Lemma 42.6.1. Let $(A, \mathfrak m)$ be a $2$-dimensional Noetherian local ring. Let $t \in \mathfrak m$ be a nonzerodivisor. Say $V(t) = \{ \mathfrak m, \mathfrak q_1, \ldots , \mathfrak q_ r\} $. Let $A_{\mathfrak q_ i} \subset B_ i$ be a finite ring extension with $B_ i/A_{\mathfrak q_ i}$ annihilated by a power of $t$. Then there exists a finite extension $A \subset B$ of local rings identifying residue fields with $B_ i \cong B_{\mathfrak q_ i}$ and $B/A$ annihilated by a power of $t$.
Proof. Choose $n > 0$ such that $B_ i \subset t^{-n}A_{\mathfrak q_ i}$. Let $M \subset t^{-n}A$, resp. $M' \subset t^{-2n}A$ be the $A$-submodule consisting of elements mapping to $B_ i$ in $t^{-n}A_{\mathfrak q_ i}$, resp. $t^{-2n}A_{\mathfrak q_ i}$. Then $M \subset M'$ are finite $A$-modules as $A$ is Noetherian and $M_{\mathfrak q_ i} = M'_{\mathfrak q_ i} = B_ i$ as localization is exact. Thus $M'/M$ is annihilated by $\mathfrak m^ c$ for some $c > 0$. Observe that $M \cdot M \subset M'$ under the multiplication $t^{-n}A \times t^{-n}A \to t^{-2n}A$. Hence $B = A + \mathfrak m^{c + 1}M$ is a finite $A$-algebra with the correct localizations. We omit the verification that $B$ is local with maximal ideal $\mathfrak m + \mathfrak m^{c + 1}M$. $\square$
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Your email address will not be published.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)