# Nonzero-divisor on the associated graded ring at the maximal ideal implies nonzero-divisor on the ring? [closed]

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 $$R$$ ?

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 $$R$$.
• what is $\overline{y}$ exactly – reuns Oct 28 '19 at 8:50
• 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 $\mathrm{gr}_{\mathfrak{m}}(R)$. – Slup Oct 28 '19 at 8:58