Partial dehomogenization and smoothness

Question

Let $P_1l_1+P_2l_2$ be a homogeneous degree $d$ polynomial in $\mathbb{C}[X_0,X_1,X_2,X_3]$ which defines a smooth surface in $\mathbb{P}^3$. Here $l_i$ are linear polynomials and $l_1 \not=\lambda l_2$ for any $\lambda \in \mathbb{C}$. Does there exist $\lambda' \in \mathbb{C}$ such that $P_1+\lambda' P_2$ defines a smooth surface in $\mathbb{P}^3$? If so can we further say that this is true for general $\lambda'$?

Answer 1

The answer seems yes, but not quite trivial. Here is an attempt.Your condition of smoothness of $P_1l_1+P_2l_2=0$ implies that $P_1=P_2=0$ is a curve $C$ and at any point of $C$ either $P_1=0$ or $P_2=0$ is smooth. Let us use the phrase $P_1,P_2$ are linearly dependent at a point $p\in C$' to mean that the images of $P_1,P_2$ in $m_p/m_p^2$ are linearly dependent, where $m_p$ is the sheaf of maximal ideal of $p\in\mathbb{P}^3$.

Let $S\subset C$ be the set of points where $P_1,P_2$ are both smooth and linearly dependent. Clearly $S$ is a locally closed subscheme. The main point to observe is that under your hypothesis, $S$ is finite. Let us assume this for a moment and finish the rest.

By Bertini, $aP_1+bP_2$ for general $a,b\in\mathbb{C}$ is smooth outside $C$. If we take $ab\neq 0$, then $aP_1+bP_2$ is smooth at points of $C$ where one of $P_1,P_2$ is smooth and the other singular. At points where both $P_i$s are smooth and linearly independent, for all choice of $a,b$ at least one of them non-zero, $aP_1+bP_2$ is smooth. The only remaining points are in $S$ and for this finite set, clearly only finitely many choices of $a,b$ (upto constant multiples) need to be avoided. This proves what you need.

To prove $S$ is finite, we proceed as follows. Let $D\subset C$ be a reduced irreducible component of the closure of $S$. Assume that $D$ is a curve and we reach a contradiction. Consider the natural map, $I_C/I_C^2\to \Omega^1_{\mathbb{P}^3}|C$. We have the natural map $\mathcal{O}_C(-d-1)\to I_C/I_C^2$ given by your smooth surface, whose image under composition in $\Omega^1|C$ is a line subbundle. If we restrict all this to $D$, we see that since on a dense subset of $D$, $P_i$'s are linearly dependent and $D$ is reduced irreducible, the image of $I_C/I_C^2|D$ in $\Omega^1|D$ is a line bundle and thus necessarily equal to the image of $\mathcal{O}_D(-d-1)$. This says that the map $\mathcal{O}_D(-d-1)\to I_C/I_C^2|D=\mathcal{O}_D(-d)^{\oplus 2}$ splits. This is absurd if $D$ is a reduced irreducible curve. So, $D$ must be finite.