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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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    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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    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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    @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