Following the Calabi Yau Bestiary for Physicsts, I have the following (basic) doubts.
We have a Hirzebruch surface $S_{\epsilon}$ which is a member of the following configuration $$ \left[\begin{array}{c||ccc} 3&1& 1\\ 1&1& 1\\ \end{array}\right] $$
$S_{\epsilon}$ is chosen to be defined by*:
$ \left\{\begin{matrix} z_{0}w_{0}+z_{1}w_{1}=0\\ z_{2}w_{0}+\left [ \sum_{i=0}^{2}a_{i}z_{i}+\varepsilon z_{3} \right ]w_{1}=0 \end{matrix}\right.$
It says that the surface is not singular in $\mathbb{P}^{3}\times \mathbb{P}^{1}$ everywhere except in the cases: (1) $a_{1}=0$, (2) $a_{2}=0$ or (3) $a_{0}=-a_{1}a_{2} \not=0$**
*Question 1: I have supposed that that equation system is an example of the configuration matrix, what will be the most general equation system (counting all the linear rescalings)?
** Question 2: Can someone show me why is the surface singular in those three cases? I have substituted that values and I do not see anything special (in fact what is the general method to see which ones are the singular points?)
This is a partial answer to (**); somebody wiser than me tell me if it should be in a comment. Find singular points by looking on affine patches and using Jacobians. On the patch $z_3=1, w_0=1$ get the matrix $$\pmatrix{z_1 & a_1z_1 \\ 1 & a_0w_1 \\ w_1 & a_1w_1 \\ 0 & 1+a_2w_1}$$ One way for this matrix to have rank $1$ is for $z_1$ to equal $0$, forcing $z_0=0$. $w_1\neq 0$ else the first and fourth rows form a nonsingular minor; and now when $a_0w_1=a_1, a_2w_1=-1$ the matrix has rank $1$. Solving these equations shows that when $-a_0=a_1a_2$ we get a singularity where $w_1=\frac{a_1}{a_0}$.