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For a Noetherian local ring $(R, \mathfrak m)$ , let $\mathrm{gr}_{\mathfrak m} (R):= \oplus_{n \ge 0} \mathfrak m^n/\mathfrak m^{n+1}$ be the associated graded ring.

If $x\in \mathfrak m/\mathfrak m^2 $ is not a zero divisor on $\mathrm{gr}_{\mathfrak m} (R)$ , then is $x$ also not a zero divisor on $R$ ?

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More general result is true.

Suppose that $x\in R$ is a non zero element in $R$ and assume that the corresponding class $\overline{x}$ in $\mathrm{gr}_{\mathfrak{m}}(R)$ is not a zero divisor. Pick $y \in R\setminus \{0\}$ such that $x\cdot y = 0$. Now $$\overline{x}\cdot \overline{y} = 0$$ where $\overline{y}$ is the class of $y$ in $\mathrm{gr}_{\mathfrak{m}}(R)$. But $\overline{y}$ is nonzero in $\mathrm{gr}_{\mathfrak{m}}(R)$ (because $y\neq 0$ in $R$ and $\bigcap_{n\in \mathbb{N}}\mathfrak{m}^n = 0$). Hence by contradiction we deduce that $x$ is not a zero divisor in $R$.

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  • $\begingroup$ what is $\overline{y}$ exactly $\endgroup$ – reuns Oct 28 '19 at 8:50
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    $\begingroup$ Since $y$ is non zero, there exists $n$ such that $y\in \mathfrak{m}^n\setminus \mathfrak{m}^{n+1}$. Then $\overline{y} = y\,\mathrm{mod}\,\mathfrak{m}^{n+1}$ is an element of $\mathrm{gr}_{\mathfrak{m}}(R)$. $\endgroup$ – Slup Oct 28 '19 at 8:58

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