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?

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i may be wrong, but intuitively, you have a map of a smooth surface S into the variety of quadrics and the pullback of the locus of singular quadrics is a smooth curve. hence the map does not meet the singular locus of the locus of singular quadrics, which is to say it meets no quadrics of corank 2. – roy smith Sep 1 '11 at 16:43

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.

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thank you! did you use anywhere the fact that $X$ is smooth? – The Chopper Aug 20 '11 at 17:00
Not really. This is the local equation around a singular fibre, and you can check that for a general choice of the $a_{ij}$ the total space $X$ is smooth. But I guess that the statement also holds when $X$ is singular – Francesco Polizzi Aug 20 '11 at 17:18