Lemma 42.68.39. Let $A$ be a Noetherian local domain of dimension $1$ with fraction field $K$. For $x \in K \setminus \{ 0, 1\} $ we have
\[ d_ A(x, 1 -x) = 1 \]
Proof. Write $x = a/b$ with $a, b \in A$. The hypothesis implies, since $1 - x = (b - a)/b$, that also $b - a \not= 0$. Hence we compute
\[ d_ A(x, 1 - x) = d_ A(a, b - a)d_ A(a, b)^{-1}d_ A(b, b - a)^{-1}d_ A(b, b) \]
Thus we have to show that $d_ A(a, b - a) d_ A(b, b) = d_ A(b, b - a) d_ A(a, b)$. This is Lemma 42.68.38. $\square$
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