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!
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.