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Given the following subsets of $\mathbb{C}^5$ $$X=\cases{z_1^2+z_2^3+z_3^4+z_4^5+z_5^6=1\\z_1z_2z_3z_4z_5=1}\\Y=\{z_1z_2+z_2z_3+z_3z_4+z_4z_5=0\}\\Z=X-(X\cap Y)$$Determine whether $Z$ is a projective or an affine set.

Since $Y$ is a hyperplane of one homogenous polynomial that is an affine set $\subseteq \mathbb{A}^5$. An isomorphism can be also defined between $Y\to Y^\prime\subseteq \mathbb{A}^5$. About $X$ both equations are not homogenous and I can't find a precise isomorphism between it to another set (either affine, or projective). How can I find if $X$ is projective?

If it's projective: $X\cap Y$ will be a hyper surface with the limitations from $X$ and $Y$. Does that mean that $Z$ is affine?

  • $Y$ is a closed subscheme of $\mathbb{A}^5$, it is not projective. If you take the same equation in $\mathbb{P}^5$ is would be projective. But your hypothesis is that $Y$ lives in $\mathbb{A}^5$. – RghtHndSd Feb 14 '15 at 16:08
  • That's because $\mathbb{C}^5\subseteq\mathbb{A}^5$? What about $X$? –  Feb 14 '15 at 16:13
  • $\mathbb{C}^5 = \mathbb{A}^5$, but in algebraic geometry $\mathbb{C}^5$ is nonstandard. – RghtHndSd Feb 14 '15 at 20:22

2 Answers2

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There is a general theorem which suits here as well.

If $X$ is an affine variety and $Y$ is a hypersurface (=affine variety that is given by 1 equation), Then $X-(X\cap Y)$ is affine.

Thus your set $Z$ is affine. Since a set which is both affine and projective is precisely finite, $Z$ is not projective.

Mike
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  • but why this theorem is correct? –  Feb 14 '15 at 19:05
  • This should be proved in a basic course on algebraic geometry; I'm sure it may be found in a relevant book. – Mike Feb 14 '15 at 20:13
  • Can you not provide a standard reference (even if it differs from the hone Danis uses)? "It's in some book" is not helpful to those who are not familiar with algebraic geometry. – RghtHndSd Feb 14 '15 at 20:24
  • It is not that hard to google "Algebraic geometry books", and a quick look at the content will give you the answer. – Mike Feb 14 '15 at 20:36
  • anyways, here: http://math.stanford.edu/~vakil/725/class9.pdf you can find a sketch of the proof. – Mike Feb 14 '15 at 20:41
  • See also here http://math.stackexchange.com/questions/538136/complements-of-hypersurfaces-in-a-projective-space-is-affine – Mike Feb 14 '15 at 20:48
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The projection on the first $5$ coordinates maps $$\tilde X=\cases{z_1^2+z_2^3+z_3^4+z_4^5+z_5^6=1\\ z_1z_2z_3z_4z_5=1\\ (z_1z_2+z_2z_3+z_3z_4+z_4z_5) \cdot z_6 =1 }$$

isomorphically onto $X \backslash Y$. A concrete form of the concise statement by @Mike:

orangeskid
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