$S^1$ 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?

 A: Let $V$ be a complex algebraic variety in some $\mathbb{C}^n$, let $I$ be the associated ideal to $V$; by definition
\begin{equation}
\mathbb{C}[V]=\mathbb{C}[x_1,\dots,x_n]_{\displaystyle/I};
\end{equation}
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:
\begin{equation}
\forall P\in V,\,\varphi(P)=q(P),\,\,\text{where}\,\,q=\varphi^{*}(x),
\end{equation}
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.
Is it all clear?
A: 
Can one just observe that polynomials with domain $\mathbb{C}$ are univariate and therefore zero sets are finite?

Yes.
Also I don't think your original question was really ambiguous -- it's perfectly comprehensible to me.
