Lemma 33.21.2. Let $K/k$ be an extension of fields. Let $X$ be a locally algebraic $k$-scheme. Set $Y = X_ K$. Let $y \in Y$ be a point with image $x \in X$. Assume that $\dim (\mathcal{O}_{X, x}) = \dim (\mathcal{O}_{Y, y})$ and that $\kappa (x)/k$ is separable. Choose an isomorphism
\[ \kappa (x)[[x_1, \ldots , x_ n]]/(g_1, \ldots , g_ m) \longrightarrow \mathcal{O}_{X, x}^\wedge \]
of $k$-algebras as in (33.21.1.1). Then we have an isomorphism
\[ \kappa (y)[[x_1, \ldots , x_ n]]/(g_1, \ldots , g_ m) \longrightarrow \mathcal{O}_{Y, y}^\wedge \]
of $K$-algebras as in (33.21.1.1). Here we use $\kappa (x) \to \kappa (y)$ to view $g_ j$ as a power series over $\kappa (y)$.
Proof.
The local ring map $\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$ induces a local ring map $\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{Y, y}^\wedge $. The induced map
\[ \kappa (x) \to \kappa (x)[[x_1, \ldots , x_ n]]/(g_1, \ldots , g_ m) \to \mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{Y, y}^\wedge \]
composed with the projection to $\kappa (y)$ is the canonical homomorphism $\kappa (x) \to \kappa (y)$. By Lemma 33.5.1 the residue field $\kappa (y)$ is a localization of $\kappa (x) \otimes _ k K$ at the kernel $\mathfrak p_0$ of $\kappa (x) \otimes _ k K \to \kappa (y)$. On the other hand, by Lemma 33.5.3 the local ring $(\kappa (x) \otimes _ k K)_{\mathfrak p_0}$ is equal to $\kappa (y)$. Hence the map
\[ \kappa (x) \otimes _ k K \to \mathcal{O}_{Y, y}^\wedge \]
factors canonically through $\kappa (y)$. We obtain a commutative diagram
\[ \xymatrix{ \kappa (y) \ar[rr] & & \mathcal{O}_{Y, y}^\wedge \\ \kappa (x) \ar[r] \ar[u] & \kappa (x)[[x_1, \ldots , x_ n]]/(g_1, \ldots , g_ m) \ar[r] & \mathcal{O}_{X, x}^\wedge \ar[u] } \]
Let $f_ i \in \mathfrak m_ x^\wedge \subset \mathcal{O}_{X, x}^\wedge $ be the image of $x_ i$. Observe that $\mathfrak m_ x^\wedge = (f_1, \ldots , f_ n)$ as the map is surjective. Consider the map
\[ \kappa (y)[[x_1, \ldots , x_ n]] \longrightarrow \mathcal{O}_{Y, y}^\wedge ,\quad x_ i \longmapsto f_ i \]
where here $f_ i$ really means the image of $f_ i$ in $\mathfrak m_ y^\wedge $. Since $\mathfrak m_ x \mathcal{O}_{Y, y} = \mathfrak m_ y$ by Lemma 33.5.3 we see that the right hand side is complete with respect to $(x_1, \ldots , x_ n)$ (use Algebra, Lemma 10.96.3 to see that it is a complete local ring). Since both sides are $(x_1, \ldots , x_ n)$-adically complete our map is surjective by Algebra, Lemma 10.96.1 as it is surjective modulo $(x_1, \ldots , x_ n)$. Of course the power series $g_1, \ldots , g_ m$ are mapped to zero under this map, as they already map to zero in $\mathcal{O}_{X, x}^\wedge $. Thus we have the commutative diagram
\[ \xymatrix{ \kappa (y)[[x_1, \ldots , x_ n]]/(g_1, \ldots , g_ m) \ar[r] & \mathcal{O}_{Y, y}^\wedge \\ \kappa (x)[[x_1, \ldots , x_ n]]/(g_1, \ldots , g_ m) \ar[r] \ar[u] & \mathcal{O}_{X, x}^\wedge \ar[u] } \]
We still need to show that the top horizontal arrow is an isomorphism. We already know that it is surjective. We know that $\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$ is flat (Lemma 33.5.1), which implies that $\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{Y, y}^\wedge $ is flat (More on Algebra, Lemma 15.43.8). Thus we may apply Algebra, Lemma 10.99.1 with $R = \kappa (x)[[x_1, \ldots , x_ n]]/(g_1, \ldots , g_ m)$, with $S = \kappa (y)[[x_1, \ldots , x_ n]]/(g_1, \ldots , g_ m)$, with $M = \mathcal{O}_{Y, y}^\wedge $, and with $N = S$ to conclude that the map is injective.
$\square$
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