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The question is as indicated in the title:

When is $\langle m+n{\Bbb Z}\rangle$ a radical ideal in ${\Bbb Z}_n$, i.e. $Rad(\langle m+n{\Bbb Z}\rangle)=\langle m+n{\Bbb Z}\rangle$?

I gathered the information in the following question I asked:

According answers to the second question, I am able to get $$ Rad(\langle m+n{\Bbb Z}\rangle)=\langle \bar{m}+n{\Bbb Z}\rangle$$ for some $\bar{m}$. To determine whether $Rad(\langle m+n{\Bbb Z}\rangle)\subset\langle m+n{\Bbb Z}\rangle$, it suffices to know when $$ \langle \bar{m}+n{\Bbb Z}\rangle\subset \langle m+n{\Bbb Z}\rangle $$ Is this equivalent to $\langle \bar{m}\rangle\subset \langle m\rangle$ in $\Bbb Z$? How can I approach the problem in the title?

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The last condition is clearly sufficient. It it not necessary because once you find a pair $m, \bar{m}$ as desired, for any natural number $r$ prime to $n\bar{m}$, you also have $$\langle \bar{m}+n\mathbb Z\rangle \subseteq \langle r{m}+n\mathbb Z\rangle =\langle {m}+n\mathbb Z\rangle$$ but $\langle \bar{m}\rangle \subseteq \langle r{m}\rangle$ is never true.

To answer your original question, the ideal generated by $m\in\mathbb Z$ in $\mathbb Z/n\mathbb Z$ is radicial if and only if all multiple prime factors in $m$ are prime to $n$. Can you prove this ?

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I don't understand you statement: "all multiple prime factors in $m$ are prime to $n$". Do you mean: if $m=p_i^{k_1}p_2^{k_2}\cdots p_m^{k_m}$, then $gcd(p_i^{k_i},n)=1$ for all the prime numbers $p_i$? – Jack Nov 8 '12 at 0:18
@Jack: sorry, I mean $\gcd(p_i, n)=1$ if $k_i\ge 2$. – user18119 Nov 8 '12 at 6:52

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