Find all of the ideals in $\mathbb{Z}/n\mathbb{Z}$.
I have a hint with the theorem of correspondence where the ideals in $\mathbb{Z}/n\mathbb{Z}$ are the ideals $r\mathbb{Z}/n\mathbb{Z}$ where $r\mid n$.
I am not sure how to apply this.
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Welcome to TeX SX! Clearly, you have to find all divisors of $n$. – Bernard Jun 24 '19 at 19:17
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4@Bernard TeX SX?????????? – user10354138 Jun 24 '19 at 19:19
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2It should say "where $ r\Bbb Z \supseteq n\Bbb Z$", which is equivalent to $,r\mid n,$ since contains = divides for principal ideals. – Bill Dubuque Jun 24 '19 at 19:19
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@Bernard Find all the divisors while not losing sight of the site for the questions! – Bill Dubuque Jun 24 '19 at 19:25
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For the quotient group(ring) $r\Bbb{Z}/n\Bbb{Z}$ to be defined, we need $n\Bbb{Z}$ to be a subgroup (subring) of $r\Bbb{Z}$. So at the very least $n\Bbb{Z} \subseteq r\Bbb{Z}$. Thus $n=ra$ for some $a \in \Bbb{Z}$. – Anurag A Jun 24 '19 at 19:27
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@user10354138: Sorry, I operate on both sites,and I had a hard day ;o) – Bernard Jun 24 '19 at 19:29
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The question is asking to find all of the ideals in the general format, that is what I am unsure of – Vkirch Jun 24 '19 at 19:41
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Also, this is a duplicate of https://math.stackexchange.com/questions/1657210/find-all-ideals-of-mathbbz-n?rq=1 – miraunpajaro Jun 24 '19 at 19:43
1 Answers
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The correspondance theorem yields that if $I$ is an ideal of $\mathbb{Z}/n\mathbb{Z}$, then there exists an unique $J$ ideal of $\mathbb{Z}$ so that $n\mathbb{Z}\subset J$ and $I=J/n\mathbb{Z}$
Since $\mathbb{Z}$ is an euclidean domain, $J=m\mathbb{Z}$ for some $m$ and $m\mid n$.
This basically gives away the answer, there are as many ideals of $\mathbb{Z}/n\mathbb{Z}$ as divisors of $n$. For example $\mathbb{Z}_6=\mathbb{Z}/6\mathbb{Z}$ has two ideals, for $6\mathbb{Z}\subset 2\mathbb{Z}$ and $6\mathbb{Z}\subset 3\mathbb{Z}$ and $6\mathbb{Z}$ has no other proper superideal (that is ideal containing it). So $\mathbb{Z}_6$ has two ideals (namely, $2\mathbb{Z}_6$ and $3\mathbb{Z}_6$)
PS: If you find this answer messy, please feel free to edit for clarification. Thank you very much
José Carlos Santos
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miraunpajaro
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