
Primary ideals in Z
In Dummit and Foote on page 682 Example 1 reads as follows:

The primary ideals in $\displaystyle \mathbb{Z} $ are 0 and the ideals $\displaystyle (p^m) $ for p a prime and $\displaystyle 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 $\displaystyle \mathbb{Z} $ are 0 and the ideals $\displaystyle (p^m) $ for p a prime and $\displaystyle m \ge 1 $.
Peter
Note: 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 $\displaystyle ab \in Q $ and $\displaystyle a \notin Q $ then $\displaystyle b^n \in Q $ for some positive integer n.
Equivalently, if $\displaystyle ab \in Q $ and $\displaystyle a \notin Q $ then $\displaystyle b \in rad \ Q $

Re: Primary ideals in Z
(4) is primary, as $\displaystyle 4=2^2$ and 2 is prime. Suppose $\displaystyle ab \in (4)$. Then $\displaystyle 4ab$. Suppose $\displaystyle a \not\in (4)$, so $\displaystyle 4\nmid a$.
But we can have two cases, either $\displaystyle 2a$ or $\displaystyle 2\nmid a$:
If $\displaystyle 2a$, as $\displaystyle 4ab$ and $\displaystyle 4\nmid a$, we get $\displaystyle 2b$, which means $\displaystyle b\in(2)$ therefore $\displaystyle b^2 \in (4)$.
If $\displaystyle 2\nmid a$, then $\displaystyle 4b$, and so $\displaystyle b\in 4$.
You can use a similar proof to prove that the primary ideals are (0) and $\displaystyle (p^m)$.