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Let $f : X \to Y$ be a morphism of schemes and $x \mapsto y$ be points at which $f$ is smooth of relative dimension $N$ and which have the same residue fields (i.e., $k(x) = k(y)$). Then does the map of stalks induce $\widehat{\mathscr{O}_{Y, y}} [[T_1, \dots, T_N]] \cong \widehat{\mathscr{O}_{X, x}}$?

For $N = 0$, this is standard (e.g., see 4.3.26 of Liu's "Algebraic geometry and arithmetic curves"). For general $N$, since there exists an open neighbour $U$ of $x$ such that $f\mathclose|_U$ is the composition of étale $U \to \mathbb{A}_Y^N$ and the projection $\mathbb{A}_Y^N \to Y$, we may assume that $X = \mathbb{A}_Y^N$. So it suffices to show:

Let $A$ be a local ring with the maximal ideal $\mathfrak{m}$, and $\mathfrak{n} = \mathfrak{m} + (T_1, \dots, T_N)$. Then $\hat{A}[[T_1, \dots, T_N]] \cong A[T_1, \dots, T_N]\widehat{\ \ }$ (the completion with respect to $\mathfrak{n}$).

How can I show it?

And it seems that the converse holds. If this is true, please suggest me its references.

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    $\begingroup$ The statement is false as stated, the map $\mathrm{Spec } \mathbf C \to \mathrm{Spec } \mathbf R$ is 'etale, but the claim clearly does not hold in that case. Though the statement is correct if $X$ and $Y$ are finite type over an alg. closed field and $x$ is a closed point. $\endgroup$ – gdb May 17 at 4:51
  • $\begingroup$ @gdb Thank you for pointing out. I edited. $\endgroup$ – k.j. May 17 at 5:02
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    $\begingroup$ Does a direct proof not work? There is a natural map from the left hand side to the completion, you can check injectivity by hand I suspect and to check surjectivity, you check it after reduction + show one can always lift (again, by hand)? $\endgroup$ – Asvin May 17 at 6:01

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