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I'm studying Fraleigh's abstract algebra(7ed), and there are little contents about algebraic geometry, just the definitions of varieties and ideals. Since I have few backgrounds about algebraic geometry, I don't know how to solve the following two exercises.

Sec28, Ex27, b. Give an example of a subset of $\mathbb{R}^2$ which is not an algebraic variety.
Ex34. Give an example of a subset $S$ of $\mathbb{R}^2$ such that $V(I(S))\neq S$.
(Here, the algebraic variety $V(S)$ in $F^n$ is the set of all common zeros in $F^n$ of the polynomial in $S$, where $S$ is a finite subset of $F[\mathbf{x}]$.)

I think that the answer of the two exercises can be same. But I don't know how to show some subset is not an algebraic variety. How can I solve it?

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    Does your definition of algebraic subvariety of $\mathbb{R}^2$ allow for non-Zariski closed subsets, like $\mathbb{R}^2 \setminus \{(0,0)\}$? What is the precise definition given in the text?2011-08-01
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    It is the precise definition given in the text. And for an ideal $I$ in $F[\mathbf{x}]$, we let $V(I)$ be the set of all common zeros of all elements of $I$. Here is no comment about Zariski topology or something, so I think there is no restriction.2011-08-01
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    Are you saying that the definition of an algebraic variety is $V(I)$, the common zero set of an ideal? Okay, then you can easily show that such a thing is closed in the usual topology on $\mathbb{R}^2$ -- polynomials are continuous, the preimage of a (closed!) point under a continuous function is closed, and an intersection of closed sets is closed. So take any non-(Euclidean-closed) subset, for instance.2011-08-01

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Hint1: A univariate polynomial can have infinitely many zeros only if it is the zero polynomial.

Hint2: Show that if the polynomial $p(x,y)$ has infinitely many zeros on the line $y=y_0$, it must be divisible by $y-y_0$.

Hint3: Show that if the conclusion of the previous hint holds for infinitely many choices of $y_0$, then $p(x,y)$ must be the zero polynomial.

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    But for $\mathbb{R}^2$, It is not the case of univariate polynomial. Inifinite many zeros on x-axis can be achieved by zeros of $y \in \mathbb{R}[x,y]$2011-08-01
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    @Gobi: Second hint: maybe you can restrict your attention to a copy of $\mathbb{R}$ -- i.e., a line -- inside $\mathbb{R}^2$?2011-08-01
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    @Gobi: I added a couple more hints to my answer.2011-08-01
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    Thanks for your hints, I found a example like $\{(x,y)|y-x\in \mathbb{Z}\}$2011-08-03
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The easiest solution to the first exercise depends on exactly what your running definition of a subvariety of $\mathbb{R}^2$ is. But no one would call a subset of affine $n$-space a subvariety unless it is at least locally Zariski-closed, i.e., the intersection of a Zariski-closed subset and a Zariski-open subset. It is not too hard to show that the complement of any countably infinite subset of $\mathbb{R}^2$ is not of this form. One could for instance proceed as follows:

  1. Show that any proper Zariski-closed subset of $\mathbb{R}^2$ has measure zero.
  2. Show that any infinite Zariski-closed subset of $\mathbb{R}^2$ is uncountable.

[On second thought, maybe 2. is overkill here. If you take your countably infinite subset to be contained in, for instance, the line $y = x$, then it is much easier than what I proposed above to see that it is not Zariski-closed, so that its complement is not Zariski-open.]

As for the second exercise, you can take any non-Zariski closed subset, so for instance any subset which is not closed in the Euclidean topology. As you say, any solution to the first exercise will certainly do here.

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    Actually I don't know about Zariski-clsoed or Zariski-open subset. Given that it is an introductory abstract algebra book, I think the definition here is just as written in the question, the common zeros. Then can it be possible to answer it with more easy concepts, without Zariski-topology and measure zero(background in analysis)?2011-08-01
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    Yes, given what you said the solution is much easier -- as I said, without having seen the text I gave an example of something which is not a variety *according to the most general definition*. See my comment above. If you want a more specific hint, try to prove that the set $\{ (n,n) \ | \ n \in \mathbb{Z} \}$ is not of the form $V(I)$.2011-08-01
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    @Gobi: by the way, I don't mean to be discouraging, but the rudiments of measure theory are relatively low-tech compared to what one has to learn to do algebraic geometry at all seriously. In other words, yes, you can definitely reason purely algebraically, but it takes some work to do so.2011-08-01