Prove $\mathbb{C}^n \setminus X$ is path connected.

Question

Let $f \in \mathbb{C}[z_{1}, . . . , z_{n}]$ be a nonzero polynomial ($n ≥ 1$) and $X = \{ z ∈ \mathbb{C}^n| f(z) = 0 \}.$ How do we prove that $\mathbb{C}^n\setminus X$ is path connected?

In one variable case the polynomial has only finitely many roots and by considering straight lines passing through any two distict points we can conclude that $\mathbb{C}^n \setminus X$ is path connected. How do we proceed for any $n$?

Answer 1

Hint: Any two points $a, b \in \mathbb{C}^n$ lie on the "line" $\{a + (b -a) z \mid z \in \mathbb{C}\}$. Use this to reduce to the one-variable case.

Answer 2

Consider $\gamma(t) = a+(b-a)t$ and look at the function $f\circ \gamma(t)\in \mathbb{C}[t]$ . Now we need a path $\delta(t)$ connecting $0,1$ such that $f\circ \gamma \circ \delta \neq 0$ i.e. a path $\delta $ in $(f\circ \gamma )^{-1}(\mathbb{C}\setminus \{0\}) $ . Observe that $(f\circ \gamma )^{-1}(\mathbb{C}\setminus \{0\}) $ is path connected being cofinite(polynomial in $\mathbb{C}[t]$ hence finite number of roots) and nonempty $(f\circ \gamma (t)\neq 0 $ for $t=0,1)$. Therefore there is such a path $\delta$, connecting $0,1$. And our required path is $\gamma\circ \delta $.