# Prove $X=(y^{2}z-x^{3}+xz^{2})\backslash\{(1,0,-1)\}$ is irreducible.

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.

• I guess if $f = y^2z - x^3 +xz^2$ were reducible, $Z(f)$ should split in at least two components, and $Z(f)$ should be singular where they meet. Since there is just one singular point, the components must meet in exactly that point. By Bezout, the components are either a degree 2 curve and a line meeting transversally at the point, or three lines all meeting eachother in the same point. I guess that gives enough restrictions on $f$. However, Bezout only works if $k$ is algebraically closed, so i hope that was indeed an assumption we could make. Commented Mar 11, 2014 at 21:46
• Yes, $k$ is algebraically closed here. This problem came well before the proof of Bezout and I feel like there must be another solution. Interesting idea though. Commented Mar 11, 2014 at 21:49

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.
• Now that I think about this, the factor theorem only applies for polynomials over a field I believe. The proof required $k[x]$ being a euclidean domain. Does it work over general commutative rings? Commented Mar 13, 2014 at 4:35
• Dear @TheNumber23, the first half of my answer works more generally over an arbitrary integral domain. The second half about factoring quadratic polynomials is true in $R[x]$, where $R$ is a UFD. So the whole argument works in the ring $R[x,y,z]$, where $R$ is a UFD, just as well. I haven't thought it through carefully, but I doubt that it can be extended in a sensible way beyond these cases. Regards Commented Mar 13, 2014 at 9:12