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 $\lambda'$?
1 Answer
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 nonzero, $aP_1+bP_2$ 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 $\mathcal{O}_C(d1)\to I_C/I_C^2$ given by your smooth surface, whose image under composition in $\Omega^1C$ 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 $I_C/I_C^2D$ in $\Omega^1D$ is a line bundle and thus necessarily equal to the image of $\mathcal{O}_D(d1)$. This says that the map $\mathcal{O}_D(d1)\to I_C/I_C^2D=\mathcal{O}_D(d)^{\oplus 2}$ splits. This is absurd if $D$ is a reduced irreducible curve. So, $D$ must be finite.

$\begingroup$ @Mohan: Thanks a lot for the answer. I understand most of it. However, could you please elaborate on the natural map from $O_C(−d−1)\to I_C/I^2_C$ induced by the degree d smooth surface $X$? Is it simply multiplication by the defining equation of $X$? $\endgroup$ Jan 24, 2013 at 6:43

$\begingroup$ Exactly. The equation of the surface of degree $d+1$ gives an inclusion $\mathcal{O}_{\mathbb{P}^3}(d1)\to I_C$ and this induces the above map. $\endgroup$– MohanJan 24, 2013 at 15:19