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I have a question about the following proof:

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How do I get that $\mathfrak a$ is reducible? I thought perhaps one can argue that $\mathfrak a \cap \mathfrak a = \mathfrak a$ is a finite intersection hence $\mathfrak a$ can't be irreducible. But this feels stupid so it must be wrong. Thanks for your help.

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Where do you read that $\mathfrak a$ is irreducible? In fact, the proof makes use of the fact that $\mathfrak a$ must be reducible or else it (trivially!) would have a primary decomposition. – Bruno Joyal Jul 23 '12 at 13:38
@Bruno Sorry, that was a typo. – Rudy the Reindeer Jul 23 '12 at 13:39
up vote 2 down vote accepted

If $\mathfrak a$ were irreducible, then it would trivially be a finite intersection of irreducible ideals (namely, $\mathfrak a = \mathfrak a$, or $\mathfrak a = \mathfrak a \cap \mathfrak a$ if you like your intersections to have more than one intersectee). Since $\mathfrak a$ is an element of the collection of all ideals which have no such representation, $\mathfrak a$ must be reducible, which further leads to the contradiction exposed in the proof.

Sometimes things are as simple as they appear!

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Heh, you're saying the argument as I give in the question is correct as it is? – Rudy the Reindeer Jul 23 '12 at 14:35
@MattN., yes, precisely. A blue ideal is trivially a finite intersection of blue ideals, where "blue" is your favorite property. – Bruno Joyal Jul 23 '12 at 14:37
I think blue is my favourite property. : ) – Rudy the Reindeer Jul 23 '12 at 14:47

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