Compute the Zariski-closure of two sets Compute the Zariski-closure of the following two sets:


*

*$X = \{(z_1, z_2) \in \mathbb{C}^2 \mid |z_1|^2 + |z_2|^2 = 1\}$ in $\mathbb{C}^2$. 

*$X = \{z \in \mathbb{C} \mid z = n \in \mathbb{N}\}$ in $\mathbb{C}$.

 A: Regarding 2: We define $Z \subset \mathbb{C}$ to be closed if $Z = V(S)$ for some $S \subset \mathbb{C}[x]$.  Since $\mathbb{C}$ is a field, it is a PID, so $S$ generates a principal ideal, i.e., $(S) = (f)$ for some $f \in \mathbb{C}[x]$.  Then every point in $V(S) = V(f)$ is a root of $f$, but $f$ has at most $\operatorname{deg}(f) = n$ distinct roots.  This shows that every proper algebraic set in $\mathbb{C}$ is finite.  Thus, the Zariski-closure of $\{z \in \mathbb{C} \;|\; z = n \in \mathbb{N} \} \subset \mathbb{C}$ must not be a proper subset, so it's $\mathbb{C}$.
A: The first set isn't Zariski closed either. A proper Zariski-closed set would be a contained in the zero locus of a bivariate polynomial. Thus (at least at non-singular points) it locally looks like a 1-dimensional complex manifold, i.e. a 2-dimensional real manifold. But this set is a 3-dimensional real manifold, and thus we don't expect it to be contained in a proper Zariski-closed subset of $\mathbb{C}^2$.
Let $P(Z_1,Z_2)$ be a polynomial that vanishes on all of $X$. Let $z_0$ be an arbitrary complex number with the property $|z_0|<1$. The univariate polynomial
$$
F(Z_2)=P(z_0,Z_2)
$$
vanishes whenever $|z_2|=\sqrt{1-|z_0|^2}$. Therefore it has infinitely many zeros. This is possible for a univariate polynomial only, if it is the zero polynomial. Thus we can conclude that $Z_1-z_0\mid P(Z_1,Z_2)$. But there are infinitely many distinct such numbers $z_0$. This implies that $\deg P(Z_1,Z_2)$ exceeds any bound unless it is the zero polynomial. So the zero polynomial is the only polynomial vanishing on $X$. Hence the Zariski-closure of $X$ is all of $\mathbb{C}^2.$
