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Brittany Murphy
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In a Noetherian local ring $(R,m)$ with a prime ideal $P\neq m$, $P^{(n)}=P^n:m^{\infty}$
@Hailong: Thanks for the comment. Could you please point me to a reference of the proof of this statement when dim$(R/P)=1$. I am unable to find any reference in the paper.
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In a Noetherian local ring $(R,m)$ with a prime ideal $P\neq m$, $P^{(n)}=P^n:m^{\infty}$
@Mohan: I am not sure I understand your comment. The author uses this formula in section 4 of this paper that I referenced above people.reed.edu/~iswanson/topology.pdf in a power series ring with 3 variables.
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In a Noetherian local ring $(R,m)$ with a prime ideal $P\neq m$, $P^{(n)}=P^n:m^{\infty}$
@Graham: I guess I am missing some hypothesis then. In section 4, in the following paper, this is the formula used for calculating symbolic powers.people.reed.edu/~iswanson/topology.pdf Is the statement true if $R$ is regular, local?
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In a Noetherian local ring $(R,m)$ with a prime ideal $P\neq m$, $P^{(n)}=P^n:m^{\infty}$