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Let $ X\rightarrow Spec(\mathbb{C}[t])=\mathbb{C}$ be a projective variety over $\mathbb{C}[t]$ (a flat family of projective varieties $X_{t}$, $t\in \mathbb{C}$), and $X_{\eta}$ be the base change via $\mathbb{C}[t] \rightarrow \mathbb{C}((t))$ by $t\mapsto t$. ($X_{\eta}$ is the formal neighborhood of $X_{0}$ in $X$.) Let $ Y\rightarrow Spec(\mathbb{C}[[t]])$ be a projective variety over $\mathbb{C}[[t]]$, $Y_{0}$ be the special fiber, and $Y_{\eta}$ be the general fiber, i.e. the base change via $\mathbb{C}[[t]] \rightarrow \mathbb{C}((t))$. Assume that $X_{\eta}$ is isomorphic to $Y_{\eta}$ as varieties over $\mathbb{C}((t))$.

The question: Is there a projective variety $Z \rightarrow Spec(\mathbb{C}[t])$ over $\mathbb{C}[t]$ such that $Z_{t}$ is isomorphic to $X_{t}$ for any $t\neq 0$, and $Z_{0}$ is isomorphic to $Y_{0}$? (A sort of surgery to replace $X_{0}$ by $Y_{0}$.)

Thanks in advance!

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    $\begingroup$ As the question stands, yes: take $Z=X_{\neq 0}\amalg Z$, where $X_{\neq 0}$ is the open subset of $X$ above $\mathbb{C}\smallsetminus\{0\} $. So you need some assumptions on $Z\rightarrow \mathbb{C}[t]$. $\endgroup$
    – abx
    Commented Jun 16, 2016 at 6:06

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That's fpqc-descent. For this to work you need additionally ample line bundles on both $X_{\ne0}$ and $Y$ and a compatible extension of the isomorphism $X_\eta\to Y_\eta$ to these line bundles. See SGA 1,Proposition 7.8.

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