Lemma 42.4.3. Let $(R, \mathfrak m)$ be a Noetherian local ring of dimension $1$. Let $a, b \in R$ be nonzerodivisors with $a \in \mathfrak m$. There exists an integer $n = n(R, a, b)$ such that for a finite ring extension $R \subset R'$ if $b = a^ m c$ for some $c \in R'$, then $m \leq n$.

Proof. Choose a minimal prime $\mathfrak q \subset R$. Observe that $\dim (R/\mathfrak q) = 1$, in particular $R/\mathfrak q$ is not a field. We can choose a discrete valuation ring $A$ dominating $R/\mathfrak q$ with the same fraction field, see Algebra, Lemma 10.119.1. Observe that $a$ and $b$ map to nonzero elements of $A$ as nonzerodivisors in $R$ are not contained in $\mathfrak q$. Let $v$ be the discrete valuation on $A$. Then $v(a) > 0$ as $a \in \mathfrak m$. We claim $n = v(b)/v(a)$ works.

Let $R \subset R'$ be given. Set $A' = A \otimes _ R R'$. Since $\mathop{\mathrm{Spec}}(R') \to \mathop{\mathrm{Spec}}(R)$ is surjective (Algebra, Lemma 10.36.17) also $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A)$ is surjective (Algebra, Lemma 10.30.3). Pick a prime $\mathfrak q' \subset A'$ lying over $(0) \subset A$. Then $A \subset A'' = A'/\mathfrak q'$ is a finite extension of rings (again inducing a surjection on spectra). Pick a maximal ideal $\mathfrak m'' \subset A''$ lying over the maximal ideal of $A$ and a discrete valuation ring $A'''$ dominating $A''_{\mathfrak m''}$ (see lemma cited above). Then $A \to A'''$ is an extension of discrete valuation rings and we have $b = a^ m c$ in $A'''$. Thus $v'''(b) \geq mv'''(a)$. Since $v''' = ev$ where $e$ is the ramification index of $A'''/A$, we find that $m \leq n$ as desired. $\square$

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