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I learned the concept radical of an ideal from this wikipedia article. I tried some examples and I found that it's not easy to find $Rad(I)$. (That article gives some examples when $R={\Bbb Z}$.) For examples, let $R$ be the ring ${\Bbb Z}_n$ and let $I=\langle m\rangle $ where $m\in {\Bbb Z}_n$. How can I find $Rad(I)$? One properties I read in the article may be useful is

$Rad(I)$ is the intersection of all the prime ideals of $R$ that contain $I$.

But how can I find all the prime ideals of ${\Bbb Z}_n$ that contains $\langle m\rangle$?

I didn't find related materials in some introduction-level abstract algebra textbooks. Some happen to give this concept in exercises, say, "show that $Rad(I)$" is an ideal. Can any one come up with some useful references for this topic?

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I assume your notation means $\mathbb{Z}_n=\mathbb{Z}/n$. If $m=p_1^{n_1}\cdots p_k^{n_k}$ is the prime decomposition then $Rad(m)=(p_1\cdots p_k) \trianglelefteq \mathbb{Z}_n$. In particular, if $m$ is squarefree, then $(m)$ is a radical ideal in $\mathbb{Z}_n$. – Ralph Nov 7 '12 at 2:11
@Ralph, doesn't $n$ play a role as well? For instance, the radical of $m \bmod n$ will contain all the nilpotent elements and for general $n$ there are nontrivial nilpotent elements in $\mathbb Z/n$. – lhf Nov 7 '12 at 2:17
If $m$ chosen is the unique one satisfying $m | n$, then what Ralph said is true. – user27126 Nov 7 '12 at 2:52
Let $g = gcd(n,m)$. Write $m=gq$ with $q,n$ coprime. Then $q$ is a unit in $\mathbb{Z}_n$ and $(m)=(g)$ in $\mathbb{Z}_n$. Now if $g=p_1^{n_1}\cdots p_k^{n_k}$ then I think we have $Rad(m)=(p_1\cdots p_k)$. – Ralph Nov 7 '12 at 3:08
up vote 2 down vote accepted

I assume your notation means $\mathbb{Z}_n=\mathbb{Z}/n$. Set $\bar{m} := m + n\mathbb{Z} \in \mathbb{Z}_n$. Let $g=gcd(n,m)$ have the prime decomposition $g=p_1^{n_1}\cdots p_k^{n_k}$ and set $g_0 := p_1\cdots p_k$. Then we have $$Rad(\bar{m})=(\overline{g_0}) \trianglelefteq \mathbb{Z}_n$$

Proof: By writing $m=gq$, $\bar{q}$ is a unit in $\mathbb{Z}_n$ and hence $(\bar{m})=(\bar{g})$.

$(\supseteq)$ Let $\bar{x} \in (\overline{g_0})$. There are $y,z \in \mathbb{Z}$ s.t. $x=g_0y+nz=:g_0w$. Choose $l > 0$ s.t. $g \mid g_0^l$ (say $g_0^l=gh$). Then $\bar{x}^l=\overline{g_0}^l\bar{w}^l=\bar{g}\bar{h}\bar{w}^l \in (\bar{g})=(\bar{m})$. Thus $\bar{x} \in Rad(\bar{m})$.

$(\subseteq)$ Let $\bar{x} \in Rad(\bar{m})$. There is $l > 0$ s.t. $\bar{x}^l \in (\bar{m})=(\bar{g})$, i.e. there is an integer $y$ with $\bar{x}^l=\bar{g}\bar{y}=\overline{gy}$. Hence there is an integer $z$ with $x^l=gy+nz=:g_0w$. Thus $p_i \mid x^l$, whence $p_i \mid x$ for all $1 \le i \le k$. Consequently $g_0=p_1 \cdots p_k \mid x$ (say $x=g_0h$) and hence $\bar{x}=\overline{g_0}\bar{h} \in (\overline{g_0})$.

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