In the second paragraph of the proof below, I don't understand why the zero ideal having primary decomposition implies that there are finitely many prime ideals in $A$ (prime ideals that happen to be minimal). I would appreciate any help.
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Let $(0) = q_1\cap q_2\cap \cdots \cap q_n$ be a primary decomposition, where each $q_i$ is primary, and $\sqrt{q_i} = p_i$ is prime. Now take a prime ideal $p$. I claim is must contain at least one of the $q_i$, and therefore also the corresponding $p_i$, and therefore either be equal to $p_i$ or non-minimal.
For contradiction, assume $q_i\not\subseteq p$ for all $i$. Then for each $i$ we can pick an element $x_i\in q_i\setminus p$. We have $x_1x_2\cdots x_n\in q_1\cap q_2\cap \cdots \cap q_n = (0)$, so therefore this product is equal to $0$, and therefore an element of $p$. But this contradicts either primality of $p$ or our choice of the $x_i$.
Arthur
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