TITLE: Is there an irreducible projective hypersurface such that its complement has zero Euler characteristic?
QUESTION [9 upvotes]: We know that, if $f=X_0X_1...X_n \in \mathbb{C}[X_0,...,X_n]$ and $Z(f)\subset \mathbb{CP}^n$, then the Euler characteristic of its complement is zero, i.e.
$$
\chi(\mathbb{CP}^n\setminus Z(f))=0.
$$
But $f$ is not irreducible.
Let $Z\subset \mathbb{CP}^n$ be a smooth, irreducible hypersurface. Then, we know that
$$
\chi(Z)=\frac{1}{d}((1-d)^{n+1}-1)+n+1,
$$
where $d$ degree of $Z$.
In particular, if $g=X_0^2+...+X_3^2 \in \mathbb{C}[X_0,...,X_3]$, we have
$$
\chi(Z(g))=\frac{1}{2}((1-2)^4-1)+4=4,
$$
then $\chi(\mathbb{CP}^3\setminus Z(g))=0$, since $\chi(\mathbb{CP}^3)=4$.
So I ask: is there an irreducible homogeneous polynomial $h \in \mathbb{C}[X_0,...,X_n]$ such that deg$h>2$ and $\chi(\mathbb{CP}^n\setminus Z(h))=0$?
Remark: this is not possible if $Z(h)$ is smooth (with deg$h>2$).

REPLY [3 votes]: Let $S = \{ x_1^j x_2^{d-j} = x_3^k x_4^{d-k} \}$ in $\mathbb{P}^3$, with $0 < j,k < d$ and $GCD(j,k,d)=1$. I claim that $S$ is irreducible and $\chi(S)=4$ and hence $\chi(\mathbb{P}^3 \setminus S)=0$. 
Let $T =\{ x_1 x_2 x_3 x_4 \neq 0 \} \subset \mathbb{P}^3$, this is isomorphic to $(\mathbb{C}^{\ast})^3$. Since $GCD(j,k,d)=GCD(j,k,d-j,d-k)=1$, the locus in $T$ where $x_1^j x_2^{d-j} x_3^{-k} x_4^{-(d-k)}=1$ is isomorphic to $(\mathbb{C}^{\ast})^2$, with Euler characteristic $0$. So $\chi(S \cap T) =0$ and thus $\chi(S) = \chi(S \setminus T)$. 
But (using the assumptions $0 < j,k < d$) the complement $S \setminus T$ is just four $\mathbb{P}^1$'s: $\{x_1=x_3=0\}$, $\{x_1=x_4=0\}$, $\{x_2=x_3=0\}$ and $\{x_2=x_4=0\}$, and this has Euler characteristic $4$. Also, $T \cap S$ is obviously irreducible and is dense in $S$, so $S$ is irreducible.