# Showing a (relatively simple) set of polynomial zeros in projective space is irreducible

I'm teaching myself a little algebraic geometry and I was hoping you could help me with an exercise. I have my head around affine spaces alright but I am having a little more trouble with projective space. The exercise I am confused about is to show that the curve given by set of zeros $W \subset \mathbb{P}^2$ of $X_0^3 X_2^2 + X_2^3 X_1^2 + X_1^3 X_0^2$ is irreducible (over field of zero characteristic).

More generally, I would like to know if there is a sensible method to approach a problem like this: my first thought was to pick some point which must be in at least 1 irreducible component, and then show any component containing that point must be the whole of W: I got nowhere with this, however. For this and in general zeros of other reasonably simple polynomials, how should I try to show irreducibility? Please keep answers simple, I haven't been doing this for very long! -K

-

Here's a more algebraic version of Luboš Motl's answer. A curve is irreducible if and only if the only closed subsets of it are finite. Thus to show that it's irreducible you can ignore finitely many points (edit: as long as their complement is dense!); in particular, you can ignore the finitely many points with $X_2 = 0$, so work on the chart $X_2 \neq 0$ with the affine equation $$x^3 + y^2 + x^2 y^3 = 0$$

where $x = \frac{X_0}{X_2}, y = \frac{X_1}{X_2}$. It suffices to show that this polynomial is irreducible. Think of it as a polynomial in $x$ with coefficients in $k[y]$. Then by the rational root theorem, it is irreducible if and only if it has a root in $k[y]$. Its constant term is $y^2$, so the only possible roots are scalar multiples of $1, y, y^2$, and by inspection none of these works (the different terms don't even all have the same degree when these are substituted). So the conclusion follows.

For the special case of projective curves defined by one homogeneous equation, here's a nice trick which doesn't seem to apply in this case, but is good to keep in mind in general: if they're not irreducible, they have components which intersect over $\bar{k}$ by Bezout's theorem. At any such intersection point, all of the partial derivatives of the defining equation vanish. Hence if the curve is geometrically smooth (smooth over $\bar{k}$) it is automatically irreducible.

-

It's a complex curve in the two-dimensional complex projective space. If I set $X_2=1$, I don't lose anything because only $X_2=0$ points are omitted and, substituting $X_2=0$ to your polynomial, we only get two points of $P^2$, $(1,0,0)$ and $(0,1,0)$ which are easily seen to be connected with some one-dimensional curves.

For $X_2=1$, the equation is $$X_0^3 + X_1^2 + X_1^3 X_0^2=0$$ It may be viewed as a cubic equation for $X_0$. So for each generic value of $X_1$, there are three possible values of $X_0$ - all of them are continuous functions of $X_1$. The only question is whether one can get from one branch to the second branch or third branch by changing $X_1$ along smooth points at which the three allowed values of $X_0$ (roots) are different.

They're clearly connected because the equation is generic. The discriminant entering Cardan's expression for the roots of the cubic equation is $$27 X_1^4 + 4 X_1^{11}$$ so whenever one makes a trip around a generic point in the $X_1$ plane where this discriminant is zero - and aside from the quadruple $X_1=0$ root, it clearly has 7 extra simple roots on a circle in the complex plane which are relevant - the three roots $X_0$ get cyclically permuted because $X_0$ depends on the third root of the discriminant.

This is not a fascinating conclusion - degenerate curves are given by very special polynomials whose measure is zero within all such polynomials. It would be more interesting to see what is the genus of this curve (Riemann surface). I guess that experienced algebraic geometers can calculate the genus within a second of looking at the equation; the degree is probably enough for them. It may be pretty large.

-
It's a plane curve of degree $5$, so it has genus $\frac{(5-1)(5-2)}{2} = 6$. Also note that the OP wants an answer over any field of characteristic $0$. –  Qiaochu Yuan May 29 '11 at 9:44
Excellent, it's you et al. whom I meant. ;-) Is that right that the result from $C$ directly extends to all fields of characteristic $0$? –  Luboš Motl May 29 '11 at 16:14
Maybe? That probably follows by model completeness. –  Qiaochu Yuan May 29 '11 at 16:24
@Qiaochu: If a scheme $X$ (not necessarily of finite type!) over an algebraically closed field $k$ is irreducible, then the same is true for any base-change to a bigger algebraically closed field (this is part of the whole "geometrically irreducible" story); note that if $X \times_k K$ is irreducible, so is $X$ (because the projection is a surjection). So Lubos Motl's result extends to any field of characteristic zero. (Note that "model completeness" won't handle arbitrary schemes, though it should handle varieties.) –  Akhil Mathew May 29 '11 at 16:35