Let $f : X \rightarrow S$ be a morphism between schemes, with $f$ locally of finite presentation, proper and flat, and $S$ the spectrum of a discrete valuation ring with closed point $s$.

Assume that the special fiber $X_s$ is smooth. Is the generic fiber smooth?

In other words: is it impossible to have a flat (algebraic) family of proper singular varieties degenerating into a smooth variety?

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Yes. Here is an argument provided by Kay Rülling, I hope that this is helpful for someone else:

It is sufficient (and necessary because $S$ has just two points: the special point $s$ and the generic point) to show that $f$ is smooth. By this lemma (or EGA IV, 4, 17.5.1), we use local finite presentation and smoothness of the special fiber to conclude that $f$ is smooth on an open $U \subset X$ containing the special fiber $f^{-1}(s)$.

Since $f$ is proper, we conclude that $U = X$, so in particular the generic fiber is smooth. Indeed, $X \setminus U$ is closed; if it were non empty, then the properness of $f$ implies that $f(X \setminus U) \subset S$ is closed and non-empty, and since $S$ is a discrete valuation ring, $f(X \setminus U)$ has to be equal to $s$ or $S$. In both cases $s \in f(X \setminus U)$, which implies that $X \setminus U$ meets $f^{-1}(s)$, hence a contradiction.


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