Proving Primary Ideals in \mathbb{Z}: Peter's Challenge

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SUMMARY

The primary ideals in the ring \(\mathbb{Z}\) are definitively identified as 0 and the ideals \((p^m)\) for any prime \(p\) and integer \(m \ge 1\), as stated in Dummit and Foote on page 682. The discussion revolves around proving that a specific ideal, denoted as (4), is not primary. The definition of a primary ideal, as provided by Dummit and Foote, states that a proper ideal \(Q\) in a commutative ring \(R\) is primary if whenever \(ab \in Q\) and \(a \notin Q\), then \(b^n \in Q\) for some positive integer \(n\). The participants confirm that (4) does not meet this criterion, thus validating its non-primary status.

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In Dummit and Foote on page 682 Example 1 reads as follows:
----------------------------------------------------------------------------------------------------------------------------The primary ideals in \mathbb{Z} are 0 and the ideals (p^m) for p a prime and m \ge 1.-----------------------------------------------------------------------------------------------------------------------------

So given what D&F say, (4) is obviously not primary.

I began trying to show from definition that (4) was not a primary from the definition, but failed to do this

Can anyone help in this ... and come up with an easy way to show that (4) is not primary?

Further, can anyone please help me prove that the primary ideals in \mathbb{Z} are 0 and the ideals (p^m) for p a prime and m \ge 1.

PeterNote: the definition of a primary idea is given in D&F as follows:

Definition. A proper ideal Q in the commutative ring R is called primary if whenever ab \in Q and a \notin Q then b^n \in Q for some positive integer n.

Equivalently, if ab \in Q and a \notin Q then b \in rad \ Q
 
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(4) is primary, and this agrees with what D&F said. Indeed (4)=(p^m) where, p=2 and m=2.
 
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Thanks economicsnerd!

Not entirely sure how I missed that ... :-(
 

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