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Suppose, $R$ is a noetherian ring. Let $P$ be a prime ideal in $R$. Let $Q$ be a $P$-primary ideal that contains $P^n$. Then does $Q$ contain $P^{(n)}$ which is the $n$th symbolic power of $P$ and is the $P$-primary component that occurs in any irredundant primary decomposition of $P^n$? I think this should be true, but I am not sure where to start. I was thinking in terms of taking an irredundant decomposition for $P^n$ and intersecting it with $Q$. This introduces two $P$ primary components, so one must be contained in the other?

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up vote 2 down vote accepted

The answer is yes. Suppose

$P^n = P^{(n)}\cap I_1\cap I_2\cap ...\cap I_r$

is a minimal primary decomposition. Since $P\subseteq Q$ we find that

$P^n = (P^{(n)}\cap Q)\cap I_1\cap I_2\cap ...\cap I_r$

is also a minimal primary decomposition. Now, primary decomposition is not unique in general, but in this case we now that the $P$-primary component is isolated because the radical of $P^n$ is $P$ (thus, the radicals of all the $I_i$ must contain $P$) and therefore $P^{(n)} \cap Q = P^{(n)}$ which means that $P^{(n)}\subseteq Q$.

If I am not mistaken, this observation is needed even for the definition of the symbolic power as "the $P$-primary component of $P^n$".

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Thank you, KotelKanim – BMI Oct 17 '11 at 13:49

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