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 $P R\_P$.



Basic elements and localizations
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.


