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Let $R$ be a complete discrete valuation ring with uniformizer $\pi$. Let $f:X\to \mathop{\mathrm{Spec}} R$ be a morphism of finite type and assume that the smooth locus of $f$ contains every $x\in X$ with $f(x)=(\pi)$. Is $f$ necessarily smooth?

Note that the assumption on the smooth locus is equivalent to the assumption that the base change of $f$ with respect to $R\to R/(\pi^n)$ is smooth for every $n\in\mathbb{N}$ (use Proposition 17.14.2 in EGA IV which states that it suffices to verify the infinitesimal lifting criterion with local Artin rings as test objects).

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No.

Take $f:X\rightarrow\mathrm{Spec}(R)$ such that $X_0:=X\times_R K$, $K=\mathrm{Frac}(R)$, possesses at least one singularity and there exists a point $x\in X\times_R R/(\pi)$ at which $f$ is smooth. Then there exists an (affine) open neighborhood $U$ of $x$ on $X$ such that $f$ is smooth at every $y\in U$. Since $X_0$ is open in $X$, the open subscheme $X_0\cup U$ is a counterexample to your conjecture.

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    Well, you could require $f$ to be closed ...2012-03-01