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Let $R$ be a Noetherian commutative ring and fix an ideal $I$. It is a classical result that $I$ admits a decomposition as the intersection of primary ideals, all of whose radicals are different and of which none can be omitted from the intersection. The radicals of these primary ideals are called the associated prime ideals of $I$. It can be shown that these are independent of the particular primary decomposition used.

My question is whether the minimal primes over $I$ and the minimal associated primes of $I$ coincide.

It should be clear that a minimal associated prime $P$ is also minimal over $I$. Otherwise, we can find a minimal prime over $I$ below the $P$-primary component of $I$ and insert this into a minimal primary decomposition of $I$ in place of the $P$-primary component.

I am, however, not so sure about the other inclusion. I think the same proof as above would work if we knew there was an associated prime above any minimal prime over $I$, but I don't know if that's the case.

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Yes, indeed. The minimal prime ideals containing $I$ and the minimal associated primes of $I$ coincide. You may find this result in Atiyah's commutative algebra. –  Ehsan M. Kermani Dec 30 '11 at 10:45

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Yes, the minimal associated primes of $I$ coincide with the absolutely minimal primes of $I$. (I'll only show the part you haven't yet done.)

Suppose that your primary decomposition is $I=Q_1\cap\cdots\cap Q_n$ and consider an absolutely minimal prime $P\supset I$.
Since $P=\sqrt P\supset \sqrt I=P_1\cap\cdots\cap P_n\supset P_1 \cdots P_n$, we must have $P\supset P_i$ for some $i$ (and of course $P_i\supset \sqrt I\supset I$).
Absolute minimality of $P$ then yields the conclusion: $P= P_i$.

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