Singularities in Projective Space

Question

I'm trying to compute singularities and their type for the following equation in $\mathbb{P}^1$.

$$P(X,Y) = ax^3 + bx^2y + cxy^2 + dy^3$$

$a$, $b$, $c$ and $d$ are nonzero. If we consider the partials:

• $d/dx = 3ax^2 + 2bxy + cy^2$
• $d/dy = bx^2 + 2cxy + 3dy^2$

We see that the only possible singularity is at $[1:1]$ provided that $c = -3a - 2b$ and $d = 2a+b$. Furthermore, a quick look at the second derivatives

• $d^2/dx^2 = 6ax + 2by$
• $d^2/dy^2 = 2cx + 6dy = -(6a + 4b)x + (12a + 6b)y$

shows that the second derivatives at $[1:1]$ are only $0$ if $b = -3a$. If this is the case, our basic curve becomes $$P(X,Y) = a(x-y)^3$$ and so it seems like we have a non-ordinary triple root at $[1:1]$ Is this correct? Furthermore, from here, I'm unsure of how precisely to work out what type of singularity we have if $b \neq -3a$. Our curve becomes $$P(X,Y) = ax^3 + bx^2y- (3a + 2b)xy^2 + (2a + b)y^3$$ but since our singularity is at $[1:1]$, we can't use the same trick as in Singular points of algebraic curve, multiplicity, ordinary?. How do I proceed?

Answer 1

Your effort has serious problems. It is not the case that the only singularities can be at $[1:1]$! For instance, for any point $[a:b]\in\Bbb P^1$, we can arrange to have singularities there: consider $(bx-ay)^3$.

The thing that's really going to help you out here is that $V(ax^3+bx^2y+cxy^2+dy^3)$ is three points counted with multiplicity. This is the same as saying that $f=ax^3+bx^2y+cxy^2+dy^3$ factors in to irreducibles in the following way:

• It doesn't. This corresponds to a smooth point defined over a cubic extension of $k$.
• An irreducible quadratic term and a linear term. This corresponds to the union of a smooth point over $k$ and a smooth point over a quadratic extension of $k$.
• A product of three distinct linear terms. This corresponds to a union of three distinct smooth points.
• A product of a linear term and the square of a different linear term. This corresponds to a smooth point and a fat point (not smooth!).
• The cube of a linear term. This corresponds to a single fat point which isn't smooth.

What's going on in your question is that your argument with the partial derivatives has lead you astray. Your calculation of the partials is correct, but the math that follows after them is not.