Lemma 33.37.9. Let $A$ be a Noetherian local ring of dimension $1$. Let $L = \prod A_\mathfrak p$ where the product is over the minimal primes of $A$. Let $a_1, a_2 \in \mathfrak m_ A$ map to the same element of $L$. Then $a_1^ n = a_2^ n$ for some $n > 0$.

Proof. Write $a_1 = a_2 + x$. Then $x$ maps to zero in $L$. Hence $x$ is a nilpotent element of $A$ because $\bigcap \mathfrak p$ is the radical of $(0)$ and the annihilator $I$ of $x$ contains a power of the maximal ideal because $\mathfrak p \not\in V(I)$ for all minimal primes. Say $x^ k = 0$ and $\mathfrak m^ n \subset I$. Then

$a_1^{k + n} = a_2^{k + n} + {n + k \choose 1} a_2^{n + k - 1} x + {n + k \choose 2} a_2^{n + k - 2} x^2 + \ldots + {n + k \choose k - 1} a_2^{n + 1} x^{k - 1} = a_2^{n + k}$

because $a_2 \in \mathfrak m_ A$. $\square$

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