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Let $(R, \mathfrak{m})$ be a local domain and $x$ is a basic element of $\mathfrak{m}$, that is $x \in \mathfrak{m} \setminus \mathfrak{m}^2$. Let $P$ be a prime ideal containing $x$. Is it true that $x$ is a basic element in $R_P$?

Edit: By the Mohan answer, the question has negative answer. In fact, I am interested in a stronger property for $x$. Suppose $\dim R = d$ and $(x_1, \ldots, x_d)$ be a minimal reduction of $\mathfrak{m}$ (here we assume that the residue field is infinite). Let $x = x_1$, we have $x$ is a basic element of $R$. Let $P$ be a prime ideal containing $x$. The question now as follows:

Question: Is it true that $x$ is a basic element in $R_P$? Is it an basic element of a minimal reduction of $P R_P$.

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No. Consider $R=k[[x,y,z]]/xy-z^2$ and $x$, which is basic. But when you localize at the prime $P=(x,z)$, $x$ is no longer basic.

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  • $\begingroup$ Thanks you! In my problem the element $x$ satisfies a stronger property. I edited the question. Could you help me. $\endgroup$ – Pham Hung Quy Sep 29 '17 at 2:23
  • $\begingroup$ @PhamHungQuy In my example, $(x,y)$ looks like a minimal reduction, since $(x,y)(x,y,z)=(x,y,z)^2$, so the same example should be a counter example. $\endgroup$ – Mohan Sep 29 '17 at 13:49

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