The Stacks project

33.21 Complete local rings

Some results on complete local rings of schemes over fields.

Lemma 33.21.1. Let $k$ be a field. Let $X$ be a locally Noetherian scheme over $k$. Let $x \in X$ be a point with residue field $\kappa $. There is an isomorphism

33.21.1.1
\begin{equation} \label{varieties-equation-complete-local-ring} \kappa [[x_1, \ldots , x_ n]]/I \longrightarrow \mathcal{O}_{X, x}^\wedge \end{equation}

inducing the identity on residue fields. In general we cannot choose (33.21.1.1) to be a $k$-algebra isomorphism. However, if the extension $\kappa /k$ is separable, then we can choose (33.21.1.1) to be an isomorphism of $k$-algebras.

Proof. The existence of the isomorphism is an immediate consequence of the Cohen structure theorem1 (Algebra, Theorem 10.160.8).

Let $p$ be an odd prime number, let $k = \mathbf{F}_ p(t)$, and $A = k[x, y]/(y^2 + x^ p - t)$. Then the completion $A^\wedge $ of $A$ in the maximal ideal $\mathfrak m = (y)$ is isomorphic to $k(t^{1/p})[[z]]$ as a ring but not as a $k$-algebra. The reason is that $A^\wedge $ does not contain an element whose $p$th power is $t$ (as the reader can see by computing modulo $y^2$). This also shows that any isomorphism (33.21.1.1) cannot be a $k$-algebra isomorphism.

If $\kappa /k$ is separable, then there is a $k$-algebra homomorphism $\kappa \to \mathcal{O}_{X, x}^\wedge $ inducing the identity on residue fields by More on Algebra, Lemma 15.38.3. Let $f_1, \ldots , f_ n \in \mathfrak m_ x$ be generators. Consider the map

\[ \kappa [[x_1, \ldots , x_ n]] \longrightarrow \mathcal{O}_{X, x}^\wedge ,\quad x_ i \longmapsto f_ i \]

Since both sides are $(x_1, \ldots , x_ n)$-adically complete (the right hand side by Algebra, Lemmas 10.96.3) this map is surjective by Algebra, Lemma 10.96.1 as it is surjective modulo $(x_1, \ldots , x_ n)$ by construction. $\square$

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. $\square$

[1] Note that if $\kappa $ has characteristic $p$, then the theorem just says we get a surjection $\Lambda [[x_1, \ldots , x_ n]] \to \mathcal{O}_{X, x}^\wedge $ where $\Lambda $ is a Cohen ring for $\kappa $. But of course in this case the map factors through $\Lambda /p\Lambda [[x_1, \ldots , x_ n]]$ and $\Lambda /p\Lambda = \kappa $.

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