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If $x_1,...,x_n$ is an $M$-sequence, then for prime ideal $P$ of $R$, can we localize the $M$-sequence to an $M_P$-sequence?

$I_PM_P$ is not $M_P$ by Nakayama's lemma. Then how can I prove after localization, $x_i$ is non-zero divisor in $M_P/(x_1,...,x_{i-1})M_P$?

user26857
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1 Answers1

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That is because localisation is an exact functor, hence multiplication by $x_i$, which is injective on $M/(x_1,\dots,x_{i-1})M$ remains injective on $M_\mathfrak p/(x_1,\dots,x_{i-1})M_\mathfrak p$ (writing $x_i$ for $\dfrac{x_i}1$ in $A_\mathfrak p$).

Bernard
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