Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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
1  
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

1 Answer 1

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 $\bar{x}=\overline{g_0}\bar{h} \in (\overline{g_0})$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.