Suppose I have a smooth 2-dimensional quadric bundle $f:X\to S$ over a surface $S$. Suppose furthermore that the discriminant locus $\Delta \subset S$ is smooth. Can I immedately conclude that the fibers of $f$ have at most isolated singularities? Why?



degeneration of quadric bundles
It is sufficient to prove that if the fibre over a point $s \in S$ is the union of two planes or a plane counted twice, then $\Delta$ is singular at $s$.


The question being local, we may assume that $S$ is a small polidisk centered at $(0,0) \in \mathbb{C}^2$ and $s=(0,0)$. Let $\mathcal{O}$ be the local ring of convergent power series centered in $(0,0)$ and $\mathfrak{m} \subset \mathcal{O}$ be its maximal ideal, namely the set of power series vanishing at $(0,0)$.


In the case where the central fibre is the union of two planes, the equation of our conic bundle is
$$2x\_0x\_1 + \sum a\_{ij}(z,w)x\_i x\_j=0,$$
where $i,j \in \{0,1,2,3\}$, $\{i,j \} \neq \{ 0,1 \}$ and $a\_{ij}(z,w)=a\_{ji}(z,w) \in \mathfrak{m} $. In fact, the fibre over $(0,0)$ is the reducible quadric $x\_0x\_1=0$.


The equation of the discriminant $\Delta$ is then given by
$$\det \left(\begin{array}{llll} a\_{00} & 1 & a\_{02} & a\_{03} \\\ 1 & a\_{11} & a\_{12} & a\_{13} \\\ a\_{02} & a\_{12} & a\_{22} & a\_{23} \\\ a\_{03} & a\_{13} & a\_{23} & a\_{33} \end{array}\right)=0.$$


Since $a\_{ij} \in \mathfrak{m}$, the explicit computation shows that the determinant above belongs to $\mathfrak{m}^2$. This precisely means that $\Delta$ is singular at $(0,0)$, proving the claim. 


In the case where the central fibre is a plane counted twice, the equation of our conic bundle is 
$$x\_0^2 + \sum a\_{ij}(z,w)x\_i x\_j=0,$$
with the $a\_{ij}(z,w)$ as above. In fact, the fibre over $(0,0)$ is the double plane $x\_0^2=0$. The proof that $\Delta$ is singular at $(0,0)$ is the same as in the previous case. 


