Finding singularities of a projective curve

Question

For $w \in \mathbb{C}$ we define the projective curve $$p(x,y,z):= x^3+y^3+z^3+wxyz.$$

Now I have to find all $w \in \mathbb{C}$ for which the projective curve $p(x,y,z)$ is singular and show that for this $w$ the curve $p(x,y,z)$ is reducible (and finding the prime factorization).

Answer 1

If you multiply $3x^2 +wyz =0$ by $x$ and similarly for the other equations, you can cancel the $xyz$ terms and conclude $x^3 = y^3 = z^3$. Thus $y=a^i x$ and $z=a^jx$, where $a$ is a primitive cube root of unity. The equations become $3 +w a^{i+j}=0,$ $3a^{2i} + wa^j = 0$ and $3a^{2j} + wa^i = 0,$ each of which lets you conclude $w = -3a^{-i-j}.$ So the curve seems to be singular iff $w = -3a^k$ for some $k$.

edit. Regarding the factorisation: You can factor $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(\text{something}).$ You can obtain the factorisations in the other cases by (e.g.) replacing $x$ by $a^kx.$

Answer 2

You can use resultants to look for common zeros of multiple polynomials. For example, if you use the first equation to eliminate $x$ from the other two, then combine those to eliminate $y$ as well, you end up with something like this:

$$q(z,w)=-27z^{16}(w - 3)^3(w + 3)^3(w^2 - 3w + 9)^3(w^2 + 3w + 9)^3 =-27z^{16}(w^6-3^6)^3$$

So you see the possible ways this can be zero depending on $w$. The case $z=0$ would need special treatment, but due to the symmetry of the whole thing, you could always rename variables so that $z$ is a non-zero one. Because $[0:0:0]$ is not a point of the projective plane.

The zeros of $q$ are all scaled versions of sixth roots of unity: $w_k=3e^{ik2\pi/6}$ for $k\in\{0,1,2,3,4,5\}$. But if you go back to the polynomials from which this was computed, not all of these values will yield valid solutions. For example, while the first two equations have a common zero for some $x$, and the the first and the third have a common zero as well, then it might still be the case that these two $x$ are different, so there is no common solution for all three. In this sense, the above gives you a superset of things to look for.

Doing the computation, one can find that there are indeed only three possible values for $w$, just like Tom Bachmann found out. Each of them comes with three distinct singular points.