MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f \in \mathbb{C}[x,y,z]$ be a polynomial which defines an isolated singularity $0 \in D:= (f=0) \subset \mathbb{C}^3$. Assume that $\mathcal{D}:= (f+tx =0) \subset \mathbb{C}^3 \times \mathbb{C}$ defines a smoothing of $D$ over a small disk. Let $g \in \mathbb{C}[x,y,z]$ be a polynomial which has a zero of order $\ge 2$ at $0$.

Question Let $\mathcal{D}' := (f+t(x+g)=0)$ be a deformation of $D$ induced by $x+g$. Is $\mathcal{D}'$ smoothing?

share|cite|improve this question
up vote 3 down vote accepted


Call $d$ the maximum between the degrees of $f$ and $g$. Let us homogenize the equations to homogeneous equations of degree $d$ in four variables. The polynomial $f + t(x + g)$ corresponds to a linear system of curves $C_t$ of degree $d$ in $\mathbb P^3$. By Bertini's theorem $C_t$ is smooth away from the base points of the linear system, for all but finitely many values of $t$. We don't care about the singularities at infinity. If $p \in \mathbb C^3$ is a base point of the linear system, then either $p$ is the origin, in which case $C_\infty$ is smooth at $p$, or is not, and in this case $C_0$ is smooth at $p$. Since being smooth at a base point is an open condition, we have have that $C_t$ is smooth at all points of $\mathbb C^3$ for all but finitely many values of $t$.

share|cite|improve this answer
Thank you very much for the clear answer. – tarosano Aug 5 '12 at 20:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.