# $S^1$ cannot be the zero locus of polynomials: is this argument correct?

Suppose I want to rigorously prove that the unit circle $S^1 \subseteq \mathbb C$ can't be the zero locus of any polynomials.

Can one just observe that polynomials with domain $\mathbb C$ are univariate and therefore zero sets are finite?

The old version of my question: Is it rigorous and correct to argue like this or do I also have to worry about polynomials in $n$ variables because a subset of $\mathbb C^n$ could contain $S^1$ when restricted to one of the dimensions?

Note: The zero locus of a polynomial is the set of roots of the polynomial and so the zero locus of a collection of polynomials is the intersection of each zero locus of each polynomial. (that is, the set of point that make every polynomial zero simultaneously)

Edit

Since the previous version of my question was not so easy to answer I would instead like to ask the following (with the hopes of increased chances of someone knowing an answer):

I started to learn basic algebraic geometry a few days ago and I read that $S^1$ is not a variety and I tried to look it up and I found for example that varieties cannot be compact but that's already more knowledge than I have about varieties. I tried to look up why an algebraic variety can't be bounded but could not find a proof.

How to prove that a complex algebraic variety is unbounded in the Euclidean metric?

I googled this of course and surprisingly could not find any proof (even though I read it somewhere so it's probably true)

If the above is too difficult:

Is there a basic (=almost or no knowledge of algebraic geometry required) way of proving that $S^1$ is not an algebraic variety?

• Do you really need complex analysis techniques to show that a polynomial in one variable with coefficients in a field has a finite number of roots? Use induction on the degree. Mar 5 '16 at 13:04
• @NajibIdrissi Or appeal to the fundamental theorem of algebra. But what does this have to do with the question? Mar 6 '16 at 1:21
• Can you clarify all the definitions involved in your question? It seems like you are actually asking what "zero locus of any polynomials" means, and we can't answer that without knowing what definition you are using. Mar 6 '16 at 1:22
• @EricWofsey I added some more information. Does it make it clearer? Or is there anything else I could add? Mar 6 '16 at 1:25
• But the circle is the zero-locus of $p(x,y) = x^2+y^2-1$. So you need to specify that you're considering only complex polynomials in $z$. Mar 8 '16 at 21:25

Let $V$ be a complex algebraic variety in some $\mathbb{C}^n$, let $I$ be the associated ideal to $V$; by definition $$\mathbb{C}[V]=\mathbb{C}[x_1,\dots,x_n]_{\displaystyle/I};$$ one can prove that the regular functions $\varphi:V\to\mathbb{C}$ are uniquely determinated by the ring homomorphisms $\varphi^{*}:f\in\mathbb{C}[x]\to f\circ\varphi\in\mathbb{C}[V]$, and in particular one can prove that: $$\forall P\in V,\,\varphi(P)=q(P),\,\,\text{where}\,\,q=\varphi^{*}(x),$$ that is any $\varphi$ is a holomorphic function with values in $\mathbb{C}$.
If $V$ is compact, then any $\varphi$ is a constant function, by Liouville's theroem; if $V$ is an infinite set, all this is a contraddiction, because not all ring homomorphisms $\varphi^{*}$ are constant.
Can one just observe that polynomials with domain $\mathbb{C}$ are univariate and therefore zero sets are finite?