Elementary algebraic geometry

Question

Let $p(z,w)=z^2+w^2-zw+1,$and $Z(p)=\{(z,w)\in\mathbb{C}\times\mathbb{C}|\,p(z,w)=0\}.$ Is this variety irreducible? Is $Z(p)$ a connected subset of $\mathbb{C}\times\mathbb{C}$ ? (in usual topology of $\mathbb{C}\times\mathbb{C}$. Im not interested in Zariski top.in this particular question)

Answer 1

HINT:

Write $$z^2 + w^2 - z w + 1 = (z - \zeta\cdot w)( z - \frac{1}{\zeta}\cdot w ) + 1$$

where $\zeta = (\frac{1}{2} + i \frac{\sqrt{3}}{2})$ is a root of $-1$ of order $3$. Now solve the system with parameter $t\in \mathbb{C}\backslash \{0\}$ \begin{eqnarray*} z - \zeta\cdot w &=& t \\ z - \frac{1}{\zeta}\cdot w &=& -\frac{1}{t} \end{eqnarray*} to get a parametrization of $Z$

\begin{eqnarray*} z& =& \frac{\zeta\cdot t^2 -1}{(1+\zeta) t} \\ w &=& \frac{ \zeta^2 (1+ t^2)}{(1+\zeta) t} \end{eqnarray*}

This variety is isomorphic to $\mathbb{C} \backslash \{0\}$.

Answer 2

The closure (in the usual topology) $\bar Z$ of your variety $Z$ in $\mathbb P^2(\mathbb C)$ is the smooth conic $z^2+w^2-zw+t^2=0$, which like all smooth conics is isomorphic to $\mathbb P^1(\mathbb C)$.
The original variety $Z$ is obtained by removing the two points at infinity (those with $t=0$), hence is isomorphic to $\mathbb P^1(\mathbb C)$ minus two points and thus isomorphic to $\mathbb C^\ast$, corroborating orangeskid's excellent computation.