# “A smooth morphism is a formal power series of $N$ invariants”

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.

• 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. – gdb May 17 at 4:51
• @gdb Thank you for pointing out. I edited. – k.j. May 17 at 5:02
• 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)? – Asvin May 17 at 6:01