one can take a smooth affine cover as in my construction and cut it down to the appropriate dimension by taking a sufficiently general collection of equations; this will be irreducible by standard arguments. $\endgroup$ May 11, 2017 at 13:47

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