Let $f:X \to \mbox{Spec } R$ be a projective morphism between irreducible Noetherian schemes. Assume that $R$ is a discrete valuation ring and its residue field is algebraically closed. Suppose now that $f$ restricted to the special fiber is smooth. Assume further that the special fiber is irreducible and smooth (over the residue field). Does it imply that $f$ is flat or smooth?
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No, it does not. As a counterexample, let $f$ be the inclusion of the closed point $Spec\ k \hookrightarrow Spec\ R$. Equally well, one could also take other irreducible projective smooth $k$-varieties for $X$.
answered Nov 22, 2013 at 1:55