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Let $k$ be an algebraically closed field and let $X=\left\{\xi_1,\cdots,\xi_N \right\}$ be a subset of $k^n$. Then $X$ is an affine variety of $k^n$ with vanishing ideal $I_X = \cap_{i=1}^N m_{\xi_i}$, where $m_{\xi_i} = (x_1-\xi_{1,i},\cdots,x_n - \xi_{n,i})$. Since the $m_{\xi_i}$ are maximal ideals we also have that $I_X = \prod_{i=1}^N m_{\xi_i}$. From this formula we see that $I_X$ is generated by $N$-fold products of factors of the form $x_i - \xi_{i,j}$, which are non-homogeneous. Hence $I_X$ is a non-homogeneous ideal.

Question: why can we view $X$ as a projective variety and what is the associated ambient projective space?

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Google for "homogenization" (e.g. en.wikipedia.org/wiki/…) and "projective closure". By the way, perhaps it's time to choose another route to algebraic geometry when you are able to manipulate ideals abstractly, without knowing why points are projective varieties ... –  Martin Brandenburg Sep 17 '13 at 22:56
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@MartinBrandenburg: I know what homogenization is and projective closure, at least on the level of definitions. Interesting...so what exactly do you mean by "another route"? What would be that route? What is your suggestion? –  Manos Sep 17 '13 at 23:05

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All what you write is correct. But you do not have to look at the affine ideal $I_X$ and ask whether it is homogeneous: we may have $$X\cong\textrm{Spec }k[x_1,\dots,x_n]/I\cong\textrm{Proj }k[y_0,\dots,y_n]/J$$ for a finite variety $X$. Here $I$ is your $I_X$ and $J$... yes, $J$ has to be homogeneous.

By the way, there is the following:

Theorem. Finite morphisms are projective.

To see that $X$ is projective, we can apply this result to the structural morphism $X\to \textrm{Spec }k$, where $X$ is your 0-dimensional variety.

Aside: a morphism is finite if and only if it is both proper and affine. This implies that whenever you have a projective variety $X$ sitting inside an affine variety, $X$ is 0-dimensional.

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Dear atricolf, it may be worth noting that the statement on projective morphisms is true in the sense of EGA II, but not in the sense of Hartshorne. However, both definitions coincide when the base is affine as in your context. –  Cantlog Sep 18 '13 at 6:43
    
@Cantlog: yes, thank you for pointing this out. Also, do not they coincide over a separated base? I am not sure –  Brenin Sep 18 '13 at 6:54
    
They coincide when the base has an invertible ample sheaf. Being separated (e.g. proper non-projective variety) is not enough. –  Cantlog Sep 18 '13 at 11:46

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