Let $(R, \mathfrak{m})$ be a noetherian local ring, and let $\hat R$ be its $\mathfrak{m}$-adic completion. Extension of scalars allows one to transform an $R$-scheme into an $\hat R$-scheme. Is this association surjective, up to isomorphism? More precisely, given a finite type scheme $X \to \mathrm{Spec} \ \hat R$, does there exist an $R$-scheme $Y$ and an $\hat R$-isomorphism $$X \cong Y \times_{\mathrm{Spec} \ R} \mathrm{Spec} \ \hat R\ \ ? $$

My interest stems from the case where $R$ is the ring of germs of holomorphic functions at the origin in $\mathbb{C}$. If $t$ is the usual coordinate on $\mathbb{C}$, then we have a canonical isomorphism $\hat R \cong \mathbb{C}[[t]]$, where the latter is the ring of formal power series in $t$. Is every formal family $X \to \mathbb{C}[[t]]$ isomorphic to an analytic family? I suppose it may be relevant that $R$ is henselian in my example (by the implicit function theorem).

**Update: ** Let $K$ be the field of fractions of $R$, and let $\hat K$ be the field of fractions of $\hat R$. Suppose that there exists a $K$-scheme $Y_K$ and a $\hat K$ isomorphism $X_{\hat K} \cong Y_K \times_{\mathrm{Spec} \ K} \mathrm{Spec \ \hat K}$. Can we conclude that there exists an $R$-scheme $Y$ as above, which necessarily has generic fiber $Y_K$? That is, if we already know the generic fiber is algebraic, can we arrange for the whole scheme to be algebraic?

`$\mathbf{C}[\![t]\!]$`

has radius of convergence 0 then it cannot arise from an analytic family of elliptic curves.) $\endgroup$4more comments