Lemma 15.123.4. Let $A \to B$ be a flat local homomorphism of Noetherian local normal domains. Let $f \in A$ and $h \in B$ such that $f = w h^ n$ for some $n > 1$ and some unit $w$ of $B$. Assume that for every height $1$ prime $\mathfrak p \subset A$ there is a height $1$ prime $\mathfrak q \subset B$ lying over $\mathfrak p$ such that the extension $A_\mathfrak p \subset B_\mathfrak q$ is weakly unramified. Then $f = u g^ n$ for some $g \in A$ and unit $u$ of $A$.
Proof. The local rings of $A$ and $B$ at height $1$ primes are discrete valuation rings (Algebra, Lemma 10.119.7). Thus the assumption makes sense (via Definition 15.111.1). Let $\mathfrak p_1, \ldots , \mathfrak p_ r$ be the primes of $A$ minimal over $f$. These have height $1$ by Algebra, Lemma 10.60.11. For each $i$ let $\mathfrak q_{i, j} \subset B$, $j = 1, \ldots , r_ i$ be the height $1$ primes of $B$ lying over $\mathfrak p_ i$. Say we number them so that $A_{\mathfrak p_ i} \to B_{\mathfrak q_{i, 1}}$ is weakly unramified. Since $f$ maps to an $n$th power times a unit in $B_{\mathfrak q_{i, 1}}$ we see that the valuation $v_ i$ of $f$ in $A_{\mathfrak p_ i}$ is divisible by $n$. Say $v_ i = n w_ i$ for some $w_ i \geq 0$. Consider the exact sequence
\[ 0 \to I \to A \to \prod \nolimits _{i = 1, \ldots , r} A_{\mathfrak p_ i}/\mathfrak p_ i^{w_ i}A_{\mathfrak p_ i} \]
defining the ideal $I$. Applying the exact functor $- \otimes _ A B$ we obtain an exact sequence
\[ 0 \to I \otimes _ A B \to B \to \prod \nolimits _{i = 1, \ldots , r} (A_{\mathfrak p_ i}/\mathfrak p_ i^{w_ i}A_{\mathfrak p_ i}) \otimes _ A B \]
Fix $i$. We claim that the canonical map
\[ (A_{\mathfrak p_ i}/\mathfrak p_ i^{w_ i}A_{\mathfrak p_ i}) \otimes _ A B \to \prod \nolimits _{j = 1, \ldots , r_ i} B_{\mathfrak q_{i, j}}/\mathfrak q_{i, j}^{e_{i, j}w_ i}B_{\mathfrak q_{i, j}} \]
is injective. Here $e_{i, j}$ is the ramification index of $A_{\mathfrak p_ i} \to B_{\mathfrak q_{i, j}}$. The claim asserts that $\mathfrak p_ i^{w_ i}B_{\mathfrak p_ i}$ is equal to the set of elements $b$ of $B_{\mathfrak p_ i}$ whose valuation at $\mathfrak q_{i, j}$ is $\geq e_{i, j}w_ i$. Choose a generator $a \in A_{\mathfrak p_ i}$ of the principal ideal $\mathfrak p_ i^{w_ i}$. Then the valuation of $a$ at $\mathfrak q_{i, j}$ is equal to $e_{i, j}w_ i$. Hence, as $B_{\mathfrak p_ i}$ is a normal domain whose height one primes are the primes $\mathfrak q_{i, j}$, $j = 1, \ldots , r_ i$, we see that, for $b$ as above, we have $b/a \in B_{\mathfrak p_ i}$ by Algebra, Lemma 10.157.6. Thus the claim.
The claim combined with the second exact sequence above determines an exact sequence
\[ 0 \to I \otimes _ A B \to B \to \prod \nolimits _{i = 1, \ldots , r} \prod \nolimits _{j = 1, \ldots , r_ i} B_{\mathfrak q_{i, j}}/\mathfrak q_{i, j}^{e_{i, j}w_ i}B_{\mathfrak q_{i, j}} \]
It follows that $I \otimes _ A B$ is the set of elements $h'$ of $B$ which have valuation $\geq e_{i, j}w_ i$ at $\mathfrak q_{i, j}$. Since $f = wh^ n$ in $B$ we see that $h$ has valuation $e_{i, j}w_ i$ at $\mathfrak q_{i, j}$. Thus $h'/h \in B$ by Algebra, Lemma 10.157.6. It follows that $I \otimes _ A B$ is a free $B$-module of rank $1$ (generated by $h$). Therefore $I$ is a free $A$-module of rank $1$, see Algebra, Lemma 10.78.6. Let $g \in I$ be a generator. Then we see that $g$ and $h$ differ by a unit in $B$. Working backwards we conclude that the valuation of $g$ in $A_{\mathfrak p_ i}$ is $w_ i = v_ i/n$. Hence $g^ n$ and $f$ differ by a unit in $A$ (by Algebra, Lemma 10.157.6) as desired. $\square$
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