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Situation: $I$ is an ideal in a polynomial ring with a primary decomposition, not necessarily minimal (minimal=irredundant). I want to minimal-ize it. For any couple of primary ideals with the same radical, I take the intersection of the two, and this is ok. Then I must avoid that one primary ideal contains the intersection of all the others. I see in some solved exercises the following method: find the radicals of the primary ideals (i.e the primes of $\sqrt{I}$), eliminate those that contains the intersection of the others and the primary ideals corresponding to the prime ideals left are effectively the minimal ones. I can't figure why this method should work since I know that: $Q_i\supseteq \bigcap_{j\neq i}Q_j \Rightarrow$ $\sqrt{Q_i} \supseteq \bigcap_{j\neq i}\sqrt{Q_j}\Leftrightarrow$ $P_i \supseteq \bigcap_{j\neq i}P_j \Leftrightarrow$ $P_i \supseteq P_j$ for some $j$, but the first implication does not invert. Then I can say for sure that if $P_i \nsupseteq P_j$ for all $j$ then $Q_i$ is minimal but if $P_i \supseteq P_j$ for some $j$ I'm not sure that $Q_i$ is not minimal. Am I wrong? Thanks.

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The process you are referring to, does not work for the following reason: a primary decomposition of an ideal can be irredundant, without necessarily all the associated primes being minimal, i.e. we might have $I=Q_1 \cap \cdots \cap Q_s$, an irredundant primary decomposition, with $\sqrt(Q_i) = P_i$, and it might be the case that $P_i \subset P_j$. The process you are describing would always give a decomposition in which the associated primes are minimal. This is too restrictive.

Here is an example: Let $k$ be a field, then in the ring $k[x,y]$ we have that $(x^2,xy) = (x) \cap (x^2,y)$ is an irredundant primary decomposition. $(x)$ is prime and so its radical is itself. $(x^2,y)$ is primary with radical the prime ideal $(x,y)$. Note that $(x,y)$ properly contains $(x)$ however $(x^2,y)$ does not contain $(x)$!

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