Smoothness of total spaces with nice singular fibers

Question

Let $f:X \to Y$ be a (flat and projective) family of hypersurfaces over a smooth curve $Y$. If every fiber of $f$ is smooth, it is clear that the total space $X$ is also smooth. I would like to know if the smoothness still holds when we also allow singularities of ordinary double points:

If $f$ has finitely many singular fibers but every singular fiber contains only one ordinary double point, is it still true that $X$ is smooth?

Here an ordinary double point means the tangent cone is nondegenerate. Since the total space of a Lefschetz pencil is always smooth (as blow-up a smooth one along a smooth locus is smooth), I guess it is reasonable to expect smoothness with those nice singular fibers.

Thanks in advance.

Answer 1

No. For instance take $Y = \mathbb{A}^1$ (with coordinate $t$) and set $$ X = \{ xy + t(xz - y^2) = 0 \} \subset \mathbb{A}^1 \times \mathbb{P}^2. $$ The fibers of $X$ over $\mathbb{A}^1 \setminus \{0\}$ are smooth, the fiber over 0 has an ordinary double point, but $X$ is singular (at that point).

Answer 2

In this post, I'd like to give an argument locally analytically classifies a family around an ordinary node on a special fiber:

Proposition: Let $f: X\to \Delta$ be a flat family and smooth over $\Delta^*$. Assume the central fiber $f^{-1}(0)$ has an ordinary node at point $x$, then a small neighborhood of $x$ in $X$ is analytically equivalent to $$\mathbb C^{n}\times \Delta\supset\{x_1^2+\cdots+x_n^2-t^k=0\} \to \Delta, \ \ \ \ (\vec x,t)\mapsto t.$$

This is because $k=1$ is semi-universal deformations of ordinary node (see p.238 of Introduction to Singularities and Deformations), therefore any other (nontrivial) deformations of an ordinary node arise from a base change via $$t\mapsto t^k$$ with $k\ge 1$. Hence, a direct consequence will be

Corollary: The total space $X$ is smooth at $x$ if and only if $k=1$.

Lastly, let's test the argument on @Sasha's example: Take affine chart $\{z=1\}$ and change the coordinate $\bar{y}=y+t,\ \bar{x}=x-y+t^2$, then the equation of the total space becomes $$\bar{x}\bar{y}-t^3=0.$$ It is $k=3$ in this case!