## Some obvious singularities

Wei Yue asked a question about an assertion that came up in lecture today. It was that if a curve $C$ in ${\bf P}^2$ has defining equation $F=0$, where $F=GH$ for two non-constant homogeneous polynomials $G$ and $H$, then $C$ is necessarily singular. The reason is the equation

$\nabla F=(\nabla G)H+G(\nabla H).$

So if $a$ is a point where $G(a)=0$ and $H(a)=0$, then $\nabla F(a)=0$. But the zero sets of $G$ and $H$ must meet (by Bezout’s theorem) and hence, the curve $C$ is singular.

$D: G=0$

and

$E: H=0$.

Since $F=GH$, we get

$C=D\cup E.$

But there must be a point $a\in D\cap E$, and this is a singular point of $C$. If you visualize $C$ as the union of two curves, you can imagine that these isn’t a single tangent line to $C$ along their intersection (and it *looks* singular there). For the easiest example, consider the union of two lines that meet at a point. (By the way, in ${\bf P}^2$ two lines *must* meet.)

Another question was about the factorization of such an $F$. That is, doesn’t $F$ factorize into linear factors anyways? The answer is no in general. The factorization we discussed in the lecture was for homogeneous polynomials in two variables. In three variables, many $F$ of large degree are *irreducible*. In fact, what we showed above is that if $F$ is reducible (in that case, we also say $C$ is reducible), then $C$ is necessarily singular.