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Please no flame, I am a beginner here... Given ANY subset of $\mathbb{C}$, is there always at least one set of polynomial equations which generates it? If yes, can we always explicitly construct at least one member of that set (of polynomial equations)?

What is the situation for $\mathbb{R}^2$ and polynomials in $x,y$? For example, $x^{2}+y^{2} = 1$ generates the unit circle in $\mathbb{R}^2$.

Frank
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  • I think it will be helpful if you post what you have done so far – Ana Galois Aug 29 '13 at 22:24
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    any polynomial equation $a_nx^n+...+a_1x+a_0=0$, where $a_i$ are complex numbers, has only a finite number of solutions (assuming it's not the trivial equation $0=0)$. – user64480 Aug 29 '13 at 22:28
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    What do you mean by "generate"? – Zev Chonoles Aug 29 '13 at 22:29
  • I think "generates it" should be "defines it" or "cuts it out" – user64480 Aug 29 '13 at 22:30
  • @user64480 Perhaps PO has in mind a SET of polynomial equations in several variables. – Maesumi Aug 29 '13 at 22:38
  • @Ana - What I have done so far? On that problem? Nothing - I am mildly suspecting this question is very hard. – Frank Aug 29 '13 at 22:52
  • In what sense $|z| = 1$ is generated by a SET of polynomials? – achille hui Aug 29 '13 at 22:59
  • $|z|=1$ is not a polynomial equation in the complex variable $z$. However, you can make it into a polynomial equation in two real variables if you identified $\mathbb{C}$ with $\mathbb{R}^2$. Then the set is given by the polynomial equation $x^2+y^2=1$. If this is what you meant, maybe your question should be about $\mathbb{R}^2$ instead of $\mathbb{C}$. – user64480 Aug 29 '13 at 23:21
  • Even in $\mathbb{R}^2$ the answer is no. Sets which are solutions of polynomials are always relatively small, or all of $\mathbb{R}^2$. As a simple example, consider the set $\mathbb{R}^2 - {(0,0)}$. It cannot be described by polynomial equations, since for any nontrivial polynomial you can think of, you can come up with at least one point other than the origin where it doesn't vanish. – user64480 Aug 29 '13 at 23:36
  • @user64480 The usage of $\mathbb{C}$ in the question is also the part that confuses me. – achille hui Aug 29 '13 at 23:39
  • @achillehui I know, I was directing my comments at the OP, not at you. Sorry if that was unclear – user64480 Aug 29 '13 at 23:47

2 Answers2

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The question is a little hard to interpret, I assume that by a set of polynomials generating a subset of $\mathbb C$ you mean that the subset of $\mathbb C$ is the set of common zeros of the polynomials in your set. Of course this means you should only be considering polynomials in a single variable, not in multiple variables as you explicitly stated. So maybe I've misinterpreted you, let me know if I have.

Now assuming I have interpreted your question correctly, what you are asking is whether the Zariski topology on $\mathbb C$ is discrete. The answer is no, the Zariski topology on $\mathbb C$ is the cofinite topology. This means that the closed sets (meaning the sets that arise as the common zeros of a set of polynomials) are exactly the finite subsets of $\mathbb C$ and $\mathbb C$ itself.

To see this first note that $\mathbb C$ is the solution set of the set of polynomials $\{0\}$. For any other set $S$, if $f \in S$ is any nonzero polynomial then $f$ has only finitely many solutions and the common zeros of $S$ must be a subset of this finite set. This shows that every closed proper set is finite. To see that every finite set is closed let $\{a_1, \ldots, a_n\}$ be a finite set of points in $\mathbb C$. Then the zeros of the set $\{f\}$, where $f = (x - a_1)\cdots(x - a_n)$, are exactly the $a_i$.

Edit: For $\mathbb R^2$ the answer is still no, not every subset of $\mathbb R^2$ is the solution set of a set of polynomials. The standard example is the set $\mathbb R^2 \setminus (0, 0)$. I claim that the zero polynomial is the only polynomial that is zero on this set. To see this let $f(x, y) = \sum_{i = 0}^ng_i(x)y^i$ be any polynomial that is zero on $\mathbb R^2 \setminus (0, 0)$. Then for any $a \neq 0$ the polynomial $f(a, y)$ is identically zero. This means each $g_i(x)$ is a polynomial with the property that $g_i(a) = 0$ when $a \neq 0$. As polynomials give continuous functions we then get $g_i(0) = 0$ so $g_i$ is the zero polynomial. Then $f$ is also the zero polynomial.

So we see that no set of polynomials has $\mathbb R^2 \setminus (0, 0)$ as their zeros because those polynomials are zero, and hence give the set $\mathbb R^2$ and not $\mathbb R^2 \setminus (0, 0)$.

More generally, any polynomial that's zero on an open set is zero, so you can't get any subset that contains an open set (open in the Euclidean topology). Also, I don't believe you can get a proper subset of positive Lebesgue measure, but you certainly can't get all subsets of Lebesgue measure $0$ so that's still not a complete characterization.

Jim
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  • Jim - I think my original question was indeed about the Zariski topology - but at the same time I identified C and R^2, which made my question ambiguous. I was confusedly wondering whether it was always possible to find a set of polynomial equations for any figure drawn in R^2. – Frank Aug 30 '13 at 00:09
  • The answer is still no in $\mathbb R^2$, I'll edit with an example – Jim Aug 30 '13 at 00:09
  • Jim - no need - in fact, it's not so mysterious, in hindsight. – Frank Aug 30 '13 at 01:57
  • One further question though - are there objects that would be "infinite" polynomial, the product of an infinite number of monomials $z-z_{i}$ ? It looks like there are, but would those have a name as mathematical entities? – Frank Aug 30 '13 at 01:59
  • Not that I know of. – Jim Aug 30 '13 at 05:17
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Adapted from my comment:

For the revised question about $\mathbb{R}^2$, the answer is still no.

First, a note: every polynomial equation $f=g$ is equivalent to $f-g=0$, thus instead of talking of the locus of solutions to a set of polynomial equations, we may talk about the locus of zeros of a set of polynomials.

Let $X=\mathbb{R}^2-\{(0,0)\}$.

Suppose for contradiction $X$ was described as the locus of zeros of a set $S$ of polynomials. If $S$ consists only of the polynomial $0$, then the locus of zeros is all of $\mathbb{R}^2$, so suppose there is a nonzero polynomial $f\in S$. Now, $f$ is a polynomial potentially in both $x$ and $y$.

Case $1$: $f$ has terms involving $x$. Then if we do the substitution $y=1$, we get a nonzero polynomial, $f_x$, which involves only $x$. Since this is a one-variable nonzero polynomial, it has only finitely many zeros. So let $a$ be a value which is not a zero of $f_x$. We see that $(a,1)$ is not a zero of $f$.

Case $2$: $f$ has terms involving $y$. Similarly to above, we can find a point $(1,b)$ which is not a zero of $f$.

In conclusion, there is a point other than the origin which is not a zero of $f$. So the locus of points where the set $S$ of polynomials vanishes is necessarily smaller than $\mathbb{R}^2-\{(0,0)\}$.

user64480
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