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 $\mathfrak m + \mathfrak m^{c + 1}M$.
$\square$

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