# If $P$ is a prime ideal in a ring, then its only associated prime is the ideal itself?

This seems obvious to me, but I don't see it explicitly mentioned, so I may be wrong here, but since a prime ideal is primary, we have a primary decomposition for a prime ideal consisting of just the prime ideal and it's radical is itself. So this must be it's only associated prime, since the set of associated primes is independent of the particular primary decomposition.

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I'm happy to put you at rest: what you write is perfectly correct. If you need a reference, check page 107 of these fine notes by Hochster: math.lsa.umich.edu/~hochster/614F10/614.pdf –  Georges Elencwajg Oct 9 '11 at 17:23
@Georges Elencwajg: Perfect, thank you. –  Sanjoy Pagg Oct 9 '11 at 17:30