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I am having trouble with classification of the radical ideals of $\mathbb{Z}$.

We know that for a commutative ring $R$ with an ideal $I$, the radical of $I$ is defined (and denoted as $\sqrt{I}$) as a set ot all elements $a\in R$ such that for some $n\in \mathbb{N}_{> 0}$ we have $a^{n}\in I$. If $\sqrt{I}=I$, then $I$ is a radical ideal.

I think we have to look at the $m\mathbb{Z}$ for $m$ prime or composite. Wikipedia says that in general the radical ideal of $m\mathbb{Z}$ is $r\mathbb{Z}$, where $r$ is the product of all prime factors of $m$. I would like also to prove this but i don't know how...

Did i understand the question correct or i have to look at some other ideals of the set of the integers? Thank you in advance!

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    $\begingroup$ Show this: $m\mid a^n$ for some $n\ge 1$ iff all primes of $m$ divide $a$. $\endgroup$ – user26857 Sep 18 '14 at 19:16
  • $\begingroup$ Thank you for the hint! Could i also say that all prime ideals (2),(3),(5)... are radicals? $\endgroup$ – Lullaby Sep 18 '14 at 20:14
  • $\begingroup$ Prime ideals are always also radical ideals. $\endgroup$ – Martin Brandenburg Sep 21 '14 at 12:50
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The radical ideals are precisely those which can be written as intersections of prime ideals.

In a PID, the prime ideals are $\{0\}$ and $(p)$ for prime elements $p$. Hence, the radical ideals are $(0)$ and $\cap_i (p_i)$ for prime elements $p_i$. If there are infinitely many of them, the intersection is $(0)$. If not, we get $(\prod_i p_i)$. Hence, the radical ideals are those ideals $(n)$ for which $n$ is square-free.

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  • $\begingroup$ Maybe. I just wanted to show this method. $\endgroup$ – Martin Brandenburg Sep 21 '14 at 14:29
  • $\begingroup$ Others may post their own solutions ;). $\endgroup$ – Martin Brandenburg Sep 21 '14 at 18:15

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