Lemma 54.11.1. Let $(A, \mathfrak m, \kappa )$ be a local ring with finitely generated maximal ideal $\mathfrak m$. Let $X$ be a scheme over $A$. Let $Y = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A^\wedge )$ where $A^\wedge $ is the $\mathfrak m$-adic completion of $A$. For a point $q \in Y$ with image $p \in X$ lying over the closed point of $\mathop{\mathrm{Spec}}(A)$ the local ring map $\mathcal{O}_{X, p} \to \mathcal{O}_{Y, q}$ induces an isomorphism on completions.
Proof. We may assume $X$ is affine. Then we may write $X = \mathop{\mathrm{Spec}}(B)$. Let $\mathfrak q \subset B' = B \otimes _ A A^\wedge $ be the prime corresponding to $q$ and let $\mathfrak p \subset B$ be the prime ideal corresponding to $p$. By Algebra, Lemma 10.96.3 we have
\[ B'/(\mathfrak m^\wedge )^ n B' = A^\wedge /(\mathfrak m^\wedge )^ n \otimes _ A B = A/\mathfrak m^ n \otimes _ A B = B/\mathfrak m^ n B \]
for all $n$. Since $\mathfrak m B \subset \mathfrak p$ and $\mathfrak m^\wedge B' \subset \mathfrak q$ we see that $B/\mathfrak p^ n$ and $B'/\mathfrak q^ n$ are both quotients of the ring displayed above by the $n$th power of the same prime ideal. The lemma follows. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)