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This is Exercise 14.3 of Matsumura's Commutative Ring Theory.

Let $(A,m)$ be a Noetherian local ring and $G=\operatorname{gr}_{m}(A)$ be the associated graded ring, generated by direct sum of $R_0=A/m$ and $R_{i} = m^{i}/m^{i+1}$.

Then, given an ideal $I \subset m$ of $A$, we may let $I^{\ast}$ be an ideal generated by the image of elements of $I$ in $G$. For example, each $a \in I$ has an $i \in \mathbb{N}$ such that $a \in m^{i}$ but $a \not\in m^{i+1}$, thus we may let $a^{\ast}:=a+m^{i+1} \in R_{i}$, therefore represent $I^{\ast}$ as $(a^{\ast}: a \in I)$.

Then the book asks to give an example such that if $G$ is an integral domain, there is an ideal $I=(a_1,\dots, a_{r})$ with $r>1$ but $I^{\ast} \neq (a_{1}^{\ast},\dots,a_{r}^{\ast})$. Could you give me a hint for construction?

I firstly worked with $k[[X,Y]]$, but do not have any result. Any idea will be appreciated.

user26857
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user124697
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An example is given in Bruns and Herzog, Exercise 4.6.12(b). Let $R=K[[X,Y,Z]]$ and $I=(X^2,XY+Z^3)$. Then $XZ^3\in I^*$, but $XZ^3\notin(X^2,XY)$.
The associated graded ring $\mathrm{gr}_{\mathfrak m}(R)$ is isomorphic to $K[X,Y,Z]$, and therefore it is and integral domain.

user26857
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