Let $X=Z(y^{2}z-x^{3}+xz^{2})\subset\mathbb{P}^{2}$ and $P=(1,0,-1)$. Prove that $X\backslash P$ is irreducible in the Zariski topology.
When char$(k)\neq 2$, I've answered this problem by showing that $y^{2}z-x^{3}+xz^{2}$ is irreducible because $X$ is a nonsingular variety in $\mathbb{P}^{2}$. $X\backslash P$ is an an open set of an irreducible set. Whence irreducible.
When char$(k)=2$, I have a problem because $X$ has a singularity at $(1,0,-1)$ and nowhere else. So I am unable to show that the polynomial $y^{2}z-x^{3}+xz^{2}$ is irreducible. I have no idea how to proceed from here. Any direction or hints would be great. Thanks.
Since no positive power $z^k$ of $z$ divides the polynomial $F(x,y,z)=y^2z-x^3+xz^2$, it is irreducible, if and only if its dehomogenization with respect to $z$ $$ F_*(x,y)=y^2-(x^3-x) $$ is irreducible.
Now consider $F_*$ as a quadratic polynomial in $y$. It is reducible, if and only if $x^3-x$ is a square in $k[x]$, which it obviously isn't, as $x^3-x$ has odd degree. Thus $F_*$ is irreducible, and therefore $F$ as well.