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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.

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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.)

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  • $\begingroup$ Thank you for your help. Can you please explain why "there is some $x$ such that there are infinitely many values of $y$ such that $(x,y)$ is in the open ball"? $\endgroup$ – Mark Mar 8 '17 at 11:16
  • $\begingroup$ Just imagine what an open ball looks like. For every $x$ value there are infinitely many $y$ values such that $(x, y)$ is a point in the ball. The important thing here though is because there are infinitely many different $x$-values amongst the points in the ball, we can also choose the value of $x$ so that it is not a common root of $P_0, P_1, \dots, P_d$. $\endgroup$ – Dylan Mar 8 '17 at 19:07
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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.

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  • $\begingroup$ Thank you. Please correct me if I do not understand completely: you're picking an arbitrary polynomial and then assuming $V$ is a variety containing the interior point $(0, 0)$, you arrive at the contradiction that $f$ is zero. I do not know how this rules out the possibility that there cannot be another interior point? $\endgroup$ – Mark Mar 7 '17 at 5:36
  • $\begingroup$ No, $V$ denotes the set in your question, which contains $(0,0)$ in its interior: there is no need to assume that: it simply does. Then I am showing that if a polynomial vanishes on $V$ then it vanishes on all of $\mathbb C^2$. That implies that if an algebraic set $X$ contains $V$, then $X$ is all of $C^2$. $\endgroup$ – Mariano Suárez-Álvarez Mar 7 '17 at 6:52
  • $\begingroup$ If you wouldn't mind, can you please explain also how any algebraic variety in $\mathbb{C}^2$ can't contain interior points? $\endgroup$ – Mark Mar 8 '17 at 6:07
  • $\begingroup$ The proof proves exactly this...If you have any interior point, you can just use a translation to assume the origin is an interior point...Or you can even use the same proof with any interior point since a polynomial coincides with its taylor series at every point. $\endgroup$ – MooS Mar 8 '17 at 6:13

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