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Let $R$ be a Noetherian local ring with maximal ideal $I$.

Suppose we have a morphism of smooth $R$-algebras $f : A\to B$ such that its reduction modulo $I^n$

$$f_n : A/I^n \to B/I^n$$

is an isomorphism for all $n\ge 1$. What can we say about $f$? Is $f$ smooth?

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No. Take for $R$ a discrete valuation ring with fraction field $K$, and for $f$ any non-smooth morphism of smooth $K$-algebras (viewed as a morphism of $R$-algebras).

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