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Background

I recently encountered this textbook exercise (from Ideals, Varieties, and Algorithms by Cox) in my abstract algebra course that I found rather interesting:

Let $R=\{(x,y)\in\mathbb{R}^2 \ | \ y>0\}$ be the upper half plane. Prove that $R$ is not an affine variety.

As an assignment, I am to come up with some exercises of my own, and I wanted to see if I can generalize this question at all, which brings us to my question.

Question

My question might seem strange, and it might not even have an answer, but what I'm wondering is this: can we extend this result (that the upper half of $\mathbb{R}^2$ is not an affine variety) further?

For example, would it be true also that

$R=\{(x,y,z)\in\mathbb{R}^3 \ | \ z>0\}$

is not an affine variety? Or more generally that

$R=\{(x_1,\dots,x_n)\in\mathbb{R}^n \ | \ x_n>0\}$

isn't an affine variety?

As always, thank you all for taking the time to help me out!

Thy Art is Math
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  • Of course it generalizes. It should be proved the same way. – D_S Sep 06 '17 at 01:32
  • An affine subvariety of $\mathbb{R}^n$ must be a closed subset of $\mathbb{R}^n$. – quasi Sep 06 '17 at 01:37
  • And (correct me if I'm wrong) all the examples I listed are not closed subsets of $\mathbb{R}^n$, right? I haven't learned about affine subvarieties but I think I understand what you're saying. – Thy Art is Math Sep 06 '17 at 01:41
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    In this context, subvariety just means a variety which is a subset of $\mathbb{R}^n$. – quasi Sep 06 '17 at 01:49
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    As to the closed property, if a polynomial $p \in \mathbb{R}^n$ is zero on a set $S \subseteq \mathbb{R}^n$, then since polynomials are continuous functions, $p$ must be zero on the closure of $S$. It follows that the set of zeros of $p$ is a closed subset of $R^{n}$. But then a variety is just the intersection of the zero sets of a set of polynomials, and an interesection of closed sets is a closed set. – quasi Sep 06 '17 at 01:52
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    More generally, if a polynomial $p \in \mathbb{R}[x_1,...,x_n]$ is zero on a nonempty open subset of $\mathbb{R}^n$, then $p$ must be identically zero, hence $p$ is zero on all of $\mathbb{R}^n$. – quasi Sep 06 '17 at 01:55
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    Thus, even a closed half-space would fail to be a variety. – quasi Sep 06 '17 at 01:58
  • Thank you for the clarification, it really made sense! – Thy Art is Math Sep 06 '17 at 02:01

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