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Consider the product $\mathbb{A}^n \times \mathbb{P}^{m}$. Let $x_i$ be affine coordinates on $\mathbb{A}^n$ and $y_j$ homogeneous coordinates on $\mathbb{P}^{m}$.

Question: Is $A=k[x_1,\dots,x_n,y_0,\dots,y_m]$ the coordinate ring of $\mathbb{A}^n \times \mathbb{P}^{m}$ "in some sense"? If yes, how can we rigorously see that and what is "that sense"?

Manos
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  • No. Your variety is not affine, so it does not have a ring of coordinate functions in any useful sense. – Mariano Suárez-Álvarez Jun 27 '14 at 20:05
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    I think the upshot is that you want to consider polynomials and ideals that are homogeneous in the $y_i$, viewed as indeterminates over $k[x_1, \dots, x_n]$. Another option would be to work out what the Segre embedding does to this open subset of $\mathbb{P}^n \times \mathbb{P}^m$, which is how Hartshorne is giving this thing the structure of a variety anyway. That sounds messy, though. – Hoot Jun 27 '14 at 20:07
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    Isn't $\mathbb{A}^n \times \mathbb{P}^m$ basically projective $m$-space over the base $\mathbb{A}^n$? I guess you could use global Proj to make sense of it. – Zhen Lin Jun 27 '14 at 20:31
  • The ring of global sections is the one of $\mathbb{A}^n$, I think. Probably you mean the homogeneous coordinate ring? But this depends on the embedding into some projective space - you should add this embedding to your question! – Martin Brandenburg Jun 27 '14 at 21:03
  • Surely this is the homogeneous coordinate ring of relative projective space, as Zhen Lin says. In particular, this should be the ring of sections $\bigoplus_{d\geq 0} H^0(\mathcal{O}(d))$, coming from the relatively ample $\mathcal{O}(1)$. – Jake Levinson Jun 28 '14 at 03:37

2 Answers2

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Your intuition is correct, this should be the coordinate ring in some sense and it is. You just need to be careful to treat the $x$ variables and the $y$ variables somewhat differently. In particular you want to treat the $y$ variables as homogeneous variables and the $x$ variables as regular affine ones.

To do this in one step: Define a grading on this ring by putting the $x$ variables in degree 0, and the $y$ variables in degree 1. Now if you take $\textbf{proj}$ with respect to this grading you get what you want.

Nate
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  • One should not confuse the coordinate ring (of an affine variety) with the homogeneous coordinate ring. They are different beasts. While it may seem ok to do so at first because there are certain parallels (I did things like this a lot when I first started), they often lead to great confusion down the road. – RghtHndSd Jun 28 '14 at 15:36
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Yes. $X=\mathbb A^n \times \mathbb P^m$ is a smooth toric variety, and thus have a homogeneous coordinate ring (in the sense of Cox). It is graded by the Chow group $A_{n+m-1}(X)$, and homogeneous ideals (w.r.t. this grading) correspond to subvarieties on $X$ just as in the projective case.

See the description here.

Fredrik Meyer
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