Why is $\{(x,y) : |x| \le 1, |y| \le1\}\subseteq \mathbb{C}^2$ not an algebraic variety? This is an example in Karen Smith's An Invitation to AG. I don't follow the explanation that is given and would appreciate if someone can explain the following:
$V=\{(x,y) \in \mathbb{C}^2 : |x| \le 1, |y| \le1\}$ is a closed set that is not an algebraic variety. This follows from the fact that no nontrivial algebraic variety in $\mathbb{C}^2$ can have interior points, since the zero set of one nonzero polynomial has no interior points.
 A: Suppose that $f\in\mathbb C[x,y]$ is a polynomial which vanishes on your set $V$. As $V$ contains an open set which contains $(0,0)$, you can easily check that $f$ and all its derivatives vanish at $(0,0)$, and then, since for example $f$ is an analytical function and coincides with the sum of its Taylor series at $(0,0),$ the function $f$ vanishes on all of $\mathbb C^2$.
This shows that the only polynomial in $\mathbb C[x,y]$ which vanishes on $V$ is the zero polynomial. As $V$ is not the zero set of the zero polymonial, $V$ is not an algebraic set.
A: Suppose that $f(x, y)$ is a polynomial. You're asking why its set of zeroes can not have any interior points.
As a preliminary observation, consider a fixed value of $x$. Substitute this value into $f$, and you'll see that all of the powers of $x$ are just coefficients for the powers of $y$, and so for a fixed value of $x$, we can view $f(x, y)$ as a polynomial in $y$. This is then either the $0$ polynomial, or it only has finitely many roots.
How often can it arise that $f(x, y)$ is the zero polynomial in $y$?
Write the polynomial $f(x, y)$ as
$$ f(x, y) = \sum_{k=0}^{d} P_k(x) y^k. $$
We see that for $f(x, y)$ to be the zero polynomial in $y$, that $x$ must be a root of each of the polynomials $P_0, P_1, \dots, P_d$. In particular, there are only finitely many possible values of $x$ for which this occurs.
We see that except possibly for a finite number of values of $x$, that for a particular value of $x$, there are only finitely many values of $y$ such that $(x, y)$ is a root of $f$.
Now suppose that the zero set of $f$ has an interior point. Then there is an open ball around this point contained in the zero set. But then there is some $x$ such that there are infinitely many values of $y$ such that $(x, y)$ is in the open ball, and hence such that $(x, y)$ is a zero of $f$. This contradicts our earlier observation. (Choose an $x$ that is not common zero of $P_0, P_1, \dots, P_d$, and look at all of the points in the open ball with that $x$ value for their $x$-coordinate.)
