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On page 52 they write "...By (4.3) we can achieve (i)..."

where (4.3) is the lemma on the previous page that states that if $q_i$ are all $p$-primary then $\bigcap_i q_i$ is $p$-primary and (i) is the property of a minimal primary decomposition that $r(q_i)$ are pairwise distinct.

I'm confused about how $r(q_i)=p$ for all $i$ helps us to get $r(q_i) \neq r(q_j)$ for all $i \neq j$. If $r(q_i)=p$ for all $i$ then the primary decomposition would only consist of one ideal because we'd have to throw all the others away to get (i). Or not?

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I think the point is that if, say, $\mathfrak q_1, \ldots, \mathfrak q_s$ are $\mathfrak p$-primary then you just replace them by the single ideal $\bigcap_{i = 1}^s \mathfrak q_i$, and what you get is still a primary decomposition. – Dylan Moreland Jun 29 '12 at 21:34
up vote 1 down vote accepted

Let $I = q_1 \cap q_2 \cap q_3 \cap q_4 \cap q_5$. Suppose $q_1, q_2$ are $p_1$-primary and $q_3, q_4, q_5$ are $p_2$-primary, where $p_1 \neq p_2$. Let $q_1' = q_1 \cap q_2$, $q_3' = q_3 \cap q_4 \cap q_5$. Then $q_1'$ is $p_1$-primary and $q_3'$ is $p_2$-primary. And $ I = q_1 \cap q_2 \cap q_3 \cap q_4 \cap q_5 = q_1' \cap q_3'$.

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