The task is to find all ideals of ${\mathbb{Z}_n}$, where $n$ is positive integer, greater than one.

My effort

Let $I$ be an ideal of ${\mathbb{Z}_n}$. It is obvious that $I$ is an additive subgroup of ${\mathbb{Z}_n}$. Consider $G$ as an additive subgroup of ${\mathbb{Z}_n}$. Then $G$ is a cyclic additive subgroup generated by $\left\langle d \right\rangle $, where $d \mid n$. We know that for a finite cyclic group of order $k$, every subgroup's order is a divisor of $k$, and there is exactly one subgroup for each divisor. It follows that all ideals of ${\mathbb{Z}_n}$ are of form $\left\langle {{d_1}} \right\rangle ,\left\langle {{d_2}} \right\rangle , \ldots \left\langle {{d_i}} \right\rangle $, where ${d_1},{d_2}, \ldots ,{d_i}$ are positive divisors or $n$.


Is my proof correct?

  • $\begingroup$ You have to show that every ideal is principal, see here. $\endgroup$ – Dietrich Burde Feb 15 '16 at 20:20

You are almost right! You have to pay attention to the fact that an ideal of $\mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z}$ is in particular a subset: it is a set of equivalences classes (in $\mathbb{Z}_n$). But $\left\langle {{d_i}} \right\rangle$ is a subset of $\Bbb Z$.

So you should write $\left\langle {{[d_i]_n}} \right\rangle$ where $[d_i]_n$ denotes the equivalence class of $d_i$, where $d_i \mid n$.

If you are conviced that all the additive subgroups of $\Bbb Z_n$ are of the form $\left\langle {{[d_i]_n}} \right\rangle$ for some $d_i \mid n$, then it just remains to show that these subgroups are actually ideals of $\Bbb Z_n$.

More precisely, you have shown that: if $I$ is an ideal of $\Bbb Z_n$, then it is of the form $I = \langle [d_i]_n \rangle = d_i\mathbb{Z}/n\mathbb{Z}$ for some $d_i \mid n$. It just remains to show the converse: if $I = \left\langle {{[d_i]_n}} \right\rangle$ for some $d_i \mid n$ then $I$ is an ideal of $\Bbb Z_n$.


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